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Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 7 Fractions Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 7 Fractions Solutions

Ganita Prakash Class 6 Chapter 7 Solutions

Class 6 Maths Ganita Prakash Chapter 7 Solutions Fractions

Question 1.
Draw a picture and write an addition statement to show:
(a) 5 times \(\frac{1}{4}\) of a roti
(b) 9 times \(\frac{1}{4}\) of a roti
Solution:
(a)

5 times \(\frac{1}{4}\) of a roti = \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)

(b)

9 times \(\frac{1}{4}\) of a roti = \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)

Question 2.
Match each fractional unit with the correct picture:

Solution:

Question 3.
On a number line, draw lines of lengths \(\frac{1}{10}\), \(\frac{3}{10}\), and \(\frac{4}{5}\).
Solution:

Question 4.
Write the fraction that gives the lengths of the lines in the respective boxes.

Solution:
\(\frac{6}{5}\), \(\frac{7}{5}\), \(\frac{8}{5}\), \(\frac{9}{5}\)

Question 5.
Figure out the number of whole units in each of the following fractions:
(a) \(\frac{8}{3}\)
(b) \(\frac{11}{5}\)
(c) \(\frac{9}{4}\)
Solution:
(a) \(\frac{8}{3}\) = 2\(\frac{2}{3}\). So, 2 whole units.
(b) \(\frac{11}{5}\) = 2\(\frac{1}{5}\). So, 2 whole units.
(c) \(\frac{9}{4}\) = 2\(\frac{1}{4}\). So, 2 whole units.

Question 6.
Are \(\frac{3}{6}\), \(\frac{4}{8}\) and \(\frac{5}{10}\) equivalent fractions ? Why?
Solution:
Yes, because all of them have the same length i.e., \(\frac{3}{6}\) = \(\frac{4}{8}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 7.
\(\frac{4}{6}\) = ___ = ____ = _____ = ____ (write as many as you can)
Solution:
\(\frac{4}{6}\) = \(\frac{2}{3}\) = \(\frac{6}{9}\) = \(\frac{8}{12}\) = \(\frac{10}{15}\)

Question 8.
Rahim mixes \(\frac{2}{3}\) litres of yellow paint with \(\frac{3}{4}\) litres of blue paint to make green paint. What is the volume of green paint he has made?
Solution:
Volume of yellow paint = \(\frac{2}{3}\) litres
Volume of blue paint = \(\frac{3}{4}\)
Volume of green paint = (\(\frac{2}{3}\) + \(\frac{3}{4}\)) litres = (\(\frac{2}{3}\) × \(\frac{4}{4}\) + \(\frac{3}{4}\) × \(\frac{3}{3}\))litres
= (\(\frac{8}{12}\) + \(\frac{9}{12}\))litres = \(\frac{17}{12}\)litres = 1\(\frac{5}{12}\) litres

Question 9.
Geeta bought \(\frac{2}{5}\) metre of lace and Shamim bought \(\frac{3}{4}\) metre of the same lace to put a complete border on a table cloth whose perimeter is 1 metre long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?
Solution:
Length of lace bought by Geeta = \(\frac{2}{5}\) metre
Length of lace bought by Shamim = \(\frac{3}{4}\) metre
∴ Total length of lace bought = (\(\frac{2}{5}\) + \(\frac{3}{4}\))metres = (\(\frac{2}{5}\) × \(\frac{4}{4}\) + \(\frac{3}{4}\) × \(\frac{5}{5}\))metres
= (\(\frac{8}{20}\) + \(\frac{15}{20}\))metres = \(\frac{23}{20}\) metres = 1\(\frac{3}{20}\) metres > 1 m
Yes, the lace will be sufficient to cover the whole border as it exceeds the perimeter of table cloth, which is 1 m long.

Question 10.
Solve the following problems:
(a) Jaya’s school is \(\frac{7}{10}\) km from her home. She takes an auto for \(\frac{2}{5}\) km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?
(b) Jeevika takes \(\frac{10}{3}\) minutes to take a complete round of the park and her friend Namit takes \(\frac{13}{4}\) minutes to do the same. Who takes less time and by bow much?
Solution:
(a) Total distance between school and home = \(\frac{7}{10}\) km
Distance travelled in auto = \(\frac{1}{2}\) km.
∴ Distance she walks daily to reach the school = (\(\frac{7}{10}\) – \(\frac{1}{2}\))km = (\(\frac{7}{10}\) – \(\frac{1}{2}\) × \(\frac{5}{5}\))km
= (\(\frac{7}{10}\) – \(\frac{5}{10}\))km = \(\frac{2}{10}\) km = \(\frac{1}{5}\) km

(b) Time taken by Jeevika = \(\frac{10}{3}\) minutes and time taken by Namit = \(\frac{13}{4}\) minutes
Now, \(\frac{10}{3}\) × \(\frac{4}{4}\) = \(\frac{40}{12}\) and \(\frac{13}{4}\) × \(\frac{3}{3}\) = \(\frac{39}{12}\)
Clearly, \(\frac{10}{3}\) > \(\frac{13}{4}\)
∴ Namit takes less time by (\(\frac{10}{3}\) – \(\frac{13}{4}\)) minutes = (\(\frac{40}{12}\) – \(\frac{39}{12}\)) minutes = \(\frac{1}{12}\) minutes

Fractions Class 6 Extra Questions

Fractions Class 6 Very Short Question Answer

Question 1.
What fraction of a year is 5 months?
Solution:
We know, number of months in a year = 12
So. 5 month is of a \(\frac{5}{12}\) year.

Question 2.
Represent \(\frac{2}{7}\), \(\frac{4}{7}\), \(\frac{6}{7}\) on a number line.
Solution:
In order to represent \(\frac{2}{7}\), \(\frac{4}{7}\), \(\frac{6}{7}\) on a number line. we divide the gap between 0 and 1. i.e. 1 unit, into 7 equal parts and take second, fourth and sixth points from 0, as shown in the figure.

Question 3.
What fraction of a dozen banana is 7 bananas?
Solution:
Number of bananas in 1 dozen = 12
So, 7 bananas is \(\frac{7}{12}\) of a dozen.

Question 4.
Represent \(\frac{1}{8}\), \(\frac{2}{8}\), \(\frac{4}{8}\) on a number line.
Solution:
In order to represent \(\frac{1}{8}\), \(\frac{2}{8}\), \(\frac{4}{8}\) on a number line, we divide the gap between 0 and 1 into 8 equal parts and take first, second and fourth points from 0, as shown in the figure.

Question 5.
Write the fraction that gives the length of the lines in the respective boxes.

Solution:

Question 6.
Write some equivalent fractions which contain all digits from 1 to 9 once only.
Solution:
\(\frac{2}{6}\) = \(\frac{3}{9}\) = \(\frac{58}{174}\),
\(\frac{2}{4}\) = \(\frac{3}{6}\) = \(\frac{79}{158}\)

Question 7.
Write three equivalent fractions of \(\frac{3}{4}\).
Solution:
Equivalent fractions of \(\frac{3}{4}\) are:
\(\frac{3}{4}\) = \(\frac{3 \times 2}{4 \times 2}\) = \(\frac{6}{8}\),
\(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 3}\) = \(\frac{9}{12}\),
\(\frac{3}{4}\) = \(\frac{3 \times 4}{4 \times 4}\) = \(\frac{12}{16}\)

Question 8.
Subtract \(\frac{3}{7}\) from \(\frac{6}{7}\).
Solution:
\(\frac{6}{7}\) – \(\frac{3}{7}\) = \(\frac{6-3}{7}\) = \(\frac{3}{7}\)

Question 9.
Subtract 8\(\frac{1}{5}\) from 12\(\frac{2}{5}\).
Solution:
8\(\frac{1}{5}\) from 12\(\frac{2}{5}\) = \(\left(\frac{12 \times 5+2}{5}\right)\) – \(\left(\frac{8 \times 5+1}{5}\right)\)
= \(\left(\frac{60+2}{5}\right)\) = \(\left(\frac{40+1}{5}\right)\)
= \(\frac{62}{5}\) – \(\frac{41}{5}\) = \(\frac{62-41}{5}\) = \(\frac{21}{5}\)

Question 10.
Rohit travelled \(\frac{200}{3}\) km by train and \(\frac{50}{3}\) km by bus. What is the total distance travelled by Rohit?
Solution:
Distance travelled by train = \(\frac{200}{3}\) km;
Distance travelled by bus = \(\frac{50}{3}\) km
∴ Total distance covered
= (\(\frac{200}{3}\) \(\frac{50}{3}\)) km = (\(\frac{200+50}{3}\))km
= \(\frac{250}{3}\) km

Question 11.
Find the difference of \(\frac{19}{24}\) and \(\frac{13}{16}\).
Solution:
We can write 24 = 2 × 2 × 2 × 3 and
16 = 2 × 2 × 2 × 2
So, LCM of 24 and 16 is 2 × 2 × 2 × 2 × 3 = 48.
Now, \(\frac{19}{24}\) = \(\frac{19 \times 2}{24 \times 2}\) = \(\frac{38}{48}\) and
\(\frac{13}{16}\) = \(\frac{13 \times 3}{16 \times 3}\) = \(\frac{39}{48}\)
Clearly, \(\frac{38}{48}\) < \(\frac{39}{48}\) ⇒ \(\frac{19}{24}\) < \(\frac{13}{16}\)
Thus, required difference
= \(\frac{13}{16}\) – \(\frac{19}{24}\) = \(\frac{39}{48}\) – \(\frac{38}{48}\)
= \(\frac{39-38}{48}\) = \(\frac{1}{48}\)

Question 12.
Subtract \(\frac{5}{9}\) from \(\frac{7}{9}\).
Solution:
\(\frac{7}{9}\) – \(\frac{5}{9}\) = \(\frac{7-5}{9}\) = \(\frac{2}{9}\)

Fractions Class 6 Short Question Answer

Question 1.
Write the following fractions as mixed fractions:
(i) \(\frac{10}{3}\)
(ii) \(\frac{12}{5}\)
(iii) \(\frac{16}{7}\)
(iv) \(\frac{11}{3}\)
(v) \(\frac{63}{4}\)
Solution:
(i) \(\frac{10}{3}\) = 3 + \(\frac{1}{3}\) = 3\(\frac{1}{3}\)

(ii) \(\frac{12}{5}\) = 2 + \(\frac{2}{5}\) = 2\(\frac{2}{5}\)

(iii) \(\frac{16}{7}\) = 2 + \(\frac{2}{7}\) = 2\(\frac{2}{7}\)

(iv) \(\frac{11}{3}\) = 3 + \(\frac{2}{3}\) = 3\(\frac{2}{3}\)

(v) \(\frac{63}{4}\) = 15 + \(\frac{3}{4}\) = 15\(\frac{3}{4}\)

Question 2.
Write the following mixed fractions into improper fractions:
(i) 4\(\frac{1}{3}\)
(ii) 2\(\frac{1}{4}\)
(iii) 7\(\frac{3}{10}\)
(iv) 12\(\frac{1}{2}\)
(v) 5\(\frac{3}{7}\)
Solution:
(i) 4\(\frac{1}{3}\) = 4 + \(\frac{1}{3}\) = \(\frac{4 \times 3+1}{3}\)
= \(\frac{12+1}{3}\) = \(\frac{13}{3}\)

(ii) 2\(\frac{1}{4}\) = 2 + \(\frac{1}{4}\) = \(\frac{2 \times 4+1}{4}\)
= \(\frac{8+1}{4}\) = \(\frac{9}{4}\)

(iii) 7\(\frac{3}{10}\) = 7 + \(\frac{3}{10}\) = \(\frac{7 \times 10+3}{10}\)
= \(\frac{70+3}{10}\) = \(\frac{73}{10}\)

(iv) 12\(\frac{1}{2}\) = 12 + \(\frac{1}{2}\) = \(\frac{12 \times 2+1}{2}\)
= \(\frac{24+1}{2}\) = \(\frac{73}{2}\)

(v) 5\(\frac{3}{7}\) = 5 + \(\frac{3}{7}\) = \(\frac{5 \times 7+3}{7}\)
= \(\frac{35+3}{7}\) = \(\frac{35}{7}\)

Question 3.
Replace ☐ in each of the following by the correct number:
(i) \(\frac{2}{5}\) = \(\frac{10}{}\)
(ii) \(\frac{3}{7}\) = \(\frac{27}{}\)
(iii) \(\frac{9}{13}\) = \(\frac{27}{}\)
(iv) \(\frac{6}{7}\) = \(\frac{}{49}\)
(v) \(\frac{5}{7}\) = \(\frac{}{35}\)
Solution:
(i) We know, 10 = 2 × 5

So, we replace ☐ by 25 to get \(\frac{2}{5}\) = \(\frac{10}{25}\)

(ii) We know, 27 = 3 × 7

So, we replace ☐ by 63 to get \(\frac{3}{7}\) = \(\frac{27}{63}\)

(iii) We know, 27 = 9 × 3

So, we replace ☐ by 39 to get \(\frac{9}{13}\) = \(\frac{27}{39}\)

(iv) We know, 49 = 7 × 7

So, we replace ☐ by 42 to get \(\frac{6}{7}\) = \(\frac{42}{49}\)

(v) We know, 27 = 3 × 7

So, we replace ☐ by 63 to get \(\frac{3}{7}\) = \(\frac{27}{63}\)

Question 4.
Find the fraction equivalent to \(\frac{30}{45}\), having:
(i) Numerator 16
(ii) Denominator 30
Solution:
We have, \(\frac{30}{45}\) = \(\frac{2 \times 15}{3 \times 15}\) = \(\frac{2}{3}\)
(i) On dividing 16 by 2, we get 8

So, the fraction with numerator 16 and equivalent to \(\frac{30}{45}\) is \(\frac{16}{24}\).

(ii) On dividing 30 by 3, we get 10

So, the fraction with denominator 30 and equivalent to \(\frac{30}{45}\) is \(\frac{20}{30}\).

Question 5.
Write the following mixed fractions as fractions:
(i) 12\(\frac{3}{10}\)
(ii) 9\(\frac{5}{8}\)
(iii) 7\(\frac{6}{11}\)
(iv) 4\(\frac{2}{9}\)
(v) 6\(\frac{3}{7}\)
Solution:
(i) 12\(\frac{3}{10}\) = 12 + \(\frac{3}{10}\) = \(\frac{12 \times 10+3}{10}\)
= \(\frac{120+3}{10}\) = \(\frac{123}{10}\)

(ii) 9\(\frac{5}{8}\) = 9 + \(\frac{5}{8}\) = \(\frac{9 \times 8+5}{8}\)
= \(\frac{72+5}{8}\) = \(\frac{77}{8}\)

(iii) 7\(\frac{6}{11}\) = 7 + \(\frac{6}{11}\) = \(\frac{7 \times 11+6}{11}\)
= \(\frac{77+6}{11}\) = \(\frac{83}{11}\)

(iv) 4\(\frac{2}{9}\) = 4 + \(\frac{2}{9}\) = \(\frac{4 \times 9+2}{9}\)
= \(\frac{36+2}{9}\) = \(\frac{38}{9}\)

(v) 6\(\frac{3}{7}\) = 6 + \(\frac{3}{7}\) = \(\frac{6 \times 7+3}{7}\)
= \(\frac{42+3}{7}\) = \(\frac{45}{7}\)

Question 6.
Replace ☐ in each of the following by the correct number:
(i) \(\frac{3}{4}\) = \(\frac{15}{}\)
(ii) \(\frac{7}{3}\) = \(\frac{}{15}\)
(iii) \(\frac{10}{7}\) = \(\frac{30}{}\)
(iv) \(\frac{11}{15}\) = \(\frac{44}{}\)
(v) \(\frac{15}{4}\) = \(\frac{}{24}\)
Solution:
(i) On dividing 15 by 3, we get 5.
Now, multiplying the numerator and denominator of \(\frac{3}{4}\) by 5, we get
\(\frac{3}{4}\) = \(\frac{3 \times 5}{4 \times 5}\) = \(\frac{15}{20}\)
So, we replace ☐ by 20.

(ii) On dividing 15 by 3, we get 5.
Now, multiplying the numerator and denominator of \(\frac{7}{3}\) by 5, we get
\(\frac{7}{3}\) = \(\frac{7 \times 5}{3 \times 5}\) = \(\frac{35}{15}\)
So, we replace ☐ by 35.

(iii) On dividing 30 by 10, we get 3.
Now, multiplying the numerator and denominator of \(\frac{10}{7}\) by 3, we get
\(\frac{10}{7}\) = \(\frac{10 \times 3}{7 \times 3}\) = \(\frac{30}{21}\)
So, we replace ☐ by 21.

(iv) On dividing 44 by 11, we get 4.
Now, multiplying the numerator and denominator of \(\frac{11}{15}\) by 4, we get
\(\frac{11}{15}\) = \(\frac{11 \times 4}{15 \times 4}\) = \(\frac{44}{60}\)
So, we replace ☐ by 60.

(v) On dividing 24 by 4, we get 6.
Now, multiplying the numerator and denominator of \(\frac{15}{4}\) by 6, we get
\(\frac{15}{4}\) = \(\frac{15 \times 6}{4 \times 6}\) = \(\frac{90}{24}\)
So, we replace ☐ by 90.

Question 7.
Find the fraction equivalent to \(\frac{18}{45}\), having:
(i) Numerator 50
(ii) Denominator 60
Solution:
We have, \(\frac{15}{45}\) = \(\frac{2 \times 9}{5 \times 9}\) = \(\frac{2}{5}\)
(i) On dividing 50 by 2, we get 25.
Now, multiplying the numerator and denominator of \(\frac{2}{5}\) by 25, we get
\(\frac{2}{5}\) = \(\frac{2 \times 25}{5 \times 25}\) = \(\frac{50}{125}\)
So, the fraction with numerator 50 and equivalent to \(\frac{18}{45}\) is \(\frac{50}{125}\).

(ii) On dividing 60 by 5, we get 12.
Now, multiplying the numerator and ‘ denominator of \(\frac{2}{5}\) by 12, we get
\(\frac{2}{5}\) = \(\frac{2 \times 12}{5 \times 12}\) = \(\frac{24}{60}\)
So, the fraction with denominator 60 and equivalent to\(\frac{18}{45}\) is \(\frac{24}{60}\).

Question 8.
Write the following fractions in the simplest form:
(i) \(\frac{45}{81}\)
(ii) \(\frac{65}{91}\)
(iii) \(\frac{441}{483}\)
Solution:
(i) We have, \(\frac{45}{81}\) = \(\frac{3 \times 3 \times 5}{3 \times 3 \times 3 \times 3}\) = \(\frac{5}{9}\)
(ii) We have, \(\frac{65}{91}\) = \(\frac{5 \times 13}{7 \times 13}\) = \(\frac{5}{7}\)
(iii) We have, \(\frac{441}{483}\) = \(\frac{3 \times 3 \times 7 \times 7}{3 \times 7 \times 23}\) = \(\frac{21}{23}\)

Question 9.
Arrange the following fractions in ascending order:
\(\frac{10}{3}\), \(\frac{21}{6}\), \(\frac{9}{2}\), \(\frac{13}{4}\), \(\frac{25}{8}\)
Solution:
Denominators of the given fractions are 3, 6, 2, 4 and 8.
The smallest common multiple of 3, 6, 2, 4 and 8 is 24.
Now, converting each fraction into equivalent fraction with 24 as its denominator, we get
\(\frac{10}{3}\) = \(\frac{10 \times 8}{3 \times 8}\) = \(\frac{80}{24}\)
\(\frac{21}{6}\) = \(\frac{21 \times 4}{6 \times 4}\) = \(\frac{84}{24}\)
\(\frac{9}{2}\) = \(\frac{9 \times 12}{2 \times 12}\) = \(\frac{108}{24}\)
\(\frac{13}{4}\) = \(\frac{13 \times 6}{4 \times 6}\) = \(\frac{78}{24}\)
\(\frac{25}{8}\) = \(\frac{25 \times 3}{8 \times 3}\) = \(\frac{75}{24}\)
We know, 75 < 78 < 80 < 84 < 108
⇒ \(\frac{75}{24}\) < \(\frac{78}{24}\) < \(\frac{80}{24}\) < \(\frac{84}{24}\) < \(\frac{108}{24}\)
⇒ \(\frac{25}{8}\) < \(\frac{13}{4}\) < \(\frac{10}{3}\) < \(\frac{21}{6}\) < \(\frac{9}{2}\)

Question 10.
There are 30 students in section A & 40 in section B of class VI. Among them, 25 students from section A 8c 34 from section B passed with distinction. Which section performed better?
Solution:
Here, we have to compare \(\frac{25}{30}\) and \(\frac{34}{40}\).
We can write 30 = 3 × 10 and 40 = 4 × 10.
The least common multiple of 30 and 40 is 3 × 4 × 10 = 120.
∴ \(\frac{25}{30}\) = \(\frac{25 \times 4}{30 \times 4}\) = \(\frac{100}{120}\) and \(\frac{34}{40}\) = \(\frac{34 \times 3}{40 \times 3}\) = \(\frac{102}{120}\)
We know, 100 < 102
⇒ \(\frac{100}{120}\) < \(\frac{102}{120}\)
⇒ \(\frac{25}{30}\) < \(\frac{34}{40}\)
So, section B performed better than section A.

Question 11.
The refractive index of stone A and stone B are \(\frac{121}{50}\) and \(\frac{58}{25}\) respectively. Which stone has greater refractive index?
Solution:
Here, we have to compare \(\frac{121}{50}\) and \(\frac{58}{25}\).
The least common multiple of 50 and 25 is 50.
∴ \(\frac{58}{25}\) = \(\frac{58 \times 2}{25 \times 2}\) = \(\frac{116}{50}\)
We know, 121 > 116
⇒ \(\frac{121}{50}\) > \(\frac{116}{50}\)
⇒ \(\frac{151}{50}\) > \(\frac{58}{25}\)
So, the refractive index of stone A is greater than the refractine index of stone B.

Question 12.
Solve the following:
(i) \(\frac{2}{9}\) + \(\frac{3}{9}\)
(ii) \(\frac{1}{7}\) + \(\frac{3}{7}\) + \(\frac{12}{7}\)
(iii) \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{2}{5}\) + \(\frac{3}{5}\)
(iv) 2\(\frac{3}{5}\) + \(\frac{2}{5}\) + 3\(\frac{1}{5}\)
Solution:
(i) \(\frac{2}{9}\) + \(\frac{3}{9}\)
= \(\frac{2+3}{9}\) = \(\frac{5}{9}\)

(ii) \(\frac{1}{7}\) + \(\frac{3}{7}\) + \(\frac{12}{7}\)
= \(\frac{1+3+12}{7}\) = \(\frac{16}{7}\)

(iii) \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{2}{5}\) + \(\frac{3}{5}\)
= (\(\frac{1+3}{4}\)) + (\(\frac{2+3}{5}\))
= \(\frac{4}{4}\) + \(\frac{5}{5}\) = 1 + 1 = 2

(iv) 2\(\frac{3}{5}\) + \(\frac{2}{5}\) + 3\(\frac{1}{5}\)
= (\(\frac{2 \times 5+3}{5}\)) + \(\frac{1}{5}\) + (\(\frac{3 \times 5+1}{5}\))
= (\(\frac{10+3}{5}\)) + \(\frac{2}{5}\) + (\(\frac{15+1}{5}\))
= \(\frac{13}{5}\) + \(\frac{2}{5}\) + \(\frac{16}{5}\)
= \(\frac{13+2+16}{5}\) = \(\frac{31}{5}\)

Question 13.
Simplify the following:
(i) 5\(\frac{1}{4}\) + \(\frac{3}{4}\) – 2\(\frac{1}{4}\)
(ii) 7\(\frac{1}{7}\) – \(\frac{13}{7}\) + 2\(\frac{2}{7}\)
Solution:
(i) 5\(\frac{1}{4}\) + \(\frac{3}{4}\) – 2\(\frac{1}{4}\) = (\(\frac{5 \times 4+2}{4}\)) + \(\frac{3}{4}\) – (\(\frac{2 \times 4+1}{4}\))
= (\(\frac{20+2}{4}\)) + \(\frac{3}{4}\) – (\(\frac{8+1}{4}\))
= \(\frac{22}{4}\) + \(\frac{3}{4}\) – \(\frac{9}{4}\)
= \(\frac{22+3-9}{4}\) = \(\frac{16}{4}\) = 4

(ii) 7\(\frac{1}{7}\) – \(\frac{13}{7}\) + 2\(\frac{2}{7}\)
= (\(\frac{7 \times 7+1}{7}\)) – \(\frac{13}{7}\) – (\(\frac{2 \times 7+2}{7}\))
= (\(\frac{49+1}{7}\)) – \(\frac{13}{7}\) – (\(\frac{14+2}{4}\))
= \(\frac{50}{7}\) – \(\frac{13}{7}\) + \(\frac{16}{7}\)
= \(\frac{50-13+16}{7}\) = \(\frac{53}{7}\)

Question 14.
Shikha ate \(\frac{1}{5}\) of the pizza and her friend Sanvi ate \(\frac{3}{5}\) of the pizza. Did they eat whole of the pizza? If not, then what fraction of the pizza is left?
Solution:
Fraction of pizza eaten by Shikha = \(\frac{1}{5}\)
Fraction of pizza eaten by Sanvi = \(\frac{3}{5}\)
Total pizza eaten by both Shikha and Sanvi
= \(\frac{1}{5}\) + \(\frac{3}{5}\) = \(\frac{4}{5}\) < 1 Fraction representing the remaining pizza = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{5-4}{5}\) = \(\frac{1}{5}\) So, \(\frac{1}{5}\) of the pizza is left.

Question 15.
Fill in the missing fractions:
(i) \(\frac{3}{8}\) + ☐ = \(\frac{5}{8}\)
(ii) \(\frac{4}{9}\) – ☐ = \(\frac{2}{9}\)
(iii) \(\frac{8}{15}\) – ☐ = \(\frac{1}{5}\)
(iv) ☐ – \(\frac{4}{10}\) = \(\frac{3}{10}\)
Solution:
Let x be the missing fraction.
(i) \(\frac{3}{8}\) + x = \(\frac{5}{8}\)
⇒ x = \(\frac{5}{8}\) – \(\frac{3}{8}\)
⇒ x = \(\frac{5-3}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)

(ii) \(\frac{4}{9}\) – x = \(\frac{2}{9}\)
⇒ \(\frac{4}{9}\) – \(\frac{2}{9}\) = x
⇒ x = \(\frac{4-2}{9}\)
⇒ x = \(\frac{2}{9}\)

(iii) \(\frac{8}{15}\) – x = \(\frac{1}{5}\)
⇒ \(\frac{8}{15}\) – \(\frac{1}{5}\) = x
⇒ \(\frac{8}{15}\) – \(\frac{3}{5}\) = x
⇒ x = \(\frac{8-3}{15}\)
⇒ x = \(\frac{5}{15}\) ⇒ x = \(\frac{1}{3}\)

(iv) x – \(\frac{4}{10}\) = \(\frac{3}{10}\)
⇒ x = \(\frac{3}{10}\) + \(\frac{4}{10}\)
⇒ x = \(\frac{3+4}{10}\)
⇒ x = \(\frac{7}{10}\)

Question 16.
Arrange the following fractions in descending order:
\(\frac{28}{9}\), \(\frac{55}{18}\), \(\frac{37}{12}\), 3, \(\frac{31}{4}\)
Solution:
Denonimators of the given fractions are 9, 18, 12, 1 and 4.
The smallest common multiple of 9, 18, 12, 1 and 4 is 36.
Now, converting each fraction into equivalent fraction with 36 as its denominator, we get
\(\frac{28}{9}\) = \(\frac{28 \times 4}{9 \times 4}\) = \(\frac{112}{36}\),
\(\frac{55}{18}\) = \(\frac{55 \times 2}{18 \times 2}\) = \(\frac{110}{36}\),
\(\frac{37}{12}\) = \(\frac{37 \times 3}{12 \times 3}\) = \(\frac{11}{36}\),
\(\frac{3}{1}\) = \(\frac{3 \times 36}{1 \times 36}\) = \(\frac{108}{36}\),
\(\frac{31}{4}\) = \(\frac{31 \times 9}{4 \times 9}\) = \(\frac{279}{36}\)
We know, 279 > 112 > 111 > 110 > 108
⇒ \(\frac{279}{36}\) > \(\frac{112}{36}\) > \(\frac{111}{36}\) > \(\frac{110}{36}\) > \(\frac{108}{36}\)
⇒ \(\frac{31}{4}\) > \(\frac{28}{9}\) > \(\frac{37}{12}\) > \(\frac{55}{18}\) > 3

Question 17.
Rahul scored 54 out of 75 marks, while Swati scored 92 out of 125. Who performed better?
Solution:
Here, we have to compare \(\frac{54}{75}\) and \(\frac{92}{125}\).
We can write, 75 = 3 × 25 and 125 = 5 × 25.
Now, the least common multiple of 75 and 125 is 3 × 5 × 25 = 375.
∴ \(\frac{54}{75}\) = \(\frac{54 \times 5}{75 \times 5}\) = \(\frac{270}{375}\) and \(\frac{92}{125}\) = \(\frac{92 \times 3}{125 \times 3}\) = \(\frac{276}{375}\)
We know 270 < 276
⇒ \(\frac{270}{375}\) < \(\frac{276}{375}\)
⇒ \(\frac{54}{75}\) < \(\frac{92}{125}\)
So, Swati performed better than Rahul.

Question 18.
The distance from Delhi to Gurugram is \(\frac{310}{15}\)km, while the distance from Delhi to Noida is \(\frac{415}{20}\) km. Which city, Gurugram or Noida, is
closer to Delhi?
Solution:
Here, we have to compare \(\frac{310}{15}\) and \(\frac{415}{20}\).
We can write, 15 = 3 × 5 and 20 = 4 × 5.
The least common multiple of 15 and 20 is 3 × 4 × 5 = 60.
∴ \(\frac{310}{15}\) = \(\frac{310 \times 4}{15 \times 4}\) = \(\frac{1240}{60}\) and \(\frac{415}{20}\) = \(\frac{415 \times 3}{20 \times 3}\) = \(\frac{1245}{60}\)
We know 1240 < 1245
⇒ \(\frac{1240}{60}\) < \(\frac{1245}{60}\)
⇒ \(\frac{310}{15}\) < \(\frac{415}{20}\)
So, Gurugram is closer to Delhi.

Question 19.
Solve the following:
(i) \(\frac{5}{11}\) + \(\frac{13}{11}\)
(ii) \(\frac{3}{12}\) + \(\frac{4}{12}\) + \(\frac{7}{12}\)
(iii) \(\frac{1}{6}\) + \(\frac{2}{6}\) + \(\frac{3}{6}\)
(iv) 4\(\frac{3}{15}\) + \(\frac{11}{15}\) + 5\(\frac{7}{15}\)
Solution:
(i) \(\frac{5}{11}\) + \(\frac{13}{11}\) = \(\frac{5+13}{11}\) = \(\frac{18}{11}\)

(ii) \(\frac{3}{12}\) + \(\frac{4}{12}\) + \(\frac{7}{12}\)
= \(\frac{3+4+7}{12}\) = \(\frac{14}{12}\)
= \(\frac{7}{6}\)

(iii) \(\frac{1}{6}\) + \(\frac{2}{6}\) + \(\frac{3}{6}\)
= \(\frac{1+2+3}{6}\) = \(\frac{6}{6}\)
= 1

(iv) 4\(\frac{3}{15}\) + \(\frac{11}{15}\) + 5\(\frac{7}{15}\)
= (\(\frac{4 \times 15+3}{15}\)) + \(\frac{11}{15}\) + (\(\frac{5 \times 15+7}{15}\))
= (\(\frac{60+3}{11}\)) + \(\frac{11}{15}\) + (\(\frac{75+7}{15}\))
= \(\frac{63}{15}\) + \(\frac{11}{15}\) + \(\frac{82}{15}\) = \(\frac{63+11+82}{15}\)
= \(\frac{156}{15}\) = \(\frac{3 \times 52}{3 \times 5}\)
= \(\frac{52}{5}\)

Question 20.
Add the following fractions:
(i) \(\frac{3}{4}\) and \(\frac{4}{3}\)
(ii) \(\frac{7}{4}\), \(\frac{2}{3}\) and \(\frac{1}{5}\)
Solution:
(i) LCM of denominators i.e., 4 and 3 is 12.
∴ \(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 3}\) = \(\frac{9}{12}\),
\(\frac{4}{3}\) = \(\frac{4 \times 4}{3 \times 4}\) = \(\frac{16}{12}\)
Now, \(\frac{3}{4}\) + \(\frac{4}{3}\) = \(\frac{9}{12}\) + \(\frac{16}{12}\)
= \(\frac{9+16}{12}\) = \(\frac{25}{12}\)

(ii) LCM of denominators i.e., 4, 3 and 5 is 60.
∴ \(\frac{7}{4}\) = \(\frac{7 \times 15}{4 \times 15}\) = \(\frac{105}{60}\),
\(\frac{2}{3}\) = \(\frac{2 \times 20}{3 \times 20}\) = \(\frac{40}{60}\),
\(\frac{1}{5}\) = \(\frac{1 \times 12}{5 \times 12}\) = \(\frac{12}{60}\)
Now, \(\frac{7}{4}\) + \(\frac{2}{3}\) + \(\frac{1}{5}\) = \(\frac{105}{60}\) + \(\frac{40}{60}\) + \(\frac{12}{60}\)
= \(\frac{105+40+12}{60}\) = \(\frac{157}{60}\)

Question 21.
Simplify: 4\(\frac{1}{5}\) – 3\(\frac{2}{3}\)
Solution:
4\(\frac{1}{5}\) – 3\(\frac{2}{3}\) = \(\frac{4 \times 5+1}{5}\) – \(\frac{3 \times 3+2}{3}\)
= \(\frac{20+1}{5}\) – \(\frac{9+2}{3}\) = \(\frac{21}{5}\) – \(\frac{11}{3}\)
L.C.M of denominator i.e. 5 and 3 is 15.
∴ \(\frac{21}{5}\) = \(\frac{21 \times 3}{5 \times 3}\) = \(\frac{63}{15}\),
\(\frac{11}{3}\) = \(\frac{11 \times 5}{3 \times 5}\) = \(\frac{55}{15}\)
Now, 4\(\frac{1}{5}\) – 3\(\frac{2}{3}\) = \(\frac{21}{5}\) – \(\frac{11}{3}\) = \(\frac{63}{15}\) – \(\frac{55}{15}\)
= \(\frac{63-55}{15}\) = \(\frac{8}{15}\)

Question 22.
Rahul, Ashok and Anshika buy a toy. Rahul gives \(\frac{3}{10}\) of the total cost, Anshika gives \(\frac{5}{10}\) of the total cost, and the remaining amount is paid by Ashok. What fraction of the total cost is paid by Ashok?
Solution:
Rahul’s share of total cost = \(\frac{3}{10}\)
Anshika’s share of total cost = \(\frac{5}{10}\)
Total share of Rahul and Anshika
= \(\frac{3}{10}\) + \(\frac{5}{10}\) = \(\frac{3+5}{10}\) = \(\frac{8}{10}\)
∴ Ashok’s share of total cost
= 1 – \(\frac{8}{10}\) = \(\frac{10}{10}\) – \(\frac{8}{10}\) = \(\frac{10-8}{10}\)
= \(\frac{2}{10}\) = \(\frac{2}{2 \times 5}\) = \(\frac{1}{5}\)

Question 23.
Find the difference of \(\frac{17}{12}\) and \(\frac{11}{18}\).
Solution:
We can write 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3.
So, LCM of 12 and 18 is 2 × 2 × 3 × 3 = 36.
∴ \(\frac{17}{12}\) = \(\frac{17 \times 3}{12 \times 3}\) = \(\frac{51}{36}\) and
\(\frac{11}{18}\) = \(\frac{11 \times 2}{18 \times 2}\) = \(\frac{22}{36}\)
We know, 51 > 22
⇒ \(\frac{51}{36}\) > \(\frac{22}{36}\)
⇒ \(\frac{17}{12}\) > \(\frac{11}{18}\)
Thus, the required difference
= \(\frac{17}{12}\) – \(\frac{11}{18}\) = \(\frac{51}{36}\) – \(\frac{22}{36}\)
= \(\frac{51-22}{36}\) = \(\frac{29}{36}\)

Fractions Class 6 Long Question Answer

Question 1.
Write a fraction to represent the shaded part in each of the following diagrams:

Solution:
(i) Number of equal parts or fractional units = 7,
Number of shaded parts = 3
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{3}{7}\)

(ii) Number of equal parts or fractional units = 9,
Number of shaded parts = 4
So, the required fraction .
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{4}{9}\)

(iii) Number of equal parts or fractional units = 8,
Number of shaded parts = 6
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{6}{8}\) = \(\frac{6 \div 2}{8 \div 2}\) = \(\frac{3}{4}\)

(iv) Number of equal parts or fractional units = 7,
Number of shaded parts = 4
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{4}{7}\)

Question 2.
Write a fraction to represent the shaded part in each of the following diagrams:

Solution:
(i) Number of equal parts or fractional units = 12,
Number of shaded parts = 8
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{8}{12}\) = \(\frac{2 \times 4}{3 \times 4}\) = \(\frac{2}{3}\)

(ii) Number of equal parts or fractional units = 16,
Number of shaded parts = 8
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{8}{16}\) = \(\frac{8 \times 1}{8 \times 2}\) = \(\frac{1}{2}\)

(iii) Number of equal parts or fractional units = 6,
Number of shaded parts = 2
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\)
= \(\frac{2}{6}\) = \(\frac{2 \times 1}{2 \times 3}\) = \(\frac{1}{3}\)

(iv) Number of equal parts or fractional units = 8,
Number of shaded parts = 3
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{3}{8}\)

(v) Number of equal parts or fractional units = 8,
Number of shaded parts = 5
So, the required fraction
= \(\frac{\text { Number of shaded parts }}{\text { Total number of fractional units }}\) = \(\frac{5}{8}\)

Question 3.
Compare the following fractions:
(i) \(\frac{7}{3}\) and \(\frac{11}{5}\)
(ii) \(\frac{4}{7}\) and \(\frac{5}{9}\)
(iii) \(\frac{26}{14}\) and \(\frac{40}{21}\)
(iv) \(\frac{13}{7}\) and \(\frac{23}{11}\)
Solution:
(i) 3 × 5 = 15 is a common multiple of 3 and 5.
∴ \(\frac{7}{3}\) = \(\frac{7 \times 5}{3 \times 5}\) = \(\frac{35}{15}\) and
\(\frac{11}{5}\) = \(\frac{11 \times 3}{5 \times 3}\) = \(\frac{33}{15}\)
We know, 35 > 33
⇒ \(\frac{35}{15}\) > \(\frac{35}{15}\)
⇒ \(\frac{7}{3}\) > \(\frac{11}{5}\)

(ii) 7 × 9 = 63 is a common multiple of 7 and 9.
∴ \(\frac{4}{7}\) = \(\frac{4 \times 9}{7 \times 9}\) = \(\frac{36}{63}\) and
\(\frac{5}{9}\) = \(\frac{5 \times 7}{9 \times 7}\) = \(\frac{35}{63}\)
We know, 35 > 33
⇒ \(\frac{36}{63}\) > \(\frac{35}{63}\)
⇒ \(\frac{4}{7}\) > \(\frac{5}{9}\)

(iii) We can write \(\frac{26}{14}\) = \(\frac{13}{7}\).
21 is a common multiple of 7 and 21.
∴ \(\frac{13}{7}\) = \(\frac{13 \times 3}{7 \times 3}\) = \(\frac{39}{21}\)
We know, 39 < 40
⇒ \(\frac{39}{21}\) > \(\frac{40}{21}\)
⇒ \(\frac{13}{7}\) > \(\frac{40}{21}\)
⇒ \(\frac{26}{14}\) > \(\frac{40}{21}\)

(iv) 7 × 11 = 77 is a common multiple of 7 and 11.
∴ \(\frac{13}{7}\) = \(\frac{13 \times 11}{7 \times 11}\) = \(\frac{143}{77}\) and
\(\frac{23}{11}\) = \(\frac{23 \times 7}{11 \times 7}\) = \(\frac{161}{77}\)
We know, 143 < 161
⇒ \(\frac{143}{77}\) > \(\frac{161}{77}\)
⇒ \(\frac{13}{7}\) > \(\frac{23}{11}\)

Question 4.
Compare the following fractions:
(i) \(\frac{17}{12}\) and \(\frac{11}{6}\)
(ii) \(\frac{13}{5}\) and \(\frac{23}{10}\)
(iii) \(\frac{6}{5}\) and \(\frac{9}{8}\)
(iv) \(\frac{23}{8}\) and \(\frac{17}{6}\)
Solution:
(i) 12 is a common multiple of 12 and 6.
∴ \(\frac{11}{6}\) = \(\frac{11 \times 2}{6 \times 2}\) = \(\frac{22}{12}\)
We know, 17 < 22
⇒ \(\frac{17}{12}\) > \(\frac{22}{12}\)
⇒ \(\frac{17}{12}\) > \(\frac{11}{6}\)

(ii) 10 is a common multiple of 5 and 10.
∴ \(\frac{13}{5}\) = \(\frac{13 \times 2}{5 \times 2}\) = \(\frac{26}{10}\)
We know, 26 < 23
⇒ \(\frac{26}{10}\) > \(\frac{23}{10}\)
⇒ \(\frac{13}{5}\) > \(\frac{23}{10}\)

(iii) 5 × 8 = 40 is a common multiple of 5 and 8.
∴ \(\frac{6}{5}\) = \(\frac{6 \times 8}{5 \times 8}\) = \(\frac{48}{40}\)
\(\frac{9}{8}\) = \(\frac{9 \times 5}{8 \times 5}\) = \(\frac{45}{40}\)
We know, 48 < 4
⇒ \(\frac{48}{40}\) > \(\frac{45}{40}\)
⇒ \(\frac{6}{5}\) > \(\frac{9}{8}\)

(iv) 24 is a common multiple of 8 and 6.
∴ \(\frac{23}{8}\) = \(\frac{23 \times 3}{8 \times 3}\) = \(\frac{69}{24}\) and
\(\frac{17}{6}\) = \(\frac{17 \times 4}{6 \times 4}\) = \(\frac{68}{24}\)
We know, 69 < 68
⇒ \(\frac{69}{24}\) > \(\frac{68}{24}\)
⇒ \(\frac{23}{8}\) > \(\frac{17}{6}\)

Question 5.
Simplify the following:
(i) 3\(\frac{7}{10}\) + \(\frac{3}{10}\) – 2\(\frac{4}{10}\)
(ii) 5\(\frac{1}{6}\) – 3\(\frac{5}{6}\) + 1\(\frac{3}{6}\)
(iii) 4\(\frac{2}{11}\) – 3\(\frac{4}{11}\) + \(\frac{10}{11}\)
(iv) 6\(\frac{3}{13}\) – 2\(\frac{4}{13}\) – 1\(\frac{9}{13}\)
Solution:
(i) 3\(\frac{7}{10}\) + \(\frac{3}{10}\) – 2\(\frac{4}{10}\)
= \(\left(\frac{3 \times 10+7}{10}\right)\) + \(\frac{3}{10}\) – \(\left(\frac{2 \times 10+4}{10}\right)\)
= \(\left(\frac{30+7}{10}\right)\) + \(\frac{3}{10}\) – \(\left(\frac{20+4}{10}\right)\)
= \(\frac{37}{10}\) + \(\frac{3}{10}\) – \(\frac{24}{10}\)
= \(\frac{37+3-24}{10}\) = \(\frac{16}{10}\) = \(\frac{2 \times 8}{2 \times 5}\)
= \(\frac{8}{5}\)

(ii) 5\(\frac{1}{6}\) – 3\(\frac{5}{6}\) + 1\(\frac{3}{6}\)
= \(\left(\frac{5 \times 6+1}{6}\right)\) – \(\left(\frac{3 \times 6+5}{6}\right)\) + \(\left(\frac{1 \times 6+3}{6}\right)\)
= \(\left(\frac{30+1}{6}\right)\) – \(\left(\frac{18+5}{6}\right)\) + \(\left(\frac{6+3}{6}\right)\)
= \(\frac{31}{6}\) – \(\frac{23}{6}\) + \(\frac{9}{6}\)
= \(\frac{31-23+9}{6}\)
= \(\frac{17}{6}\)

(iii) 4\(\frac{2}{11}\) – 3\(\frac{4}{11}\) + \(\frac{10}{11}\)
= \(\left(\frac{4 \times 11+2}{11}\right)\) – \(\left(\frac{3 \times 11+4}{11}\right)\) + \(\frac{10}{11}\)
= \(\left(\frac{44+2}{11}\right)\) – \(\left(\frac{33+4}{11}\right)\) + \(\frac{10}{11}\)
= \(\frac{46}{11}\) – \(\frac{37}{11}\) + \(\frac{10}{11}\)
= \(\frac{46-37+10}{11}\)
= \(\frac{19}{11}\)

(iv) 6\(\frac{3}{13}\) – 2\(\frac{4}{13}\) – 1\(\frac{9}{13}\)
= \(\left(\frac{6 \times 13+3}{13}\right)\) – \(\left(\frac{2 \times 13+4}{13}\right)\) – \(\left(\frac{1 \times 13+9}{13}\right)\)
= \(\left(\frac{78+3}{13}\right)\) – \(\left(\frac{26+4}{13}\right)\) – \(\left(\frac{13+9}{13}\right)\)
= \(\frac{81}{13}\) – \(\frac{30}{13}\) – \(\frac{22}{13}\)
= \(\frac{81-30-22}{13}\)
= \(\frac{29}{13}\)

Question 6.
Simplify the following:
(i) \(\frac{3}{11}\) – \(\frac{7}{9}\) + \(\frac{13}{3}\)
(ii) 5\(\frac{2}{3}\) – 2\(\frac{1}{4}\) + 3\(\frac{1}{6}\)
(iii) 1\(\frac{5}{7}\) – \(\frac{5}{9}\) + 6
(iv) 8\(\frac{1}{4}\) – 2\(\frac{5}{6}\) + 1\(\frac{2}{3}\)
Solution:
(i) \(\left(\frac{3 \times 9}{11 \times 9}\right)\) – \(\left(\frac{7\times 11}{9 \times 11}\right)\) + \(\left(\frac{13 \times 33}{3 \times 33}\right)\)
= \(\frac{29}{13}\) – \(\frac{29}{13}\) + \(\frac{29}{13}\)
[∵ L.C.M of 11, 9 and 3 is 99.]
= \(\frac{27}{99}\) – \(\frac{77}{99}\) + \(\frac{429}{99}\)
= \(\frac{29-77+429}{99}\)
= \(\frac{379}{99}\)

(ii) 5\(\frac{2}{3}\) – 2\(\frac{1}{4}\) + 3\(\frac{1}{6}\)
= \(\left(\frac{5 \times 3+2}{3}\right)\) – \(\left(\frac{2 \times 4+1}{4}\right)\) + \(\left(\frac{3 \times 6+1}{6}\right)\)
= \(\left(\frac{15+2}{3}\right)\) – \(\left(\frac{8+1}{4}\right)\) + \(\left(\frac{18+1}{6}\right)\)
= \(\frac{17}{3}\) – \(\frac{9}{4}\) + \(\frac{19}{6}\)
= \(\left(\frac{17 \times 4}{3 \times 4}\right)\) – \(\left(\frac{9 \times 3}{4 \times 3}\right)\) + \(\left(\frac{19 \times 2}{6 \times 2}\right)\)
[∵ LCM of 3, 4 and 6 is 12.]
= \(\frac{68}{12}\) – \(\frac{27}{12}\) + \(\frac{38}{12}\)
= \(\frac{68-27+38}{12}\) = \(\frac{79}{12}\)

(iii) 1\(\frac{5}{7}\) – \(\frac{5}{9}\) + 6
= \(\left(\frac{1 \times 7+5}{7}\right)\) – \(\frac{5}{9}\) + \(\frac{6}{1}\)
= \(\left(\frac{7+5}{7}\right)\) – \(\frac{5}{9}\) + \(\frac{6}{1}\)
= \(\frac{12}{7}\) – \(\frac{5}{9}\) + \(\frac{6}{1}\)
= \(\left(\frac{12 \times 9}{7 \times 9}\right)\) – \(\left(\frac{5 \times 7}{9 \times 7}\right)\) + \(\left(\frac{6 \times 63}{1 \times 63}\right)\)
[∵ LCM of 7, 9 and 1 is 63.]
= \(\frac{108}{63}\) – \(\frac{35}{63}\) + \(\frac{378}{63}\)
= \(\frac{108-35+378}{63}\) = \(\frac{451}{63}\)

(iv) 8\(\frac{1}{4}\) – 2\(\frac{5}{6}\) + 1\(\frac{2}{3}\)
= \(\left(\frac{8 \times 4+1}{4}\right)\) – \(\left(\frac{2 \times 6+5}{6}\right)\) + \(\left(\frac{1 \times 3+2}{3}\right)\)
= \(\left(\frac{32+1}{4}\right)\) – \(\left(\frac{12+5}{6}\right)\) + \(\left(\frac{3+2}{3}\right)\)
= \(\frac{33}{4}\) – \(\frac{17}{6}\) + \(\frac{5}{3}\)
= \(\frac{33 \times 3}{4 \times 3}\) – \(\frac{17 \times 2}{6 \times 2}\) + \(\frac{5 \times 4}{3 \times 4}\)
[∵ LCM of 4, 6 and 3 is 12.]
= \(\frac{99}{12}\) – \(\frac{34}{12}\) + \(\frac{20}{12}\)
= \(\frac{99-34+20}{12}\) = \(\frac{85}{12}\)

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 6 Constructions and Tilings MCQ improves accuracy in objective exams.

MCQ on Constructions and Tilings Class 7

Constructions and Tilings MCQ Class 7

Class 7 Maths Constructions and Tilings MCQ

Question 1.
If we bisect a line segment AB of 9 cm, then what will be the length of each part?
(a) 5 cm
(b) 4 cm
(c) 4.5 cm
(d) 3 cm
Solution:
(c) 4.5 cm
When we bisect a line segment, the length of each part is half of the length of given line segment.
Given, length of the line segment AB = 9 cm After bisecting AB,
Length of each part = (Length of the line segment AB) ÷ 2 = 9 cm ÷ 2 = 4.5 cm.

Question 2.
Two lines are said to be perpendicular if angle between them is:
(a) 45°
(b) 90°
(c) 60°
(d) 30°
Solution:
(b) 90°
We know, two lines are said to be perpendicular if angle between them is 90°.

Question 3.
To construct the perpendicular bisector of a line segment PQ, the compass opening must be:
(a) smaller than half of PQ
(b) more than half of PQ
(c) equal to the length of PQ
(d) equal to half of PQ
Solution:
(b) more than half of PQ
To construct the perpendicular bisector of a line segment PQ, the compass opening must be more than half of PQuestion

Question 4.
If we bisect an angle of a rectangle, we get two angles each of:
(a) 45°
(b) 30°
(c) 60°
(d) 15°
Solution:
(a) 45°
We know that the measure of each interior angle of a rectangle is 90°.
Therefore, if we bisect an angle of a rectangle, we get two angles each of 45°.

Question 5.
What angle would you get if you bisect an exterior angle of an equilateral triangle?
(a) 90°
(b) 60°
(c) 45°
(d) 30°
Solution:
(b) 60°
We know that at each vertex of an equilateral triangle, the measure of exterior angle is 120°.

Therefore, if we bisect an exterior angle of an equilateral triangle, we get two angles measuring 60° each.

Question 6.
If YA = AB and AC = BC, then measure of ∠XYC equals:

(a) 60°
(b) 45°
(c) 30°
(d) None of these
Solution:
(c) 30°
Here, the arc from Y cuts YZ and YX at points A and B respectively and YD bisects ∠XYZ.
As arc radius is same, YA = YB.
∴ YA = YB = AB [Given, YA = AB]
Thus, ∆ABY is an equilateral triangle.
∴ ∠AYB = 60° ⇒ ∠XYC = 60° ÷ 2 = 30°

Question 7.
While constructing the perpendicular bisector of a line segment PQ, if the compass opening is exactly half of PQ, then drawn arcs from P and Q:
(a) do not intersect
(b) intersect at one point
(c) intersect at two points
(d) intersect at four points
Solution:
(b) intersect at one point
While constructing the perpendicular bisector of a line segment PQ, if the compass opening is exactly half of PQ, then drawn arcs from P and Q intersect at one point only.

Question 8.
Which of the following shapes can tile a plane on its own without leaving any gaps or overlaps?
(a) Circle
(b) Oval
(c) Square
(d) Regular pentagon
Solution:
(c) Square
A regular polygon can tile the plane only if its interior angle divides 360° exactly.
Circles and ovals leave gaps, and regular pentagons also fail to tessellate. Squares, however, fit together perfectly and cover the plane without gaps or overlaps.

Question 9.
A 5 × 5 square grid consists of 25 unit squares. Can it be completely covered using 2 × 1 dominoes?
(a) Yes, because 25 squares can be covered by dominoes in different ways.
(b) Yes, because dominoes can be placed both horizontally and vertically.
(c) No, because each domino covers 2 squares, and 25 is an odd number.
(d) None of the above
Solution:
(c) No, because each domino covers 2 squares, and 25 is an odd number.
Each 2 × 1 domino covers exactly 2 unit squares. To completely cover a grid, the total number of squares must be divisible by 2.

Here, the 5 × 5 grid has 25 unit squares. Since 25 is odd, it is not divisible by 2, so it is impossible to cover the entire grid with 2 × 1 dominoes.

Question 10.
Which of the following shapes does not tessellate (tile) a plane on its own?
(a) Equilateral triangle
(b) Square
(c) Hexagon
(d) Circle
Solution:
(d) Circle
Among the given shapes, only equilateral triangles, squares, and regular hexagons can tile the plane by themselves as their interior angle divides 360° exactly.

Question 11.
In a tangram set, which piece has all its sides equal in length?
(a) Square
(b) Parallelogram
(c) Small triangle
(d) Large triangle
Solution:
(a) Square
The tangram set consists of different shapes such as triangles, a square, and a parallelogram. A square has four sides of equal length.

A parallelogram has only opposite sides equal. The small triangles and the large triangles are right-angled isosceles triangles, which have only two sides equal, not all.

Question 12.
The tiling pattern used by bees in honeycombs is based on which of the following shapes?
(a) Squares
(b) Triangles
(c) Hexagons
(d) Pentagons
Solution:
(c) Hexagons
Bees construct honeycombs using hexagonal cells.

Constructions and Tilings Class 7 Fill in the Blanks

Question 1.
If two line segments cut each other at right angles, then they are _________ .
Solution: perpendicular
If two line segments cut each other at right angles, then they are perpendicular.

Question 2.
A straight angle can be bisected into _______ right angles.
Solution: two
We know that a straight angle measures 180°.
When a straight angle is bisected, it is divided into two equal angles, each measuring 90°.
Thus, a straight angle can be bisected into two right angles.

Question 3.
If ray OR bisects ∠POQ of measure 85°, the then they are measure of ∠POR is _______ .
Solution: 42.5°
Given, ∠POQ = 85°
And, ray OR bisects ∠POQ.
Therefore, ∠POR = ∠POQ ÷ 2 = 85° ÷ 2 = 42.5°

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 5 Connecting the Dots MCQ improves accuracy in objective exams.

MCQ on Connecting the Dots Class 7

Connecting the Dots MCQ Class 7

Class 7 Maths Connecting the Dots MCQ

Question 1.
The daily temperatures (in °C) for a week are 30, 27, 29, 35, 34, 32, 30. What is the mean temperature?
(a) 31°C
(b) 32°C
(c) 30°C
(d) 29.5°C
Solution:
(a) 31°C
Given, daily temperatures (in °C) for a week are: 30, 27, 29, 35, 34, 32, 30
We know, Mean = \(\frac{Sum of observations}{Number of observations}\)
∴ Mean temperature (in °C)
= \(\frac{30+27+29+35+34+32+30}{7}\)
= \(\frac{217}{7}\) = 31

Question 2.
The mean of the first 15 natural numbers is:
(a) 5
(b) 7.5
(c) 8
(d) 9.5
Solution:
(c) 8
Sum of first 15 natural numbers
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 +10 + 11 + 12 + 13 + 14 + 15 = 120
Mean of first 15 natural numbers
= \(\frac{Sum of first 15 natural numbers}{15}\)
= \(\frac{120}{15}\) = 8

Question 3.
The following dot plot shows the hours of sleep taken by a child on different days.

The range of the number of hours of sleep is:
(a) 4
(b) 8
(c) 5
(d) 10
Solution:
(c) 5
We know, range of data = Highest value – Lowest value
In the given dot plot, minimum hours
= 5 and maximum hours = 10
∴ Range = Maximum hours – Minimum hours 10 – 5 = 5

Question 4.
The table shows the number of flowers that bloom in a garden in 7 days.

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
No. of flowers	5	4	6	7	3	5	2

What is the median number of flowers?
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(d) 5
Given, the number of flowers that bloom in garden for 7 days are: 5, 4, 6, 7, 3, 5, 2
Arranging the data in ascending order, we get: 2, 3, 4, 5, 5, 6, 7
Here, number of observations, n = 7, is odd.
∴ Median = \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= \(\left(\frac{7+1}{2}\right)^{\text {th }}\) observation = 4^th observation
⇒ Median = 5

Question 5.
The difference between mean and median of the given dot plot showing number of trips taken by different families during an year, is:

(a) 0.33
(b) 3.33
(c) 0.0033
(d) None of these
Solution:
(b) 3.33
Writing down the data from the given dot plot, we get the values:
10, 10, 20, 20, 30, 30, 30, 30, 40, 40, 40, 40, 50, 50, 60
Total number of observations (n) =15 For mean,
Sum = (10 × 2) + (20 × 2) + (30 × 4) + (40 × 4) + (50 × 2) + (60 × 1)
= 20 + 40 + 120 + 160 + 100 + 60 = 500
Mean = \(\frac{500}{15}\) = 33.33
For median, (data is already in ascending order)
Since n = 15 is odd, median = \(\left(\frac{15+1}{2}\right)^{\text {th }}\) = 8^th value of the data
The 8^th value of the data is 30.
∴ Median = 30
Now, Difference = mean – median = 33.33 – 30 = 3.33]

Question 6.
Mean of 10 numbers is 20 and the mean of 5 numbers is 8. Then, the mean of all the 15 numbers is:
(a) 10
(b) 15
(c) 16
(d) 17
Solution:
(c) 16
Given, mean of 10 numbers = 20
∴ Sum of 10 numbers = Mean × 10
= 20 × 10 = 200
Also, mean of 5 numbers = 8
∴ Sum of 5 numbers = Mean × 5
= 8 × 5 = 40
Now, sum of all 15 numbers = Sum of 10 numbers + Sum of 5 numbers
= 200 + 40 = 240
∴ Mean of all 15 numbers
= \(\frac{Sum of 15 numbers}{15}\) = \(\frac{240}{15}\) = 16

Question 7.
If the median of \(\frac{x}{5}, x, \frac{x}{4}, \frac{x}{2} \text { and } \frac{x}{3}\) (where x > 0) is 8, then the value of x would be:
(a) 24
(b) 8
(c) 15
(d) 10
Solution:
(a) 24
Given, the observations are: \(\frac{x}{5}, x, \frac{x}{4}, \frac{x}{2}, \frac{x}{3}\)
To find the median, we first arrange-them in ascending order.
Since x > 0, the larger the denominator, the smaller the value.
Thus arranging the observations in ascending order, we get: \(\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x\)
Here, number of observations, n = 5, is odd.
∴ Median = \(\left(\frac{5+1}{2}\right)^{\text {th }}\) observation
= 3^rd observation = \(\frac{x}{3}\)
According to the question,
Median = 8 ⇒ \(\frac{x}{3}\) = 8
⇒ r = 8 × 3 = 24

Question 8.
A cricket player scored 220 runs in 5 matches but did not bat in one of the innings. What is the mean score of the player?
(a) 36.67
(b) 38
(c) 55
(d) 45.67
Solution:
(c) 55
Given, the player scored 220 runs in 5 matches in total and did not bat in one inning.
∴ Mean = \(\frac{Total score}{Number of innings played}\)
= \(\frac{220}{4}\) = 55

Question 9.
The mean of five numbers is 18. If the number 32 is removed, what is the mean of the remaining four numbers?
(a) 12.5
(b) 14.5
(c) 13
(d) 16.5
Solution:
(b) 14.5
Given, mean of 5 numbers = 18
∴ Sum of 5 numbers = 5 × 18 = 90
As, number 32 is removed,
Sum of remaining 4 numbers = 90 – 32 = 58
∴ Mean of remaining four numbers
= \(\frac{58}{4}\) = 14.5

Question 10.
A data set contains 11 values arranged in ascending order. The 6th value is 42. Which of the following must be true?
(a) The mean is 42.
(b) The median is 42.
(c) All values are dose to 42.
(d) There is no outlier in the data.
Solution:
(b) The median is 42.
For an odd number of observations, the median is the middle value.
Here, the number of observations is 11 which is odd.
∴ Median = \(\left(\frac{11+1}{2}\right)^{\mathrm{th}}\) value = 6^th value = 42

Question 11.
The median of prime numbers between 20 and 50 is:
(a) 31
(b) 37
(c) 41
(d) 43
Solution:
(b) 37
The prime numbers between 20 and 50 are:
23, 29, 31, 37, 41, 43, 47 (In ascending order) Here, number of observations, n = 7, is odd.
Median = \(\left(\frac{n+1}{2}\right)^{\mathrm{th}}\) observation
= \(\left(\frac{7+1}{2}\right)^{\mathrm{th}}\) observation = 4^th observation
Thus, median = 37

Question 12.
A data set has mean = 50 and median = 70. Which of the following is most likely to be true?
(a) Presence of higher end outliers
(b) Presence of lower end outliers
(c) Data is balanced
(d) None of these
Solution:
(b) Presence of lower end outliers
Given, for a data set, mean = 50 and median = 70.
In this data set, the mean is much lower than the median (50 < 70).
The median tells us where the middle of the data is.
The mean is sensitive to outliers and gets pulled in their direction.
Because the mean has been pulled down to 50 (below the median), there must be very small numbers (lower-end outliers) dragging the average down.

Question 13.
The following bar graph shows the rainfall at six selected locations (A, B, C, D, E and F) in certain months.

Which of the following statements is correct?
(a) July rainfall exceeds August rainfall by 100 cm at each location.
(b) September rainfall exceeds August rainfall by 50 cm at each location.
(c) July rainfall is lower than August rainfall at each location.
(d) None of the above
Solution:
(c) July rainfall is lower than August rainfall at each location.
From the bar graph, comparing July, August, and September rainfalls at each location A to F, we observe:

• At every location, the August bar is higher than the July bar, i.e. July rainfall never exceeds August rainfall.
• Also, July rainfall is lower than August rainfall in each location.
• September rainfall is not exactly 50 cm more than August at each location.

Question 14.
A bar graph represents the daily water consumption (in litres) of a household.

Which conclusion is most logically sound?
(a) Mean = Median
(b) Mean > Median
(c) Median > Mean
(d) Mean and Median cannot be compared
Solution:
(b) Mean > Median
From the graph, we observe that the water consumption on days 1, 2, 3 and 4 are the same (bars of same height), while the water consumption on day 5 is the highest (longest bar).

As the water consumption on day 5 is higher than the water consumption on other days, it will pull the mean upward.
However, the median water consumption will lie in middle.
Hence, Mean > Median.

Question 15.
A bar graph uses the scale: 1 unit =10 children on y-axis.
If the scale is changed to 1 unit = 20 children, then:
(a) the data values change
(b) the order of bars changes
(c) the height of bars changes
(d) the graph becomes incorrect
Solution:
(c) the height of bars changes
If scale is changed to 1 unit = 20 children, the height of bars will reduce in size, but the data values and order of bars remain unchanged.
So, only the height of bars will change.

Question 16.
The given double bar graph represents the average rainfall of two cities in 5 months.

Which city sees increase in rainfall each month from January to May?
(a) City 1
(b) City 2
(c) Both city 1 and city 2
(d) Neither city 1 nor city 2
Solution:
(a) City 1
From the given bar graph:
Rainfall in City 1 increases steadily from January to May.
Rainfall in City 2 decreases from January to February, then increases from February to March to April and then again decreases in May.

Question 17.
The data given below shows the time (in minutes), taken by nine students to go to school from their homes. Which statement about the data is/are true?
115, 65, 225, 75, 115,65,155,125,115
(i) Median is 115.
(ii) The observation 65 is lowest observation.
(iii) The range of the data is 160.
(iv) The range of the data is 90.
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (i) and (iii) only
(c) (ii) and (iv) only
(d) (iv) only
Solution:
(a) (i), (ii) and (iii)
Arranging the data in ascending order, we get 65, 65, 75, 115, 115, 115, 125, 155, 225
Here, number of observations, n = 9, is odd.
∴ Median = \(\left(\frac{9+1}{2}\right)^{\text {th }}\) observation
= \(\left(\frac{10}{2}\right)^{\text {th }}\) observation = 5^th observation =115
In the given data set, lowest observation = 65 and highest observation = 225.
So, the range of data set
= Highest value- Lowest value
= 225 – 65 = 160
Thus, statements (i), (ii) and (iii) are true.

Question 18.
The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls.

Based on the dot plots, which of the following statements are true?
(i) The data varies more for the girls than for the
(ii) The median number of pockets for the boys is more than that for the girls.
(iii) The mean number of pockets for the girls is more than that for the boys.
(iv) The maximum number of pockets for boys is greater than that for the girls.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (ii) and (iv)
(d) (i) and (iii)
Solution:
(a) (i) and (ii)
For boys: 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6
Range = highest value-lowest value = 6 – 3 = 3
For girls: 0, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 6
Range = highest value-lowest value = 6 – 0 = 6
As the range for girls is higher than the range for boys, the data for girls is more variable than that for boys.
For median:
Number of boys, n = 12, which is even
∴ Median number of pockets for the boys
= \(\frac{\left(\frac{12}{2}\right)^{\text {th }} \text { value }+\left(\frac{12}{2}+1\right)^{\text {th }} \text { value }}{2}\)
= \(\frac{6^{\text {th }} \text { value }+7^{\text {th }} \text { value }}{2}\)
= \(\frac{5+5}{2}=\frac{10}{2}\) = 5
Now, number of girls =13, which is odd.
∴ Median number of pockets for the girls
= \(\left(\frac{13+1}{2}\right)^{\text {th }}\) observation = 7^th observation
= 4
Clearly, the median number of pockets for the boys is more than that for the girls.
For mean:
Total number of pockets for boys
= 3 + 4 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6
= 3 + 16 + 25 + 12 = 56
∴ Mean number of pockets for boys
= \(\frac{\text { Total number of pockets }}{\text { Number of boys }}=\frac{56}{12}\) ≈ 4.67
Also, total number of pockets for girls
= 0 + 2 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4 + 4 + 5 + 6
= 2 + 12 + 20 + 11 = 45
∴ Mean number of pockets for girls
= \(\frac{\text { Total number of pockets }}{\text { Number of girls }}=\frac{45}{13}\) ≈ 3.46
Clearly, the mean number of pockets for the girls is less than that for the boys.
Also, maximum number of pockets for both boys and girls is the same, i.e. 6.
Thus, statements (i) and (ii) are true.

Question 19.
A student’s weekly step counts (in thousands) are: 5, 5, 6, 7, 7, 8, 20
Which of the following will be largely affected by the removal of the outlier?
(i) Mean
(ii) Median
(iii) Range
(iv) None of these
Choose the correct answer from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iv) only
Solution:
(c) (i) and (iii)
Observing the given set of data, we see that most numbers are clustered between 5 and 8. The number 20 is much higher than the others and acts as an outlier.
With the presence of an outlier:
Mean steps = \(\frac{5+5+6+7+7+8+20}{7}=\frac{58}{7}\)
= 8.29
The data is already arranged in ascending order.
So, median = 7
Range of data = 20 – 5 = 15
On removing the outlier:
Mean steps: \(\frac{5+5+6+7+7+8}{6}=\frac{38}{6}\) = 6.33
Now, number of observations = 6 is even, so median is average of two middle values.
∴ Median = \(\frac{6+7}{2}=\frac{13}{2}\) = 6.5
Range of data = 8 – 5 = 3
Difference in mean = 8.29 – 6.33 = 1.96
Difference in median = 7 – 6.5 = 0.5
Difference in range = 15 – 3 =12
Therefore, both the mean and the range are largely affected by outliers.

Question 20.
Consider the given dot plot (for 20 students) representing the data of number of pairs of socks owned by a student.

Which of the following statements are correct?
(i) Total 77 pairs of socks are represented in thegiven dot plot.
(ii) Only one student has the highest number of pairs of socks.
(iii) Number of students that own more than 5 pairs is 4.
(iv) Range of the data is 4.
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (i) and (ii) only
(c) (ii) and (iv) only
(d) (iv) only
Solution:
(a)
The total number of pairs of socks
= 0 + 1 + (4 × 2) + (2 × 3) + (5 × 4) + (3 × 5) + (3 × 6) + 9
= 0 + 1 + 8 + 6 + 20 + 15 + 18 + 9 = 77
Since there is only one dot corresponding to 9
i. e. the highest number of pairs of socks, only one student has the highest number of pairs of socks.
The number of students that own more than 5 pairs is 4.
Range of data = Highest value – Lowest value = 9 – 0 = 9
Thus, statements (i), (ii) and (iii) are correct.

Connecting the Dots Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct ansiyer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): If two data sets have the same mean, they must have the same sum.
(R): Mean = \(\frac{Sum of observations}{Number of observations}\)
Solution:
(d) A is false but R is true.
We know, Mean = \(\frac{Sum of observations}{Number of observations}\)
Thus, Reason (R) is true.

For two data sets to have the same mean, the ratio of their sum to the number of observations must be equal. However, if the number of observations (n) is different for each set, their sums will also be different even if the mean is the same. For example:

Set A: 10, 10 (Sum = 20, Number of observations = 2, Mean =10)
Set B: 10, 10, 10 (Sum = 30, Number of observations = 3, Mean = 10)
Both have the same mean, but different sums. Thus, the Assertion (A) is false.

Question 2.
(A): If a data set has an outlier at the higher end, then the mean is more likely to be greater than the median.
(R): The outlier at the higher end does not affect the mean.
Solution:
(c) A is true but R is false.
We know that, if in a dataset, the outlier lies at the higher end, the mean is pulled upward.

Because of this, the mean becomes greater than the middle value, i.e. the median.
Thus, Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): An inappropriate scale on a column graph can make the data interpretation misleading.
(R): The scale decides the value represented by each unit length on the axis.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
The height of the bars of column graph is entirely dependent on the chosen scale. Since the scale determines the value represented by each unit on the axis, an inappropriate scale leads to confusion while interpreting the graph.

Thus, both the Assertion (A) and the Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Connecting the Dots Class 7 Fill in the Blanks

Question 1.
If there are even number of observations in a data set, then median is average of _____ middle values.
Solution: two
When number of observations in a data set is even, there exists two middle values.
Thus, if there are even number of observations in a data set, then median is average of two middle values.

Question 2.
Mean of first five multiples of 2 is _________ .
Solution: 6
The first five multiples of 2 are 2, 4, 6, 8, 10.
∴ Mean = \(\frac{Sum of first five multiples of 2}{5}\)
= \(\frac{2+4+6+8+10}{5}=\frac{30}{5}\) = 6
Thus, mean of first five multiples of 2 is 6.

Question 3.
A question is said to be statistical only if its answer is expected to show _________.
Solution: variability
A question is said to be statistical only if its answer is expected to show variability.

Question 4.
The mecpan Qf first 15 odd natural numbers is ________ .
Solution: 15
The first 15 odd numbers in ascending order are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 Here, the number of observations, n = 15, is odd.
∴ Median = \(\left(\frac{15+1}{2}\right)^{\text {th }}\)
Observation = 8^th observation =15
Thus, the median of the first 15 odd natural numbers is 15.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 6 Perimeter and Area Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 6 Perimeter and Area Solutions

Ganita Prakash Class 6 Chapter 6 Solutions

Class 6 Maths Ganita Prakash Chapter 6 Solutions Perimeter and Area

Question 1.
A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
Solutions:
Given, length of rectangle = 5 cm and breadth of rectangle = 3 cm
We know that perimeter of rectangle = 2 × (length + breadth) = 2 × (5 + 3) = 16 cm
Now, if we bend the wire to form a square, the total length of the wire (16 cm) will be divided equally among the four sides of the square.
So, each side of the square = \(\frac{\text { Perimeter }}{4}\) = \(\frac{16}{4}\) = 4 cm

Question 2.
A piece of string is 36 cm long. What will be the length of each side, if it is used to form:
(a) Square
(b) Equilateral Triangle
(c) Regular Hexagon
Solutions:
Given, total length of string = 36 cm
(a) For a square. each side of the square = \(\frac{\text { Total length of string }}{4}\) = \(\frac{36}{4}\) = 12 cm
(b) For a triangle with all si(les of edila! length. each side of the triangle = \(\frac{\text { Total length of string }}{3}\) = \(\frac{36}{3}\) = 12 cm
(c) For a hexagon wit h all sides oI’egual length, each sidle oÍ the hexagon = \(\frac{\text { Total length of string }}{6}\) = \(\frac{36}{6}\) = 6 cm

InText Questions

Question 1.
Deep Dive: In races, usually there is a common finish line for all the runners. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is of 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing fine after they run for 350 m. Mark the starting points of the runner on the inner track as ‘A’ and the runner on the outer track as ‘B’.

Solution:
For the inner square track, each side is 100 m.
For the outer square track, each side is 150 m.
So, inner track perimeter = 4 × side = 4 × 100 m = 400 m
Outer track perimeter = 4 × side = 4 × 150 m = 600 m

Both runners start at different points and run along their respective square tracks, possibly completing full or partial laps. The finish line is fixed at the middle of one side.
The total race distance is 350 m for both runners.
Therefore, each runner must start 350 m before the finish line, in the direction of the run.

Runner A runs on the inner track (400 m) and runner B runs on the outer track (600 m).
Now, we calculate how far before the finish line each runner should start:
Start Point A: 400 m – 350 m = 50 m
Thus, position A is 50 m behind the finish line (in the direction of the run) on the inner track.
Start Point B: 600 m – 350 m = 250 m
Thus, position B is 250 m behind the finish line (in the direction of the run) on the outer track.

Thus, mark point A on the inner track, 50 m behind the flag (finish line), and point B on the outer track, 250 m behind the flag, both measured in the direction of the run.

Question 2.
Look at the figures below and guess which one of them has a larger area.

Solution:
We can estimate the area of any simple closed shape by using a sheet of squared paper or graph paper where every square measures 1 unit X 1 unit or 1 square unit.
To estimate the area, we can trace the shape onto a piece of transparent paper and place the same on a piece of squared or graph paper and then follow the below conventions:

1. Count all the full squares inside the figure. Area of each is 1 square unit.
2. For squares that are more than half filled, count them as 1 square unit.
3. For squares that are exactly half filled, count them as \(\frac{1}{2}\) square unit.
4. Ignore squares that are less than half filled.

Now, adding a square grid, where every square measures 1 unit × 1 unit, we get the following figure.

In figure (a): Number of full squares inside the figure = 31
Number of squares that are more than half filled = 13
Area of figure (a) = 31 × 1 + 13 × 1
= 31 + 13 = 44 sq. units
In figure (b): Number of full squares inside the figure = 16
Number of squares that are more than half filled = 20
Area of figure (b) = 16 × 1 + 20 × 1
= 16 + 20 = 36 sq. units
Thus, figure (a) has a larger area.

Question 3.
Use your understanding from, previous grades to calculate the area of any closed figure using grid paper and

(a) Find the area of triangle BAD. _________
(b) Find the area of triangle ABE. _________
Solution:
From the given figure,

Area of rectangle A BCD = 20 sq. units
Area of rectangle AFED = 12 sq. units
Area of rectangle FBCE = 8 sq. units
(a) Area of triangle BAD
= \(\frac{1}{2}\) × Area of rectangle ABCD
= \(\frac{1}{2}\) × 20 sq. units = 10 sq. units

(b) Area of triangle ABE
= Area of triangle AFE + Area of triangle BEE
= \(\frac{1}{2}\) × Area of rectangle AFED + \(\frac{1}{2}\) × Area of rectangle FBCE
= \(\frac{1}{2}\) × 12 sq. units + \(\frac{1}{2}\) × 8 sq. units
= 6 sq. units + 4 sq. units = 10 sq. units

Question 4.
Using 9 unit squares, solve the following.
(a) What is the smallest perimeter possible?
(b) What is the largest perimeter possible?
(c) Make a figure with a perimeter of 18 units.
(d) Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?
Solution:
(a) The smallest perimeter possible is 12 units.
This occurs when the 9 unit squares are arranged to form a 3 × 3 square.

Side length = 3 units
Perimeter = 4 × 3 = 12 units

(b) The largest perimeter possible is 20 units.
This occurs when the 9 unit squares are arranged in a single straight line to form 1 × 9 or 9 × 1 rectangle.

Perimeter of rectangle = 2 × (9 + 1) = 20 units

(c) A figure with a perimeter of 18 units is given alongside:

(d) Yes, we can make multiple different shapes for each of the three perimeter values 12, 18 and 20 units as long as the number of external sides (edges not shared with another square) adds up to the correct perimeter.

Reasoning: Each square has 4 edges.

Every time two squares are placed adjacent to each other, they share an edge, reducing the total perimeter by 2 units (1 edge from each square).

By changing how the 9 squares are joined linear, block, zig-zag, L-shape, etc. we can vary how many edges are shared and hence control the perimeter.

The number of shared sides determines the final perimeter, not the specific shape.

Question 5.
Let’s do something tricky now! We have^ a figure below having perimeter 24 units.
Without calculating all over again, observe, think and find out what will be the change in the perimeter if a new square is attached as shown on the right.

Experiment placing this new square at different places and think what the change in perimeter will be. Can you place the square so that the perimeter: (a) increases; (b) decreases; (c) stays the same?
Solution:
When a new square is attached to an existing figure:
If it shares one full side with the existing figure, the perimeter increases by 2 units. (Because one new side is hidden in the shared boundary, and 3 new sides are exposed.
So, net change = + 3 – 1 = + 2).

If it shares two sides (like being inserted into a corner), the perimeter stays the same. (2 new sides added, but 2 sides of the previous figure are now internal and not counted: + 2 – 2 = 0).

If it shares three sides (nestled into a concave corner), the perimeter decreases by 2 units. (Only 1 side added is exposed, 3 sides of the previous figure are hidden: + 1 – 3 = – 2).

(a) Perimeter increases when the new square is added such that only one side touches the original figure.
(b) Perimeter decreases when the square is placed within a corner, sharing three sides with the original figure.
(c) Perimeter stays the same when the square is added in such a way that two sides are shared with the original figure (like being placed into an edge corner).

Perimeter and Area Class 6 Extra Questions

Perimeter and Area Class 6 Very Short Question Answer

Question 1.
Find the perimeter of the given quadrilateral.

Solution:
We know that the perimeter of a polygon is the sum of the lengths of its all sides.
So, the perimeter of given quadrilateral
= AB + BC + CD + DA
= 5 cm + 3 cm + 7.5 cm + 4.5 cm
= 20 cm

Question 2.
Find the perimeter of the following figure:

solution:
We know that the perimeter of a closed figure is the sum of the lengths of its all sides.
So, the perimeter of given figure
= PQ + QR + RS + ST + TU + UP
= 6.4 cm + 2.5 cm + 2.6 cm + 2.6 cm + 2.5 cm + 6.4 cm
= 23 cm

Question 3.
Find the perimeter of a rectangle whose length and breadth are 1.25 m and 75 cm, respectively.

Solution:
Given, length of the rectangle = 1.25 m = 125 cm [∵ 1 m = 100 cm]
Breadth of the rectangle = 75 cm We know,
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (125 cm + 75 cm) = 2 × 200 cm = 400 cm

Question 4.
Find the perimeter of the following figures:


Solution:
We know that the perimeter of a polygon is the sum of the lengths of its all sides.
(i) Perimeter = 2 cm + 7 cm + 2 cm + 7 cm = 18 cm
(ii) Perimeter = 4 cm + 4 cm + 3 cm = 11 cm
(iii) Perimeter = 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm = 18 cm
(iv) Perimeter = 3 cm + 5 cm + 4 cm + 2 cm = 14 cm

Question 5.
Find the perimeter of the following figure:

Solution:
We know that the perimeter of a closed figure is the sum of the lengths of its all sides.
So, the perimeter of given figure = LM + MN + NO + OP + PCI + QR + RS + SL
= 2.9 cm + 2.9 cm + 1.4 cm + 5.5 cm + 2.4 cm + 2.4 cm + 5.5 cm + 1.4 cm
= 24.4 cm

Question 6.
The lengths of two sides of a triangle are 15 cm and 21 cm. The perimeter of the triangle is 46 cm. Find the length of its third side.
Solution:
Given, the perimeter of a triangle is 46 cm.
Length of first side =15 cm

Length of second side = 21 cm
We know, the perimeter of a triangle
= Sum of the lengths of all sides of the triangle
⇒ 46 cm = Length of first side + Length of second side + Length of third side
⇒ 46 cm =15 cm + 21 cm 4 Length of third side
⇒ 46 cm = 36 cm 4 Length of third side
⇒ Length of third side = 46 cm – 36 cm = 10 cm

Question 7.
If the perimeter of a regular heptagon is 63 cm, find the length of its one side.
Solution:
Given, perimeter of a regular heptagon is 63 cm.
We know, number of sides in a regular heptagon = 7
And, perimeter of a regular polygon = Number of sides × Length of one side
So, the perimeter of given regular heptagon
= 7 × Length of one side
⇒ 63 cm = 7 × Length of one side
⇒ Length of one side = \(\frac{63}{7}\) cm = 9 cm

Question 8.
A rectangular carpet has an area of 72 sq. m and a width of 6 m. What is its length?
Solution:
Given, width (breadth) of a rectangular carpet = 6 m
And, area of a rectangular carpet = 72 sq. m
We know, area of the rectangular carpet = Length × breadth
⇒ 72 sq. m = Length × 6 m
⇒ Length = \(\frac{72}{6}\) = 12 m
Thus, the length of the rectangular carpet is 12 m.

Question 9.
If rectangle ABCD has area 42 sq. units, then find the area of triangle ABC, cut along the diagonal of the rectangle ABCD.
Solution:
Given, the area of rectangle ABCD is 42 sq. units.
We know that if a rectangle is cut along one of its diagonals, then the area of each resulting triangle is half the area of the rectangle.
∴ Area of triangle ABC
= \(\frac{1}{2}\) × Area of rectangle ABCD
= \(\frac{1}{2}\) × 42 sq. units = 21 sq. units

Perimeter and Area Class 6 Short Question Answer

Question 1.
The perimeter of a rectangle is 72 cm and its breadth is 8 cm. Find its length.

Solution:
Given, perimeter of the rectangle = 72 cm
Breadth of the rectangle = 8 cm
We know,
Perimeter of the rectangle = 2 (Length + Breadth )
⇒ 72 cm = 2(Length + 8 cm)
⇒ 2(Length + 8 cm) = 72
⇒ Length + 8 cm = \(\frac{72}{2}\) = 36 cm
⇒ Length = 36 cm – 8 cm = 28 cm

Question 2.
In the adjoining figure, the length of each side is 2.75 cm. Find the perimeter of the figure.

Solution:
Given figure is a polygon and the length of its each side is 2.75 cm.
Also, the number of sides is 10.
So, the perimeter of given polygon
= 10 × Length of one side
= 10 × 2.75 cm
= 27.5 cm

Question 3.
If the perimeter of a regular pentagon is 55 cm, find the length of its one side.

Solution:
Given, perimeter of a regular pentagon is 55 cm.
We know, the number of sides in a regular pentagon is 5.
And, perimeter of a regular polygon
= Number of sides × Length of each side
So, the perimeter of regular pentagon = 5 × Length of each side
⇒ 55 cm = 5 × Length of each side
⇒ Length of each side = \(\frac{55}{5}\) cm = 11 cm

Question 4.
The lid of a rectangular box of length 45 cm and breadth 30 cm is sealed all around with tape. Find the required length of tape.

Solution:
Given, length of the rectangular box is 45 cm.
Breadth of the rectangular box is 30 cm.
And, the lid of the rectangular box is sealed all around with tape.
Therefore, length of the tape
= Perimeter of the rectangle
= 2(Length + Breadth)
= 2(45 cm + 30 cm)
= 2 × 75 cm = 150 cm

Question 5.
A wire is in the shape of an equilateral triangle of side 26 cm. It is rebent into the shape of rectangle whose length is 23 cm, find its breadth.
Solution:
Given, length of the side of an equilateral triangle is 26 cm.
Therefore, length of the wire = Perimeter of the equilateral triangle of side 26 cm
= 3 × 26 cm [∵ Number of sides in an equilateral triangle is 3.]
= 78 cm

Given that the wire is rebent into the shape of rectangle whose length is 23 cm.
∴ Perimeter of the rectangle = Length of wire
⇒ 2(Length + Breadth) = 78 cm
⇒ 2(23 cm + Breadth) = 78 cm
⇒ 23 cm + Breadth = \(\frac{78}{2}\) = 39 cm
⇒ Breadth = 39 cm – 23 cm = 16 cm

Question 6.
Priya went to a rectangular field 160 m long and 110 m wide. She took 4 complete rounds on its boundary. Find the distance covered by her.

Solution:
Given, length of the rectangular field = 160 m
Breadth of the rectangular field = 1 10 m
∴ Distance covered by Priya in one round
= Perimeter of the rectangular field
= 2(Length + Breadth)
= 2(160 m + 110 m)
= 2 × 270 m = 540 m
Since she took 4 complete rounds on its boundary,
Distance covered by Priya in four rounds
= 4 × Distance covered by Priya in one round
= 4 × 540 m = 2160 m
Thus, Priya covers 2160 m distance in 4 rounds.

Question 7.
In below figure, the length of each side is 3.25 cm. Find the perimeter of the figure.

Solution:
In the given figure, number of sides is 14.
Also, the length of each side is 3.25 cm.
So, the perimeter of given polygon = 14 × Length of one side = 14 × 3.25 cm = 45.5 cm

Question 8.
The perimeter of a rectangle is 56 cm and its length is 20 cm. Find its breadth.
Solution:
Given, perimeter of the rectangle = 56 cm

Length of the rectangle = 20 cm
We know, perimeter of the rectangle = 2 (Length 4- Breadth)
⇒ 56 cm = 2(20 cm + Breadth)
⇒ 28 cm = 20 cm + Breadth
⇒ Breadth = 28 cm – 20 cm = 8 cm

Question 9.
A rectangular piece of land measures 0.95 km by 0.65 km. Each side of the land is to be fenced with 5 rows of wires. Find the required length of the wire.
Solution:
Given, length of the rectangular piece of land = 0.95 km

Breadth of the rectangular piece of land = 0.65 km
∴ Perimeter of the rectangular piece of land
= 2(Length + Breadth)
= 2(0.95 km + 0.65 km)
= 2 × 1.60 km = 3.20 km
Since each side of the land is to he fenced with 5 rows of wires, length of the wire is five times the perimeter of the land.
∴ Required length of wire = 5 × 3.20 km = 16 km

Question 10.
Find the perimeter of a rectangular field whose length is 345 cm and which has an area equal to 56925 sq. cm.
Solution:
Given, length of rectangular field = 345 cm and area of the field = 56925 sq. cm
We know, area of rectangular field
= Length × Breadth
⇒ 56925 sq. cm = 345 cm × Breadth
⇒ Breadth = \(\frac{56925}{345}\) = 165 cm
So, breadth of the field is 165 cm.
Hence,
perimeter of the field = 2 (Length + Breadth)
= 2 (345 cm + 165 cm) = 2 × 510 cm = 1020 cm
Thus, the perimeter of the rectangular field is 1020 cm.

Question 11.
In the figure, find the area of the path which is 2.5 m wide all around.

Solution:
Given, length of outer rectangle = 70 m
Breadth of outer rectangle = 44 m
As the path is 2.5 m wide,
Length of inner rectangle = 70 m – 2.5 m – 2.5 m
= 65 m
Breadth of inner rectangle = 44 m – 2.5 m – 2.5 m
= 39 m
Now, the area of path = Area of outer rectangle – Area of inner rectangle
= 70 m × 44 m – 65 m × 39 m
= 3080 sq. m – 2535 sq. m = 545 sq. m
Thus, the area of the path is 545 sq. m

Question 12.
Four square tiles of side 2.5 m each are placed together without any gaps. What is the total area covered?
Solution:
Given, the side of a square tile is 2.5 m.
We know, area of a square = Side × Side
∴ Area of a square tile = 2.5 m × 2.5 m
= 6.25 sq. m
As four square tiles are placed together without any gaps,
Total covered area = 4 × Area of one square tile
= 4 × 6.25 sq. m = 25 sq. m
Thus, total area covered by the four square tiles of side 2.5 m is 25 sq. m.

Question 13.
A floor measuring 15 m by 9 m is covered with a carpet of size 12 m × 8 m. Find the area that remains uncovered.
Solution:
Given, length of the floor = 15 m,
Breadth of the floor = 9 m,
Length of the carpet = 12 m
and breadth of the carpet = 8 m
Therefore, area of the floor
= Length of the floor × Breadth of the floor
= 15 m × 9 m = 135 sq. m
And, area of the carpet
= Length of the carpet × Breadth of the carpet
= 12 m × 8 m = 96 sq. m
Thus, area that remains uncovered = Area of the floor – Area of the carpet
= 135 sq. m – 96 sq. m = 39 sq. m

Question 14.
Find the area of the given shaded region (figure alongside) drawn on a square paper, taking the area of each square as 1 cm².

Solution:
The figure drawn on square paper contains 35 complete squares and 9 half squares.
∴ Area of given closed figure = Area of 35 complete squares + Area of 9 half squares
= 35 × 1 cm² + 9 × \(\frac{1}{2}\) cm²
= 35 cm² + 4.5 cm² = 39.5 cm²

Perimeter and Area Class 6 Long Question Answer

Question 1.
A wire is 39 cm long. What will be the length of each side if the wire is used to form
(i) an equilateral triangle?
(ii) a regular hexagon?
Solution:
Given, a wire is 39 cm long.
(i) The wire is used to form an equilateral triangle.
We know that there are 3 equal sides in an equilateral triangle.
∴ Perimeter of equilateral triangle = Length of wire
⇒ 3 × Length of side = 39 cm
⇒ Length of side = \(\frac{39}{3}\) cm = 13 cm

(ii) The wire is used to form a regular hexagon.
We know that there are 6 equal sides in a regular hexagon.
∴ Perimeter of regular hexagon = Length of wire
⇒ 6 × Length of side = 39 cm
⇒ Length of side = \(\frac{39}{6}\) cm = 6.5 cm

Question 2.
A farmer has a rectangular field of length 350 m and breadth 175 m. He wants to fence it with 4 rounds of rope as shown in the figure. If cost of rope is ₹ 4.5 per metre, then find the cost of fencing the field.

Solution:
Given, length of the rectangular field = 350 m
Breadth of the rectangular field = 175 m
∴ Perimeter of the rectangular field = 2(Length + Breadth)
= 2(350 m + 175 m)
= 2 × 525 m = 1050 m
Since the farmer wants to fence the field with 4 rounds of rope, length of the rope is four times the perimeter of the field.
∴ Required length of rope = 4 × 1050 m = 4200 m
Given, the cost of 1 m rope is ₹ 4.5.
∴ Total cost of 4200 m rope = ₹ 4.5 × 4200
= ₹ 18,900
Thus, the cost of fencing the field is ₹ 18,900.

Question 3.
Priya runs around a square field of side 90 m. Rishi runs around a rectangular field of length 120 m and breadth 75 m. Who covers more distance and by how much in one round?

Solution:
Given, Priya runs around a square field of side 90 m.
∴ Distance covered by Priya in one round
= Perimeter of the square field
= 4 × length of a side of the square field
= 4 × 90 m = 360 m
And, Rishi runs around a rectangular field of length 120 m and breadth 75 m.
∴ Distance covered by Rishi in one round
= Perimeter of the rectangular field
= 2(Length + Breadth)
= 2(120 m + 75 m) = 2 × 195 m = 390 m
So, difference in the distance covered in one round = 390 m – 360 m = 30 m
Thus, Rishi covers more distance by 30 m in one round.

Question 4.
A rectangle, having side lengths 19 cm and 13 cm, is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
Solution:
Given, length of the rectangle is 19 cm.
Breadth of the rectangle is 13 cm.

Therefore, length of the wire
= Perimeter of the rectangle
= 2(Length + Breadth)
= 2(19 cm + 13 cm)
= 2 × 32 cm = 64 cm
Given that the wire is rebent into the shape of a square.
∴ Perimeter of the square = Length of wire
⇒ 4 × Length of side = 64 cm
⇒ Length of side = \(\frac{64}{4}\) cm = 16 cm

Question 5.
A wire is 39 cm long. What will be the length of each side if the wire is used to form
(i) a square?
(ii) a regular pentagon?
Solution:
Given, a wire is 39 cm long.
(i) The wire is used to form a square.
We know that there are 4 equal sides in a square.
∴ Length of wire = 4 × Length of side
⇒ 39 cm = 4 × Length of side
⇒ Length of side = \(\frac{39}{4}\) cm = 9.75 cm

(ii) The wire is used to form a regular pentagon.
We know that there are 5 equal sides in a regular pentagon.
∴ Length of wire = 5 × Length of side
⇒ 39 cm = 5 × Length of side
⇒ Length of side = \(\frac{39}{5}\) cm = 7.8 cm

Question 6.
The length of a rectangular field is four times its breadth. A man runs around it 4 times and covered a distance of 5 km. What is the length of the field?
Solution:
Given, the length of a rectangular field is four times its breadth.
∴ Length = 4 × Breadth ……. (i)
And, a man runs around the field 4 times and covered a distance of 5 km.
∴ Distance covered in 4 rounds
= 4 × Perimeter of the field
⇒ 5 km = 4 × 2 (Length + Breadth)
⇒ 5000 m = 8 × (4 × Breadth + Breadth) [From (i)] [∵ 1 km = 1000 m]
⇒ 5000 m = 8 × 5 × Breadth
⇒ 5000 m = 40 × Breadth
⇒ Breadth = \(\frac{5000 \mathrm{~m}}{40}\) = 125 m
Substituting I he value of breadth in (I). we gel
Length = 4 × 125 m = 500 m
Thus, the length of the field is 500 m.

Question 7.
Seema runs 6 times around a rectangular park with length 70 m long and breadth 45 m while Ramesh runs 5 times around a square park of side 65 m. Who covers more distance and by how much?

Solution:
Given, Seema runs around a rectangular park of length 70 m and breadth 45 m.
∴ Distance covered by Seema in one round
= Perimeter of the rectangular park
= 2(Length + Breadth)
= 2(70 m + 45 m) = 2 × 115 m = 230 m
So, distance covered by Seema in 6 rounds
= 6 × Distance covered by Seema in one round
= 6 × 230 m = 1380 m
And Ramesh runs around a square park of side 65 m.
∴ Distance covered by Ramesh in one round
= Perimeter of the square park
= 4 × length of a side of the square park
= 4 × 65 m = 260 m
So, distance covered by Ramesh in 5 rounds
= 5 × Distance covered by Ramesh in one round
= 5 × 260 m = 1300 m
So, difference in the distance covered
= 1380 m- 1300 m = 80 m
Thus, Seema covers 80 m more distance than Ramesh.

Question 8.
Match the closed figure given in Column I with their perimeter given in Column II.


Solution:
We know that the perimeter of a polygon/closed figure is the sum of the lengths of its all sides.
(A) Perimeter = 40 cm + 40 cm + 40 cm + 40 cm = 160 cm
(B) Perimeter = 30 cm + 60 cm + 30 cm + 60 cm = 180 cm
(C) Perimeter = 35 cm + 35 cm + 35 cm = 105 cm
(D) Perimeter = 30 cm + 20 cm + 25 cm + 24 cm = 99 cm
(E) Perimeter = 31 cm + 40 cm + 20 cm + 22 cm + 40 cm = 153 cm
(F) Perimeter = 15 cm + 28 cm + 15 cm + 2 cm + 25 cm + 10 cm + 5 cm + 40 cm = 140 cm
Thus, (A) – (t), (B) – (s), (C) – (p), (D) – (q), (E) – (u), (T) – (r)

Question 9.
Rishi wants to cover the floor of a room 4 m wide & 8 m long by square tiles. If each square tile is of side 0.4 m, then find the number of tiles required.
Solution:
Given, length of the room = 8 m and breadth of the room = 4 m
As the room is in rectangular shape,
Area of the room = Length × Breadth
= 8 m × 4 m = 32 sq. m
And, side length of each square tile = 0.4 m
∴ Area of each tile = Side × Side
= 0.4 m × 0.4 m = 0.16 sq. m
Now, number of tiles = \(\frac{\text { Area of floor of the room }}{\text { Area of one tile }}\)
= \(\frac{32}{0.16}\) = 200
Thus, the number of required tiles is 200.

Question 10.
A square and a rectangle have equal area. If the side of the square is 36 cm and the length of the rectangle is 54 cm, then find:
(i) the breadth of the rectangle)
(ii) the perimeter of die rectangle.
Solution:
(i) Given, side of the square
= 36 cm and length of the rectangle
= 54 cm
∴ Area of the square = Side × Side
= 36 cm × 36 cm = 1296 sq. cm
As area of the rectangle = area of the square
⇒ Length × Breadth = 1296 sq. cm
⇒ 54 cm × Breadth = 1296 sq. cm
⇒ Breadth = \(\frac{1296}{54}\) = 24 cm
Thus, the breadth of the rectangle is 24 cm.

(ii) We know, perimeter of the rectangle
= 2 (Length + Breadth)
= 2 (54 cm + 24 cm)
= 2 × 78 cm = 156 cm

Question 11.
Find the area of the given figure.

Solution:
To find the required area, we divide the enclosed region of given figure into rectangles and squares as shown in the figure.

Part I is a rectangle of length 200 m and breadth 100 m.
Part II is a square of side length 100 m.
Part III is a rectangle of length 400 m and breadth 200 m.
Now, area of part I = 200 m × 100 m = 20000sq. m
Area of part II = 100 m × 100 m = 10000 sq.m
And area of part III = 400 m × 200 in = 80000 sq. m
Total area of given closed figure
= 20000 sq. m + 10000 sq. m + 80000 sq. m
= 110000 sq.m

Question 12.
In the following figure, find the missing value of the area of a region.

Solution:
We name the given figure as shown.
Here,
breadth of the rectangle II = 7 m
And area of the rectangle II = 63 sq. m
∴ Length of rectangle II × Breadth of rectangle
II = 63 sq. m
⇒ Length of rectangle II × 7 m = 63 sq. m

⇒ Length of rectangle II = \(\frac{63}{7}\) = 9 m
= IG = HD
So, JC = IC – IJ = 9m – 5m = 4 m
GD = GH + HD = 2m + 9m = 11m
Now, area of rectangle III = 33 sq. m
∴ Length of rectangle III × Breadth of rectangle III = 33 sq. m
⇒ GD × Breadth of rectangle III = 33 sq. m
⇒ 11 m × Breadth of rectangle III = 33 sq. m
⇒ Breadth of rectangle III = \(\frac{33}{11}\) = 3 m
= FG = DE
Given, BE = 14 m
⇒ BC + CD + DE = 14 m
⇒ BC + 7 m + 3 m = 14 m
⇒ BC = 14 m – 7 m – 3 m = 4 m = AJ
As AJ = JC = 4 m, ABCJ is a square.
∴ Area of square I = AJ × JC = 4 × 4
= 16 sq. m.

Question 13.
A square and a rectangle have equal area. If the side of the square is 27 cm and length of rectangle is 81 cm, then find:
(i) the breadth of the rectangle.
(ii) the perimeter of the rectangle.
Solution:
(i) Given, side of the square = 27 cm
and length of the rectangle = 81 cm
∴ Area of the square = Side × Side
= 27 cm × 27 cm
= 729 sq. cm
As area of the rectangle = Area of the square
⇒ Length × Breadth = 729 sq. cm
⇒ 81 cm × Breadth = 729 sq. cm
⇒ Breadth = \(\frac{729}{81}\) = 9 cm
Thus, the breadth of the rectangle is 9 cm.

(ii) We know, perimeter of the rectangle
= 2 (Length + Breadth)
= 2 (81 cm + 9 cm)
= 2 × 90 cm = 180 cm

Question 14.
If the perimeter of the square is thrice the perimeter of a triangle whose sides are 3 cm, 4 cm and 5 cm, then find the area of the square.
Solution:
Given, sides of a triangle are 3 cm, 4 cm and 5 cm.
∴ Perimeter of the triangle
= 3 cm + 4 cm + 5 cm = 12 cm
As the perimeter of the square is thrice the perimeter of a triangle,
Perimeter of the square = 3 × Perimeter of the triangle
⇒ 4 × Side = 3 × 12 cm
⇒ Side = \(\frac{3 \times 12}{4}\) = 9 cm 4
Now, area of the square = Side × Side
= 9 cm × 9 cm
= 81 sq. cm
Thus, the area of the square is 81 sq. cm.

Question 15.
Find the area of the given figure (a) alongside.

Solution:
To find the required area, we divide the enclosed region of given figure into rectangles as shown in the figure.
Part I is a rectangle of length 10 m and breadth 2 m.
Part II is a rectangle of length 6 m and breadth 2 m.
Part III is a rectangle of length 6 m and breadth 2 m.
Part IV is a rectangle of length 10 m and breadth 2 m.
Now, area of part I = 10 m × 2 m = 20 sq. m
Area of part II = 6 m × 2 m = 12 sq. m
Area of part III = 6 m × 2 m = 12 sq. m
And area of part IV = 10m × 2m = 20 sq. m
Total area of given dosed figure
= 20 sq. m + 12 sq. m + 12 sq. m + 20 sq. m
= 64 sq. m

Question 16.
In the following figure, find the missing value of the area of a region.

Solution:
To find the required area, we divide the enclosed region of given figure into squares and rectangles as shown in the figure.

Now, breadth of the rectangle I = 3 cm
And area of the rectangle I = 21 sq. cm
∴ Length of rectangle I × Breadth of rectangle I
= 21 sq. cm
⇒ Length of rectangle I × 3 cm = 21 sq. cm
⇒ Length of rectangle I = \(\frac{21}{3}\) = 7 cm
So, Breadth of rectangle II = 7 cm – 4 cm
= 3 cm
Now, area of rectangle II = 27 sq. cm
∴ Length of rectangle II × Breadth of rectangle
II = 27 sq. cm
⇒ Length of rectangle II × 3 = 27 sq. cm
⇒ Length of rectangle II = \(\frac{27}{3}\) = 9 cm
So, breadth of region III = 9 cm – 4 cm
= 5 cm
Now, area of square III
= Side length × Side length
= 5 cm × 5 cm = 25 sq. cm

Question 17.
In the following figure, find the missing value of the area of a region.

Solution:
We name the given figure as shown.

Area of the rectangle I is 17 sq. cm, which can be written as 1 cm × 17 cm or 17 cm × 1 cm.
And area of the rectangle II is 51 sq. cm, which can be written as 3 cm × 17 cm or 17 cm × 3 cm.
Front figure, we can see that the breadth of the rectangles I and II are equal.
Therefore,
Length of rectangle I = 1 cm
Breadth of rectangle I = 17 cm
Length of rectangle II = 3 cm
Breadth of rectangle II = 17 cm
For rectangle III, length = 1 cm
Now, area of rectangle III = 30 sq. cm [Given]
⇒ Length of rectangle III × Breadth of rectangle III = 30 sq. cm
⇒ 1 cm × Breadth of rectangle III = 30 sq. cm
⇒ Breadth of rectangle III = 30 cm
So, length of rectangle IV = 3 cm and breadth of rectangle IV = 30 cm
∴ Area of rectangle IV = 3 cm × 30 cm
= 90 sq. cm

Question 18.
Find the area of the given figure.

Solution:
To find the required area, we divide the enclosed region of given figure into rectangles and squares as shown in the figure.
Part I is a square of side 6 cm.
Part II is a rectangle of length 6 cm and breadth 2 cm.
Part III is a square of side 6 cm.
Part IV is a rectangle of length 7 cm and breadth 2 cm.
Now, area of part I = 6 cm × 6 cm = 36 sq. cm

Area of part II = 6 cm × 2 cm = 12 sq. cm
Area of part III = 6 cm × 6 cm = 36 sq. cm
And area of part IV = 7 cm × 2 cm = 14 sq. cm
Total area of given closed figure
= 36 sq. cm + 12 sq. cm + 36 sq. cm + 14 sq. cm
= 98 sq. cm

Perimeter and Area Class 6 Case Based Questions

Question 1.
Priya and Rishi start running along the rectangular tracks as shown in the figure. Rishi runs along the outer track. Priya runs along the inner track. Now, they are wondering who ran more.

Based on the above information, answer the following questions:
(i) Find the distance covered by Priya in one round.
(ii) Find the distance covered by Rishi in one round.
(iii) Find out who ran the longer distance when Rishi runs 6 rounds and Priya runs 7 rounds.
Solution:
(i) Given, Priya runs along the rectangular track of length 130 m and breadth 80 m.
∴ Distance covered by Priya in one round
= Perimeter of the inner rectangular track
= 2(Length + Breadth)
= 2(130 m + 80 m) = 2 × 210 m = 420 m

(ii) Given, Rishi runs along the rectangular track of length 150 m and breadth 100 m.
∴ Distance covered by Rishi in one round
= Perimeter of the outer rectangular track
= 2(Length + Breadth)
= 2(150 m + 100 m) = 2 × 250 m = 500 m

(iii) Distance covered by Rishi in 6 rounds
= 6 × Distance covered by Rishi in one round
= 6 × 500 m = 3000 m
And, distance covered by Priya in 7 rounds
= 7 × Distance covered by Priya in one round
= 7 × 420 m = 2940 m

So, difference in the distance covered = 3000 m – 2940 m = 60 m
Thus, Rishi covers 60 m more distance than Priya.

Question 2.
Look at the plan of a house build on a rectangular plot as shown in the below figure.

Based on the above information, answer the following questions:
(i) Find the missing measurements.
(ii) Compare areas of the small bedroom & kitchen. Which one has a bigger area? By how much?
(iii) Compare areas of small bedroom & drawing room. Which one has a bigger area? By how much?
Solution:
(i) For store room, length = 8 ft and breadth = 7 ft
∴ Area of store room = 8 ft × 7 ft = 56 sq. ft

For toilet:
Length = Length of store room = 8 ft
And breadth = 6 ft
∴ Area of toilet = 8 ft × 6 ft = 48 sq. ft

For master bedroom:
Length = 15 ft
Breadth = 26 ft – Breadth of store room – Breadth of toilet
= 26 ft – 7 ft – 6 ft = 13 ft
∴ Area of master bedroom = 15 ft × 13 ft
= 195 sq. ft

For entrance:
Breadth = 8 ft
And, Area = 80 sq. ft
⇒ Length × Breadth = 80 sq. ft
⇒ Length × 8 ft = 80 sq. ft
⇒ Length = \(\frac{80}{8}\) = 10 ft

For drawing room:
Breadth = Breadth of master bedroom = 13 ft
Length = 38 ft – Length of master bedroom – Length of entrance
= 38 ft – 15 ft – 10 ft = 13 ft
∴ Area of drawing room = 13 ft × 13 ft = 169 sq. ft

For hall:
Breadth = Length of entrance =10 ft
Length = 26 ft – Breadth of entrance = 26 ft – 8 ft = 18 ft
∴ Area of hall = 18 ft × 10 ft = 180 sq. ft

For kitchen:
Breadth = 8 ft
Length = 26 ft – Breadth of drawing room
= 26 ft – 13 ft = 13 ft
∴ Area of kitchen = 13 ft × 8 ft = 104 sq. ft

For small bedroom:
Length = Length of kitchen = 13 ft
Breadth = 38 ft – Length of toilet – Breadth of kitchen – Breadth of hall ,
= 38 ft – 8 ft – 8 ft – 10ft = 12ft
∴ Area of small bedroom = 13 ft × 12 ft = 156 sq. ft

(ii) Area of small bedroom =156 sq. ft
Area of kitchen =104 sq. ft
Difference of areas = 156 sq. ft – 104 sq. ft = 52 sq. ft
Thus, the small bedroom has 52 sq. ft more area than the kitchen.

(iii) Area of small bedroom = 156 sq. ft
Area of drawing room = 169 sq. ft
Difference of areas = 169 sq. ft – 156 sq. ft = 13 sq. ft
Thus, the drawing room has 13 sq. ft. area more than small bedroom.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 4 Another Peek Beyond the Point MCQ improves accuracy in objective exams.

MCQ on Another Peek Beyond the Point Class 7

Another Peek Beyond the Point MCQ Class 7

Class 7 Maths Another Peek Beyond the Point MCQ

Question 1.
Which digit is in the thousandths place in the number 48.50736?
(a) 5
(b) 0
(c) 7
(d) 3
Solution:
(c) 7
In 48.50736, on the right side of the decimal point, the place values of digits are tenths (5), hundredths (0), thousandths (7) and so on.

Question 2.
Which of the following is the correct expansion of 32.406?
(a) 30 + 2 + 0.4 + 0.006
(b) 30 + 2 + 0.04 + 0.006
(c) 30 + 2 + 0.4 + 0.06
(d) 30 + 0.4 + 0.006
Solution:
(a) 30 + 2 + 0.4 + 0.006
The expanded form of 32.406 is given as
32.406 = 3 × 10 + 2 × 1 + 4 × \(\frac{1}{10}\) + 0 × \(\frac{1}{100}\) + 6 × \(\frac{1}{1000}\)
= 30 + 2 + 4 × 0.1 + 0 × 0.01 + 6 × 0.001
= 30 4- 2 + 0.4 + 0.006

Question 3.
The value of 0.385 + 4.07 + 12 is:
(a) 16.455
(b) 16.450
(c) 16.40
(d) 15.455
Solution:
(a) 16.455
Adding after aligning the decimal points in given numbers:

Question 4.
Which of the following is equal to 4.8 × 2.5?
(a) 10.2
(b) 12.0
(c) 11.75
(d) 9.5
Solution:
(b) 12.0
Multiplying the decimal numbers after converting them into fractions:
48 × 25 = \(\frac{48}{10} \times \frac{25}{10}=\frac{48 \times 25}{10 \times 10}=\frac{1200}{100}\) = 12

Question 5.
Cost of a pencil is ₹7.25. What is the cost of 12 pencils?
(a) ₹87
(b) ₹88.50
(c) ₹90
(d) ₹72.50
Solution:
(a) ₹87
Cost of one pencil = ₹7.25
∴ Cost of 12 pencils = 12 × 7.25
= (12 × 7) + (12 × 0.25)
= 84 + 12 × \(\frac{1}{4}\) = ₹87

Question 6.
7 + \(\frac{5}{10}+\frac{3}{100}+\frac{1}{1000}\) =
(a) 0.7531
(b) 75.31
(c) 7.531
(d) 0.7531
Solution:
(c) 7.531
7 + \(\frac{5}{10}+\frac{3}{100}+\frac{1}{1000}\)
= 7 + 0.5 + 0.03 + 0.001 = 7.531

Question 7.
A car covers a distance of 14.75 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
(a) 147.5 km
(b) 1475 km
(c) 1.475 km
(d) 14.075 km
Solution:
(a) 147.5 km
The distance covered by car in one litre of petrol = 14.75 km
The distance covered by car in 10 litres of petrol = 14.75 × 10 = 147.5 km

Question 8.
A shop packs sugar in 0.375 kg packets. If one full box contains 24 such packets, then the total weight of one full box is:
(a) 9.0 kg
(b) 8.75 kg
(c) 9.100 kg
(d) 8.25 kg
Solution:
(a) 9.0 kg
Weight of 1 packet of sugar = 0.375 kg
Since a box contains 24 packets of sugar,
Weight of one box = 0.375 kg × 24
= \(\frac{375}{1000} \times 24=\frac{9000}{1000}\)
= 9.0 kg

Question 9.
48.6 ÷ 10 =
(a) 4.86
(b) 0.486
(c) 48.06
(d) 486
Solution:
(a) 4.86
48.6 ÷ 10 = \(\frac{486}{10} \div 10=\frac{486}{10} \times \frac{1}{10}=\frac{486}{100}\) = 4.86

Question 10.
Which of the following division gives a quotient less than the dividend?
(a) 12.5 ÷ 0.5
(b) 7.29 ÷ 9
(c) 4.8 ÷ 0.08
(d) 18.6 ÷ 0.3
Solution:
(b) 7.29 ÷ 9
We know that in decimal division, if divisor is greater than 1, the quotient is less than the dividend.
Now, 9 > 1, so on dividing 7.29 by 9 we will get a quotient (0.81) less than 7.29.

Question 11.
Which of the following statements are correct with respect to the multiplication of the decimal number 8.879 by 89?
(i) 8.879 × 89 = 8.87 × 89 + 0.009 × 89
(ii) 8.879 × 89 = 8 × 89 + 0.8 × 89 + 0.79 × 89
(iii) 8.879 × 89 = 8 × 89 + 0.8 × 89 + 0.07 × 89 + 0.009 × 89
(iv) 8.879 × 89 = 8 × 89 + 0.879 × 89
Choose the correct option from the following:
(a) Only (i), (ii) and (iii)
(b) Only (iii)
(c) Only (i), (iii) and (iv)
(d) All of the above
Solution:
(c) Only (i), (iii) and (iv)
The expanded form of 8.879 is:
8.879 = 8 × 1 + 8 × \(\frac{1}{10}\) + 7 × \(\frac{1}{100}\) + 9 × \(\frac{1}{1000}\)
= 8 + 0.8 + 0.07 + 0.009
Now, 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = 8 × 89 + 0.8 × 89 + 0.07 × 89 + 0.009 × 89
or 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = (8.87 + 0.009) × 89 = 8.87 × 89 + 0.009 × 89
or 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = (8 + 0.8 + 0.079) × 89 = 8 × 89 + 0.8 × 89 + 0.079 × 89
or 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = (8 + 0.879) × 89 = 8 × 89 + 0.879 × 89
∴ The statements (i), (iii) and (iv) are correct.

Question 12.
Which of the following multiplications of decimal numbers are correct?
(i) 0.5 × 0.05 = 0.025
(ii) 1.2 × 100 = 1200
(iii) 2.5 × 4 = 10
(iv) 0.1 × 0.1 = 0.1
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iii)
(d) (ii), (iii) and(iv)
Solution:
(a) (i) and (iii) only
Statements (i) and (iii) are correct as 25
0.5 × 0.05 = \(\frac{5}{10} \times \frac{5}{100}=\frac{25}{1000}\) = 0.025 and
2.5 × 4 = 2 × 4 + 0.5 × 4 = 8 + 2 = 10
But statements (ii) and (iv) are incorrect as 1.2 ×x 100 = 120 and 0.1 × 0.1 = 0.01

Question 13.
Which of the following years are not leap years?
(i) 2024
(ii) 2100
(iii) 2000
(iv) 1800
Choose the correct option from the following:
(a) (ii) and (iv)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Solution:
(a) (ii) and (iv)
A year divisible by 4 is a leap year; a year divisible by 100 is not a leap year; however, a year divisible by 400 is a leap year.
The year 2024 is divisible by 4 and not divisible by 100, so it is a leap year.
The year 2100 is divisible by 100 but not divisible by 400, so it is not a leap year.
The year 2000 is divisible by 400, so it is a leap year.
The year 1800 is divisible by 100 but not divisible by 400, so it is not a leap year. Therefore, the years that are not leap years are 2100 and 1800.

Another Peek Beyond the Point Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): 32 × 3 tenths = 32.3
(R): Moving the decimal point to the right makes the number bigger.
Solution:
(d) A is false but R is true.
32 × 3 tenths = 32 × \(\frac{3}{10}=\frac{32 \times 3}{10}\)
= \(\frac{96}{10}\) = 9.6
So, Assertion (A) is false.
We know that multiplying a number by 10 shifts the decimal point one place to the right, making the number bigger.
For example, 1.2 × 10 = 12
So, Reason (R) is true.

Question 2.
(A): When 56.3 is multiplied by 1000, the result is 56300.
(R): When multiplying a decimal number by 10, 100, or 1000, the decimal point shifts to the left by as many places as there are zeroes in the multiplier.
Solution:
(c) A is true but R is false.
We know that, when multiplying a decimal number by 10, 100, or 1000, the decimal point shifts to the right (not left) by as many places as there are zeroes with one.

By using the above logic, when 56.3 is multiplied by 1000, the result is 563,00.
∴ Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): The area of a square with a side length of 0.5 cm is 2.5 sQuestion cm.
(R): The area of a square of side a is calculated using the formula a × a.
Solution:
(d) A is false but R is true.
We know that the area of a square of side a is calculated using the formula a × a.
So, if the side length, a = 0.5 cm, then
area = a × a = 0.5 cm × 0.5 cm = 0.25 sq. cm
∴ Assertion (A) is false, but Reason (R) is true.

Question 4.
(A): The year 3100 is a leap year.
(R): A century year that is not divisible by 400 is not a leap year.
Solution:
(d) A is false but R is true.
Since 3100 is a century year, it must be divisible by 400 to be a leap year.
But 3100 is not divisible by 400, so it is not a leap year
Therefore, Assertion (A) is false, but Reason (R) is true.

Another Peek Beyond the Point Class 7 Fill in the Blanks

Question 1.
In the blanks below, fill ‘<’, ‘>’ or ‘ = ’ after analysing the expressions on the LHS and RHS.
(i) 17.89 × 5.9 ______ 185.2 × 0.57
(ii) 5.6 × 12.72 _____ 89.04 × 0.8
Solution: <, =
(i) LHS = 17.89 × 5.9
= \(\frac{1789}{100} \times \frac{59}{10}=\frac{105551}{1000}\) = 105.551
RHS = 185.2 × 0.57 = \(\frac{1852}{10} \times \frac{57}{100}=\frac{105564}{1000}\)
= 105.564
The whole number parts of both are equal i.e. 105.
Also, the digit in the tenths place is the same for both i.e. 5.
However, the digit in the hundredths place of the LHS i.e. 5 is less than that of the RHS i.e. 6.
Since 5 < 6, we conclude:
105.551 < 105.564
So, 17.89 × 5.9 < 185.2 × 0.57.

(ii) LHS = 56 × 12 72 = \(\frac{56}{10} \times \frac{1272}{100}=\frac{71232}{1000}\)
= 71.232
RHS = 8904 × 08 = \(\frac{8904}{100} \times \frac{8}{10}=\frac{71232}{1000}\)
= 71.232
Since all digits match, we conclude:
5.6 × 12.72 = 89.04 × 0.8

Question 2.
When a decimal number is multiplied by 10000, the decimal point shifts to the right by ______ places.
Solution: four
We know that the number of zeroes in the multiplier (10, 100, 1000, …) determines how many places the decimal point shifts to the right.
Since the multiplier (10000) hag 4 zeroes, the decimal point shifts four places to the right.

Question 3.
6.7 × 10 + 6.7 × 100 + 0.67 × 1000 = _______
Solution: 1407
6.7 × 10 + 6.7 × 100 + 0.67 × 1000
= 67 + 670 + 670 = 1407

Question 4.
Evaluate the following using the information 256 × 12 = 3072.
(i) 25.6 × 1.2 = _______
(ii) 307.2 ÷ 1.2 = _______
(iii) 30.72 ÷ 25.6 = ______
(iv) 0.256 × 0.12 = _____
Solution: 30.72, 256, 1.2, 0.03072
(i) 25.6 × 1.2 = \(\frac{256}{10} \times \frac{12}{10}\)
= \(\frac{256 \times 12}{100}=\frac{3072}{100}\) = 30.72

(ii) 307.2 ÷ 1.2 = \(\frac{3072}{10} \div \frac{12}{10}\)
= \(\frac{3072}{10} \times \frac{10}{12}=\frac{256 \times 12}{12}\) = 256

(iii) 30.72 ÷ 25.6 = \(\frac{3072}{100} \div \frac{256}{10}\)
= \(\frac{256 \times 12}{100} \times \frac{10}{256}=\frac{12}{10}\) = 1.2

(iv) 0.256 × 0.12 = \(\frac{256}{1000} \times \frac{12}{100}\)
= \(\frac{256 \times 12}{100000}=\frac{3072}{100000}\) = 0.03072

Question 5.
Fill in the blanks:
(i) 5.5 km = _____ m
(ii) 125.5 mL = _________ L
(iii) 14.5 cm = ______ mm
(iv) 68 g = _______ kg
Solution: 5500, 0.1255, 145, 0.068
(i) As 1 km = 1000 m
∴ 5.5 km = 5.5 × 1000 = 5500 m

(ii) As 1 mL = \(\frac{1}{1000}\) L
∴ 125.5 mL = \(\frac{125.5}{1000}\) = 0.1255 LHS

(iii) As 1 cm = 10 mm
∴ 14.5 cm = 14.5 × 10 = 145 mm

(iv) As 1 g = \(\frac{1}{1000}\) kg
∴ 68 g = \(\frac{68}{1000}\) = 0.068 kg

Go through BSE Odisha Class 8 Science Solutions Chapter 10 Light: Mirrors and Lenses Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 10 Question Answer

Class 8 Science Ch 10 Light: Mirrors and Lenses Question Answer

Class 8 Science Chapter 10 Light: Mirrors and Lenses Question Answer

Probe and Ponder Questions

Question 1.
Can we make mirrors which can give enlarged or diminished images?
Answer:
Yes, we can make mirrors that produce ehnlarged or diminished images. Specially, concave mirrors can form both enlarged and diminished images depending on the object’s distance from the mirror, while convex mirrors always produce diminished images. Concave mirrors give magnified images when the object is close (between the pole and the focal point) and diminished images when the object is far (beyond the centre of curvature). Convex mirrors always form virtual, erect and similar images.

Question 2.
On side-view mirrors of vehicles, there is a warning that says “Objects in mirror are closer than they appear”. Why is this warning written there?
Answer:
The warning “Objects in mirror are closer than they appear” is written on the side-view mirrors of vehicles because these mirrors are typically convex. Convex mirrors make objects appear smaller and therefore seem farther away than they really are. The warning reminds drivers to allow extra distance when judging how near other vehicles or objects are.

Question 3.
Why is there a curved line on some reading glasses?
Answer:
The curved line on reading glasses is due to the convex shape of the lens. These lenses are thicker in the middle and thinner at the edges, causing light rays to converge (bend inward) and improve near vision for individuals with presbyopia. This curvature is essential for correcting the vision by focusing light properly on the retina.

Question 4.
Share your questions ………….
Answer:
Here are some fun questions you might think of from the chapter:

• How does a magnifying glass make things look bigger?
• Why do spoons act like funny mirrors?
• Can lenses in our eyes change shape?
• What happens if you use a concave mirror to focus sunlight?

InText Questions

Question 1.
How can we distinguish between concave and convex mirrors? (Page 155)
Answer:
To distinguish between concave and convex mirrors, observe how they reflect light and the images they form. Concave mirrors curve inward, causing parallel light rays to converge at a focal point and can form both real (inverted) and virtual (erect) images depending on object distance.

Convex mirror curve outward, causing parallel light rays to diverge and always form virtual, erect and diminished images. A simple test is to hold the mirror near an object and move it; a concave mirror can magnify objects when close while a convex mirror always makes them appear smaller.

Question 2.
Do you remember learning about the use of telescope in the chapter ‘Beyond Earth’ in Curiosity, Grade 6? (Page 156)
Answer:
Yes, some laws govern image formation by mirrors, including plane, concave and convex mirrors. These laws are based on the laws of reflection, which state that the angle of incidence (the angle at which light hits the mirror) is equal to the angle of reflection (the angle at which light bounces off the mirror). Additionally, the incident ray, the reflected ray, and the normal (an imaginary line perpendicular to the mirror’s surface at the point of incidence) all lie in the same plane.

Question 3.
We have observed images formed by three types of mirrors-plane, concave, and convex. But are there any laws which govern the image formation? (Page 157)
Answer:
Yes, the two laws of reflections which govern the image formation.

Question 4.
Are laws of reflection applicable to spherical mirrors also? (Page 160)
Answer:
Yes, the laws of reflection apply to spherical mirrors. The same rules-angle of incidence equals angle of reflection and the incident ray, reflected ray and normal lie in the same plane – hold at every point on a spherical mirror’s surface. Using these laws and simple geometry we can predict where an image will form for concave and convex spherical mirrors.

Question 5.
We explored the images of an object formed by curved mirrors. But how do objects look when viewed through transparent materials with curved surfaces? (Page 162)
Answer:
When we view the printed letters by a magnifying glass which is a convex lens, it appear bigger in size. In case of concave lens the image is smaller than object.

Question 6.
What changes can be seen in the objects when viewed through lenses? (Page 163)
Answer:
Image formed by a convex lens can be enlarged, diminished or of the same size as the object and it may be erect or inverted, depending upon the distance of the object from the lens. But image formed by a concave lens is always erect and diminished in size.

Question 7.
Do lenses also converge or diverge the light beam? Can it also burn a paper ? (Page 163)
Answer:
Yes, lenses can both converge and diverge light beams. Convex (convering) lenses bring parallel rays together at a focus, while concave (diverging) lenses cause parallel rays to spread out as if they came from a virtual focus.

Question 8.
Since a convex lens converges a light beam, can it also burn a paper? (Page 164)
Answer:
Yes. A convex lens can concentrate sunlight to a small spot and raise the temperature there enough to scorch or even ignite paper. This is the same principle used in a magnifying glass to focus sunlight.

Question 9.
Where all are the lenses used ? (Page 165)
Answer:
Lenses are used in many devices such as eyeglasses, cameras, microscopes, telescopes, projectors, magnifying glasses and medical instruments. Different lens types (convex or concave) are chosen depending on whether the light needs to be converged or diverged.

Light: Mirrors and Lenses Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 166-169)

Question 1.
What is the angle made by the reflected ray with the mirror?

(i) 40°
(ii) 50°
(iii) 45°
(iv) 60°
Answer:
(ii) If the incident ray makes a 40° angle with the normal, then :
Angle of incidence (i) =40°
According to the law of reflection,
Angle of reflection (r) = Angle of incidence
(i) = 40°
Now, Angle between reflected ray and mirror = 90° angle of reflection
= 90°-40°=50°

Question 2.
The figure shows three different situations where a light ray falls on a mirror:

(i) The light ray falls along the normal.
(ii) The mirror is tilted, but the light ray still falls along the normal to the tilted surface.
(iii) The mirror is tilted, and the light ray falls at an angle of 20° from the normal.
Draw the reflected ray in each case (Use a ruler and protractor for accurate drawing). What is the angle of reflection in each case?
Answer:
(i) Light ray falls along the normal Angle of incidence = 0°
Angle of incidence =0°
Angle of reflection =0°
The ray retraces its path. It reflects straight back along the same line.

(ii) The mirror is tilted, but the light ray still falls along the normal to the tilted surface.
Even though the mirror is tilted, the ray is still perpendicular to the surface.
Angle of incidence = 0°
Angle of reflection = 0°
The ray again retraces its path – the tilt doesn’t affect the reflection if the ray is normal to the surface.

(iii) The mirror is tilted, and the light ray falls at 20° from the normal.
Angle of incidence =20°
By the law of reflection:
Angle of reflection =20°
The reflected ray will make a 20° angle with the normal, on the opposite side.

Question 3.
In the figure, the cap of a sketch pen is placed in front of three types of mirrors.

Match each image with the correct mirror.

Answer:

Image	Mirror
(i) (ii) (iii)	Convex mirror Concave mirror Plane mirror

Question 4.
In the Figure, the cap of a sketch pen is placed behind a convex lens, a concave lens, and a flat transparent glass piece- all at the same distance.

Match each image with the correct type of lens or glass.

Answer:

Image	Lens/glass type
(i) (ii) (iii)	Convex lens Concave lens Flat transparent glass piece

Question 5.
When the light is incident along the normal on the mirror, which of the following statements is true:
(i) Angle of incidence is 90°
(ii) Angle of incidence is 0°
(iii) Angle of reflection is 90°
(iv) No reflection of light takes place in this case
Answer:
(ii) When light hits a surface at a 90° angle (perpendicular), it is considered to be incident normally.
Angle of incidence = Angle of reflection: In reflection, the angle of the reflected light is always equal to the angle of the incident light. Since the light is hitting the mirror normally (at a 0° angle of incidence), the reflected light will also be at a 0° angle of reflection.

Question 6.
Three mirrors-plane, concave, and convex, are placed in the figure. Based on the images of the graph sheet formed in the mirrors, identify the mirrors and write their names above the mirrors.

Answer:

Question 7.
In a museum, a woman walks towards a large convex mirror (Figure). She will see that :
(i) her erect image keeps decreasing in size.
(ii) her inverted image keeps decreasing in size.
(iii) her inverted image keeps increasing in size, and eventually it becomes erect and magnified.
(iv) her erect image keeps increasing in size.

Answer:
The correct answer is (i) her erect image keeps decreasing in size. Because a convex mirror always forms a virtual, erect, and diminished image.

Question 8.
Hold a magnifying glass over the text and identify the distance at which you can see the text bigger than they are written. Now move it away from the text. What do you notice? Which type of lens is a magnifying glass?
Answer:
A magnifying glass uses a convex lens to make text appear larger. When we hold the magnifying glass close to the text and then slowly move it away, the image of the text will first appear larger and sharper, then progressively get larger and more blurred, and finally invert and appear smaller again.

Question 9.
Match the entries in Column I with those in Column II.

Answer:

Column I	Column II
(i) Concave mirror	(a) Spherical mirror with a reflecting surface that curves inwards.
(ii) Convex mirror	(b) It forms an image which is always erect and diminished in size.
(iii) Convex lens	(c) An object placed behind it may appear inverted at some distance.
(iv) Concave lens	(d) The object placed behind it always appears diminished in size.

Question 10.
The following question is based on Assertion/Reason.
Assertion: Convex mirrors are preferred for observing the traffic behind us.
Reason: Convex mirrors provide a significantly larger view area than plane mirrors.
Choose the correct option:
(i) Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct, but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
Answer:
(i) Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion.

Question 11.
In the Figure, note that O stands for object, M for mirror, and I for image.

Which of the following statements is true?
(i) Figure (a) indicates a plane mirror, and Figure (b) indicates a concave mirror.
(ii) Figure (a) indicates a convex mirror and Figure (b) indicates a concave mirror.
(iii) Figure (a) indicates a concave mirror and Figure (b) indicates a convex mirror.
(iv) Figure (a) indicates a plane mirror, and Figure (b) indicates a convex mirror.
Answer:
(ii) Figure (a) indicates a convex mirror and Figure (b) indicates a concave mirror.

Question 12.
Place a pencil behind a transparent glass tumbler (Figure a). Now fill the tumbler halfway with water (Figure b). How does the pencil appear when viewed through the water? Explain why its shape appears changed.

Answer:
When we place a pencil behind a transparent glass tumbler and fill the tumbler halfway with water, the pencil will appear to be bent or broken at the point where it enters the water. The submerged portion of the pencil might also appear slightly thicker or shifted from its actual position. This phenomenon is called refraction of light.

Why it Happens
Change in medium: Light travels at different speeds in different materials (media). Air is an optically rarer medium, meaning light travels faster in it, compared to water, which is an optically denser medium where light travels more slowly.

Bending of Light (Refraction): When light rays from the pencil travel from the water (denser medium) into the air (rarer medium) and then into our eyes, their speed changes. This change in speed causes the light rays to bend or deviate from their original path. Specifically, when light goes from a denser medium to a rarer medium, it bends away from the normal (an imaginary line perpendicular to the surface at the point where light enters the new medium).

Apparent Position: Because our brains interpret light rays as traveling in straight lines, the bending of light at the water-air interface makes the submerged part of the pencil appear to be at a different location than its actual position, creating the illusion of a bent or broken pencil.

The extent of this bending depends on the angle at which we view the pencil and the difference in optical density between the two media (water and air). If we view the pencil straight from above, we won’t observe the bending as the light rays would be traveling perpendicular to the surface, and refraction would be minimal.

Class 8 Science Chapter 10 Question Answer

Activity 1.

Let us explore
Aim: To study the image of an object formed by a curved reflecting surface.
Materials Required: A highly polished large stainless steel spoon.

Procedure:

• Take a highly polished large stainless steel spoon and hold its curved surface close to your face. Can you see your image in it?
• You will notice that the image of your face is different from the image you see in a plane mirror.
• Now, observe the image, by moving spoon slowly away from your face. Do you observe any change in the image?
• Repeat the same steps by flipping the spoon.

Observations:

• When we looked at the inner side of spoon which is curved inwards, we observed that the image was inverted.
• When we looked at the outer side of the spoon which is bulges outwards, the image of our face was erect but smaller in size.

Inferences:

• Curved mirrors are a part of hollow glass sphere.
• The reflecting surface of the spherical mirror may be curved inwards or outwards.

Activity 2.

Let us distinguish

Aim: To distinguish between concave and convex mirror.
Materials Required: A concave mirror, a convex mirror.

Procedure:

• Keep a concave and a convex mirrors on a table whose reflecting surfaces facing upwards.
• Now look at them from the side, keeping your eye at their level, so that you can identify whether the reflecting surface is curved inwards or outwards.

Observations:

• The Reflecting surface of one mirror is curved inwards.
• The Reflecting surface of other mirror is curved outwards.

Inference:

Concave mirror	Convex mirror
1. Reflecting surface is curved inwards.	1. Reflecting surface is curved outwards.
2. A thin layer of aluminium is coated on outer curved surface.	2. A thin layer of aluminium is coated on inner curved surface.
3. It forms both type of images: erect and inverted.	3. It forms only erect images at all distances.

Activity 3.

Let us explore
Aim: To study the characteristics of images formed by spherical mirrors.
Materials Required: A concave mirror, a convex mirror, two small wooden blocks, a small toy or some other object.

Procedure:

• Take two mirrors and hold them side by side in an upright position on a table.
• Keep the object in front of them at a small distance (3-4 cm away) as shown in Figure (a).
• What kind of images do you see in each mirror? Are the images of the same size as the object? Are they erect? Do you see lateral inversion in the images?
• Note down your observation.
• Now, move the object gradually from the mirror.
• What changes do you see in the images in both the mirrors? Do the images become smaller or larger? Do they continue to be erect? Again, note down your observations.
• Repeat the above steps with each mirror separately.

Observations:

1. When object is close to the mirror.

Concave	Convex
(i) Enlarged size (Larger)	(i) Diminished size (Smaller)
(ii) Erect	(ii) Erect
(iii) Lateral inversion is seen.	(iii) Lateral inversion is seen.

2. When object is moved away:

Concave	Convex
(i) The size decreases as the distance increases.	(i) Smaller at all distance.
(ii) Inverted	(ii) Remains erect
(iii) Lateral inversion is seen.	(iii) Lateral inversion is seen.

Inference :
For a Convex Reflecting Surface:

• An erect and smaller image is formed.
• The image remains erect and smaller at all distances.

For a Concave Reflecting Surface:

• At smaller distance, the image is erect and enlarged.
• At larger distances, the image formed is inverted. The size of the image decreases as the distance from the spoon increases.

Conclusion: Common in Both: Lateral inversion (left-right reversal) is observed in the images.

Activity 4.

Let us experiment
Aim: To verify the first laws of reflection.
Materials Required: A drawing board, a white sheet of paper, a comb, a torch, a strip of plane mirror.

Procedure:

• Fix a white sheet of paper on a drawing board.
• Take a comb and close all its openings except one in the middle.
• Hold the comb perpendicular to the sheet of paper.
• Throw light from a torch through the opening of the comb from one side. With slight adjustment of the torch and the comb you will see a ray of light along the paper on the other side of the comb.
• Keep the comb and the torch steady.
• Place a strip of plane mirror in the path of the light ray.
• What do you observe ?
  After striking the mirror, the ray of light is reflected in another direction.
• After striking the mirror, the ray of light is reflected in another direction. The light ray, which strikes any surface, is called the incident ray. The ray that comes back from the surface after reflection is known as the reflected ray.
• Daw lines showing the position of the plane mirror, the incident ray and the reflected ray on the paper. Remove the mirror and the comb. Draw a line making an angle of 90° to the line representing the mirror at the point where the incident ray strikes the mirror. This line is known as the normal to the reflecting surface at that point.
• The angle between the normal and incident ray is called the angle of incidence (∠i). The angle between the normal and the reflected ray is known as the angle of reflection (∠r).
• Measure the angle of incidence and the angle of reflection.
• Repeat the activity several times by changing the angle of incidence.

Observations:

S.No	Angle of incidence (∠i)	Angle of Reflection (∠r)
1.	30°	30°
2.	35°	35°
3.	45°	45°
4.	50°	50°
5.	60°	60°

Inference: Angle of incidence is always equal to the angle of reflection. This is known as the law of reflection.
Incident ray of light: Ray of light which falls on a polished surface.
Reflected ray of light: Ray of light which gets reflected from a polished surface.
Normal is a line at right angle to the reflecting surface at the point of incidence.
Angle of incidence: Angle made by incident ray with the normal.
Angle of reflection: Angle made by reflected ray with the normal.

Laws of Reflection: There are two laws of reflection :
Law 1: The angle of reflection (∠r) is equal to the angle of incidence (∠i)
Law 2: The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.

Activity 5.

Let us experiment

Aim: To verify the second law of reflection, i.e. the Incident ray, Reflected ray and the normal all lie in the same plane.
Materials Required: A mirror strip, holder, ray box.

Procedure:

• Fix a white sheet of paper on a drawing board in such a way that a small portion of it project a little beyond the edge of the drawing board.
• Place the plane mirror strip on the sheet of paper and hold it vertically with a mirror stand.
• Throw a ray of light on the mirror using a ray box.
• Look at the reflected ray. The reflected ray extends the projected portion of the paper.
• Make a cut in the middle of the projected portion of the sheet.
• Bend that portion of the projected sheet on which the reflected light falls.
• The reflected ray of light is not seen on the bent portion of the sheet.

Inference: The entire sheet fixed on to the drawing board represents a plane. The incident, reflected ray and the normal lie in the plane of paper.

Activity 6.

Let us explore
Aim: To observe the reflection of multiple parallel rays fall on the plane mirror and spherical mirrors.
Materials Required: A plane mirror, a concave mirror, a convex mirror, stands for mirrors, a torch, a comb, a paper clip.

Procedure:

• Arrange the setup as mentioned in Activity 4 again.
• Instead of a single slit, you have to leave many opening of the comb uncovered so that you may obtain multiple parallel beams of light.
• Now, we will fall these multiple parallel beams of light upon the plane mirror, concave mirror, and convex mirror, respectively. Observe the reflected beams. Is your observation same as what is shown in figure (b), (c) and (d)?

Observations:

Case I: Multiple reflected beams are also parallel [Figure (b)].
Case II: When multiple beams of light fall upon a concave mirror, the multiple refiected beams get closer, that is, they converge [Figure (c)].
Case III: In the case of a convex mirror, the multiple reflected beams spread, that is, they diverge [Figure (d)].

Inference:

• Spherical mirror also follows the laws of reflection.
• A concave mirror converge (get closer) the multiple beam of light after reflection.
• A convex mirror diverge (spread) the multiple beam of light after reflection.
• But in a plane mirror multiple reflected beam of light remains parallel.

Activity 7.

Let us explore

Aim: To show that concave mirror is a converging mirror.
Materials Required: A concave mirror, a sheet of thin paper or newspaper.

Procedure:
1. Take a concave mirror and move its reflecting surface towards the Sun. Let the light of the Sun be reflected by the mirror on the sheet of paper.

2. We should adjust the distance of the paper so that you can get a sharp bright spot on it see in figure.
3. You must hold the mirror and the sheet of paper contineously for a few minutes. Does the paper start to burn producing smoke?
Observation: The bright spot is formed on the paper and paper get ignited after few minutes.

Inference:

• Bright spot is due to concentration of reflected light from the sun. i.e., reflected light converged at a point after reflection.
• This produce sufficient heat at this point which can ignite the paper.
• Bright spot is actually the image of the sun.

Precaution:

• You should always perform this activity under the guidance of a teacher or an adult.
• We should not look towards the Sun or into the mirror reflecting the Sun.
• We should focus the reflected light only on a piece of paper, not towards anyone’s face or eyes.

Activity 8.

Let us explore
Aim: To read the printed letters through glass/plastic strip.
Materials Required: A flat strip of glass or clear plastic, such as a flat scale, few drops of oil, dropper, water and a paper or book.

Procedure:

• Take a glass or plastic strip.
• Spread few drops of oil on it and rub it to leave a very thin coating. We can also use wax instead of oil.
• Now with the help of a dropper or our finger, put a few drop of water on the oiled/waxed spot. (The oil/wax will help the water form a nice round drop.)
• Observe the water drop. What is the shape of its surface? Is it flat or curved inward or curved outward?
• Now, observe the paper underneath the glass/plastic strip. The paper or book should be directly under the water drop (see figure).
• Now, look down through the water drop at the text below. Do you find some change in the size of the letters just below the water drop? Do they look enlarged or smaller?

Observation:

• The surface of the water drop is curved outside.
• The letters under the water drop appear than the letters nearby.

Inference:

• The curved drop of water acts like a simple lens.
• This is similar to magnifying glass which is also a lens that helps in reading small print making the letters appear larger.

Activity 9.

Let us experiment
Aim: To show the formation of images by a lens when the objects are placed at different distances.
Materials Required: A convex lens, a concave lens, a lens holder or stand and a smaller object.

Procedure:
(A) For a convex lens

• Take a convex lens and mount it on a lens stand. Place it on a table.
• Place a lighted candle at a known distance (say 50 cm) from the lens.
• Place a white paper screen on the other side of the lens.
• Move the screen towards or away from the lens to get a sharp image of the candle flame.
• Since the image is obtained on the screen, therefore the image formed is real. Note down the positions of the lens, the candle, and that of the screen.
• Now move the candle by 10 cm towards the lens and try to obtain its image on the screen by moving the screen. Note down the positions of the lens, the candle and the screen.
• Repeat the experiment by changing the position of the candle and that of the screen.

Observation:

Inference:

• Convex lens forms a real and inverted image of the object when placed away from the lens.
• Convex lens forms virtual and erect image when the object is placed very near to the lens.

Procedure:

• Mount a concave lens on a lens stand and keep it on a table.
• Place a lighted candle at a known distance (say 50 cm) from the lens.
• Place white paper sheet on the other side of the lens.
• Move the screen towards or away from the lens and look for the image of the candle flame. Image of the candle flame is not seen on the screen.
• Now fix the position of the screen and move the candle towards the lens. Image of the candle flame is not seen on the screen.
• Remove the screen and look into the concave lens from that side. An erect virtual and smaller image is seen through the concave lens.
• Look into the concave lens for different distances from the lens and record your observations.
  (i) When the object is far off from the lens.
  (ii) When the object is nearer to the lens.

Observations and Inference:

• The image formed by a concave lens cannot be obtained on a screen. Therefore, the images formed by a concave lens are virtual.
• The image formed by a concave lens for all distances from the lens is erect and smaller in size than the object.

Activity 10.

Let us investigate
Aim: To show that convex lens is a congerging lens and concave lens is a diverging lens.
Materials Required: A thin transparent glass plale, a convex lens, a concave lens, a torch and a comb, a paper clip, two books of same size, white sheet of paper.

Procedure:

• Hold a comb upright with a paper clip to obtain multiple parallel beams of light.
• With the help of two books placed adjacent to each other, fix the glass plate or lens upright in between them see in figure and spread paper sheets on both books.
• Now pass the multiple parallel beams of light through the thin glass plate, convex lens, and concave lens one by one as shown in figure. Does the parallel beam of light pass through without any change in its direction in all three cases?
• Note down and analyse your observations.

Observation:

• The beam of light passes through the thin glass place without any deviation (i.e., no change in its path).
• In the convex lens, beam of light converges after passing through it (i.e., come closer).
• In concave lens, beam of light diverges after passing through it (i.e., beam spread).

Inference: A convex lens is called a converging lens and a concave lens is called a diverging lens. We draw the following figures for above Activity-10.

Activity 11.

Let us investigate
Aim: To show that a convex lens converges sunlight at a point and can burn a piece of paper.
Materials Required: A convex lens, a sheet of thin paper or newspaper.

Procedure:

• Hold a comb upright with a paper clip to obtain multiple parallel beams of light.
• With the help of two books placed adjacent to each other, fix the glass plate or lens upright in between them see in figure and spread paper sheets on both books.
• Now pass the multiple parallel beams of light through the thin glass plate, convex lens, and concave lens one by one as shown in figure. Does the parallel beam of light pass through without any change in its direction in all three cases?
• Note down and analyse your observations.

Observation:

• The beam of light passes through the thin glass place without any deviation (i.e., no change in its path).
• In the convex lens, beam of light converges after passing through it (i.e., come closer).
• In concave lens, beam of light diverges after passing through it (i.e., beam spread).

Inference:

• Bright spot is due to concentration of reflected light from the sun. i.e., reflected light converged at a point after reflection.
• This produce sufficient heat at this point which can ignite the paper.
• Bright spot is actually the image of the sun.

Precautions: Do not look at the Sun directly or through the lens as it may damage your eyes.

Light: Mirrors and Lenses Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
State the characteristics of the image formed by a plane mirror.
Answer:

• Plane mirror forms an erect image.
• It forms a virtual image.
• Size of the image is same as that of the object.
• Image gets formed at the same distance behind the mirror as the object stands in front of it.
• Image formed is a laterally inverted image. i.e. right hand side of the object seems to be the left hand side and vice versa.

Question 2.
What is a virtual image ? Give one situation where a virtual image is formed.
Answer:
The image which cannot be taken on a screen is called a virtual image. When some object is placed very close to the concave mirror we do not get any image of that object on the white screen placed in front of the mirror. Such image is called a virtual image.

Question 3.
What is the difference between virtual images produced by concave, plane and convex mirrors ?
Answer:
Virtual image produced by concave mirror is magnified, that produced by plane mirror is of the same size and the virtual image produced by convex mirror is diminished.

Question 4.
State one way in which the image formed in a convex mirror is similar to that in plane mirror and one way in which it is different.
Answer:
Similarity: Both convex mirror and plane mirror always form a virtual and erect image of an object.
Difference: Convex mirror always forms an image which is smaller than the object but a plane mirror always forms an image which is of the same size as the object.

Question 5.
Which type of mirror is used as a shaving mirror ? Support your answer with reason.
Answer:
Concave mirrors are used as shaving mirrors to see a large image of the face. This is because when the face is held within the focus of a concave mirror, then an enlarged image of the face is seen in the concave mirror. This helps in making a smooth shave.

Long Answer Type Questions

Question 1.
How can you distinguish between plane mirror, convex mirror and concave mirror by merely looking at the image formed in each case ?
Answer:
To distinguish between a plane mirror, a convex mirror and a concave mirror, the given mirror is held near the face and the image is seen.

• If the image is upright and same size as the object and, so it does not change in size when the mirror is moved, then the mirror is a plane mirror.
• If the image is upright magnified and it becomes inverted when the mirror is moved away from the face, then the mirror is a concave mirror.
• If the image is upright and diminished and it remains upright when the mirror is moved away from the face, then the mirror is a convex mirror.

Question 2.
Write the uses of solar concentrator.
Answer:

• Solar concentrators are devices which can concentrate sunlight to a small area with the help of mirrors and lenses.
• The concentrated sunlight is utilised to heat a liquid to produce steam which can be used to generate electricity or for providing heat for various purposes, such as large scale cooking or for solar furnaces.
• Solar furnaces are even used for melting steel! Do you remember learning in an earlier chapter, about electric furnaces for melting steel?

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow :

If the reflecting surface of a spherical mirror is concave, it is called a concave mirror. If the reflecting surface is convex, then it is a convex mirror. The inner surface of a spoon acts like a concave mirror, while its outer surface acts like a convex mirror. The image of an object formed by a plane mirror cannot be obtained on a screen. Let us investigate if it is also true for the image formed by a concave mirror. The image formed by a plane mirror could not be obtained on a screen. Such an image is called a virtual image.

(i) Which of the following can be used to form a real image ?
(a) Concave mirror only
(b) Plane mirror only
(c) Convex mirror only
(d) Both concave and convex mirrors
Answer:
(a) Concave mirror only

(ii) Which of the statement is ture :
(a) If reflecting surface is convex then this a covex mirror
(b) Inner surface of spoon acts like concave
(c) Outer surface is spoon acts like convex mirror
(d) All of these
Answer:
(d) All of these

(iii) Which of the statement is false :
(a) The image formed by plane mirror can be obtained on screen
(b) Plane mirror image can not be obtained on screen
(c) An inverted image can be obtained by plan mirror
(d) None of these
Answer:
(b) Plane mirror image can not be obtained on screen

(iv) A spherical mirror having reflecting surface curved outward is a :
(a) plane mirror
(b) concave mirror
(c) convex mirror
(d) either concave or convex
Answer:
(c) convex mirror

Picture Based Questions

I. Observe the following picture and answer the questions.

(i) Angle of incidence is equal to angle of reflection. This is one of the :
(a) law of refraction
(b) normal law
(c) both (a) and (b)
(d) angle of reflection
Answer:
(d) angle of reflection

(ii) The angle between the normal and the reflected ray is called :
(a) angle of refraction
(b) angle
(c) angle of incidence
(d) angle of reflection
Answer:
(d) angle of reflection

(iii) Define angle of incidence?
Answer:
The angle between the normal and the incident ray is called angle of incidence.

Light: Mirrors and Lenses Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
A virtual image: ………..
(a) can be formed on the screen
(b) cannot be formed on the screen
(c) is formed only by the plane mirror
(d) is formed only by the convex mirror
Answer:
(b) cannot be formed on the screen

Question 2.
Which of the following would you prefer to use while reading small letters found in a dictionary ?
(a) A concave mirror
(b) A concave lens
(c) A convex mirror
(d) A convex lens
Answer:
(d) A convex lens

Question 3.
The image of an object formed by a plane mirror is:
(a) virtual
(b) real
(c) diminished
(d) upside down
Answer:
(a) virtual

Question 4.
A diverging mirror is:
(a) a plane mirror
(b) a convex mirror
(c) a concave mirror
(d) none of the above
Answer:
(b) a convex mirror

Question 5.
The image formed by spherical mirror is virtual. The mirror will be:
(a) concave
(b) convex
(c) either concave or convex
(d) none of the above
Answer:
(c) either concave or convex

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.

(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): When the object is placed very close to the lens, the image formed is virtual, erect and magnified.
Reason (R): This happens because the convex lens can form real and inverted image when the object place very close.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

2. Assertion (A): The light ray, which strikes any surface, is called the incident ray.
Reason (R): The ray that comes back from the surface after reflection is known as the reflected ray.
Answer:
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.

Fill in the blanks

1. An image that cannot be obtained on a screen is called ………….
Answer:
virtual image

2. Light travels in ………… lines.
Answer:
straight

3. Convex mirrors are ………… in the middle than at the edges whereas concave lenses are ………… in the middle than at the edges.
Answer:
thicker, thinner

4. The rear view mirror/side mirror in automobiles is a ………… mirror.
Answer:
convex

5. The inner surface of a steel spoon acts as a ………… mirror.
Answer:
concave

True or False

1. A concave lens can be used to produce an enlarged and erect image.
Answer:
False

2. A convex lens always produces a real image.
Answer:
False

3. The sides of an object and its image formed by a concave mirror are always interchanged.
Answer:
True

4. An object can be seen only if it emits light.
Answer:
False

5. A concave mirror can not be used as a magnifying mirror.
Answer:
False

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 3 Finding Common Ground MCQ improves accuracy in objective exams.

MCQ on Finding Common Ground Class 7

Finding Common Ground MCQ Class 7

Class 7 Maths Finding Common Ground MCQ

Question 1.
The correct prime factorisation of 168 is:
(a) 2 × 2 × 2 × 21
(b) 2 × 4 × 3 × 7
(c) 2 × 2 × 6 × 7
(d) 2 × 2 × 2 × 3 × 7
Solution:
(c) 2 × 2 × 6 × 7
Prime factorisation of number 168 using division method:

Thus, 168 = 2 × 2 × 2 × 3 × 7
Hence, the prime factorisation of 168 is 2 × 2 × 2 × 3 × 7

Question 2.
If the number 198 can be written as product of prime numbers as 198 = 2 × 3 × 11 × ______. The
missing prime factor is:
(a) 1
(b) 2
(c) 3
(d) 11
Solution:
(c) 3

Prime factorisation of number 198 using division method:
Thus, 198 = 2 × 3 × 3 × 11
Hence, the missing prime factor is 3.

Question 3.
The multiple of 5 × 3 × 7 is:
(a) 41
(b) 420
(c) (5 × 3 × 7) × 2
(d) Both (b) and (c)
Solution:
(d) Both (b) and (c)
The multiple of 5 × 3 × 7 is a number that contains all its factors.
Here, (5 × 3 × 7) × 2 and (5 × 3 × 7) × 4 i.e. 420, are multiples of 5 × 3 × 7.

Question 4.
21 mango trees, 42 apple trees and 56 orange trees have to be planted in rows such that each row contains the same number of trees of one variety only. Minimum number of rows in which the trees may be planted is:
(a) 3
(b) 15
(c) 17
(d) 20
Solution:
(c) 17
Maximum number of plants in each row
= HCF of 21, 42 and 56 = 7

Number of rows of mango trees = 21 ÷ 7 = 3
Number of rows of apple trees = 42 ÷ 7 = 6
Number of rows of orange trees = 56 ÷ 7 = 8
∴ Required number of rows = 3 + 6 + 8
= 17

Question 5.
All the factors of the number 96 are:
(a) 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48
(b) 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
(c) 1, 2, 4, 8, 16, 32, 64
(d) 1, 2, 3, 6, 8, 12, 24, 48, 96
Solution:
(b) 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Prime factorisation of 96 is given as
96 = 2 × 2 × 2 × 2 × 2 × 3
Taking combinations of prime factors, all factors of 96 are:
1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Question 6.
The HCF of numbers 144, 216 and 360 is:
(a) 72
(b) 24
(c) 36
(d) 12
Solution:
(a) 72
Prime factorisations of 144, 216 and 360 are:
144 = 2 × 2 × 2 × 2 × 3 × 3
216 = 2 × 2 × 2 × 3 × 3 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
∴HCF (144, 216, 360) = 2 × 2 × 2 × 3 × 3 = 72

Question 7.
The LCM of 57 and 84 is:
(a) 1584
(b) 1596
(c) 4688
(d) 4788
Solution:
(b) 1596
Using prime factorisation method, we have
57 = 3 × 19 and
84 = 2 × 2 × 3 × 7
∴ LCM = 2 × 2 × 3 × 7 × 19 = 1596

Question 8.
The LCM and HCF of 117, 693 and 819 are:
(a) LCM = 9007, HCF = 9
(b) LCM = 9009, HCF = 7
(c) LCM = 9009, HCF = 9
(d) LCM = 9007, HCF = 7
Solution:
(c) LCM = 9009, HCF = 9
Prime factorisations of 117, 693 and 819 are:
117 = 3 × 3 × 13
693 = 3 × 3 × 7 × 11
819 = 3 × 3 × 7 × 13
Thus, HCF = 3 × 3 = 9
The LCM is the product of the highest occurrence of each prime factor, i.e.
LCM = 3 × 3 × 7 × 11 × 13 = 9009

Question 9.
The HCF of two numbers is 40, and their LCM is 8400. If one of the numbers is 240, what is the other number?
(a) 1200
(b) 1400
(c) 1600
(d) 1800
Solution:
(b) 1400
Let the unknown number be x.
We know,
Product of two numbers = HCF × LCM
⇒ 240 × x = 40 × 8400
⇒ x = \(\frac{40 \times 8400}{240}=\frac{8400}{6}\) = 1400

Question 10.
The HCF of two numbers is 8. Which one of the following can never be their LCM?
(a) 24
(b) 48
(c) 56
(d) 60
Solution:
(d) 60
We know, LCM of numbers is a multiple of their HCF.
Since 60 is not a multiple of 8, it can never be the LCM.

Question 11.
There are some sweets in a jar. The sweets can be put into boxes of 3 or 7 with no sweets left over. When the sweets are put into boxes of 20, there are 5 sweets left over. What is the smallest possible number of sweets in the jar?
(a) 48
(b) 105
(c) 125
(d) 165
Solution:
(b) 105
If the sweets can be packed in boxes of 3 or 7 with no remainder, then the number of sweets in the jar is a multiple of LCM of 3 and 7, i.e. 21.
Clearly, the required number is multiple of 21 which leaves a remainder of 5 when divided by
20, i.e. 105.

Question 12.
Which of the following statements are correct?
(i) Prime factorisation of 120 is 2 × 2 × 2 × 3 × 5.
(ii) 18 is a factor of 120.
(iii) Prime factorisation of 126 is 2 × 3 × 3 × 11.
(iv) 42 is a factor of 126.
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Solution:
(a) (i) and (iv)
(i) Prime factorisation of 120 is given as:
120 = 2 × 2 × 2 × 3 × 5
∴ (i) is correct.

(ii) Prime factorisation of 18 is 2 × 3 × 3
For 18 to be a factor of 120, the product 2 × 3 × 3 must be the part of the prime factorisation of 120.
Since 3 × 3 does not appear in the prime factorisation of 120, 18 is not a factor of 120.
∴ (ii) is not correct.

(iii) Prime factorisation of 126 is given as:

Thus, 126 = 2 × 3 × 3 × 7
∴ (iii) is not correct.

(iv) All factors of 126 are 1,2, 3,6, 7,9, 14, 18, 21, 42, 63 and 126.
∴ (iv) is correct.

Question 13.
A rectangular plot measures 168 metres by 196 metres. Find which of the following correctly represent the side length (in metres) of possible square tile that can be used to pave the entire plot without cutting any tiles.
(i) 3
(ii) 14
(iii) 28
(iv) 6
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Solution:
(d) (ii) and (iii)
The possible square tiles that can be used to pave the entire plot without cutting any tiles includes HCF of the dimensions of the rectangular plot and its all factors.
HCF of 168 and 196 using common factor method:

∴ HCF (168, 196) = 2 × 2 × 7
All the factors of 28 are 1, 2, 4, 7, 14 and 28.
Thus, 14 and 28 are the possible tile sizes to cover the floor completely.

Question 14.
Which of the following statements are correct?
(i) 14 and 20 have exactly three common factors.
(ii) The LCM of 14 and 20 is 140.
(iii) 4 × 5 × 7 is a multiple of both 14 and 20.
(iv) 40 is a factor of LCM (4, 5, 10).
Choose the correct option from the following:
(a) (i) and (iv) only
(b) (i), (iii) and (iv)
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Solution:
(c) (ii) and (iii) only
(i) As 14 = 2 × 7; 20 = 2 × 2 × 5
Common factors of 14 and 20 are 1 and 2 only.

(ii) We have, 14 = 2 × 7; 20 = 2 × 2 × 5
LCM of 14 and 20 = 2 × 2 × 5 × 7 = 140

(iii) As, 14 = 2 × 7 and 20 = 2 × 2 × 5
4 × 5 × 7 is the LCM of 14 and 20.
Thus, it is a multiple of both 14 and 20.

(iv) Prime factorisations of 4, 5 and.10 are:
4 = 2 × 2; 5 = 5; 10 = 2 × 5
The LCM is the product of the highest occurrence of each prime factor, i.e.
LCM (4, 5, 10) = 2 × 2 × 5 = 20
So, 40 is a multiple of 20 (not a factor).
Hence, statements (ii) and (iii) are correct.

Finding Common Ground Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The HCF of numbers 31 and 43 is 1.
(R): HCF of two prime numbers is always 1.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that a prime number has two factors only i.e. 1 and the number itself.
∴ HCF of two prime numbers is always 1.
Here, 31 and 43 are prime numbers.
∴ HCF(31, 43) = 1
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The HCF of numbers 313 and 314 is 1.
(R): The HCF of any two consecutive numbers is always 1.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, the HCF of two consecutive natural numbers is always 1 because they have no common factor other than 1.
∴ HCF (313, 314) = 1
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): Two traffic lights change simultaneously every 60 seconds and 90 seconds. They will both change together again after 6 minutes.
(R): The time after which repeating events coincide is a multiple of the LCM.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Prime factorisations of 60 and 90 are:
60 = 2 × 2 × 3 × 5; 90 = 2 × 3 × 3 × 5
LCM is the product of the highest occurrence of each prime factor, i.e.
LCM (60, 90) = 2 × 2 × 3 × 3 × 5 = 180
So, both traffic lights will change together again after 180 seconds, i.e. 3 minute.
Thus, both lights will change together again after 3 minutes, 6 minutes, 9 minutes… (Multiples of 3).
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): If two numbers are co-prime, their HCF is 1 and their LCM is equal to their product.
(R): Co-prime numbers have no common factors except 1.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
When two numbers are co-prime, they do not their highest common factor (HCF) is 1.

Since there are no common factors to be shared by two numbers, all the prime factors of both numbers, taken to their highest occurrence, are included while finding the LCM. As a result, the LCM becomes equal to the product of the two numbers.

Hence, both Assertion (A) and Reason (R) are true, and Reason (R)

Finding Common Ground Class 7 Fill in the Blanks

Question 1.
The HCF (336, 378) has the prime factorisation: 2 × 3 × ________ .
Solution: 7
Prime factorisations of 336 and 378 using division method:
336 = 2 × 2 × 2 × 2 × 3 × 7
378 = 2 × 3 × 3 × 3 × 7

∴ HCF (336, 378) = 2 × 3 × 7

Question 2.
Three mugs contain 145 mL, 235 mL and 495 mL of water respectively. The maximum capacity of a mug that can exactly measure the water of three mugs is _______ mL.
Solution: 5
We have, 145 = 5 × 29 ; 235 = 5 × 47 ;
495 = 5 × 9 × 11
Required capacity = HCF of 145, 235 and 495 = 5
Hence, the maximum capacity of a mug that can exactly measure the water of three mugs is 5 mL.

Question 3.
546 = 2 × 3 × ________ × 13.
Solution: 7
Prime factorisation of 546 is given as

Thus, 546 = 2 × 3 × 7 × 13

Question 4.
If the LCM of x and y is 720, their HCF is 12, and x is 144, then y is _____ .
Solution: 60
We know, product of two numbers = HCF × LCM
⇒ x × y = HCF × LCM
⇒ 144 × y = 12 × 720
⇒ y = \(\frac{12 \times 720}{144}\) ⇒ y = \(\frac{720}{12}\) = 60

Question 5.
_______ is the smallest number which, when divided by 12 and 18, leaves a remainder of 5 in each case.
Solution: 41
Required smallest number = LCM(12, 18) + 5
The LCM of 12 and 18 using prime factorisation method:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
The LCM is the product of the highest occurrence of each prime factor.
Hence, LCM (12, 18) = 2 × 2 × 3 × 3 = 36
Thus, the smallest number = 36 + 5 = 41

Question 6.
If the product of two numbers is 600 and their HCF is 10, then their LCM is ______.
Solution: 60
Let two numbers be A and B, then according to question A × B = 600 and HCF =10.
Using the property: A × B = HCF × LCM
⇒ 600 = 10 × LCM
⇒ LCM = \(\frac{600}{10}\) = 60

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 5 Prime Time Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 5 Prime Time Solutions

Ganita Prakash Class 6 Chapter 5 Solutions

Class 6 Maths Ganita Prakash Chapter 5 Solutions Prime Time

Question 1.
Who am I?
(a) I am a number less than 40. One of my factors is 7. The sum of my digits is 8.
(b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
Solution:
(a) The numbers less than 40 whose one of the factor is 7 are 7, 14, 21, 28, 35. Out of these numbers the sum of digits of 35 is 8.
Hence, the number less than 40 whose one of the factors is 7 and sum of digits equals to 8 is 35.

(b) The number less than 100 whose two factors 3 and 5 are 15, 30, 45, 60, 75, 90. Out of these numbers, the number 45 has one of its digits 1 more than the other.
Hence, the number less than 100 whose two factors are 3 and 5 and one of its digits is 1 more than the other is 45.

Question 2.
In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions.

Solution:

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Common factors of 24, 48 and 72 are 1, 2, 3, 4, 6, 8, 12, 24.
So, in multiples of 6 and 8, common multiples are 24, 48 and 72.
Other possibilities of the numbers are 3 and 8, 3 and 24, 4 and 6, 4 and 24, 6 and 24, etc.

Question 3.
Find the smallest number that is a multiple of all the numbers from 1 to 10.
Solution:
To find the smallest number that is a multiple of all numbers from 1 to 10, we need to determine least common multiple (LCM) of the numbers from 1 to 10.
Prime factorisation of numbers from 1 to 10.
1 = 1; 2 = 2; 3 = 3; 4 = 2 × 2; 5 = 5; 6 = 2 × 3; 7 = 7; 8 = 2 × 2 × 2; 9 = 3 × 3; 10 = 2 × 5
LCM (1 to 10) = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 Thus, the required smallest number is 2520.

Question 4.
Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Now, 2 + 3 = 5 (multiple of 5)
2 + 13 = 15 (multiple of 5)
7 + 13 = 20 (multiple of 5)
Thus, three pairs are (2, 3), (2, 13) and (7, 13).

Question 5.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3 . Find such pairs of prime numbers up to 100.
Solution:
Pairs of prime numbers having same digits upto 100 are:
17 and 71 (both have digits 1 and 7);
37 and 73 (both have digits 3 and 7)
79 and 97 (both have digits 7 and 9)

Question 6.
Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.
Solution:
Twin primes between 1 and 100 are
(3, 5) → 5 – 3 = 2;
(5, 7) → 7 – 5 = 2;
(17, 19) → 19- 17 = 2;
(29, 31) → 31 – 29 = 2;
(59, 61) → 61 – 59 = 2;
(71, 73) → 73 – 71 = 2

Question 7.
Observe that 3 is a prime number and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 give another prime? Find atleast five such examples.
Solution:
Five such examples are:
2 × 2+ 1 = 5 (prime number)
2 × 5 + 1 = 11 (prime number)
2 × 11 + 1 = 23 (prime number)
2 × 23 + 1 = 47 (prime number)
and 2 × 29 + 1 = 59 (prime number)

Question 8.
What is the smallest number whose prime factorisation has
(a) three different prime numbers?
(b) four different prime numbers?
Solution:
(a) The smallest three prime numbers are 2, 3 and 5.
Thus, the smallest number with exactly three different prime factors = 2 × 3 × 5 = 30

(b) The smallest four prime numbers are 2, 3, 5 and 7.
Thus, the smallest number with exactly four different prime factors = 2 × 3 × 5 × 7 = 210

Question 9.
The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
Prime factorisation of two numbers are 2 × 3 × 7 and 3 × 7 × 11.
Both the numbers have common factors 3 and 7. Therefore, they are not co-prime.
Since the prime factors of one number are not the prime factors of another number, therefore one of them does not divide the other.

Question 10.
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Solution:
The smallest 4-digit palindrome is 1001 .
Checking divisibility by 4:
1001 (not divisible by 4);
1111 (not divisible by 4);
1221 (not divisible by 4)
1331 (not divisible by 4);
1441 (not divisible by 4);
1551 (not divisible by 4);
1661 (not divisible by 4);
1771 (not divisible by 4);
1881 (not divisible by 4);
1991 (not divisible by 4);
2002 (not divisible by 4);
2112 (divisible by 4)
Therefore, die smallest 4-digit palindromic number divisible by 4 is 2112 .
The largest 4-digit palindrome is 9999.
Checking divisibility by 4:
9999 (not divisible by 4);
9669 (not divisible by 4);
9339 (not divisible by 4);
9009 (not divisible by 4):
9889 (not divisible by 4);
9559 (not divisible by 4);
9229 (not divisible by 4);
8998 (not divisible by 4);
9779 (not divisible by 4);
9449 (not divisible by 4);
9119 (not divisible by 4);
8888 (divisible by 4)
Therefore, the largest 4 -digit palindromic number divisible by 4 is 8888.

Question 11.
The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Solution:
If a number is divisible by 8 and 5, then it will also be divisible by 2, 4 and 10.
Divisibility by 8 ensures it is always divisible by 2 and 4.
Divisibility by 5 ensures it ends with 0 or 5. But since the number also needs to be divisible by 8 then it must end with 0.
A number ending with 0 is also divisible by 10.
Therefore, 8 and 5 are the required numbers.

InText Questions

Question 1.
Look at the table below. What do you notice?

(i) Is there anything common among the shaded numbers?
(ii) Is there anything common among the circled numbers?
(iii) Which numbers are both shaded and circled? What are these numbers called?
Solution:
(i) All shaded numbers are multiples of 3.
(ii) All circled numbers are multiples of 4.
(iii) Numbers (both shaded and circled) are 36, 48 and 60.
They are called ‘common multiples of 3 and 4’.

Question 2.
Grumpy and Jumpy are playing treasure finding game. Treasures are kept on two numbers. Jumpy gets the treasures only if he is able to reach both the numbers with the same jump size. Also, a jump size of 1 is not allowed.
Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures? Check if these pairs are safe:
(a) 15 and 39
(b) 4 and 15
(c) 18 and 29
(d) 20 and 55
Solution:
Grumpy should place the treasures at co-prime numbers.
(a) Pair (15 and 39) is not safe because 15 and 39 are not co-prime as 3 is their common factor.
(b) Pair (4 and 15) is safe because 4 and 15 are co-prime.
(c) Pair (18 and 29) is safe because 18 and 29 are co-prime.
(d) Pair (20 and 55) is not safe because 20 and 55 are not co-prime as 5 is their common factor.

Question 3.
Observe the following thread art. The first diagram has 12 pegs and the thread is tied to every fourth peg (we say that the thread-gap is 4). The second diagram has 13 pegs and the thread- gap is 3. What about the other diagrams? Observe these pictures, share and discuss your findings in class. In some diagrams, the thread is tied to every peg. In some, it is not. Is it related to the two numbers (the number of pegs and the thread- gap) being co-prime?

Make such pictures for the following:
(a) 15 pegs, thread-gap of 10
(b) 10 pegs. thread-gap of’ 7
(c) 14 pegs, thread-gap of 6
(d) 8 pegs, thread -gap of 3
Solution:
Yes, when two numbers are co-prime, the thread is tied to every peg.

Question 4.
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe?
Solution:
Numbers between 330 and 340 that are divisible by 4 are 332 and 336
Numbers between 1730 and 1740 that are divisible by 4 are 1732 and 1736
Numbers between 2030 and 2040 that are divisible by 4 are 2032 and 2036
We observe that if number formed by last two digits of a number is divisible by 4, then the number is divisible by 4.

Question 5.
Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe?
Solution:
Numbers between 120 and 140 that are divisible by 8 are 128 and 136.
Numbers between 1120 and 1140 that are divisible by 8 are 1128 and 1136.
Numbers between 3120 and 3140 that are divisible by 8 are 3128 and 3136.
We observe that if last three digits of a number is divisible by 8, then the number is divisible by 8.

Question 6.
Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

Solution:
(i)

7	5	3	105
2	5	2	20
2	5	3	30
28	125	18

(ii)

2	2	2	8
3	5	7	105
5	7	2	70
30	70	28

Prime Time Class 6 Extra Questions

Prime Time Class 6 Very Short Question Answer

Question 1.
Write all factors of each of the following numbers:
(i) 24
(ii) 32
Solution:
(i) We can write 24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
∴ 1,2,3, 4, 6, 8, 12 and 24 are the factors of 24.

(ii) We can write 32 = 1 × 32 = 2 × 16 = 4×8
1, 2, 4, 8, 16 and 32 are the factors of 32.

Question 2.
Write first five multiples of each of the following numbers.
(i) 9
(ii) 13
Solution:
(i) To obtain first five multiples of 9, we multiply it by 1,2, 3, 4 and 5 respectively.
9 × 1 = 9;
9 × 2 = 18;
9 × 3 = 27;
9 × 4 = 36;
9 × 5= 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45.

(ii) To obtain first five multiples of 13, we multiply it by 1, 2, 3, 4 and 5 respectively.
13 × 1 = 13;
13 × 2 = 26;
13 × 3 = 39;
13 × 4 = 52;
13 × 5 = 65
Hence, the first five multiples of 13 are 13, 26, 39, 52 and 65.

Question 3.
Show that 17 is a factor of 170017 without actual division.
Solution:
We can write
170017 = 170000 + 17 = 17 × 10000 + 17 × 1 = 17 × (10000 + 1) = 17 × 10001
Hence, 17 is a factor of 170017.

Question 4.
Write any three numbers that are multiples of 15 but not of 30.
Solution:
Multiples of 15 are 15, 30, 45, 60, 75, 90, … Multiples of 30 are 30, 60, 90, …
Thus, three numbers that are multiples of 15 but not of 30 are 15, 45 and 75.

Question 5.
Find the common factors of 16 and 28.
Solution:
We can write 16 = 1 × 16 = 2 × 8 = 4 × 4
∴ The factors of 16 are 1, 2, 4, 8 and 16.
And, 28 = 1 × 28 = 2 × 14 = 4 × 7
∴ The factors of 28 are 1, 2, 4, 7, 14 and 28.
Hence, the common factors of 16 and 28 are 1, 2 and 4.

Question 6.
Write five prime triplets such that the difference between the greatest and the smallest prime number is 6.
Solution:
{7, 11, 13}, (11, 13, 17}, {13, 17, 19}, {17, 19, 23} and {37, 41, 43}.

Question 7.
Write all prime numbers less than 100.
Solution:
The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Question 8.
Write five pairs of co-prime numbers.
Solution:
Two numbers are said to be co-prime if they do not have a common factor other than 1. Thus, five pairs of co-primes are (2, 3), (4, 5), (5, 6), (9, 10), (10, 21).

Question 9.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Solution:
We can write 10000 = 100 × 100
= (2 × 2 × 5 × 5) × (2 × 2 × 5 × 5)
= (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5)
= 16 × 625

Question 10.
Check the divisibility 100100 by 10:
Solution:
Since 100100 ends with the digit 0, it is divisible by 10.

Question 11.
Check the divisibility 412030 by 10:
Solution:
Since 412030 ends with the digit 0, it is divisible by 10.

Question 12.
Check the divisibility 718005 by 10:
Solution:
Since 718005 does not end with the digit 0, it is not divisible by 10.

Question 13.
Check the divisibility 123456 by 4:
Solution:
The number formed by the last two digits of 123456 is 56.
Since 56 = 4 × 14, it is divisible by 4.
∴ 123456 is divisible by 4.

Question 14.
Check the divisibility 725976 by 4:
Solution:
The number formed by the last two digits of 725976 is 76.
Since 76 = 4 × 19, it is divisible by 4.
∴ 725976 is divisible by 4.

Question 15.
Check the divisibility 985442 by 4:
Solution:
The number formed by the last two digits of 985442 is 42.
Clearly, 42 is not divisible by 4.
∴ 985442 is not divisible by 4.

Question 16.
Give an example of a number which is divisible by 2 but not by 4
Solution:
6

Question 17.
Give an example of a number which is divisible by 4 but not by 8.
Solution:
20

Question 18.
Give an example of a number which is divisible by both 4 and 8 but not by 32.
Solution:
48

Question 19.
Express each of the following as a sum of three unique prime numbers.
(i) 10
(ii) 25
(iii) 35
(iv) 49
Solution:
(i) 10 = 2 + 3 + 5
(ii) 25 = 5 + 7 + 13
(iii) 35 = 5 + 7 + 23
(iv) 49 = 13 + 17 + 19

Question 20.
Express each of the following as a sum of two unique prime numbers:
(i) 15
(ii) 30
(iii) 54
(iv) 80
Solution:
(i) 15 = 2 + 13
(ii) 30 = 7 + 23
(iii) 54 = 13 + 41
(iv) 80 = 19 + 61

Question 21.
Express each of the following as a sum of twin primes:
(i) 12
(ii) 36
(iii) 60
(iv) 84
Solution:
(i) 12 = 5 + 7
(ii) 36 = 17 + 18
(iii) 60 = 29 + 31
(iv) 84 = 41 + 43

Question 22.
Check the divisibility of the following numbers by 5:
(i) 235965
(ii) 783120
(iii) 415812
Solution:
(i) Since 235965 ends with the digit 5, it is divisible by 5.
(ii) Since 783120 ends with the digit 0, it is divisible by 5.
(iii) Since 415812 does not end with the digits 0 or 5, it is not divisible by 5.

Prime Time Class 6 Short Question Answer

Question 1.
Write all factors of each of the following numbers:
(i) 60
(ii) 420
Solution:
(i) We can write
60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15
= 5 × 12 = 6 × 10
∴ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 are the factors of 60.

(ii) We can write
420 = 1 × 420 = 2 × 210 = 3 × 140 = 4 × 105 = 5 × 84 = 6 × 70 = 7 × 60 = 10 × 42
= 12 × 35 = 14 × 30 = 15 × 28 = 20 × 21
∴ 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210 and 420 are the factors of 420.

Question 2.
The product of two numbers is 42. Their sum is 17. What are the numbers?
Solution:
We can write 42 = 1 × 42 = 2 × 21 = 3 × 14 = 6 × 7.
∴ 1, 2, 3, 6, 7, 14, 21 and 42 are the factors of 42.
Out of all the above factors, only 3 and 14 add up to a total of 17.
Hence, the required numbers are 3 and 14.

Question 3.
Find the smallest number that is a multiple of all the numbers from 1 to 10 except 7.
Solution:
We know, 8 is a multiple of 1, 2, 4 and 8.
9 is a multiple of 1, 3 and 9.
8 × 9 is a multiple of 6.
5 × 8 × 9 is a multiple of 5 and 10.
[As 5 × 8 = 40 is a multiple of 5 and 10.]
∴ Required smallest number is 5 × 8 × 9 = 360.

Question 4.
Find the common factors of 12, 18 and 24.
Solution:
We can write 12 = 1 × 12 = 2 × 6 = 3 × 4
∴ The factors of 12 are 1, 2, 3, 4, 6 and 12.
18 = 1 × 18 = 2 × 9 = 3 × 6
∴ The factors of 18 are 1, 2, 3, 6, 9 and 18.
24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
∴ The factors of 24 are 1,2, 3, 4, 6, 8, 12 and 24.
Hence, the common factors of 12, 18 and 24 are 1, 2, 3 and 6.

Question 5.
Find the common multiples of 6, 9 and 12.
Solution:
The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
The multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, ……
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, ………
Hence, the common multiples of 6, 9 and 12 are 36, 72, 108, … .

Question 6.
A number is divisible by both 6 and 14. By which other number will that number be always divisible?
Solution:
Since the number is divisible by 6 and 14, the number is a common multiple of 6 and 14. So, the number will always be divisible by the smallest common multiple of 6 and 14.
Now, the multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, … .
And, the multiples of 14 are 14, 28. 42, 56, 70, 84,
So, the common multiples of 6 and 14 are 42, 84, 126,….
∴ The smallest common multiple of 6 and 14 is 42.
Hence, 42 is the required number.

Question 7.
How many numbers between 1 and 100 have exactly three factors?
Solution:
As we know, prime numbers have exactly two factors, 1 and itself. So, squares of prime numbers will have exactly three factors.
2² = 4 has factors 1, 2 and 4.
3² = 9 has factors 1,3 and 9. .
5² = 25 has factors 1,5 and 25.
7² = 49 has factors 1, 7 and 49.
Thus, four numbers i.e. 4, 9, 25 and 49 between 1 and 100 have exactly three factors.

Question 8.
The product of two numbers is 36. Their difference is 9. What are the numbers?
Solution:
We can write 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6.
∴ 1, 2, 3, 4, 6, 9, 12, 18 and 36 are the factors of 36.
The difference between factors 12 and 3 is 9.
Hence, the required numbers are 3 and 12.
Note: The difference between factors 18 and 9 is also 9, but their product is not equal to 36.

Question 9.
Without actual division, show that 13 is a factor of each of the following numbers:
(i) 1313
(ii) 13013
(iii) 131313
Solution:
(i) We can write
1313 = 1300 + 13 = 13 × 100 + 13× 1 = 13(100 + 1) = 13 × 101
Hence, 13 is a factor of 1313.

(ii) We can write
13013 = 13000 + 13 = 13 × 1000 + 13 × 1 = 13(1000 + 1) = 13 × 1001
Hence, 13 is a factor of 13013.

(iii) We can write
131313 = 130000 + 1300 + 13 = 13 × 10000 + 13 × 100 + 13 × 1
= 13(10000 + 100 + 1) = 13 × 10101
Hence, 13 is a factor of’ 131313.

Question 10.
Find the common factors of:
(i) 12 and 18
(ii) 35 and 63
(iii) 60 and 210
Solution:
(i) We can write 12 = 1 × 12 = 2 × 6 = 3 × 4
∴ The factors of 12 are 1, 2, 3, 4, 6 and 12.
And, 18 = 1 × 18 = 2 × 9 = 3 × 6
∴ The factors of 18 are 1,2, 3, 6, 9 and 18.
Hence, the common factors of 12 and 18 are 1, 2, 3 and 6.

(ii) We can write 35 = 1 × 35 = 5 × 7
∴ The factors of 35 are 1, 5, 7 and 35.
And, 63 = 1 × 63 = 3 × 21 = 7 × 9
∴ The factors of 63 are 1. 3, 7, 9, 21 and 63.
Hence, the common factors of 35 and 63 are 1 and 7.

(iii) We can write 60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10
∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
210 = 1 × 210 = 2 × 105 = 3 × 70 = 5 × 42 = 6 × 35 = 7 × 30 = 10 × 21 = 14 × 15
∴ The factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
Hence, the common factors of 60 and 210 are 1, 2, 3, 5, 6, 10, 15 and 30.

Question 11.
Find first three common multiples of:
(i) 10 and 20
(ii) 25 and 50
(iii) 40 and 60
Solution:
(i) The multiples of 10 are 10, 20, 30, 40, 50, 60,
The multiples of 20 are 20, 40, 60, 80, …
Hence, the first three common multiples of 10 and 20 are 20, 40 and 60.

(ii) The multiples of 25 are 25, 50, 75, 100, 125, 150, …
The multiples of 50 are 50, 100, 150, 200, …
Hence, the first three common multiples of 25 and 50 are 50, 100 and 150.

(iii) The multiples of 40 are 40; 80, 120, 160, 200, 240, 280, 320, 360 …
The multiples of 60 are 60, 120, 180, 240, 300, 360 …
Hence, the first three common multiples of 40 and 60 are 120, 240 and 360.

Question 12.
A number is divisible by both 8 and 12. By which other numbers will that number be always divisible?
Solution:
Since the number is divisible by 8 and 12, the number is a common multiple of 8 and 12. So, the number will always he divisible by the smallest common multiple of 8 and 12.
Now, the multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72 …
The multiples of 12 are 12, 24, 36, 48, 60, 72, …
So, the common multiples of 8 and 12 are 24, 48, 72, …
∴ The smallest common multiple of 8 and 12 is 24.
Hence, the number is divisible by 24 and hence by all factors of 24, i.e., 1, 2, 3, 4, 6, 8, 12 and 24.

Question 13.
Write all pairs of twin primes between 1 and 100.
Solution:
The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83,89 and 97.
The required pairs of twin primes are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61) and (71, 73).

Question 14.
Write all pairs of prime numbers less than 30 whose sum is a multiple of 5.
Solution:
The prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Now, 2 + 3 = 5, 2 + 13 = 15, 2 + 23 = 25, 3 + 7=10,
3 + 17 = 20, 7 + 13 = 20, 7 + 23 = 30, 11 + 19 = 30,
11 + 29 = 40, 13 + 17 = 30, 17 + 23 = 40
∴ The required pairs are (2, 3), (2, 13), (2, 23), (3, 7), (3, 17), (7, 13), (7, 23), (1 1, 19), (11, 29), (13, 17) and (17, 23).

Question 15.
Determine the prime factorisation of the following numbers:
(i) 3094
(ii) 5082
(iii) 10010
Solution:
(i) The prime factorisation of 3094 is 2 × 7 × 13 × 17.
(ii) The prime factorisation of 5082 is 2 × 3 × 7 × 11 × 11.
(iii) The prime factorisation of 10010 is 2 × 5 × 7 × 11 × 13.

Question 16.
Find prime factorisation of the following numbers without multiplying first:
(i) 48 × 60
(ii) 56 × 75
(iii) 81 × 25
Solution:
(i) Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3
Prime factorisation of 60 = 2 × 2 × 3 × 5
Thus, prime factorisation of 48 × 60 is 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5.

(ii) Prime factorisation of 56 = 2 × 2 × 2 × 7
Prime factorisation of 75 = 3 × 5 × 5
Thus, prime factorisation of 56 × 75 is 2 × 2 × 2 × 3 × 5 × 5 × 7.

(iii) Prime factorisation of 81 = 3 × 3 × 3 × 3
Prime factorisation of 25 = 5 × 5
Thus, prime factorisation of 81 × 25 is 3 × 3 × 3 × 3 × 5 × 5.

Question 17.
The first number has prime factorisation 2 × 5 × 7 and the second number has prime factorisation 5 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
Since 5 and 7 are the common prime factors of two numbers, they are not co-prime.
Clearly, 2 is a prime factor of first number but not a prime factor of second number.
Hence, second number is not divisible by first number.
And, 1 1 is a prime factor of second number but not a prime factor of first number.
Hence, first number is not divisible by second number. So, none of them divide the other number.

Question 18.
Check the divisibility of the following numbers by 8:
(i) 6245826
(ii) 2727272
(iii) 2491664
Solution:
(i) The number formed by the last three digits of 6245826 is 826.
Clearly, 826 is not divisible by 8.
∴ 6245826 is not divisible by 8.

(ii) The number formed by the last three digits of 2727272 is 272.
Since 272 = 8 x 34, it is divisible by 8.
∴ 2727272 is divisible by 8.

(iii) The number formed by the last three digits of 2491664 is 664.
Since 664 = 8 X 83, it is divisible by 8.
∴ 2491664 is divisible by 8.

Prime Time Class 6 Long Question Answer

Question 1.
Who am I?
(i) I am a number less than 50. One of my factors is 11. The sum of my digits is 8.
(ii) I am an even number less than 100. One of my factors is 7. One of my digits is 1 more than the other.
Solution:
(i) Numbers less than 50 and having 11 as a factor are
11 × 1 = 11, 11 × 2 = 22, 11 × 3 = 33, 11 × 4 = 44
Out of these numbers, only 44 is the number whose sum of digits is 8.
Hence, the required number is 44.

(ii) Even numbers less than 100 and having 7 as a factor are
7 × 2 = 14, 7 × 4 = 28, 7 × 6 = 42, 7 × 8 = 56, 7 × 10 = 70, 7 × 12 = 84, 7 × 14 = 98
Out of these numbers, only 56 and 98 are the numbers whose one digit is 1 more than the other.
Hence, the required number is either 56 or 98.

Question 2.
Using prime factorisation, check whether the following pairs of numbers are co-prime or not:
(i) 154 and 195
(ii) 132 and 225
(iii) 357 and 286
Solution:
(i) Prime factorisations of 154 and 195 are as follows:
154 = 2 × 7 × 11 and 195 = 3 × 5 × 13
Since there are no common prime factors, 154 and 195 are co-prime.

(ii) Prime factorisations of 132 and 225 are as follows:
132 = 2 × 2 × 3 × 11 and 225 = 3 × 3 × 5 × 5
Since 3 is a common prime factor, 132 and 225 are not co-prime.

(iii) Prime factorisations of 357 and 286 are as follows:
357 = 3 × 7 × 17 and 286 = 2 × 11 × 13
Since there are no common prime factors, 357 and 286 are co-prime.

Question 3.
Using prime factorisation method, determine whether the first number is divisible by the second number or not:
(i) 420 and 105
(ii) 693 and 78
(iii) 990 and 90
Solution:
(i) Prime factorisations of 420 and 105 are as follows:
420 = 2 × 2 × 3 × 5 × 7 and 105 = 3 × 5 × 7
Clearly, all prime factors of 105 are prime factors of 420 and prime factorisation of 105 is included in the prime factorisation of 420.
We can write, 420 = (3 × 5 × 7) × 2 × 2 = 105 × 4
Hence, 420 is divisible by 105.

(ii) Prime factorisations of 693 and 78 are as follows:
693 = 3 × 3 × 7 × 11 and 78 = 2 × 3 × 13 Clearly, 13 is a prime factor of 78 but not a prime factor of 693.
Hence, 693 is not divisible by 78.

(iii) Prime factorisations of 990 and 90 are as follows:
990 = 2 × 3 × 3 × 5 × 11 and 90 = 2 × 3 × 3 × 5
Clearly, all prime factors of 90 are prime factors of 990 and prime factorisation of 90 is included in the prime factorisation of 990.
We can write, 990 = (2 × 3 × 3 × 5) × 11 = 90 × 11
Hence, 990 is divisible by 90.

Question 4.
Using prime factorisation, check whether the following pairs of numbers are co-prime or not:
(i) 308 and 585
(ii) 396 and 450
(iii) 1071 and 1430
Solution:
(i) Prime factorisations of 308 and 585 are as follows:
308 = 2 × 2 × 7 × 11 and 585 = 3 × 3 × 5 × 13
Since there are no common factors in these prime factorisations, 308 and 585 are co-prime.

(ii) Prime factorisations of 396 and 450 are as follows:
396 = 2 × 2 × 3 × 3 × 11 and 450 = 2 × 3 × 3 × 5 × 5
Since 2 and 3 are common factors in these prime factorisations, 396 and 450 are not co-prime.

(iii) Prime factorisations of 1071 and 1430 are as follows:
1071 = 3 × 3 × 7 × 17 and 1430 = 2 × 5 × 11 × 13
Since there are no common factors in these prime factorisations, 1071 and 1430 are co-prime.

Question 5.
Using prime factorisation method, determine whether the first number is divisible by the second number or not:
(i) 1260 and 315
(ii) 7623 and 198
(iii) 3780 and 126
Solution:
(i) Prime factorisations of 1260 and 315 are as follows:
1260 = 2 × 2 × 3 × 3 × 5 × 7 and 315 = 3 × 3 × 5 × 7
Clearly, all prime factors of 315 are prime factors of 1260 and prime factorisation of 315 is included in the prime factorisation of 1260.
We can write, 1260 = (3 × 3 × 5 × 7) × 2 × 2 = 315 × 4
Hence, 1260 is divisible by 315.

(ii) Prime factorisations of 7623 and 198 are as follows:
7623 = 3 × 3 × 7 × 11 × 11 and 198 = 2 × 3 × 3 × 11
Clearly, 2 is a prime factor of 198 but not a prime factor of 7623.
Hence, 7623 is not divisible by 198.

(iii) Prime factorisations of 3780 and 126 are as follows:
3780 = 2 × 2 ×3 × 3 × 3 × 5 × 7 and 126 = 2 × 3 × 3 × 7
Clearly, all prime factors of 126 are prime factors of 3780 and prime factorisation of 126 is included in the prime factorisation of 3780.
We can write,
3780 = (2 × 3 × 3 × 7) × 2 × 3 × 5 = 126 × 30
Hence, 3780 is divisible by 126.

Question 6.
State whether the following statements are true or false. Give reasons.
(i) If a number is divisible by 4, then it must be divisible by 8.
(ii) If a number is divisible by 8, then it must be divisible by 4.
(iii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(iv) The sum of two consecutive odd numbers is always divisible by 4.
(v) Sum of two even numbers gives a multiple of 4.
(vi) Sum of two odd numbers gives a multiple of 4.
Solution:
(i) False.
For example, 28 is divisible by 4 but not by 8.

(ii) True.
If a number is divisible by 8, then it must be divisible by 4 because 4 is a factor of 8.

(iii) False.
For example, let us take 14 and 6. Their sum, 14 + 6 = 20, is divisible by 4. But neither 14 nor 6 are divisible by 4.

(iv) True.
Let 2n + 1 and 2n + 3 be two consecutive odd numbers, where n = 0, 1,2,3, … .
Then, their sum,
(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), is divisible by 4.
For example, 3 and 5 are two consecutive odd numbers, whose sum is 8 and 8 is divisible by 4.

(v) False.
For example, 12 and 6 are even numbers but their sum 18 is not a multiple of 4.

(vi) False.
For example, 13 and 5 are odd numbers but their sum 18 is not a multiple of 4.

Question 7.
Solve the prime puzzle given below:

Solution:
385 = 5 × 7 × 11;
190 = 2 × 5 × 19;
42 = 2 × 3 × 7;
285 = 3 × 5 × 19;
154 = 2 × 7 × 11;
70 = 2 × 5 × 7
∴ Solution of given prime puzzle is as follows:

Question 8.
Solve the prime puzzle given below:

Solution:
Prime factorisations of the given numbers are as follows:
231 = 3 × 7 × 11;
130 = 2 × 13 × 5;
170 = 2 × 5 × 17;
102 = 3 × 2 × 17;
182 = 2 ×7 × 13;
275 = 5 × 5 × 11
∴ Solution of given prime puzzle is as follows:

Question 9.
Solve the prime puzzle given below:

Solution:
Prime factorisations of the given numbers are as follows:
286 = 2 × 11 × 13;
30 = 2 × 3 × 5;
70 = 2 × 5 × 7;
42 = 2 × 3 × 7;
110 = 2 × 5 × 11;
130 = 2 × 5 × 13
∴ Solution of given prime puzzle is as follows:

Prime Time Class 6 Case Based Questions

Question 1.
A Greek mathematician Eratosthenes, who lived around 2200 years ago, gave a method to list the prime numbers. The method is called the Sieve of Eratosthenes. In the given figure, encircled numbers arc prime numbers.

Based on the above information, answer the following questions:
(i) How many prime numbers are there between 1 and 100?
(ii) Which digits can never appear in the units place of a prime number?
(iii) Which digit most frequently appears in the units place of prime numbers less than 100?
(iv) Write live consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
(i) There are 25 prime numbers between 1 and 100.
(ii) 0, 4, 6 and 8 never appear in the unit place of a prime number.
(iii) Digit 3 appears 7 times in the units place of prime numbers less than 100.
(iv) 24, 25, 26, 27, 28 or 32, 33, 34, 35, 36 or 48, 49, 50, 51, 52 or 54, 55,56,57, 58 or 62, 63, 64, 65, 66 or 74, 75, 76, 77, 78 or 84, 85, 86, 87, 88
or 90, 91,92, 93, 94 or 91,92, 93, 94, 95 or 92, 93, 94, 95,96

Question 2.
The school gardening club has 210 marigold plants and 150 sunflower plants. The club members want to create equal rows for each type of flower such that each row contains the same number of one kind of plant and no plants are left unused.
Based on the above information, answer the following questions:
(i) Can they arrange the marigold plants in rows of 2, 5 or 7?
(ii) What is the largest number of plants per row they can use for each type so that all plants are used and plants in each row are equal?
(iii) They decide to pack the flowers into minimum flower boxes, and each box must contain the same number of plants. Which number less than 10 can they choose so that both 210 and 150 divide evenly by it?
(iv) Perform prime factorisation of 150 and 210 and find the common prime factors.
Solution:
(i) It is given that the school gardening club has 210 marigold plants. For 210 marigold plants to be arranged in rows of 2, 5 or 7, 210 must be divisible by 2, 5 or 7.
As 210 ends with 0, it is divisible by 2 and 5 both.
Now, 210 ÷ 7 = 30. Thus, 210 is also divisible by 7.
Thus, the marigold plants can be arranged in rows of 2, 5 or 7.

(ii) We need the largest number of plants of the same type per row such that

• Each row has the same number of plants.
• There are no leftover plants.
• Thus, we are looking for the largest number that can:
• Evenly divide 210 (so marigold plants are used up completely)
• Evenly divide 150 (so sunflower plants are used up completely)

This number will be the highest common factor of 210 and 150.
Now, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
And, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150.
The highest common factor is 30. Thus, if we use 30 plants in each row we get 210 ÷ 30 = 7 rows of marigold plant and 150 ÷ 30 = 5 rows of sunflower plant and no plants are left.

(iii) It is given that each flower box must contain equal number of plants and the number must be less than 10. Thus, we need to find the common factors of 150 and 210 which are less than 10.
Now, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
And, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150.
Thus, the common factors less than 10 are 1,2, 3, 5 and 6. ‘
For minimum boxes, the club members must place maximum plants in each box.
Thus, club members can keep 6 plants in each box, so that the flower boxes are minimum and there are equal plants in each box.

(iv) Prime factorisation of 210 = 2 × 3 × 5 × 7.
Prime factorisation of 150 = 2 × 3 × 5 × 5.
Thus, the common prime factors of 210 and 150 are 2, 3 and 5.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 2 Operations with Integers MCQ improves accuracy in objective exams.

MCQ on Operations with Integers Class 7

Operations with Integers MCQ Class 7

Class 7 Maths Operations with Integers MCQ

Question 1.
If sum of two integers is – 5, and one of the integers is 2, then the other integer is:
(a) -2
(b) -3
(c) -7
(d) -8
Solution:
(c) -7
Given, sum of two integers is – 5 and one of the integers is 2.
Let the other integer be 6.
Then, 2 + 6 = – 5 ⇒ b = – 5 – 2 = – 7

Question 2.
A pair of integers that give a product of – 20 is:
(a) -4, -5,
(b) 4, -3
(c) -5, 4
(d) -2, -10
Solution:
(c) -5, 4
We know, ( + ) × (-) = (-) or (-) × (+) = (-)
Thus, one integer must be negative and other must be positive to get product as – 20.
Also, we know that 4 ⇒ 5 = 20.
Taking 5 as negative, we get 4 ⇒ (- 5) = – 20
Thus, the pair of integers is -5 and 4.

Question 3.
For a pair of numbers, if the sum is 10 and the difference is 4, then least number in the pair is:
(a) 5
(b) 6
(c) 7
(d) 3
Solution:
(d) 3
Let a and b be the two numbers. Then,
a + b = 10 …(i)
and a – b = 4 …(ii)
Now, (a + b) + (a – b) = 10 + 4
[On adding (i) and (ii) ]
⇒ 2 × a = 14 ⇒ a = \(\frac{14}{2}\) = 7
Substituting a = 7 in a + b = 10, we get
7 + b = 10 ⇒ b = 10 – 7 = 3
Hence, the numbers are 7 and 3, and the least number in the pair is 3.

Question 4.
The temperature at 12 noon at a certain place was 15°C. If it decreases at the rate of 3°C per hour at what time will it be – 12°C?
(a) 11 am
(b) 11 pm
(c) 10 am
(d) 9 pm
Solution:
(d) 9 pm
Given, temperature at 12 noon, i.e. 12 pm at certain place = 15°C
And, change in temperature in 1 hour = – 3°C
Suppose, – 12°C is the temperature x hours after 12 noon.
∴ x × (-3) = (- 12)- 15
⇒ x × (-3) = – 27 ⇒ x = 9
Thus, time after 9 hours
= 12 pm + 9 hours= 9 pm

Question 5.
What term is used to describe a pair of a positive and a negative token that cancel out each other in the token model?
(a) Additive inverse
(b) Opposite pair
(c) Zero pair
(d) Neutral set
Solution:
(c) Zero pair
A pair of a positive (green) and a negative (red) token is called zero pair.

Question 6.
(-3) × (-6) =
(a) 18
(b) -18
(c) 12
(d) -12
Solution:
(a) 18
We know (-) × (-) = ( + )
Thus, (- 3) × (- 6) = 3 × 6 = 18

Question 7.
If both integers are same, then sum = ____ × integer.
(a) 2
(b) 0
(c) 3
(d) 4
Solution:
(a) 2
Let both integers be a.
Then, sum = a + a = 2 × a
Hence, sum = 2 × integer

Question 8.
If -400 × 3 × 3 + 567 = – 3033, then 43 – 3600 + 567 = ?
(a) – 2990
(b) -2980
(c) -3000
(d) -2970
Solution:
(a) – 2990
Given, – 400 × 3 × 3 + 567 = – 3033
⇒ (- 3600) + 567 = – 3033
Adding 43 to both sides, we get
43 + (- 3600) + 567 = 43 + (- 3033)
⇒ 43 – 3600 + 567 = 43 – 3033 = – 2990

Question 9.
If the sum and difference for a pair of integers is the same, one of the integers, is:
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(d) 0
Let the first integer be a and the second integer be 6.
We are given that, sum = difference
∴ a + b = a – b
⇒ a + b – a + b = 0 ⇒ 2b = 0 ⇒ b = 0
Thus, one of the integers is 0.

Question 10.
What will be the sign of product if we multiply 37 negative integers and 63 positive integers?
(a) Sometimes positive
(b) Always negative
(c) Never positive
(d) Can’t be determined
Solution:
(b) Always negative
We know, product of odd number of negative integers is negative and product of any number of positive integers is positive.
∴ Product of 37 (odd) negative integers and 63 positive integers is always negative.

Question 11.
The value of (- 30) ÷ 10 is:
(a) 3
(b) -3
(c) 1
(d) 0
Solution:
(b) -3
(-30) ÷ 10 = -(30 + 10)
[∵ (-a) ÷ b = -(a ÷ b)]
= -(3) = -3

Question 12.
The value of 0 ÷ (- 12) is:
(a) 4
(b) -12
(c) 12
(d) 0
Solution:
(d) 0
0 ÷ (- 12) = -(0 ÷ 12)
[∵ a ÷ (-b) = -(a ÷ b)]
= -(0) = 0

Question 13.
An integer which when multiplied by – 9 gives 117, is:
(a) 13
(b) -1053
(c) -13
(d) 1053
Solution:
(c) -13
Let the required integer be x.
Then, x × (- 9) = 117
⇒ x = 117 ÷ (-9) = -(117 ÷ 9) = – 13
Thus, the required integer is – 13.

Question 14.
If 11 × (a + 4) = 11 × (- 3) + 11 × 4, then a is :
(a) -1
(b) -2
(c) – 3
(d) – 4
Solution:
(c) -3
11 × (a + 4) = 11 × a + 11 × 4
[Using distributive property of multiplication over addition]
Given, 11 × (a + 4) = 11 × (- 3) + 11 × 4
⇒ 11 × a + 11 × 4 = 11 × (-3) + 11 × 4
On comparing, we get a = (- 3).

Question 15.
The value of (-36) ÷ (-9) is:
(a) 4
(b) – 9
(c) 6
(d) -2
Solution:
(a) 4
(-36) ÷ (-9) = 36 ÷ 9[∵ (-a) ÷ (-b) = a ÷ b]
= 4

Question 16.
The value of [(- 7) + (- 5)] ÷ [(- 4) + 1] is:
(a) -7
(b) -5
(c) 7
(d) 4
Solution:
(d) 4
[(- 7) + (- 5)] ÷ [(- 4) + 1]
= (-7 – 5) ÷ (-4 + 1)
= (-12) ÷ (-3)
= 12 ÷ 3 [∵ (-a) ÷ (-b) = a ÷ b]
= 4

Question 17.
The value of [8 × (-5) + 15 × 2 + 2] ÷ [1 × (-4)] is:
(a) -15
(b) – 5
(c) 7
(d) 2
Solution:
(d) 2
[8 × (- 5) + 15 × 2 + 2] ÷ [1 × (- 4)]
= (-40 + 30 + 2) ÷ (- 4)
= (-40 + 32) ÷ (- 4)
= (- 8) ÷ (- 4)
= 8 ÷ 4 [∵ (-a) ÷ (-b) = a ÷ b]
= 2

Question 18.
Which of the following expressions are equal to -30?
(i) -20 – (-5 × 2)
(ii) (-6 × 10) + (6 × 5)
(iii) (-2 × 5) + (-4 × 5)
(iv) (-6) × 5
Choose the correct option from the following:
(a) (ii) and (iv) only
(b) (iii) and (iv) only
(c) (ii), (iii) and (iv)
(d) (i), (ii) and (iv)
Solution:
(c) (ii), (iii) and (iv)
(i) -20 – (-5 × 2) = -20 – (-10) = -20 + 10 = -10
(ii) (-6 × 10) + (6 × 5) = -60 + 30 = – 30
(iii) (-2 × 5) + (-4 × 5) = – 10 + (- 20)
= -10 – 20 = -30
(iv) (-6) × 5 = – 30
Thus, expressions (ii), (iii) and (iv) are equal to – 30.

Question 19.
Which of the following expressions result in a negative integer?
(i) (-2) × (-3)
(ii) 10 + 3 × (-7)
(iii) (-5) × 2 + 10
(iv) (-5) × 7
Choose the correct option from the following:
(a) (i) and (iv)
(b) (iii) and (iv)
(c) (i) and (ii)
(d) (ii) and (iv)
Solution:
(d) (ii) and (iv)
(i) (- 2) × (- 3) = 6, which is a positive integer.
(ii) 10 + 3 × (- 7) = 10 + (- 21) = 10 – 21
= -11, which is a negative integer.
(iii) (-5) × 2 + 10 = -10 + 10 = 0, which is neither a positive nor a negative integer.
(iv) (-5) × 7 = -35, which is a negative integer.
Thus, the expressions (ii) and (iv) result in a negative integer.

Question 20.
Which of the following expressions are equal to -25?
(i) – 20 – (5 × 7)
(ii) (-7 × 1) + (- 3 × 6)
(iii) (-5 × 10) + (5 × 5)
(iv) (-5) × 8
Choose the correct option from the following:
(a) (ii) and (iii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (ii) and (iii)
(i) -20 – (5 × 7) = – 20 – 35 = – 55 + – 25
(ii) (-7 × 1) + (-3 × 6) = -7 + (-18) = -7 – 18 = -25
(iii) (-5 × 10) + (5 × 5) = -50 + 25 = – 25
(iv) (-5) × 8 = -(5 × 8) = -40 ≠ -25
Thus, expressions (ii) and (iii) are equal to – 25.

Question 21.
Which of the following expressions result in odd integer?
(i) (-7) × (-3)
(ii) 10 + 3 × (- 2)
(iii) (-4) × 2 + 8
(iv) (-3) × 9
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(d) (i) and (iv)
(i) (-7) × (-3) = 21,
which is an odd integer.
(ii) 10 + 3 × (- 2) = 10 + (- 6) = 4, which is an even integer.
(iii) (-4) × 2 + 8 = -8 + 8 = 0, which is an even integer.
(iv) (- 3) × 9 = – 27, which is an odd integer.
Thus, the expressions (i) and (iv) result in odd integers.

Question 22.
Which of the following expressions are correct?
(i) 72 ÷ (- 8) = – 9
(ii) 36 ÷ (-12) = – 3
(iii) (-78) ÷ (-2) = 38
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (ii) and (iii) only
(c) (i) and (ii) only
(d) (i) only
Solution:
(c) (i) and (ii) only
(i) 72 ÷ (- 8) = – (72 ÷ 8) = -(9) = – 9
(Correct)
(ii) 36 ÷ (-12) = -(36 ÷ 12) = -3 (Correct)
(iii) (-78) ÷ (-2) = 78 ÷ 2 = 39 (Incorrect)

Operations with Integers Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The additive inverse of (-12) × (-3) + 4 is -40.
(R): The additive inverse of a is – a.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
(-12) × (- 3) + 4 = 36 + 4 = 40
We know, the additive inverse of a is – a.
∴ The additive inverse of 40 is – 40.
Hence, the additive inverse of (-12) × (-3) + 4 is – 40.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The value of (-3) × (-4) × 5 × 2 × (-1) is negative.
(R): The product of odd number of negative integers is positive.
Solution:
(c) A is true but R is false.
We know that the product of odd number of negative integers is always negative.
In product (-3) × (-4) × 5 × 2 × (-1), there are 3 (odd) negative integers.
So, the product is negative.
Thus, Assertion (A) is true, but Reason (R) is false

Operations with Integers Class 7 Fill in the Blanks

Question 1.
60 + 40 + ______ = 0
Solution: -100
Let the required integer be a.
Then, 60 + 40 + a = 0 ⇒ 100 + a = 0
We know that the sum of an integer and its additive inverse is 0.
∴ a = – 100
Thus, 60 + 40 + (-100) = 0.

Question 2.
(-1) × _____ = – 43
Solution: 43
We know, (-1) × a = -a, for all integers a.
Thus, (-1) × 43 = -43

Question 3.
(-9) × (-5) × 6 × (-3) = _______
Solution: -810
The product of odd number of negative integers is negative.
∴ (-9) × (-5) × 6 × (-3)
= -(9 × 5 × 6 × 3) = -810

Question 4.
If we multiply 17 positive integers and 92 negative integers, then the sign of the product is _________ .
Solution: positive
We know that the product of even number of negative integers is always positive.
Final product = (Product of 17 positive integers) × (Product of 92 negative integers)
= (Positive) × (Positive) = Positive
Thus, if we multiply 17 positive integers and 92 negative integers, then the sign of the product is
positive.

Question 5.
When two negative integers are added, we get a ______ integer.
Solution: negative
When two negative integers are added, we get a negative integer.

Question 6.
Find:
(i) -13 + 42 = _____
(ii) 47 + (-26) = ______
(iii) 91 – 19 = ______
(iv) -38 – (-29) = _____
Solution: 29, 21, 72, -9
(i) -13 + 42 = 29
(ii) 47 + (-26) = 21
(iii) 91 – 19 = 72
(iv) -38 – (-29) = – 9

Question 7.
(7 + 3) + (5 + 2) + ( _____ + 6)
= (3 + 6) + (5 + 7) + (2 + 9)
Solution: 9
RHS = (3 + 6) + (5 + 7) + (2 + 9)
= 3 + 6 + 5 + 7 + 2 + 9
= (7 + 3) + (5 + 2) + (6 + 9)
= (7 + 3) + (5 + 2) + (9 + 6) = LHS
∴ (7 + 3) + (5 + 2) + (9 + 6)
= (3 + 6) + (5 + 7) + (2 + 9)

Question 8.
-19 + 23 – 8 + _____ = 12
Solution: 16
Let the required number be x. Then
-19 + 23 – 8 + x = 12 ⇒ 23 – 27 + x = 12
⇒ -4 + x = 12 ⇒ x = 12 + 4= 16
∴ -19 + 23 – 8 + 16 = 12

Question 9.
(-8) × (-9) + (-3) × (-3) × (-3) = _________
Solution: 45, 9, -36
(-8) × (- 9) + (- 3) × (- 3) × (- 3)
= (8 × 9) – (3 × 3 × 3) = 72 – 27 = 45

Question 10.
(-4) × (10 – 5 – 4 + 8) = (-4) × ____ = ______
Solution: 9, -36
(-4) × (10 – 5 – 4 + 8) = (-4) × 9 = – 36

Question 11.
When one of the multiplier or the multiplicand is positive and the other is negative, their product is _______ .
Solution: negative
We know that, when one of the multiplier or the multiplicand is positive and the other is negative, their product is negative.

Question 12.
When both the multiplier and the multiplicand are negative, the product is __________ .
Solution: positive
When both the multiplier and the multiplicand are negative, the product is positive.

Question 13.
Fill in the blanks keeping in view the properties of multiplication and division of integers:
(i) (-2) × 5 = _____ × (-2)
(ii) (-4) × [(____) + (2)] = (-4) × (-5) + (______) × (2)
(iii) 100 × [(____) × (-45)] = [ ____ × (-4)] × (-45)
(iv) (-15) ÷ (_____) = (15) ÷ (-3)
Solution: 5, -5, -4, -4, 100, 3
(i) (-2) × 5 = 5 × (- 2)
[Using commutative law of multiplication]
(ii) (-4) × [(-5) + (2)] = (-4) × (- 5) + (-4) × (2)
[Using distributive law of multiplication]
(iii) 100 × [(-4) × (-45)] = [100 × (-4)] × (-45)
[Using associative law of multiplication]
(iv) (-15) ÷ (3) = (15) ÷ (-3)

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 4 Data Handling and Presentation Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 4 Data Handling and Presentation Solutions

Ganita Prakash Class 6 Chapter 4 Solutions

Class 6 Maths Ganita Prakash Chapter 4 Solutions Data Handling and Presentation

Question 1.
Shri Nilesh is a teacher. He decided to bring sweets to the class to celebrate the new year. The sweets shop nearby has jalebi, gulab jamun, gujiya, barfi, and rasgulla. He wanted to know the choices of the children. He wrote the names of the sweets on the board and asked each child to tell him their preference. He put a tally mark ‘ | ’ for each student and when the count reached 5, he put a line through the previous four and marked it as

Complete the table to help Shri Nilesh to purchase the correct numbers of sweets.
(a) How many students chose jalebi? _______
(b) Barfi was chosen by _______ students?
(c) How many students chose gujiya? _______
(d) Rasgulla was chosen by _______ students?
(e) How many students chose gulab jamun? _______
Solution:

(a) It is clear from the table, Jalebi was chosen by 6 students.
(b) It is clear from the table, Barfi was chosen by 3 students.
(c) Gujiya was chosen by 13 students.
(d) Rasgulla was chosen by 7 students.
(e) Gulab jamun was chosen by 9 students.

Question 2.
Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to purchase kites from him. The number of kites he sold to these six shopkeepers are given below.

Shopkeeper	Number of Kites sold
Chaman	250
Rani	300
Rukhsana	100
Jasmeet	450
Jetha Lai	250
Poonam Ben	700

Prepare a pictograph using the symbol to represent 100 kites.
Answer the following questions:
(a) How many symbols represent the kites that Rani purchased?
(b) Who purchased the maximum number of kites?
(c) Who purchased more kites, Jasmeet or Chaman?
(d) Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why?
Solution:
Required pictograph is given below.

(a) Since Rani purchased 300 kites, it is clear from the pictograph, 3 symbols represent the kites that Rani purchased.

(b) The shopkeeper who purchased the maximum number of kites is the one with the most symbols in pictograph. It is clear from the pictograph that Poonam Ben purchased the maximum number of kites (7 × 100 = 700 kites).

(c) Number of kites purchased by Jasmeet = 450 Number of kites purchased by Chaman = 250 Jasmeet purchased more kites.

(d) Number of kites purchased by Poonam Ben = 700 Number of kites purchased by Rani = 300
Hence, Rukhsana is correct as 700 kites is more than double of 300 kites (300 × 2 = 600).

Question 3.
Samantha visited a tea garden and collected data of the insects and critters she saw there. Here is the data she collected —

Help her prepare a bar graph representing this data.
Solution:
To prepare a bar graph representing the given data, follow the point given below.
(i) Draw a horizontal line labeled, ‘Insects and Critters’ with each type of insect or critter evenly spaced.

(ii) Draw a vertical line labeled ‘Number of insects and critters seen’ with numbers starting from 0 up to the maximum number seen (in this case, 10) evenly spaced.

(iii) For each insect or critter, draw a bar that rises to the number seen. For example, the bar of caterpillars should reach up to 10.
Required bar graph is given alongside.

Question 4.
Chinu listed the various means of transport that passed across the road in front of his house from 9 a.m. to 10 a.m.

(a) Prepare a frequency distribution table for the data.
(b) Which means of transport was used the most?
(c) If you were there to collect this data, how could you do it? Write the steps or process.
Solution:
(a) Frequency distribution table for the given data is given below:

Means of transport	Bike	Car	Bus	Auto rickshaw	Bicycle	Bullock Cart	Scooter
Frequency	13	6	4	8	8	2	9

(b) The bike has the highest frequency in the table, indicating it was the most common means of transport observed.

(c) To collect this data, you could follow the steps given below. ;
Observation Timeframe: I will choose a specific timeframe, such as 9 a.m. to 10 a.m., to observe the road traffic.
Recording Data: 1 will use a tally chart or counting app to record the type of transport passing by during that hour.
Categorisation: Then I will organise the data into categories (e.g., bike, car, scooter, bus, etc.).
Final Count: After the observation period, I will get the total the number of occurrences for each category.
Analysis: Finally, I prepare a frequency distribution table based on the recorded data for analysis.

Question 5.
The number of girl students in each class of a school is depicted by the pictograph.

Observe this pictograph and answer the following questions.
(a) Which class has the least number of girl students?
(b) What is the difference between the number of girls in Class 5 and 6?
(c) If two more girls were admitted in Class 2, how would the graph change?
(d) How many girls are there in Class 7?
Solution:
(a) The pictograph shows the least number of symbols for Class 8.
Hence, Class 8 has the least number of girl students.

(b) Girl students in Class 5 = 2.5 × 4 = 10 Girl students in Class 6 = 4 × 4 = 16
Hence, there are 6 more girls in Class 6 than in Class 5.

(c) Here, adding 2 more girls would increase the count by 2 and requires a half additional symbol.
The pictograph would show half additional symbol for class 2, assuming each symbol represent 4 girls.

(d) Girl students in Class 7 = 3 × 4 = 12.

Data Handling and Presentation Class 6 Extra Questions

Data Handling and Presentation Class 6 Very Short Question Answer

Question 1.
Define the following terms:
(i) Observations
(ii) Data
Solution:
(i) Observations are the individual pieces of information we collect.
For example, if we ask five friends their ages, each friend’s age is one observation.

(ii) Any collection of facts, numbers, measures, observations or other descriptions of things that convey information about those things is called data.

Question 2.
Define the following terms:
(i) Tabulation of data
(ii) Tally marks
Solution:
(i) Tabulation of data is the process of systematically arranging data into rows and columns within a table to make it easier to understand, compare and analyse.

(ii) Tally marks are short vertical lines used to count items. Each time we count one item, we draw one line (|). After four lines (| | | |), we draw a oblique line (\) across them to show five.

Question 3.
Define the following terms:
(i) Raw data
(ii) Array
Solution:
(i) Raw data is the information as we first collect it (original form), before we sort or arrange it.
For example, writing ages on a sheet of paper as friends tell you, without putting them in any order.

(ii) Organising the data by arranging it in ascending or descending order is called an array.
For example, Meenal arranged the shoe sizes of the students in ascending order as follows:
3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7

Question 4.
Define the following terms:
(i) Frequency of observation
(ii) Frequency distribution
Solution:
(i) Frequency of observation tells us how many times a particular value appears in the data. For example, if the age ‘12’ appears three times in the list, then the frequency of age ‘12’ is 3.
(ii) Frequency distribution is a table that shows each value (or group of values) and how often it appears (frequency). For example:

Age	Frequency
10	2
11	1
12	3

Question 5.
A die was thrown 20 times and the following outcomes were noted:
1, 2, 5, 1, 6, 2, 3, 4, 2, 4, 2, 5, 6, 6, 6, 2, 3, 1, 1, 5
Represent the above data in the form of frequency distribution.
Solution:

Outcome	Frequency
1	4
2	5
3	2
4	2
5	3
6	4

Question 6.
Enlist 1 advantage and 1 disadvantage of presentation of data through pictograph.
Answer:
1 advantage and 1 disadvantage of pictograph are as follows:

Advantage	Disadvantage
1. Pictorial representation makes it easier to understand the data.	1. Drawing a pictograph is time consuming.

Data Handling and Presentation Class 6 Short Question Answer

Question 1.
Given below is the data showing the number of children in 15 families of a colony.
2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
Arrange the above data in ascending order and then form a tally table.
Solution:
Given data (number of children in 15 families) is as follows: 2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
(i) Data in ascending order: 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4.
(ii) Given data in tabular form is shown below:

Question 2.
The number of cricket bat and ball pairs, sold by a shopkeeper during a week are given below:

Day	Mon.	Tue.	Wed.	Thu.	Fri.	Sat.
Number of bat and ball pair sold	12	9	15	15	3	27

Decide the scaling and draw a pictograph to represent the given data.
Solution:

Question 3.
Total number of cows in five villages are as follows:

Village	A	B	C	D	E
Number of cows	40	60	20	100	120

Decide the scale and draw a pictograph to represent the given data.
Solution:

Question 4.
Following pictograph shows the data of the number of students who like a particular sport.

Answer the following questions:
(i) How many students like cricket?
(ii) How many more students like basketball than volleyball?
Solution:
(i) Number of symbols for cricket = 4
Students who like cricket = 4 × 3= 12

(ii) Number of symbols for basketball = 3
∴ Students who like basketball = 3 × 3 = 9
Number of symbols for volleyball = 2
∴ Students who like volleyball = 2 × 3 = 6
Difference = 9 – 6 = 3
Hence, 3 more students like basketball than volleyball.

Question 5.
The bar graph below shows the production of Kharif and Rabi crops (in tons) in 5 different states in India. Draw a table that represents the data in the bar graph.

Solution:
Table for the given bar graph can be created as follows:

States	Rabi crops (in tons)	Kharif crops (in tons)
West Bengal	30	60
Karnataka	45	40
Madhya Pradesh	20	54
Tamil Nadu	36	22
Uttar Pradesh	50	42

Data Handling and Presentation Class 6 Long Question Answer

Question 1.
The weights of newborn babies (in Kg) in a hospital on a particular day are as follows:
2.1, 2.5, 3.5, 2.7, 2.9, 3.1, 2.6, 2.5, 2.8, 2.3, 2.9, 3.3, 3.4, 2.7, 2.8
Answer the following questions:
(i) How many babies are born on that day?
(ii) Arrange the above data in descending order.
(iii) Determine the highest weight.
(iv) Determine the lowest weight.
(v) Determine the range of weight.
(vi) How many babies weigh below 2.5 kg?
Solution:
(i) There are 15 babies born on that day.

(ii) Writing the weights from highest to lowest:
3.5, 3.4, 3.3, 3.1, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.3, 2.1

(iii) The highest weight is the first weight in the descending order of weights.
Highest weight = 3.5 kg

(iv) The lowest weight is the last weight in the descending order of weights.
Lowest weight = 2.1 kg

(v) We know, range = highest value – lowest value
∴ Range of weight = highest weight – lowest weight
= 3.5 kg – 2.1 kg = 1.4 kg

(vi) The weights below 2.5 kg are 2.1 kg and 2.3 kg. So, the required number of babies is 2.

Question 2.
Study the following pictograph:

Answer the following questions:
(i) In which city was the rainfall maximum and how much?
(ii) In which city was the rainfall minimum and how much?
(iii) Which cities had rainfall of more than 50 cm?
Solution:
(i) Chennai has three full icons, which is maximum.
∴ Rainfall in Chennai = 3 × 50 cm = 150 cm

(ii) Delhi has one full icon, which is minimum.
∴ Rainfall in Delhi = 1 × 50 cm = 50 cm.

(iii) Mumbai has 2 full icons.
∴ Rainfall in Mumbai = 2 × 50 cm = 100 cm (which is greater than 50 cm)
Hyderabad has 1 full and 1 half icons.
∴ Rainfall in Hyderabad = 50 cm + 25 cm = 75 cm (which is greater than 50 cm) Chennai has 3 full icons.
∴ Rainfall in Chennai = 50 × 3 = 150 cm (which is greater than 50 cm)
Kolkata has 2 full and 1 half icons.
∴ Rainfall in Kolkata =100 cm + 25 cm =125 cm (which is greater than 50 cm)
Hence, Mumbai, Hyderabad, Chennai, and Kolkata had rainfall of more than 50 cm.

Question 3.
In a shop, there are 5 different types of toys. The number of each toy is given in the table below. Construct a bar graph representing this data. Use appropriate scaling.

Toys Name	Blocks	Cars	Dolls	Balls	Bikes
Number of Toys	40	48	60	35	30

Solution:
The required bar graph can be easily created using the following steps:
Step 1: On a graph sheet draw two mutually perpendicular lines, a horizontal and a vertical line.

Step 2: Label horizontal line as ‘Toys name’ & write names of toys from the table with equal gaps between them: Blocks, Cars, Dolls, Balls, Bikes. Label vertical line as ‘Number of Toys’.

Step 3: Scaling: Choose a suitable scale to show the number of toys.
Here, we take: 1 unit (1 big division) = 10 toys. Each big division is further divided into 10 sub-divisions, representing 1 toy.

Step 4: Calculate the height of each bar:
10 toys = 1 large divisions
∴ 1 toy = \(\frac{1}{10}\) large divisions
= 1 sub-division
Now, the height of the bar for Blocks
= 40 × (\(\frac{1}{10}\)) = 4 large divisions
The height of the bar for Cars
48 × (\(\frac{1}{10}\)) = 4 large divisions and 8 sub-divisions
The height of the bar for Dolls = 60 × (\(\frac{1}{10}\)) = 6 large divisions
The height of the bar for Balls = 35 × (\(\frac{1}{10}\)) = 3 large divisions and 5 sub-divisions
The height of the bar for Bikes = 30 × (\(\frac{1}{10}\))

Step 5: Now draw vertical bars for each toy using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 4.
The zookeeper in Delhi is preparing a presentation for higher officers on the number of animals in the zoo. He intends to create a bar graph for better visualisations. The number of each animal is given below in the table. Construct a bar graph for the table.

Animals	Number of Animals
Lion	8
Elephant	12
Gorilla	9
Zebra	15
Giraffe	7
Cheetah	6

Solution:
Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as ‘Animals’ and write the names of the animals from the table with equal gaps between them: Lion, Elephant, Gorilla, Zebra, Giraffe, Cheetah Label the vertical lines ‘ Number of Animals’.

Step 3 (Scaling) : Choose a suitable scale to show the number of animals.
Here, we take: 1 unit (1 large division)
= 2 animals
∴ 1 animal = \(\frac{1}{2}\) unit = 1 sub-division
This scale helps us draw the graph clearly.

Step 4: Calculate the height of each bar using the scale:
Height of the bar for the Lion = 8 ÷ 2 = 4 large divisions
Height of the bar for the Elephant = 12 ÷ 2 = 6 large divisions
Height of the bar for the Gorilla = 9 ÷ 2 = 4 large divisions and 1 subdivision
Height of the bar for the Zebra = 15 ÷ 2 = 7 large divisions and 1 sub-division
Height of the bar for the Giraffe = 7 + 2 = 3 large divisions and 1 subdivision
Height of the bar for the Cheetah = 6 + 2 = 3 large divisions

Step 5: Now draw vertical bars for each animal using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 5.
The following bar graph shows the revenue of your country from exports of various items in 1 year. (1 unit = 10 crore rupees)

(a) Write the information given by the bar graph in a table.
(b) What is the difference between the maximum revenue and the minimum revenue?
(c) What is the total revenue from exports?
Solution:
(a) The required table is shown below:

Items	Revenue (Rupees in crores)
Electronics	30
Foodgrains	70
Livestock	25
Metals	50
Softwares	80

(b) From the table, it can be observed that the maximum revenue was generated by the export of software, i.e. 80 crore rupees.
Also, the minimum revenue was generated by the exports of livestock, i.e. 25 crore rupees.
∴ The difference between the revenue from software and livestock = 80 – 25 = 55 crore

(c) Total revenue from exports can be calculated by summing up the revenues from all the commodities.
∴ Total revenue from exports = 30 + 70 + 25 + 50 + 80 = 255 crore rupees

Question 6.
The table shows how much money Imran’s family spends each month on different items:

Items	Expenditure (in ₹)
House rent	3,000
Food	3,400
Education	800
Electricity	400
Transport	600
Miscellaneous	1,200

Answer the following questions:
(a) Represent the given data in the form of a bar graph
(b) On which item does the Imran’s family spend the most and the second most?
(c) Is the cost of electricity about one half of the cost of education?
Solution:
(a) To represent this data in the form of a bar graph, here are the steps:

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘items’ with equal spacing between them and mark the corresponding expenditures along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length = ₹ 200, and then mark and write the corresponding values (₹ 200, ₹400, etc.) representing each unit length and calculate the heights of the bars for various items as shown below:

House rent	3000 ÷ 200	15 units
Food	3400 ÷ 200	17 units
Education	800 ÷ 200	4 units
Electricity	400 ÷ 200	2 units
Transport	600 ÷ 200	3 units
Miscellaneous	1200 ÷ 200	6 units

Here is the bar graph that we obtained based on the above steps.

(b) From the given data, we can say that on food Imran’s family spend the most and on house rent they spend the second most.

(c) Yes. As the cost of electricity is ₹ 400 and the cost of eduction is ₹ 800, we can say that the cost of electricity is about one-half the cost of education.

Question 7.
Your school is collecting data on various transportation options that students used to opt. The table below shows the data collected by you. Prepare a bar graph to be given to your school principal, also use appropriate scaling.

Transportation	Number of Students
Bus	200
Cycle	50
Walk	80
Auto	120
Car	40

Solution:
The required bar graph can be easily created using the following steps:

Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as‘Transportation’ and write the names from the table with equal gaps between them: Bus, Cycle, Walk, Auto, Car.
Label the vertical line as ‘Number of Students’.

Step 3: (Scaling): Choose a suitable scale to show the number of students.
Here, we take:
1 unit (1 big division) = 20 students (This helps us fit large numbers like 200 easily on the graph.)

Step 4: Calculate the height of each bar:
= 200 × (\(\frac{1}{20}\)) = 10 large divisions
Height of the bar for the cycle
= 50 × (\(\frac{1}{20}\))= 2 large divisions and 5 sub-divisions
Height of the bar for the walk
= 80 × (\(\frac{1}{20}\)) = 4 large divisions
Height of the bar for the auto
= 120 × (\(\frac{1}{20}\)) = 6 large divisions
Height of the bar for the car
= 40 × (\(\frac{1}{20}\)) = 2 large divisions

Step 5: Now draw vertical bars for each mode of transportation using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 8.
The table below displays the number of bicycles produced at a factory from 1998 to 2002.

Year	1998	1999	2000	2001	2002
No. of Bicycles	800	600	900	1100	1200

Answer the following questions:
(a) Create a bar graph to represent this information. Select your preferred scale.
(b) Which year had the greatest number of bicycles produced?
(c) Which year had the least number of bicycles produced?
Solution:
(a) To represent this data in the form of a bar

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘years’ with equal spacing between them and mark the corresponding ‘No. of bicycles’ along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length =100 bicycles, and then mark and write the corresponding values (100 bicycles, 200 bicycles, etc.) representing each unit length and calculate the heights of the bars for various years as shown below:

1998	800 ÷ 100	8 units
1999	600 ÷ 100	6 units
2000	900 ÷ 100	9 units
2001	1100 ÷ 100	11 units
2002	1200 ÷ 100	12 units

(b) The greatest number of bicycles produced in 2002.
(c) The least number of bicycles produced in 1999.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 8 Working with Fractions MCQ improves accuracy in objective exams.

MCQ on Working with Fractions Class 7

Working with Fractions MCQ Class 7

Class 7 Maths Working with Fractions MCQ

Question 1.
The value of \(\frac{5}{6}\) of 30 is:
(a) 20
(b) 25
(c) 30
(d) 35
Solution:
(b) 25
We know, ‘of’ means multiplication.
∴ \(\frac{5}{6} \text { of } 30=\frac{5}{6} \times 30\) = 5 × 5 = 25

Question 2.
What is the value of \(\frac{5}{8}\) × 64 ?
(a) 35
(b) 40
(c) 45
(d) 50
Solution:
(b) 40
\(\frac{5}{8} \times 64\) = 5 × 8 = 40

Question 3.
The value of \(3 \frac{1}{2}+\frac{4}{3}-2 \frac{2}{5}\) is:
(a) \(\frac{89}{30}\)
(b) \(\frac{91}{30}\)
(c) \(\frac{87}{30}\)
(d) \(\frac{73}{30}\)
Solution:
(d) \(\frac{73}{30}\)
We can write, \(3 \frac{1}{2}=\frac{7}{2} \text { and } 2 \frac{2}{5}=\frac{12}{5}\)
∴ \(3 \frac{1}{2}+\frac{4}{3}-2 \frac{2}{5}=\frac{7}{2}+\frac{4}{3}-\frac{12}{5}=\frac{105}{30}+\frac{40}{30}-\frac{72}{30}\)
[∵ LCM of 2, 3 and 5 is 30.]
= \(\frac{105+40-72}{30}=\frac{73}{30}\)

Question 4.
The cost of one notebook is \(\frac{3}{4}\) rupees. What is the cost of 32 such notebooks?
(a) ₹20
(b) ₹22
(c) ₹24
(d) ₹26
Solution:
(c) ₹24
Given, the cost of one notebook is \(\frac{3}{4}\) rupees.
Therefore, the cost of 32 notebooks
= \(32 \times \frac{3}{4}\) = 8 × 3 = ₹24

Question 5.
The value of \(\frac{11}{18}+\left(\frac{7}{15} \times \frac{1}{4}\right)\) is:
(a) \(\frac{3}{180}\)
(b) \(\frac{56}{180}\)
(c) \(\frac{131}{180}\)
(d) \(\frac{9}{180}\)
Solution:
(c) \(\frac{131}{180}\)
Given, \(\frac{11}{18}+\left(\frac{7}{15} \times \frac{1}{4}\right)=\frac{11}{18}+\frac{7}{60}=\frac{110}{180}+\frac{21}{180}\)
[∵ LCM of 18 and 60 is 180]
= \(\frac{110+21}{180}=\frac{131}{180}\)

Question 6.
The reciprocal of \(\frac{8}{7}\) is:
(a) \(\frac{1}{8}\)
(b) \(\frac{7}{8}\)
(c) \(\frac{1}{7}\)
(d) \(\frac{8}{7}\)
Solution:
(b) \(\frac{7}{8}\)
The reciprocal of \(\frac{8}{7} \text { is } \frac{7}{8}\).

Question 7.
The value of \(5 \frac{2}{3} \div 1 \frac{5}{6}\) is:
(a) \(3 \frac{1}{11}\)
(b) \(2 \frac{2}{3}\)
(c) \(1 \frac{1}{6}\)
(d) \(3 \frac{2}{5}\)
Solution:
(a) \(3 \frac{1}{11}\)
We have, \(5 \frac{2}{3} \div 1 \frac{5}{6}\)
= \(\frac{17}{3} \div \frac{11}{6}=\frac{17}{8} \times \frac{6}{11}\)
[As \(5\frac{2}{3}=\frac{17}{3} \text { and } 1 \frac{5}{6}=\frac{11}{6}\)]
= \(\frac{17 \times 2}{1 \times 11}=\frac{34}{11}=3 \frac{1}{11}\)

Question 8.
\(\left(\frac{18}{6} \div \frac{3}{9}\right)+\left(\frac{21}{7} \div \frac{6}{4}\right)\) =
(a) 9
(b) 11
(c) 12
(d) 10
Solution:
(b) 11
We have, \(\left(\frac{18}{6} \div \frac{3}{9}\right)+\left(\frac{21}{7} \div \frac{6}{4}\right)\)
= \(\left(\frac{38}{6} \times \frac{3}{8}\right)+\left(\frac{21}{7} \times \frac{4}{6}\right)\)
= \(\frac{3 \times 3}{1 \times 1}+\frac{9 \times 2}{1 \times 3}\) = 9 + 2 = 11

Question 9.
The reciprocal of \(2 \frac{3}{4}\) is:
(a) \(\frac{11}{4}\)
(b) \(\frac{13}{4}\)
(c) \(\frac{4}{11}\)
(d) \(\frac{4}{13}\)
Solution:
(c) \(\frac{4}{11}\)
We have \(2\frac{3}{4}=\frac{(2 \times 4)+3}{4}=\frac{8+3}{4}=\frac{11}{4}\)
Now, reciprocal of \(\frac{11}{4} \text { is } \frac{4}{11}\)

Question 10.
The value of \(\frac{7}{8} \div 4\) is:
(a) \(\frac{7}{2}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{4}{7}\)
(d) \(2\frac{1}{4}\)
Solution:
(b) \(\frac{7}{32}\)
We have, \(\frac{7}{8} \div 4=\frac{7}{8} \times \frac{1}{4}=\frac{7 \times 1}{8 \times 4}=\frac{7}{32}\)

Question 11.
The value of \(4 \frac{3}{5} \div 2 \frac{1}{4}\) is:
(a) \(2\frac{2}{45}\)
(b) \(3\frac{1}{5}\)
(c) \(1\frac{4}{9}\)
(d) \(2\frac{1}{4}\)
Solution:
(a) \(2\frac{2}{45}\)
We have, \(4 \frac{3}{5} \div 2 \frac{1}{4}=\frac{23}{5} \div \frac{9}{4}=\frac{23}{5} \times \frac{4}{9}\)
[∵ \(4 \frac{3}{5}=\frac{23}{5}, 2 \frac{1}{4}=\frac{9}{4}\)
= \(\frac{23 \times 4}{5 \times 9}=\frac{92}{45}=2 \frac{2}{45}\)

Question 12.
Which of the following is/are correct?
(i) \(36 \div \frac{3}{4}=48\)
(ii) \(20 \div 2 \frac{1}{2}=8\)
(iii) \(\frac{5}{6} \div \frac{1}{3}=\frac{5}{2}\)
(iv) \(3 \frac{1}{2} \div \frac{7}{4}=2\)
Choose the correct option from the following:
(a) (i), (ii) and (iii) only
(b) (i), (iii) and (iv) only
(c) (ii), (iii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Solution:
(d) (i), (ii), (iii) and (iv)

Question 13.
Which of the following is/are correct?
(i) \(36 \div \frac{3}{4}=48\)
(ii) \(45 \div 5 \frac{5}{6}=6\)
(iii) \(\frac{7}{9} \div \frac{1}{3}=\frac{7}{2}\)
(iv) \(3 \frac{1}{2} \div \frac{7}{4}=2\)
Choose the correct option from the following:
(a) (i), (ii) and (iv)
(b) (ii), (iii) and (iv)
(c) (i) and (iv) only
(d) (i), (ii) and (iii)
Solution:
(c) (i) and (iv) only
(i)
= 48 – correct

(ii)
= \(\frac{9 \times 6}{7}=\frac{54}{7} \neq 6\) = Incorrect

(iii) – Incorrect

(iv) – Correct

A Tale of Three Intersecting Lines Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The product of \(\frac{2}{5} \text { and } \frac{3}{4} \text { is } \frac{6}{20}\)
(R): To multiply two fractions, we take the LCM of the denominators and then add the numerators.
Solution:
(c) A is true but R is false.
To multiply two fractions, we do not take the LCM or add numerators. Instead, we multiply numerators and denominators directly.
∴ \(\frac{2}{5} \times \frac{3}{4}=\frac{2 \times 3}{5 \times 4}=\frac{6}{20}\)
Thus, Assertion (A) is true, but Reason (R) is false.

Question 2.
(A): \(\frac{4}{3}\) of 15 is 20.
(R): In \(\frac{4}{3}\) of 15 , ‘of’ means ‘multiplication’.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
In \(\frac{4}{3}\) of 15 , ‘of ’ means ‘multiplication’.
Thus, \(\frac{4}{3}\) of 15 = \(\frac{4}{3} \times 15\) = 4 × 5 = 20
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): The product of \(\frac{2}{3}, \frac{3}{4} \text { and } \frac{1}{2} \text { is } \frac{1}{4}\)
(R): To multiply three fractions, we take the LCM of the denominators and then add the numerators.
Solution:
(c) A is true but R is false.
To multiply three fractions, we do not take the LCM or add the numerators.
Instead, we multiply the numerators and multiply the denominators directly.
∴ \(\frac{2}{3} \times \frac{3}{4} \times \frac{1}{2}=\frac{2 \times 3 \times  1}{3 \times 4 \times 2}=\frac{1}{4}\)
Therefore, Assertion (A) is true, but Reason (R) is false.

Question 4.
(A): The product of two improper fractions is not smaller than any of the two fractions.
(R): Multiplication of two improper fractions gives a result that is greater than both the fractions or equal to either of them.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
An improper fraction has its numerator equal to or greater than the denominator (e.g. \(\frac{5}{4}, \frac{7}{3},\) 1, etc.). Its value is greater than or equal to 1.
Thus, the product of two improper fractions is equal to either of them or greater than each fraction.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

A Tale of Three Intersecting Lines Class 7 Fill in the Blanks

Question 1.
The product of two proper fractions is found by multiplying the _______ of both and the _____ of both.
Solution: numerators, denominators
The product of two proper fractions is found by multiplying the numerators of both and the
denominators of both.

Question 2.
Priya is making lemonade. She uses \(\frac{2}{3}\) of a lemon for one glass. If she makes \(\frac{3}{4}\) of a glass, she will use ______ of a lemon.
Solution: \(\frac{1}{2}\) of a glass
Given, lemon used for 1 full glass is \(\frac{2}{3}\) and glass prepared = \(\frac{3}{4}\) of a glass
∴ Required lemon = \(\frac{3}{4} \times \frac{2}{3}=\frac{3 \times 2}{4 \times 3}=\frac{6}{12}=\frac{1}{2}\) = of a lemon.

Question 3.
The product of \(\frac{5}{8}\) and 96 is _______ .
Solution: 60
Product of \(\frac{5}{8}\) and 96 is \(\frac{5}{8} \times 96\) = 5 × 12 = 60

Question 4.
Ravi is painting a wall. He uses \(\frac{3}{5}\) of a bucket of paint for one wall. If he paints \(\frac{2}{3}\) of a wall, he wil use ______ of a bucket.
Solution: \(\frac{2}{5}\)
Given, paint required to paint 1 wall = \(\frac{3}{5}\) of a bucket
Paint required to paint \(\frac{2}{3}\) of a wall
= \(\frac{2}{3} \times \frac{3}{5}=\frac{2}{5}\) of a bucket

Question 5.
Complete the following:

Solution:

Question 6.
If the product of two numbers is ________ the numbers are called reciprocal of each other.
Solution: 1
If the product of two numbers is 1, then the numbers are called reciprocal of each other.

Question 7.
Reciprocal of \(\frac{5}{11}\) is ______.
Solution: \(\frac{11}{5}\)
Reciprocal of \(\frac{5}{11} \text { is } \frac{\mathbf{1 1}}{\mathbf{5}}\)

Question 8.
\(\frac{1}{5}\) is ________ of 5.
Solution: reciprocal or multiplicative inverse
\(\frac{1}{5}\) is reciprocal or multiplicative inverse of 5.

Question 9.
\(\frac{8}{5} \div \frac{9}{}=\frac{8}{5} \times \frac{7}{9}\)
Solution: \(\frac{9}{7}\)
\(\frac{8}{5} \div \frac{9}{7}=\frac{8}{5} \times \frac{7}{9}\)