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.]
= \(\frac{99}{12}\) – \(\frac{34}{12}\) + \(\frac{20}{12}\)
= \(\frac{99-34+20}{12}\) = \(\frac{85}{12}\)