Class 7

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 3 A Peek Beyond the Point MCQ improves accuracy in objective exams.

MCQ on A Peek Beyond the Point Class 7

A Peek Beyond the Point MCQ Class 7

Class 7 Maths A Peek Beyond the Point MCQ

Question 1.
How many tenths make a unit?
(a) \(\frac{1}{10}\)
(b) 10
(c) 100
(d) 1
Solution:
(b) 10
We know, 1 unit = 10 one-tenths.

Question 2.
The value of \(4 \frac{4}{10}+7 \frac{3}{10}\) is:
(a) \(7 \frac{7}{10}\)
(b) \(11 \frac{7}{10}\)
(c) \(10 \frac{1}{10}\)
(d) \(7 \frac{17}{10}\)
Solution:
(b) \(11 \frac{7}{10}\)
We can write, \(4 \frac{4}{10}=\frac{44}{10} \text { and } 7 \frac{3}{10}=\frac{73}{10}\)
Now, \(4 \frac{4}{10}+7 \frac{3}{10}=\frac{44}{10}+\frac{73}{10}\)
= \(\frac{117}{10}=\frac{110}{10}+\frac{7}{10}=11+\frac{7}{10}=11 \frac{7}{10}\)

Question 3.
How is one-hundredth represented in decimal form?
(a) 0.01
(b) 0.001
(c) 1.0
(d) 0.1
Solution:
(a) 0.01
One-hundredth represents 100 parts of a unit, i.e. \(\frac{1}{100}=\frac{01}{100}\)
Here, number of zeros in denominator is 2.
So, putting decimal point in the numerator 2 places to the left, we get 0.01.
Thus, the decimal representation of one- hundredth is 0.01.

Question 4.
The place value of 5 in the decimal 11.485 is:
(a) Five-tenths
(b) Five-hundredths
(c) Five-thousandths
(d) Five-ones
Solution:
(c) Five-thousandths
The expanded form of 11.485 is:
11.485 = 10 × 1 + 1 × 1 + \(\left(4 \times \frac{1}{10}\right)\) + \(\left(8 \times \frac{1}{100}\right)+\left(5 \times \frac{1}{1000}\right)\)
Thus, the place value of 5 in the decimal 11.485 is five-thousandths, i.e. 0.005.

Question 5.
The decimal form of 729 hundredths is:
(a) 729
(b) 72.9
(c) 7.29
(d) None of these
Solution:
(c) 7.29
The decimal form of 729 hundredths is \(\frac{729}{100}\) = 7.29.

Question 6.
The decimal representation of \(\frac{3}{4}\) is:
(a) 0.75
(b) 0.85
(c) 0.65
(d) 0.95
Solution:
(a) 0.75
We have, \(\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}\) = 0.75

Question 7.
The place value of the digit 7 in the number 3.476 is:
(a) 7 ones
(b) 7 tenths
(c) 7 hundredths
(d) 7 thousandths
Solution:
(c) 7 hundredths
3.476 can he represented in the decimal place value chart as:

Hence, the place value of the digit 7 in the number 3.476 is 7 hundredths, i.e. 0.07.

Question 8.
If a number line is divided into 10 equal parts between 3 and 4, then the decimal number at the 7^th division is:
(a) 3.07
(b) 3.7
(c) 3.17
(d) 37
Solution:
(b) 3.7
The number line, divided into 10 equal parts between 3 and 4, is given as:

Thus, the decimal number at the 7^th division is 3.7.

Question 9.
The greatest decimal number among the following is:
(a) 1.2
(b) 1.02
(c) 1.22
(d) 1.21
Solution:
(c) 1.22
The given decimal numbers are: 1.2, 1.02, 1.22, 1.21
Converting them to like decimals, we get 1.20, 1.02, 1.22, 1.21

Now, as the whole number part of all the decimal numbers is the same, i.e. 1, we compare the digits after the decimal point from left to right.

As 1.02 has 0 at the tenths place while other decimals have 2, 1.02 is the smallest.
Now, comparing the hundredths place of 1.20, 1.22, 1.21, we get
1.22 > 1.21 > 1.20 (As 2 >1 > 0)
Thus, 1.22 is the greatest among the given numbers.

Question 10.
Which of the following sequence is decreasing?
(a) 4.5, 4.55, 4.6
(b) 2.3, 2.2, 2.1
(c) 3.1, 3.0, 3.1
(d) 1.01,2.001,3.0001
Solution:
(b) 2.3, 2.2, 2.1
A sequence is decreasing if each term is smaller
than the one before it.
2.3, 2.2, 2.1 → each term decreases by 0.1 i.e. (2.3 > 2.2 > 2.1).
Thus, 2.3, 2.2, 2.1 is the decreasing sequence.

Question 11.
How can we express 0.5 hours in minutes?
(a) 50 minutes
(b) 45 minutes
(c) 25 minutes
(d) 30 minutes
Solution:
(d) 30 minutes
We know, 1 hour = 60 minutes
Thus, 0.5 hours = 0.5 × 60 = \(\frac{5}{10}\) × 60
= 30 minutes

Question 12.
Which of the following pairs of decimal numbers are equivalent?
(i) 0.5 and 0.500
(ii) 0.05 and 0.0500
(iii) 1.06 and 1.060
(iv) 0.06 and 0.600
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iii) only
(c) (i), (ii) and (iii)
(d) (i) and (iii) only
Solution:
(c) (i), (ii) and (iii)
We know that adding trailing zeros after the last non-zero digit of a decimal does not change its value. Thus,
(i) 0.5 and 0.500 are equivalent.
(ii) 0.05 and 0.0500 are equivalent.
(iii) 1.06 and 1.060 are equivalent.
(iv) 0.06 and 0.600 are not equivalent as 0.06 represents 6-hundredths, while 0.600 represents 6-tenths.

Question 13.
Which of these are correct comparisons of decimals?
(i) 0.70 = 0.7
(ii) 0.701 > 0.7
(iii) 0.65 < 0.605
(iv) 0.506 < 0.56
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iv)
(d) (i), (iii) and (iv)
Solution:
(c) (i), (ii) and (iv)
(i) 0.70 and 0.7 are equal in value, i.e. 7 tenths. Thus, this comparison is correct.
(ii) 0.701 is greater than 0.7 because it has a 1 in the thousandths place, while 0.7 has 0. Thus, this comparison is correct.
(iii) 0.65 is greater than 0.605, since 0.65 = 0.650. So, this comparison is incorrect.
(iv) 0.506 < 0.56, since 0.506 = 506 thousandths, and 0.56 = 560 thousandths. Thus, this comparison is correct.

Question 14.
Which of the following statements are true?
(i) The decimal number with the greater whole number part is greater.
(ii) Milligrams are used for measuring heavier weights like bags of rice, etc. while kilograms are used for very light weights like medicine dosage, etc.
(iii) 3.5 feet means 3 feet and 5 inches.
(iv) 2.4 hours means 2 hours 24 minutes.
Choose the correct option from the following:
(a) (ii) and (iv)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (i) and (iii)
Solution:
(c) (i) and (iv)
The decimal number with greater whole number part is greater. For example, 7.1 > 6.99 because 7 > 6.
Kilograms are used for heavier things (like bags of rice) while milligrams are used for very tiny weights (like medicine).
Since 1 foot =12 inches, 0.5 feet = 6 inches. So, 3.5 feet = 3 feet 6 inches.
0.4 hours = 0.4 × 60 = 24 minutes. So, 2.4 hours = 2 hours 24 minutes.
Thus, statements (i) and (iv) are true.

A 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): Fractional part of a number is always less than 1.
(R): Decimal places represent values like
\(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000},\) and so on. These are all parts of a whole.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Decimal places represent values like \(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000},\)
and so on. These are a parts of a whole, they are less than 1.
That’s why fractional part of a number is always less than 1.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): \(3 \frac{6}{100}\) is greater than \(3 \frac{6}{10}\).
(R): \(\frac{6}{100}\) has a larger denominator than \(\frac{6}{10}\).
Solution:
(d) A is false but R is true.
We can write \(3 \frac{6}{100}=\frac{306}{100}\) and \(3 \frac{6}{10}=\frac{36}{10}=\frac{360}{100}\)
Clearly, 360 > 306 ⇒ \(\frac{360}{100}>\frac{306}{100}\)
⇒ \(3 \frac{6}{10}>3 \frac{6}{100}\)
Thus, Assertion (A) is false.
Now, as 100 > 10, \(\frac{6}{100}\) has larger denominator than \(\).
Thus, Reason (R) is true.

Question 3.
(A): If we subtract 47.38 from 89.62, the whole number parts differ by 42, and the actual result is more than 41 and less than 43.
(R): When subtracting two decimal numbers, the result lies between one less than and one more than the difference of their whole number parts.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Whole number part of 89.62 = 89
Whole number part of 47.38 = 47
Difference of whole number parts
= 89 – 47 = 42
Actual difference: 89.62 – 47.38 = 42.24
Now, (42 – 1) < 42.24 < (42 + 1) i.e.
41 < 42.24 < 43.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): 38 paise = ₹0.38
(R): One rupee =100 paise
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that 1 rupee = 100 paise.
So, 1 paisa = \(\frac{1}{100}\) rupee
Therefore, 38 paise = ₹ \(\frac{38}{100}\) = ₹0.38
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

A Peek Beyond the Point Class 7 Fill in the Blanks

Question 1.
The length \(2\frac{7}{10}\) cm is read as _____.
Solution: two and seven- tenths centimetres
The length \(2\frac{7}{10}\) cm is read as two and seven- tenths centimetres.

Question 2.
Kashvi cuts a 1-metre ribbon into 10 equal pieces. The length of one piece is ______ metre.
Solution: \(\frac{1}{10}\) or one-tenth
Given, length of ribbon is 1 metre and number of equal pieces is 10.
∴ Length of each piece = \(\frac{1}{10}\) metre
Thus, the length of one piece is \(\frac{1}{10}\) or one-tenth metre.

Question 3.
Fill in the blanks:
(i) __________ mm = 42.9 cm
(ii) 826 cm = ________ m
(iii) _____ g = 9.5 kg
(iv) 1834 mm = _____ cm
(v) 1749 g = ____ kg
(vi) ₹7 = _____ paise
(vii) ₹ 18.75 = ______ paise
(viii) ₹ _______ = 4250 paise
Solution: 429, 8.26, 9500, 183.4, 1.749, 700, 1875, 42.50
(i) We know that 1 cm = 10 mm.
Therefore, 42.9 cm = 42.9 × 10 mm
= \(\frac{429}{10}\) × 10 mm = 429 mm

(ii) We know that 1 cm = \(\frac{1}{100}\) m
Therefore, 826 cm = \(\frac{826}{100}\) m = 0.86 m

(iii) We know that 1 kg = 1000 g.
Therefore, 9.5 kg = 9.5 × 1000 g 95
= \(\frac{95}{10}\) × 1000 g = 9500 g

(iv) We know that 1 mm = \(\frac{1}{10}\) cm.
Therefore, 1834 mm = \(\frac{1834}{10}\) cm = 183.4 cm

(v) We know that 1 g = \(\frac{1}{1000}\) kg.
Therefore, 1749 g = \(\frac{1749}{100}\) kg = 1.749 kg

(vi) We know that 1 rupee = 100 paise.
Therefore, 7 rupees = 7 × 100 paise
= 700 paise

(vii) We know that 1 rupee =100 paise.
Therefore, ₹ 18.75 = 18.75 × 100 paise
= \(\frac{1875}{100}\) × 100 = 1875 paise

(viii) We know that 1 paise = \(\frac{1}{100}\) rupee
4250 paise = \(\frac{4250}{100}\) rupees = 42.50 rupees

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 2 Arithmetic Expressions MCQ improves accuracy in objective exams.

MCQ on Arithmetic Expressions Class 7

Arithmetic Expressions MCQ Class 7

Class 7 Maths Arithmetic Expressions MCQ

Question 1.
The value of expression 8 + 7 + 6 is:
(a) 21
(b) 19
(c) 20
(d) 17
Solution:
(a) 21
We have, 8 + 7 + 6 = 15 + 6 = 21

Question 2.
Compare and choose the correct option to fill the box: 24 + 16 ☐ 56 – 16.
(a) >
(b) <
(c) =
(d) None of these
Solution:
(c) =
Given, 24 + 16 ☐ 56 – 16
Left hand side (LHS) = 24 + 16 = 40
Right hand side (RHS) = 56 – 16 = 40
Clearly, LHS = RHS So, 24 + 16 = 56 – 16

Question 3.
Compare and choose the correct option to fill the box: 4 × 3 ☐ 78 ÷ 6.
(a) >
(b) <
(c) =
(d) None of these
Solution:
(b) <
Left hand side (LHS): 4 × 3 = 12
Right hand side (RHS): 78 ÷ 6 = 13
Clearly, 12 < 13 ⇒ 4 × 3 < 78 ÷ 6

Question 4.
In a sweet shop, Meena packed 12 laddoos in one box and 4 in another. Kunal packed 10 laddoos in one big box and none in the second. Who packed more laddoos?
(a) Meena
(b) Kunal
(c) Both packed the same number
(d) Depends on the sweetness
Solution:
(a) Meena
Laddoos packed by Meena = 12 + 4= 16
Laddoos packed by Kunal = 10 + 0= 10
Thus, Meena packed more laddoos.

Question 5.
The number of terms in the expression 40 – 7 × 3 + 8 ÷ 2 – 4is:
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(c) 4
We have, 40 – 7 × 3 + 8 ÷ 2 – 4

So, there are 4 terms in the given expression.

Question 6.
The value of the expression 100 – {25 – (10 × 5)} is:
(a) 125
(b) 115
(c) 120
(d) 130
Solution:
(a) 125
We have, 100 – {25 – (10 × 5)}
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
Thus, 100 – {25 – (10 × 5)}
= 100 – 25 + (10 × 5)
= 100 – 25 + 50 = 150 – 25 = 125

Question 7.
The value of the expression 240 – [60 – (30 – 8 ÷ 2) + 10] is:
(a) 192
(b) 194
(c) 196
(d) 198
Solution:
(c) 196
240 – [60 – (30 – 8 ÷ 2) + 10]
= 240 – [60 – (30 – 4) + 10]
= 240 – [60 – 26 + 10]
= 240 – 44 = 196

Question 8.
Which of the following is equal to the expression ‘53 – 17 + 4’?
(a) 53 – (17 + 4)
(b) 53 + (17 – 4)
(c) 53 – (17 – 4)
(d) 4 + (17 – 53)
Solution:
(c) 53 – (17 – 4)
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
So, 53 – (17 – 4) = 53 – 17 + 4

Question 9.
Which of the following expressions are the arithmetic expressions?
(i) 12 + (2 × 4 – 6)
(ii) 7x + 2y – 3
(iii) 18 ÷ 3 + 7
(iv) 4xy – 11
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii), (iii) and (iv)
(d) None of these
Solution:
(a) (i) and (iii) only
We know, an arithmetic expression is a
mathematical sentence that contains numbers and operations like addition (+), subtraction (-), multiplication (×) or division (÷).
So, 12 + (2 × 4 – 6) and 18 ÷ 3 + 7 are arithmetic expressions.
And, 7x + 2y – 3 and 4xy – 11 are not arithmetic expressions because x and y are not numbers.

Question 10.
Which of the following are correct?
(i) The signs ‘>’, ‘=’ and ‘<’ are used to compare the values of two expressions.
(ii) 276- 19 < 275- 18 (iii) 366 – 20 = 380 – 34 (iv) 490 – 20 > 396 – 30
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
(i) Yes, the signs ‘>’, ‘=’ and ‘<’ are used to compare the values of two expressions.
So, (i) is correct.

(ii) Left-hand side (LHS): 276 – 19 = 257
Right-hand side (RHS): 275 – 18 = 257
Clearly, 276- 19 = 275- 18.
So, (ii) is incorrect.

(iii) Left-hand side (LHS): 366 – 20 = 346
Right-hand side (RHS): 380 – 34 = 346
Clearly, 366 – 20 = 380 – 34.
So, (iii) is correct.

(iv) Left-hand side (LHS): 490 – 20 = 470
Right-hand side (RHS): 396 – 30 = 366
Clearly, 470 > 366
⇒ 490 – 20 > 396 – 30
So, (iv) is correct.

Arithmetic Expressions 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): 110 – 27 > 112 – 27
(R): When the same number is subtracted from two different numbers, the difference is greater for the larger number.
Solution:
(d) A is false but R is true.
We know, when the same number is subtracted
from two different numbers, the difference is greater for the larger number.
Since 112 is larger than 110,
110 – 27 < 112 – 27 Therefore, Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): 568 + 247 > 572 + 248
(R): If a < b and c < d, then a + c < b + d, where a, b, c and d are numbers.
Solution:
(d) A is false but R is true.
If a < b and c < d, then a + c < b + d, where a,
b, c and d are numbers.
Clearly, 568 < 572 and 247 < 248
So, 568 + 247 < 572 + 248
Therefore, Assertion (A) is false, but Reason (R) is true.

Question 3.
(A): 5 – 2 ≠ 5 + (-2)
(R): Subtracting a number is the same as adding its additive inverse.
Solution:
(d) A is false but R is true.
We know that subtracting a number is the same
as adding its additive inverse.
∴ 5 – 2 = 5 + (- 2) as – 2 is additive inverse of 2.
Therefore, Assertion (A) is false and Reason (R) is true.

Question 4.
(A) : 16 – (- 7 + 19) = 16 – 7 + 19
(R) : On removing the brackets preceded by a plus “ + ’ sign, the signs of all the terms inside the brackets remain same.
Solution:
(d) A is false but R is true.
We know, on removing the brackets preceded by a ‘+’ sign, the signs of all the terms inside the brackets remains same.
But in expression ‘16 – (-7 + 19)’, brackets are preceded by minus sign.
Therefore, 16 – (- 7 + 19) = 16 + 7 – 19
Therefore, Assertion (A) is false, but Reason (R) is true.

Arithmetic Expressions Class 7 Fill in the Blanks

Question 1.
Fill in the blanks with the correct sign: ‘<’ , ‘>’ or ‘=’:
(i) 111 – 28 _____ 85
(ii) 55 ÷ 11 ____ 5
Solution: <, =
(i) LHS: 111 – 28 = 83
RHS: 85
Clearly, 83 < 85 ⇒ 111 – 28 < 85

(ii) LHS: 55 ÷ 11 = 5
RHS: 5
Clearly, LHS = RHS ⇒ 55 – 11 = 5

Question 2.
Fill in the blanks to make the expressions equal on both sides of the ‘ = ’ sign:
(i) 19 + 6 = ____ + 9
(ii) 9 × _____ = 72 ÷ 2
Solution: 16, 4
(i) 19 + 6 = 16 + 9
[∵ 19 + 6 = 25 and 16 + 9 = 25]
(ii) 9 × 4 = 72 ÷ 2
[∵ 72 ÷ 2 = 36 and 9 × 4 = 36]

Question 3.
In the blanks below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out, and not by evaluating the expressions.
(i) (12 – 7) × 36 _____ (7 – 12) × 36
(ii) 20 + 6 × 15 _____ (20 + 6) × 15
Solution: <, <
(i) LHS : (12 – 7) × 36 = Positive number × 36 → Positive number
RHS : (7 – 12) × 36 = Negative number × 36 → Negative number
We know, a positive number is always greater than a negative number.
Thus, LHS > RHS
Hence, (12 – 7) × 36 > (7 – 12) × 36

(ii) RHS : (20 + 6) × 15

Thus, LHS < RHS
Hence, 20 + 6 × 15 < (20 + 6) × 15

Question 4.
Fill in the blanks with numbers and boxes with operation signs such that the expressions on both sides are equal.
(i) 21 + (___ ☐ ____) = 21 + 12 – 5
(ii) 32 – (3 + 8) = 32 ☐ 3 – _____
(iii) 25 – (11 ☐ 7) = 25 – 11 + 7
(iv) 19 – (17 – 12) = 19 ☐ 17 ☐ 12
Solution: 12 – 5, 8, -, -, +
(i) 21 + ( 12 – 5) = 21 + 12 – 5
(ii) 32 – (3 + 8) = 32 – 3 – 8
(iii) 25 – (11 – 7) = 25 – 11 + 7
(iv) 19 – (17 – 12) = 19 – 17 + 12

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 1 Large Numbers Around Us MCQ improves accuracy in objective exams.

MCQ on Large Numbers Around Us Class 7

Large Numbers Around Us MCQ Class 7

Class 7 Maths Large Numbers Around Us MCQ

Question 1.
Which of the following numbers has 7 at the lakhs place?
(a) 7,50,812
(b) 70,15,902
(c) 5,74,210
(d) 52,08,701
Solution:
(a) 7,50,812
In 7,50,812, the digit 7 is at lakhs place.
In 70,15,902, the digit 7 is at ten lakhs place.
In 5,74,210, the digit 7 is at ten thousands place.
In 52,08,701, the digit 7 is at hundreds place.

Question 2.
(20 × 10,000) + (53 × 1,000) + (1 × 100) + (2 × 1) represents the number:
(a) 53,20,102
(b) 20,53,102
(c) 2,53,102
(d) 5,20,301
Solution:
(c) 2,53,102
(20 × 10,000) + (53 × 1,000) + (1 × 100) + (2 × 1) = 2,00,000 + 53,000 + 100 + 2
= 2,53,102

Question 3.
The correct way of writing ‘sixty five lakh nine thousand four hundred twenty two’ in numbers is:
(a) 65,90,422
(b) 65,09,422
(c) 6,59,422
(d) 65,40,922
Solution:
(b) 65,09,422
Sixty five lakh nine thousand four hundred twenty two = 65,09,422

Question 4.
10 million is equal to:
(a) 100 crore
(b) 1 lakh
(c) 100 lakh
(d) 10,000 lakh
Solution:
(c) 100 lakh
We know,
1 million = 10 lakh
Thus, 10 million =10 × 10 lakh = 100 lakh

Question 5.
The estimated difference of 1,76,346 and 3,865 rounded off to the nearest thousand is:
(a) 1,70,000
(b) 1,72,000
(c) 1,90,000
(d) 1,54,000
Solution:
(b) 1,72,000
1,76,346 is rounded off to nearest thousands as
1,76,000. [∵ At hundreds place, 3 < 5] 3,865 is rounded off to nearest thousands as 4.000. [∵ At hundreds place, 8 > 5]
Thus, estimated difference = 1,76,000 – 4,000
= 1,72,000

Question 6.
The product 974 × 95 will likely have:
(a) 3 digits
(b) 4 digits
(c) 6 digits
(d) 5 digits
Solution:
(d) 5 digits
Rounding 974 to nearest hundreds we get
1,0 and rounding 95 to nearest tens we get 100.
Thus, estimated product = 1,000 × 100
= 1,00,000.
So, the product will likely have five digits and it will be less than 1,00,000 as the numbers are rounded up and 1,00,000 is the smallest 6-digit number.
Verification: 974 × 95 = 92,530, which has five digits.

Question 7.
The estimated sum of 5,47,86,326 and 45,56,767 rounded off to the nearest lakhs is:
(a) 5,94,00,000
(b) 59,40,000
(c) 54,90,000
(d) 9,47,000
Solution:
(a) 5,94,00,000
5,47,86,326 is rounded off to nearest lakhs as
5,48,0,000. [∵ At ten thousands place, 8 > 5]
45,56,767 is rounded off to nearest lakhs as
46,0,000. [∵ At ten thousands place, 5 = 5]
Thus, estimated sum
= 5,48,00,000 + 46,00,000 = 5,94,00,000

Question 8.
The product 105 × 93 will likely have:
(a) 3 digits
(b) 4 digits
(c) 6 digits
(d) 5 digits
Solution:
(b) 4 digits
Rounding off 105 to nearest hundreds, we get 100 and rounding off 93 to nearest tens we get 90,
Thus, estimated product = 100 × 90 = 9,000
So, the product will likely have four digits and it will be close to 9,000.
Verification: 105 × 93 = 9,765, which has four digits.

Question 9.
1 billion is equal to:
(i) 100 crore
(ii) 100 million
(iii) 100 lakh
(iv) 10,000 lakh
Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Solution:
(c) (i) and (iv)
We know, 1 million = 10 lakh
And, 1 billion = 1,000 million
Thus, 1 billion = 1,000 × 10 lakh
= 10,000 lakh =100 crore [∵ 1 crore = 100 lakh]

Question 10.
In which of the following numbers the comma is placed incorrectly according to the Indian Number System?
(i) 23,56,76,145
(ii) 123,98,34,67
(iii) 780,444,112
(iv) 56,12,34,889
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)
We know that while writing a number in Indian Number System, the commas are placed in 3-2-2-2… pattern from right to left.

In 23,56,76,145 and 56,12,34,889, the commas are placed in 3-2-2-2… pattern from right to left. They are written correctly according to Indian Number System.

However, in 123,98,34,67 the commas are not placed in 3-2-2-2… pattern from right to left.
Also, in 780,444,112 the commas are placed according to the International Number System,
i. e. in 3-3-3… pattern from right to left.

Thus, the commas are not placed correctly according to the Indian Number System in 123,98,34,67 and 780,444,112.

Large Numbers Around Us 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): 1,000 notes of ₹100 are needed to make ₹ 1,00,000.
(R): Place value helps us break large numbers into smaller, meaningful parts.
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
As ₹1 ,00,000 ÷ 100 = 1,000,
1,000 notes of ₹100 are needed to make ₹1,00,000.
Thus, Assertion (A) is true.
And we know that place value helps us to express larger numbers into smaller, meaningful parts.
Thus, Reason (R) is true, but Reason (R) is not the correct explanation of Assertion (A).

Question 2.
(A): The digit 6 in the number 9,645,123 has a place value of 600,000 in the International Number System.
(R): The place value of a digit is the product of the digit and its position value.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that the place value of a digit in the number system is found by multiplying the
digit with the value of the position it occupies.

Thus, in 9,645, 123, the place value of 6 is 6 × 100,000 = 600,000.
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is tile correct explanation of Assertion (A).

Question 3.
(A): When rounding 3,42,765 to the nearest lakhs, we get 4,00,000.
(R): Rounding to the nearest lakhs depends on the digit in the ten thousands place.
Solution:
(d) A is false but R is true.
When rounding 3,42,765 to nearest lakhs, we check the digit on the just right of lakhs place, i.e. the digit at the ten thousands place.
In 3,42,765, 4 is at ten thousands place and 4 < 5. So, the digit at the lakhs place, i.e. 3 remains unchanged.
Hence, if 3,42,765 is rounded off to nearest lakhs, we get 3,00,000.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 4.
(A): If 98 × 102 is calculated, then the product will be very close to 10,000.
(R): Both numbers 98 and 102 are close to 100.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
98 is rounded off to nearest tens as 100 and 102 is rounded off to nearest hundreds as 100. Thus, the product is estimated as 100 × 100 = 10,000.
This means that the actual product will be close to 10,000.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Large Numbers Around Us Class 7 Fill in the Blanks

Question 1.
In one lakh, 1 is follwed by _____________________ zeros.
Solution: five
We know, 1 lakh = 1,00,000, where the digit 1 is followed by five zeros.
∴ In one lakh, 1 is followed by five zeros.

Question 2.
In the Indian Number Svswm, the correct way to write the number 9625084 is __________.
Solution: 96,25,084
In the Indian Number System, commas are placed in a 3-2-2-2… pattern from the right to left. This helps to separate the digits into hundreds, thousands, lakhs, crores, and so on.
Therefore, in the Indian Number System, the correct way to write the number 9625084 is 96,25,084.

Question 3.
The smallest 7—digit number is _______ lakh.
Solution: ten
The smallest 7-digit number is 10,00,000 i.e. ten lakh.

Question 4.
The largest 8—digit number is _________
Solution: 9,99,99,999
The largest 8-digit number is 9,99,99,999.

Question 5.
The place value of 5 in 7,005,380 is _________.
Solution: 5000
Given number is 7,005,380, which is represented in International Number System.

Thus, the place value of 5 in 7,005,380 is 5 thousand = 5 × 1000 = 5000.

Question 6.
The expanded form of the number 76,70,905 is _________.
Solution: (7 × 10,00,000) + (6 × 1,00,000) + (7 × 10,000) + (9 × 100) + (5 × 1)
The expanded form of the number 76,70,905 is
(7 × 10,00,000) + (6 × 1,00,000) + (7 × 10,000) + (9 × 100) + (5 × 1).

Question 7.
In a crore, 1 is followed by ________ zeros.
Solution: seven
We know that 1 crore = 1,00,00,000, where 1 is followed by seven zeros.
∴ In a crore, 1 is followed by seven zeros.

Question 8.
If 728 is rounded off to the nearest hundreds, we get ________.
Solution: 700
In 728, the tens digit is 2 and 2 < 5. So, 7 at the hundreds place remains unchanged while rounded off to the nearest hundreds.
Thus, if 728 is rounded off to the nearest hundreds, we get 700.

MCQ Questions for Class 7 Maths with Answers

Class 7 Maths MCQ Chapter Wise

Class 7 Maths MCQ Questions Part 1

• Chapter 1 Large Numbers Around Us Class 7 MCQ
• Chapter 2 Arithmetic Expressions Class 7 MCQ
• Chapter 3 A Peek Beyond the Point Class 7 MCQ
• Chapter 4 Expressions using Letter Numbers Class 7 MCQ
• Chapter 5 Parallel and Intersecting Lines Class 7 MCQ
• Chapter 6 Number Play Class 7 MCQ
• Chapter 7 A Tale of Three Intersecting Lines Class 7 MCQ
• Chapter 8 Working with Fractions Class 7 MCQ

Ganita Prakash Class 7 Part 2 MCQ

• Chapter 1 Geometric Twins Class 7 MCQ
• Chapter 2 Operations with Integers Class 7 MCQ
• Chapter 3 Finding Common Ground Class 7 MCQ
• Chapter 4 Another Peek Beyond the Point Class 7 MCQ
• Chapter 5 Connecting the Dots Class 7 MCQ
• Chapter 6 Constructions and Tilings Class 7 MCQ
• Chapter 7 Finding the Unknown Class 7 MCQ

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 5 Parallel and Intersecting Lines Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 5 Parallel and Intersecting Lines Solutions

Ganita Prakash Class 7 Chapter 5 Solutions

Class 7 Maths Ganita Prakash Chapter 5 Solutions Parallel and Intersecting Lines

Question 1.
List all the linear pairs and vertically opposite angles you observe in the given figure:

Solution:
We know that adjacent angles formed by two intersecting lines, are called linear pairs. Linear pairs always add up to 180°.
And opposite angles formed by two intersecting lines, are called vertically opposite angles. Vertically opposite angles are always equal to each other.

Question 2.
Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper.
(a) Did you find it challenging to draw sorne of them?
(b) Which ones?
(c) How did you do ii?

Solution:

(a) Yes, some line segments are a litle more difficult to draw than others.
(b) Line segments e,f. h and g.
(c) Lines parallel to a given line segment are drawn by keeping thern equidistant from the given line segment.

Question 3.
In the figure, which line is parallel to line a — line b or line c? How do you decide this?

Solution:
Line c is parallel to line a because the corresponding dots on these lines are equidistant from each other. So, they do not intersect, no matter how far they are extended.

Question 4.
Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method.

Solution:
Tools needed: Ruler, Set-squares (right-angled triangle), Pencil Steps of Construction:
Step 1: Place the set square so that one side is along the line l.
Step 2: Hold the ruler against the other side of the set square (the ruler won’t move).
Step 3: Slide the set square along the ruler until one side reaches point A.
Step 4: Draw a line along the edge of the set square through point A.
Step 5: This new line is parallel to line l and passes through point A.

Question 5.
Find the angles marked below.


Solution:
(i) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain a = 48°.

(ii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain b = 52°.

(iii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain c = 81°.

(iv) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain d = 99°.

(v) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain e = 69°.

(vi) Since the sum of interior angles on the same side of a transversal intersecting a pair of parallel lines is always equal to 180°,f + 132° = 180° ⇒ f = 180°- 132° ⇒ f = 48°

(vii) Since corresponding angles formed by a transversal intersecting a pair of parallel sides are equal, we obtain g = 122°.

(viii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain h = 75°.

(ix) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain i = 54°.

(x) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain j = 97°.

Question 6.
In the figures below, what angles do x and y stand for?

Solution:
(i) The lines and angles are marked as shown in the figure.

Line m is parallel to line n and line a is a transversal.
∴ ∠2 = 65° + ∠1 [∵ Corresponding angles]
⇒ 90° = 65° + ∠1
⇒ ∠1 = 90°- 65°
⇒ ∠1 = 25°
And, ∠1 = x = 25° [∵ Vertically opposite angles]
Also, line m is parallel to line n and line b is a transversal.
∴ ∠1 +y = 180° [∵ Sum of co-interior angles = 180°]
⇒ 25° + y = 180° [∵ ∠1 = 25°]
⇒ y = 180° – 25° ⇒ y = 155°
Thus, the values of x and y are 25° and 155°, respectively.

(ii) The lines and angles are marked as shown in the figure.

Line a is parallel to line b and line d is a transversal.
∴ ∠2 = 53° [∵ Alternate interior angles]
Also, line a is parallel to line b and c is a transversal.
∴ ∠1 + ∠2 = 78° [∵ Alternate interior angles]
⇒ ∠1 + 53° = 78° [∵ ∠2 = 53°]
⇒ ∠1 = 78°- 53°
⇒ ∠1 = 25°
Therefore, ∠1 = x = 25° [∵ Vertically opposite angles]

Question 7.
What is the measure of ∠NOP in the figure?

[Hint: Draw lines parallel to LM and PQ through points N and O.]
Solution:
Lines parallel to LM and PQ through points N and O are drawn and angles are marked as shown in the figure.

Since LM is parallel to EF and MN is a transversal,
∠1 = 40° [∵ Alternate interior angles]
Now, ∠1 + ∠2 = 96°
⇒ 40° + ∠2 = 96° [∵ ∠1 = 40°]
⇒ ∠2 = 96° – 40°
⇒ ∠2 = 56°
Since EF is parallel to GH and AT is a transversal,
∠2 = ∠3 = 56° [∵ Alternate interior angles]
Also, GH is parallel to PQ and OP is a transversal.
∴ ∠4 = 52° [∵ Alternate interior angles]
So, a = ∠3 + ∠4
⇒ a = 56° + 52° ⇒ a = 108°
Thus, ∠NOP = 108°.

InText Questions

Question 1.
Can two straight lines intersect at more than one point?
Solution:
No, two straight lines cannot intersect at more than one point. If two lines intersect at more than one point, then they are coincident lines.

Question 2.
Take a plain square sheet of paper (use a newspaper for this).

(i) How would you describe the opposite edges of the sheet? They are ________ to each other.
(ii) How would you describe the adjacent edges of the sheet? The adjacent edges are ________ to each other. They meet at a point. They form right angles.
(iii) Fold the sheet horizontally in half. A new line is formed (see figure).
How many parallel lines do you see now?
(iv) Make one more horizontal fold in the folded sheet. How many parallel lines do you see now?
(v) What will happen if you do it once more? How many parallel lines will you get? Is there a pattern? Check if the pattern extends further, if you make another horizontal fold.
(vi) Make a vertical fold in the square sheet. This new vertical line is _________ to the previous horizontal lines.
(vii) Fold the sheet along a diagonal. Can you find a fold that creates a line parallel to the diagonal line?
Solution:
(i) They are parallel to each other.
(ii) The adjacent edges are perpendicular to each other.
(iii) We see three parallel horizontal lines — the top edge, the fold and the bottom edge. The new horizontal line is perpendicular to the vertical edges of paper.
(iv) On folding the paper horizontally one more time, we see five parallel horizontal lines.
(v) On folding the paper horizontally once more, we will get nine parallel lines. The number of horizontal parallel lines follow the sequence:
1^st fold → 3 lines
2^nd fold → 5 lines
3^rd fold → 9 lines and so on
So, after each fold, the number of horizontal lines increases as folding doubles the sections and add extra fold lines. The pattern continues as we fold more.

(vi) This new vertical line is perpendicular to the previous horizontal lines.

(vii) Yes, we can make a fold parallel to the diagonal by folding the sheet in the same slanting direction at equal angles or by folding a smaller triangle inside the square.

Parallel and Intersecting Lines Class 7 Extra Questions

Parallel and Intersecting Lines Class 7 Very Short Question Answer

Question 1.
In the given figure, write the vertically opposite angles of ∠AOD and ∠AOC.

Solution:
We know that when two lines intersect, the angles opposite to each other are called vertically opposite angles. They are formed without sharing a common arm and are always equal.
Thus, in the given figure, ∠BOC is vertically opposite angle of ∠AOD and ∠BOD is vertically opposite angle of ∠AOC.

Question 2.
Find the complement of each of the following angles:
(i) 28°
(ii) 66°
(iii) 75°
(iv) 80°
Solution:
We know that complement of angle x° is (90° – x°).
(i) Complement of 28° = 90° – 28° = 62°
(ii) Complement of 66° = 90° – 66° = 24°
(iii) Complement of 75° = 90° – 75° = 15°
(iv) Complement of 80° = 90° – 80° = 10°

Question 3.
Find the supplement of each of the following angles:
(i) 28°
(ii) 95°
(iii) 130°
(iv) 155°
Solution:
We know that supplement of angle x° is (180° – x°).
(i) Supplement of 28° = 180° – 28° = 152°
(ii) Supplement of 95° = 180° – 95° = 85°
(iii) Supplement of 130° = 180° – 130° = 50°
(iv) Supplement of 155° = 180° – 155° = 25°

Question 4.
Find the complement of each of the following angles:
(i) 32°
(ii) 72°
(iii) 68°
(iv) 20°
Solution:
We know that complement of angle x° is (90 – x)°.
(i) Complement of 32° = 90° – 32° = 58°
(ii) Complement of 72° = 90° – 72° = 18°
(iii) Complement of 68° = 90° – 68° = 22°
(iv) Complement of 20° = 90° – 20° = 70°

Question 5.
Find the supplement of each of the following angles:
(i) 25°
(ii) 92°
(iii) 142°
(iv) 165°
Solution:
We know that supplement of angle x° is (180 – x)°.
(i) Supplement of 25° = 180° – 25° = 155°
(ii) Supplement of 92° = 180° – 92° = 88°
(iii) Supplement of 142° = 180° – 142° = 38°
(iv) Supplement of 165° =180° – 165° = 15°

Question 6.
In the given figure, write the angles that form a linear pair with ∠AOD and with ∠BOC.

Solution:
We know that the adjacent angles formed by two lines intersecting each other, are called linear pair of angles. Linear pairs always add up to 180°.
In the given figure, ∠AOC and ∠BOD form a linear pair of angles with ∠AOD.
∠AOC and ∠BOD form a linear pair of angles with ∠BOC.

Question 7.
In the given figure, find the value of a.

Given,
∠SOR = 3a + 20° and ∠ROT = a
From the given figure, we can say that ∠SOR and ∠ROT form a linear pair.
∴ ∠SOR + ∠ROT = 180°
⇒ 3a + 20° + a = 180°
⇒ 4a = 160° ⇒ a = 40°

Question 8.
Find the value of x in the following figure, if AB is parallel to CD.

Solution:
Given, AB is parallel CD.
Since x and 3x are co-exterior angles on the same side of transversal, they add up to 180°.
∴ x + 3x = 180° ⇒ 4x = 180°
⇒ x = \(\frac{180^{\circ}}{4}\) ⇒ x = 45°

Question 9.
Find the value of x in the following figure, if AB is parallel to CD.

Solution:
Given, AB is parallel CD.
Since x and 2x are interior angles on the same side of transversal, they add up to 180°.
∴ x + 2x = 180°
⇒ 3x = 180° ⇒ x = 60°

Parallel and Intersecting Lines Class 7 Short Question Answer

Question 1.
Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 42°, 48°
(ii) 85°, 95°
(iii) 30°, 60°
(iv) 135°, 45°
Solution:
We know that if the sum of the measures of two angles is 90°, then the angles are called complementary angles and if the sum of the measures of two angles is 180°, then the angles are called supplementary angles.
(i) The given angles are 42° and 48°.
Now, sum of given angles = 42° + 48° = 90°
Thus, the given angles are complementary angles.

(ii) The given angles are 85° and 95°.
Now, sum of given angles = 85° + 95° = 180°
Thus, the given angles are supplementary angles.

(iii) The given angles are 30° and 60°.
Now, sum of given angles = 30° + 60° = 90°
Thus, the given angles are complementary angles.

(iv) The given angles are 135° and 45°.
Now, sum of given angles = 135° + 45° = 180°
Thus, the given angles are supplementary angles.

Question 2.
In the given figure, find the value of x.

Solution:
Given, ∠AOB = 47°, ∠BOC = x, ∠COD = 83°, ∠DOE = 92° and ∠EOA = 75°
Now, as ∠AOB, ∠BOC, ∠COD, ∠DOE and ∠EOA are angles at a point, they add up to 360°.
∴ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
⇒ 47° + x + 83° + 92° + 75°= 360°
⇒ 297° + x = 360°
⇒ x = 360° – 297° = 63°
Thus, the value of x is 63°.

Question 3.
In the given figure, line AB is parallel to line DG. Find the value of x + y.

Solution:
Given, line AB is parallel to line DG.
∴ ∠ACE = ∠CEG
[∵ Alternate interior angles are equal]
⇒ x = 80° [∵ ∠ACE = 80° and ∠CEF = x]
Since ∠GFH and ∠EFH form linear pair, they add up to 180°.
∴ ∠GFH + ∠EFH = 180°
⇒ 150° + y = 180° [∵ ∠GFH = 150°]
⇒ y = 180° – 150° = 30°
∴ x + y = 80° + 30° = 110°

Question 4.
In the given figure, l is parallel to m. Find the value of x and y.

Solution:
Given, l is parallel to m and n is transversal to l and m.
Since 2x and y are vertically opposite angles, they are equal.
∴ y = 2x …(i)
Since 4x and y are interior angles on the same side of the transversal, they add up to 180°.
∴ 4x + y = 180°
⇒ 4x + 2x = 180° [From (i)]
⇒ 6x = 180° ⇒ x = 30°
Substituting the value of x in (i), we get
y = 2 × 30° = 60°

Parallel and Intersecting Lines Class 7 Long Question Answer

Question 1.
Which lines appear to be perpendicular to each other in the given figure?

Solution:
When two lines intersect and the angles formed are 90° (i.e. all four angles are equal), the lines are said to be perpendicular to each other.

In the given figure, since line i is parallel to line d, perpendicular to any of these lines is also perpendicular to other. Therefore, lines b, c, h and /are perpendicular to lines i and d.

Similarly, lines d and i are perpendicular to lines b, c, h andf.

Question 2.
In the given figure, AB is parallel to CD and AD is parallel to BC. Find the values of x, y and z.

Solution:
Given, AB || CD, AD ||
BC and ∠BAD = 70°.
As AD || BC and AB is transversal, ∠ABC and ∠BAD are co-interior angles.
∴ ∠ABC + ∠BAD = 180°
⇒ x + 70° = 180° [∵ ∠BAD = 70°]
⇒ x = 180°- 70° = 110°
Now, as AB || DC and AD is transversal, ∠ADC and ∠BAD are co-interior angles.
∴ ∠ADC + ∠BAD = 180°
⇒ z + 70° = 180° [∵ ∠BAD = 70°]
⇒ z = 180°- 70° = 110°
Also, as AD || BC and DC is transversal, ∠ADC and ∠BCD are co-interior angles.
∴ ∠ADC + ∠BCD = 180°
⇒ 110°+ y = 180° [∵ ∠ADC =110°]
⇒ y = 180°- 110° = 70°

Question 3.
In the following figure, find the value of each marked angle.

Solution:
Given, line l || m || n.
Let point P lie on line l, point R lie on line p and the point of intersection of lines p and l be Q, as shown in the figure.

Since ∠PQR and 32° are vertically opposite angles, they are equal.
∴ ∠PQR = 32°
As l || m, ∠b and ∠PQR are interior angles on the same side of transversal p.
∴ ∠b + ∠PQR = 180°
⇒ ∠b + 32° = 180° [∵ ∠PQR = 32°]
⇒ ∠b = 180°- 32° = 148°
As l || n, ∠a and ∠PQR are corresponding angles.
∴ ∠a = ∠PQR
⇒ ∠a = 32° [∵ ∠PQR = 32°]

Question 4.
In the following figure, AB is parallel to CD and AD is parallel to BC. Find the values of x, y and z.

Solution:
Given, AB || CD, AD || BC, ∠DAC = 45° and ∠BAC = 30°.
As AB || CD anddC is transversal, ∠DAC and ∠ACB are alternate interior angles.
∴ ∠ACB = ∠DAC ⇒ x = 45°
[∵ ∠DAC = 45° and ∠ACB = x]
As AB || CD and AC is transversal, ∠BAC and ∠ACD are alternate interior angles.
∴ ∠ACD = ∠BAC ⇒ y = 30°
[∵ ∠BAC = 30° and ∠ACD = y]
As AD || BC and AB is transversal. ∠DAB and ∠ABC are co-interior angles.
∴ ∠DAB +∠ABC = 180°
⇒ ∠DAC + ∠CAB + ∠ABC = 180°
[∵ ∠DAB = ∠DAC + ∠CAB]
⇒ 45° + 30° + z = 180°
[∵ ∠DAC = 45°, ∠CAB = 30° and ∠ABC = z]
⇒ z = 180°- 75° = 105°

Parallel and Intersecting Lines Class 7 Case Based Questions

Question 1.
In the given figure, two straight lines PQ and RS intersect each other at O such that ∠POT = 75°.

Based on the above information, answer the following questions:
(i) Find the value of b.
(ii) Find the value of a.
(iii) Find the value of c.
Solution:
(i) Given, ∠ROP = 4b, ∠POT = 75°, ∠TOS = b
Here, ∠ROS = 180°
[Since ∠ROS is a straight angle]
⇒ ∠ROP + ∠POT + ∠TOS = 180°
⇒ 4b + 75° + b = 180°
⇒ 5b + 75° = 180°
⇒ 5b = 180° – 75° = 105°
b = \(\frac{105^{\circ}}{5}\) = 21°
Thus, the value of b is 21°.

(ii) Since ∠ROP and ∠QOS are vertically opposite angles, they are equal.
∴ ∠QOS = ∠ROP
⇒ a = 4b
⇒ a = 4 × 21° [∵ b = 21°]
⇒ a = 84°
Thus, the value of a is 84°.

(iii) Since ∠QOS and ∠QOR form a linear pair, they add up to 180°.
∴ ∠QOS + ∠QOR = 180°
⇒ a + 2c = 180°
⇒ 84° + 2c =180° [∵ a = 84°]
⇒ 2c = 180° – 84° = 96°
c = \(\frac{96^{\circ}}{2}\) = 48°
Thus, the value of c is 48°.

Question 2.
In a class, a teacher asked a student to draw three lines on the board. The student draws the lines on the board as shown.

The line l is perpendicular to line n and ∠1 = 75°.
Based on the above information answer the following questions:
(i) What is the measure of ∠6?
(ii) What is the measure of ∠5?
(iii) What is the measure of ∠3?
(iv) What is the sum of the measure of ∠2 and ∠1?
Solution:
(i) Given, line l is perpendicular to line n.
∴ ∠1 + ∠6 = 90°
⇒ 75° + ∠6 = 90° [∵ ∠1 = 75°]
⇒ ∠6 = 90° – 75° = 15°

(ii) Given, line l is perpendicular to line n.
∴ ∠5 = 90°

(iii) We have, ∠6 = 15°
Since ∠3 and ∠6 are vertically opposite angles, they are equal.
∴ ∠3 = ∠6 = 15°

(iv) Given, line l is perpendicular to line n. So, ∠2 is equal to 90°.
∴ ∠1 + ∠2 = 75° + 90° = 165°

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 4 Expressions using Letter Numbers Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 4 Expressions using Letter Numbers Solutions

Ganita Prakash Class 7 Chapter 4 Solutions

Class 7 Maths Ganita Prakash Chapter 4 Solutions Expressions using Letter Numbers

Question 1.
One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expressions(s) describe the total amount in rupees earned that day?
(a) 30x + 20y
(b) (30 + 20) × (x + y)
(c) 20x + 30)
(d) (20 + 30) × (x + y)
(e) 30x – 20y
Solution:
Cost of one plate of Jowar roti = ₹30
Cost of x plates of Jowar roti = ₹30x
Cost of one plate of Pulao = ₹20
Cost of y plate of Pulao = y
So, the expression for the total amount earned that day = 30x + 20y
Hence, the correct answer is option (a).

Question 2.
Write formulas for the perimeter of:
(i) triangle with all sides equal.
(ii) a regular pentagon.
(iii) a regular hexagon.
Solution:
(i) Let side length of triangle be a. Then,
Perimeter of triangle with all sides equal = a + a + a = 3a = 3 × side

(ii) Let side length of a regular pentagon be a. Then,
Perimeter of the regular pentagon = a + a + a + a + a = 5a = 5 × side

(iii) Let side length of a regular hexagon be a. Then,
Perimeter of the regular hexagon = a + a + a + a + a + a = 6a = 6 × side

Question 3.
Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter- number to denote the length in metres of the other pipe.
Solution:
Length of pipe Munirathna has = 20 m
Length of another pipe Munirathna wants to join = k m
∴ Combined length of the pipe = (20 + k) m

Question 4.
What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5?
Complete the following table:

Solution:

Question 5.
In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

Solution:
Date to the left to w will be 1 less than w.
Date to the right to w will be 1 more than w.
Since there are 7 clays in a week,
Date above w will be 7 less than w.
Date in the diagonally left cell to w will be 8 less than w.
Date in the diagonally right will to w will be 6 less than w.
Thus, the expressions in other five blank cells of the grid are as shown:

Question 6.
A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.
(1) Write an expression describing how far away the snail is from its starting position.
(ii) What can we say about the snail’s movement if d > u?
Solution:
(i) During the day the snail climbs up ‘u’ cm.
During the night the snail slips down ‘d’ cm.
So, the net distance covered in one day is (u – d) cm.
So, in 10 days and 10 nights the net distance covered by the snail = 10(u – d) cm.
Hence, the expression describing how far away the snail is from it starting position is 10(u – d).

(ii) If d > u, snail slips down more than it climbs.
It means the snail will never reach the top.

Question 7.
Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometres would Radha have cycled after 3 weeks?
Solution:
In first week, Radha cycles 5 km every day.
So, she cycled 5 × 7 = 35 km in first week.
In second week, Radha cycles (5 + z) km every day.
In third week, she cycles 5 + z + z = (5 + 2z) km every day.
So, she cycled (5 + 2z) × 7 = (35 + 14z) km in third week.
Thus, number of kilometres Radha cycled in 3 weeks
= 35 + (35 + 7z) + (35 + 14z)
= (35 + 35 + 35) + (7z + 14z) = (105 + 21z)km

Question 8.
In the following figure, observe how the expression it w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

Solution:
[(w + 2) – 5] × 3 = [w – 3] × 3 = 3w – 9
[(w + 2) – 8] – 4 = [w – 6] – 4 = w – 10
[(w + 2) – 4] × 3 = [w – 2] × 3 = 3w – 6

Question 9.
A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.
(i) What is the algebraic expression for the time taken to travel from Yahapur to Yahapur?
(ii) If t = 4, what is the time taken to travel from Yahapur to Vahapur?
Solution:
(i)
Let the time taken to travel from one station to another station = t
So, time taken to travel from Yahanpur to Vahapur = 4t
As there are three stoppages between these two stations and the train stops for 2 minutes at each stoppage, total time taken during stoppages = 2 × 3 = 6 minutes.
So, the algebraic expression for total time taken (in minutes) is (4t + 6).

(ii) From (i), the algebraic expression for total time (in minutes) taken from Yahanpur to Vahapur is (4t + 6). So, the time taken to travel from Yahapur to Vahapur = 4 × 4 + 6 = 16 + 6 = 22 minutes.

Question 10.
Simplify the following expressions:
(i) 3a + 9b – 6 + 8a – 4b – 7a + 16
(ii) 3(3a – 3b) – 8a – 4b – 16
(iii) 2(2x – 3) + 8x + 12
(iv) 8x – (2x – 3) + 12
(v) 8h – (5 + 7h) + 9
(vi) 23 + 4(6m – 3n) – 8n – 3m – 18
Solution:
(i) 3a + 9b – 6 + 8a – 4b – 7a + 16 = (3a + 8a – 7a) + (9b – 4b) + (- 6 + 16) = 4a + 5b + 10
(ii) 3(3a – 3b) – 8a – 4b – 16 = 9a – 9b – 8a – 4b – 16 = (9a – 8a) + (-9b – 4b) – 16 = a – 13b – 16
(iii) 2(2x – 3) + 8x + 12 = 4x – 6 + 8x + 12 = (4x + 8x) + (-6 + 12) = 12x + 6
(iv) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12 = 6x + 15
(v) 8h – (5 + 7h) + 9 = 8h – 5 – 7h + 9 = 8h – 7h – 5 + 9 = h + 4
(vi) 23 + 4(6m – 3n) – 8n – 3m – 18 = 23 + 24m – 12n – 8n – 3m – 18
= 24m – 3m – 12w – 8n + 23 – 18 = 21m – 20n + 5

Question 11.
Add the expressions given below:
(i) 4d – 7c + 9 and 8c – 11 +9d
(ii) – 6f + 19 – 8s and – 23 + 13f + 12s
(iii) 8d – 14c + 9 and 16c – (11 + 9d)
(iv) 6f – 20 + 8s and 23 – 13f – 12s
(v) 13m – 12n and 12n – 13m
(vi) – 26m + 24n and 26m – 24n
Solution:
(i) (4d – 7c + 9) + (8c – 11 + 9d) = (4d + 9d) + (- 7c + 8c) + (9 – 11) = 13d + c – 2
(ii) (-6f + 19 – 8s) + (-23 + 13f + 12s) = (-6f + 13f) + (-8s + 12s) + (19 – 23) = 7f + 4s – 4
(iii) (8d – 14c + 9) + [16c – (11 + 9d)] = (8d – 14c + 9) + (16c – 11 – 9d)
= (8d – 9d) + (-14c + 16c) + (9 – 11) = -d + 2c – 2
(iv) (6f – 20 + 8s) + (23 – 13f – 12s) = (6f – 13f) + (8s – 12s) + (-20 + 23) = -7f – 4s + 3
(v) (13m – 12n) + (12n – 13m) = (13m – 13m) + (- 12n + 12n) = 0
(vi) (-26m + 24n) + (26m – 24n) = (-26m + 26m) + (24n – 24n) = 0

Question 12.
Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

Solution:
Step 1 (0 fold): We get 0 + 2 = 2 pieces
Step 2 (1 fold): We get 1 + 2 = 3 pieces
Step 3 (2 folds): We get 2 + 2 = 4 pieces
In the same way, if rope is folded 10 times and cut, we get 10 + 2 = 12 pieces.
In the same way, when the rope is folded r times and cut, we get r + 2 pieces.

Question 13.
Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

Solution:
Step 1: To make 1 square, we need 4 matchsticks.
Step 2: To make 2 squares, we need 4 + 3 = 7 matchsticks.
Step 3: To make 3 squares, we need 4 + 3 + 3 = 10 matchsticks.
So, to make w squares, we need —
4 + (w – 1) × 3 = 4 + 3 (w – 1) = 4 + 3w – 3 = (3w + 1) matchsticks.
To make 10 squares, substituting w = 10, we get
Number of required matchsticks = 3(10) + 1 = 30 + 1 = 31

Question 14.
Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

Solution:
Number of squares in step 1 = 5
Number of squares in step 2 = 5 + 4 = 9
Number of squares in step 3 = 5 + 4 + 4 = 5 + 2 × 4 = 5 + 8 = 13
Thus, number of squares in step 4 = 5 + 3 × 4 = 5 + 12 = 17
Number of squares in step 10 = 5 + 9 × 4 = 5 + 36 = 41
And, number of squares in step 50 = 5 + 49 × 4 = 5 + 196 = 201
So, the general formula for number of squares in step n = 5 + (n – 1) × 4 = 5 + 4 (n – 1) = 5 + 4n – 4 = 4n + 1
Number of vertices in step 1 = 16
Number of vertices in step 2 = 16 + 12 = 28
Number of vertices in step 3 = 16 + 12 + 12 = 16 + 2 × 12 = 16 + 24 = 40
Number of vertices in step n = 16 + (n – 1) × 12 = 16 + 12n – 12 = 12n + 4

InText Questions

Question 1.
Shabnam is 3 years older than Aftab. When Aftab’s age is 10 years, Shabnam’s age will be 13 years. Now, Aftab’s age is 18 years, what will Shabnam’s age be?
Solution:
Shabnam’s age = Aftab’s age + 3
When .Aftab’s age = 18 years,
Shabnam’s age = 18 + 3 = 21 years

Question 2.
Find the values of the following arithmetic expressions:
(i) 23 – 10 × 2
(ii) 83 + 28 – 13 + 32
(iii) 34 – 14 + 20
(iv) 42 + 15 – (8 – 7)
(v) 68 – (18 + 13)
(vi) 7 × 4 + 9 × 6
(vii) 20 + 8 × (16 – 6)
Solution:
(i) 23 – 10 × 2 = 23 – 20 = 3
(ii) 83 + 28 – 13 + 32
= (83 – 13) + (28 + 32)
= 70 + 60 = 130
(iii) 34 – 14 + 20 = (34 – 14) + 20 = 20 + 20 = 40
(vi) 42 + 15 – (8 – 7) = 42 + 15 – 1
= 42 + 14 = 56
(v) 68 – (18 + 13) = 68 – 31 = 37
(vi) 7 × 4 + 9 × 6 = 28 + 54 = 82
(vii) 20 + 8 × (16 – 6) = 20 + 8 × 10
= 20 + 80 = 100

Question 3.
Some simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained.
1. Observe each of them and identify if there is a mistake.
2. If you think there is a mistake, try to explain what might have gone wrong.
3. Then, correct it and give the value of the expression.

Solution:
(1) If a = – 4, then 10 – a = 6 is wrong.

(2) If d = 6, then 3d = 36 is wrong.
As 10 – a = 10 – (-4) = 10 + 4 = 14
As 3d = 3 × d = 3 × 6 = 18
So, if a = -4, then 10 – a = 14 is correct.
So, if d = 6, then 3d = 18 is correct.

(3) Ifs = 7, then 3s – 2 = 15 is wrong.

(4) If r = 8, then 2r + 1 = 29 is wrong.
As 3s – 2 = 3 × 7 – 2 = 21 – 2 = 19
As 2r + 1 = 2 × 8 + 1 = 16 + 1 = 17
So, if s = 7, then 3s – 2 = 19 is correct.
So, if r = 8, then 2r + 1 = 17 is correct.

(5) If j = 5, then 2j = 10 is correct.

(6) If m = – 6, then 3(m + 1) = 19 is wrong.
As 2j = 2 × 5 = 10
As 3(m + 1) = 3 × (- 6 + 1) = 3 × (- 5) = – 15
So, if m = – 6, then 3(m + 1) = – 15 is correct.

(7) If f = 3, g = 1, then 2f – 2g = 2 is wrong.

(8) If t = 4, b = 3, then 2t + b = 24 is wrong.
As 2f – 2g = 2 × 3 – 2 × 1 = 6 – 2 = 4
As 2t + b = 2 × 4 + 3 = 8 + 3 = 11
So, if f = 3, g = 1, then 2f – 2g = 4 is correct.
So, if t = 4, b = 3, then 2t + b = 11 is correct,

(9) If h = 5, n = 6, then h – (3 – n) = 4 is wrong.
As h – (3 – n) = 5 – (3 – 6) = 5 – (- 3) = 5 + 3 = 8
So, if h = 5, n = 6, then h – (3 – n) = 8 is correct.

Question 4.
Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.

(i) If c = ₹50, find the total amount earned by the sale of pencils.
(ii) Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Solution:
Total money earned by the shopkeeper
= Money earned on day 1 + Money earned on day 2 + Money earned on day 3
= 5c + 4 d + 3c + 6d + 10c + d = 18c + 11d

(i) Total amount earned by the sale of pencils
= 5c + 3c + 10c = 18c
= 18 × ₹50 = ₹900

(ii) Given, the price per eraser is d.
Total money earned by selling erasers
= 4d + 6d + d = 11 d

Question 5.
Fill in the blanks below by replacing the letter- numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.

Solution:

We see that the values of 5u and 5 + u are not equal for different values of u. So, the expressions 5w and 5 + u are not equal.

Question 6.
Are the expressions 10y – 3 and 10(y – 3) equal?
After filling in the two diagrams (given below), do you think the two expressions are equal?

Solution:

Since the values of 10v – 3 and 10(y – 3) are not equal for different values of y, the expressions 10y – 3 and 10(y – 3) are not equal.

Question 7.
Find the formulas of the number machines below and write the expression for each set of inputs.

Solution:
(i) The formula for the number machine is “sum of first number and second number minus two” and the expression is a + b – 2.
The expression for each set of inputs are:
5 + 2 – 2 = 5, 8 + 1 – 2 = 7,
9 + 11 – 2 = 18, 10 + 10 – 2 = 18
and a + b – 2

(ii) The formula for the number machine is “product of first number and second number plus one” and the expression is a × b + 1.
The expression for each set of inputs are:
4 × 1 + 1= 5, 6 × 0 + 1 = 1,
3 × 2 + 1 = 7, 10 × 3 + 1 = 31
and a × b + 1 = ab + 1

Question 8.
Somjit noticed a repeating pattern along the border of a saree.

Use this to find what design appears at positions 99, 122 and 148.
Solution:
We can see the design A, B and C appear at the positions 3n – 2, 3n – 1 and 3n, respectively.
For 99, the remainder on division by 3 is 0,’i.e. it is a multiple of 3.
So, at position 99, design C will appear.
For 122, the remainder on division by 3 is 2, i.e. it is 1 less than a multiple of 3, i.e. 3n – 1.
So, at position 122, design B will appear.
For 148 , the remainder on division by 3 is 1, i.e. it is 2 less than a multiple of 3, i.e. 3n – 2.
So, at position 148, design A will appear.

Expressions using Letter Numbers Class 7 Extra Questions

Expressions using Letter Numbers Class 7 Very Short Question Answer

Question 1.
Simplify the expression 7(u – 2v) + 2v.
Solution:
Using the distributive property, this expression can be simplified as
7(u – 2v) + 2v = 7u – 7 × 2v + 2v
= 7u – 14v + 2v = 7u – 12v
[Adding like terms together]

Question 2.
What is the sum of the numbers in the given picture (unknown values are denoted by letter- numbers)?

Solution:
Adding row wise, we get Sum of numbers
= (5 × 4) + (2g + 2h) + (2g + 2h) + (5 × 4)
= 20 + 2g + 2 h + 2g + 2h + 20
= (20 + 20) + (2g + 2g) + (2h + 2 h)
= 40 + 4g + 4 h

Expressions using Letter Numbers Class 7 Short Question Answer

Question 1.
Add the numbers in each picture below. Write their corresponding expressions and simplify them.

Solution:
(i) (x + y + y + x + y + y + x + y + y + x + y + y) + (4 × 6)
= 4x + 8y + 24

(ii) 2 × 3y + 4 × 2 + 2 × (-4x) + 4 × (-8)
= 6y + 8 – 8x – 32
= 6y – 8x – 24

(iii) 4 × (- 3n) + 6 × 7m
= -12m + 42m
= 42m – 12n

Question 2.
Simplify each of the following expressions:
(i) e + e + e + f + f
(ii) m + m – (m – n) – n – n
(iii) l – 7m + l – (m – 1)
(iv) 2x – x – (x + x)
(v) (3u – 2v) – (2u – 3v)
Solution:
(i) e + e + e + f + f = 3e + 2f
(ii) m + m – (m – n) – n – n = 2m – (m – n) – 2n
= 2m – m + n – 2n
= m – n

(iii) l – m + l – (m – l) = l + l – m – (m – l)
= 2l – m – m + l
= 3l – 2m

(iv) 2x – x – (x + x) = 2x – x – 2x = -x

(v) (3u – 2v) – (2u – 3v) = 3u – 2v – 2u + 3v
= 3u – 2v – 2v + 3u
= u + v

Expressions using Letter Numbers Class 7 Long Question Answer

Question 1.
Simplify the following expressions:
(i) 2x – 3y + 7x + y – 12
(ii) 4(x – 2y) + 3x – 5y + 1
(iii) 7 + 3h – 2g – (5h – 4g)
(iv) 3g + 9h – (5 + 6h – 2g)
(v) 5(2u + 7v – 2) – 3(8u – 10v + 4)
(vi) 17 – 13a + 19b – 2(9 + 7a – 3b)
Solution:
(i) 2x – 3y + 7x + y – 12
= 9x – 2y – 12

(ii) 4(x – 2y) + 3x – 5y + 1
= 4x – 8y + 3x – 5y + 1
= 7x – 13y + 1

(iii) 7 + 3h – 2g – (5h – 4g)
= 7 + 3h – 2g – 5h + 4g
= 7 – 2h + 2g

(iv) 3g + 9h – (5 + 6h – 2g)
= 3g + 9h – 5 – 6h + 2g
= 5g + 3h – 5

(v) 5(2u + 7v – 2) – 3(8u – 10v + 4)
= 10u + 35v – 10 – 24u + 30v – 12
= – 14u + 65v – 22

(vi) 17 – 13a + 19b – 2(9 + 7a – 3b)
= 17 – 13a + 19b – 18 – 14a + 6b
= – 27a + 25b – 1

Question 2.
Add the expressions given below:
(i) 3d+ 6c – 7 and 3c – 4d + 2
(ii) -8g + 3h + 2 and 6g – 4h – 12
(iii) 11k + l + 10 and k – (11l – 10)
(iv) 21m – 7n and -17m + 10n + 13
(v) 5(x + 7y) – 3 and 2x + 3y – 2
(vi) 7a + 2b + 9 and 3a + 8b + 1
Solution:
(i) (3d + 6c – 7) + (3c – 4d + 2)
= 6c + 3c + 3d – 4d – 7 + 2 = 9c – d – 5

(ii) (-8g + 3h + 2) + (6g – 4h – 12)
= -8g + 6g + 3h – 4h + 2 – 12
= – 2g – h – 10

(iii) (11k + l + 10) + {k – (11l – 10)}
= 11k + l + 10 + k = 11l + 10
= 11k + k + l – 11l + 10 + 10
= 12k – 10l + 20

(iv) (21m – 7n) + (-17m + 10n + 13)
= 21m – 17m – 7n + 10n + 13
= 4m + 3n + 13

(v) 5(x + 7y) – 3 + (2x + 3y – 2)
= 5x + 35y – 3 + 2x + 3y – 2
= 5x + 2x + 35y + 3y – 3 – 2
= 7x + 38y – 5

(vi) (7a + 2b + 9) + (3a + 8b + 1)
= 7a + 3a + 2b + 8b + 9 + 1
= 10a + 10b + 10

Expressions using Letter Numbers Class 7 Case Based Questions

Question 1.
A movie theatre has 6 columns of seats (labelled A to F) and the seats are arranged in endless rows, starting from the front. Each seat is numbered sequentially row wise, beginning with Seat 1 in Column A of Row 1, moving left to right.

Based on the above information, answer the following questions:
(i) In the seating layout, imagine selecting any 2 × 3 block of seats (2 rows and 3 columns), like the one shown in the picture.

If the seat in the top middle of selected block is numbered ‘x’ write expressions to represent the seat numbers in the other five blank positions of the block.

(ii) Look at the group of seats arranged in the shape.

Find the sum of all the seat numbers in this shape. Then compare this total with the number
at the centre i.e. 11.
Try this again with a different set of numbers arranged in the same shape and write your observation.
Solution:
(i) Number to the left of Y will be 1 less than x.
Number to the right of V will be 1 more than x.
Since there are 6 seats in one row,
Number below x will be 6 more than x.
Number in the diagonally left cell to ‘x’ will be 5 more than x for one less than x + 6).
Number in the diagonally right cell to ‘x’ will be 7 more thain x (or one more than x + 6).
Thus, the expressions in other five blank s positions of the block are as shown:

(ii) Sum of all numbers
= 4 + 6 + 11 + 16 + 18 = 55 = 5 × 11
The sum is 5 times the number in the centre.
Now, let the number at the centre be 14, then the shape is given below:

Sum of all the numbers
= 7 + 9 + 14 + 19 + 21
= 70 = 5 × 14
Again, the sum is 5 times the number in the centre.
Now, let the number at the centre be 9, then the shape is given below:

Sum of all the numbers
= 2 + 4 + 9 + 14 + 16 = 45 = 5 × 9
Again, the sum is 5 times the number in the centre.

Question 2.
Observe the picture below. It shows a growing pattern of huts made using toothbrushes.
In Step 1, there is 1 hut.
In Step 2, there are 2 huts.
In Step 3, there are 3 huts and this pattern continues.

Based on the above information, answer the following questions:
(i) What is the general rule to find the number of toothbrushes in step n?
(ii) How many toothbrushes will be there in step 9, step 37 and step 54?
Solution:
(i) We can see that the number of toothbrushes increases by 3 at each step.

We can write general rule to find the number of toothbrushes in step n as:
4 + (n – 1) × 3 = 4 + 3n – 3 = 3n + 1,
where n = 1, 2, 3, …

(ii) Number of toothbrushes in step 9
= 3 × 9 + 1 = 27 + 1 = 28
Number of toothbrushes in step 37
= 3 × 37 + 1 = 111 + 1 = 112
Number of toothbrushes in step 54
= 3 × 54 + 1 = 162 + 1 = 163

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

Class 7 Maths Chapter 6 Number Play Solutions

Ganita Prakash Class 7 Chapter 6 Solutions

Class 7 Maths Ganita Prakash Chapter 6 Solutions Number Play

Question 1.
Arrange the stick figure cutouts in order of their heights, ensuring that each child (from left to right) states the number of children standing ahead who are taller than them.


(i) 0, 1 , 1 , 2, 4, 1 , 5
(ii) 0, 0, 0, 0, 0, 0, 0
(iii) 0, 1, 2, 3, 4, 5, 6
(iv) 0, 1,0, 1, 0, 1, 0
(v) 0, 1, 1, 1, 1, 1, 1
(vi) 0,0,0,3,3,3,3
Solution:
(i) The required arrangement is FCBGADF:

(ii) The required arrangement is AECGBDF.

(iii) The required arrangement is FDBGCEA

(iv) The required arrangement is EAGCDBF

(v) The required arrangement is FAECGBD

(vi) The required arrangement is BDFAECG

Question 2.
We know that:
(i) even + even = even
(ii) 0dd + 0dd = 0dd
(iii) even + odd = odd
Similarly, find out the parity for the below scenarios:
(iv) even – even = ___________
(v) 0dd – 0dd = _______
(vi) even – odd = ________
(vii) odd – even = ________
Solution:
(iv) even – even
Example: 6 – 2 = 4 → even
8 – 4 = 4 → even
Parity of result = even
∴ even – even = even

(v) odd – odd
Example: 7 – 3 = 4 → even
9 – 5 = 4 = 4 → even
Parity of result = even
∴ old – old = even

Question 3.
How many different magic squares can be made using numbers 1-9?
Solution:
Using the numbers 1-9, there is exactly one unique magic square (excluding rotations and reflections).

If transformations like rotations are allowed, then there are 8 variations of this magic square.

Question 4.
Create a magic square using numbers 2-10. What strategy would you use for this? Compare it with the magic squares made using 1-9.
Solution:
The numbers 2-10 are 9 consecutive numbers, just like 1-9, but increased up by 1.
Strategy: Start with the classic 1-9 magic square, and add 1 to each number.

Original: After adding 1 to each.
The magic square for numbers 2-10 would have a different magic sum (18) compared to numbers 1-9 (15). The structure remains similar, but the values are shifted up to 1.

Question 5.
Take a magic square, and
(i) increase each number by 1
(ii) double each number
In each case, is the resulting grid also a magic square?
How do the magic sums change in each case?
Solution:
Original:

(i) After increasing each number by 1:

This is still a magic square.
New magic sum = 15 + 3 × 1 = 18

(ii) After doubling each number

Still a magic square
New magic sum = 15 × 2 = 30

In case (i), adding a constant to every number → magic sum (for 3×3 grid) is increased by three times of that constant.
In case (ii), multiplying all by a constant → magic sum multiplied by that constant.

Question 6.
Using the generalised form, find a magic square if the centre number is 25.
Solution:
The generalised form of a magic square is given below.

If the centre number is 25, then m = 25.
Substituting m = 25 in generalised form of a magic square, we get

Question 7.
Write the result obtained by—
(i) adding 1 to every term in the generalised form.
(ii) doubling every term in the generalised form.
Solution:
The generalised form of a magic square is given below.

(i) Adding one to every term:

(ii) Doubling every term:

Question 8.
Create a magic square whose magic sum is 60.
Solution:
A 3 × 3 magic square’s sum is 3 × middle element.
So, for a sum is 60, the middle element should be \(\frac{60}{3}\) = 20 .
To get a magic sum of 60, we will multiply the original magic square by 4 i.e.

Question 9.
Is it possible to get a magic square by filling nine non-consecutive numbers?
Solution:
Yes, it is possible.
Justification: Let us consider the two magic squares with magic sum 45.

9 consecutive numbers
and

9 non-consecutive numbers

Question 10.
A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
Solution:
Dorjee toggles the switch 77 times.

Each toggle changes the state of the bulb (ON to OFF or OFF to ON). Starting from ON.

An odd number of toggles will leaves the bulb OFF and an even number of toggles will leave the bulb ON. Since 77 is odd, after 77 toggles, the bulb will be OFF.

Question 11.
Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and column sums.

Solution:
Let’s label the cells as:

So the constraints are:

• Row 1 (A, B, C): sum is odd
• Row 2 (D, E, F): sum is even
• Column 1 (A, D): sum is even
• Column 2 (B, E): sum is even
• Column 3 (C, F): sum is odd

We’ll track parities only (o or e), not actual numbers.
Row 1: A = o, B = e, C = e, then o + e + e = odd
Column 1 (A, D) – e means A must be paired with D as odd to get the sum as even.
So, if A = o, D = o, then o + o = even
Similarly, if B = e, E = e, then e + e = even
Again, if C = e, F = o, then e + o = odd
So, the 6 boxes with 3 odd numbers and 3 even numbers can be filled as follows:

Question 12.
Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.
Solution:
It is given that

• The magic square is 3 × 3.
• The magic sum is 0.
• All numbers in the square cannot be zero, we can use negative numbers as needed.

So, we will use the numbers (- 4) to 4 to create a magic square whose magic sum is 0.
The required magic square is given below.

Question 13.
Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?
Solution:
Given numbers are 987 and 1597.
In the Virahanka sequence, each number is the sum of the two preceding numbers.
The next two numbers are:
987 + 1597 = 2584 and 1597 + 2584 = 4181
The previous two numbers are:
1597 – 987 = 610 and 987 – 610 = 377
The sequence is …, 377, 610, 987, 1597, 2584, 4181,…

Question 14.
What is the parity of the 20th term of the Virahanka sequence.
Solution:
Consider the Virahanka sequence given below:
1,2, 3, 5, 8, 13, 21,34, 55, 89, 144, 233, 377, 610, 987, …
Let us observe the pattern of odd/even in Virahanka sequence.
Here 1 → odd;
2 → even;
3 → odd
5 → odd;
8 → even;
13 → odd
21 → odd;
34 → even;
55 → odd
So parity cycle: odd, even, odd, (repeats every 3 terms)
So the parity of 20th term in Virahanka sequence is even.

Question 15.
Solve the following cryptarithm:

Solution:
Here, in TAT, letter T is a hundreds place.
So, T = 1.
⇒ A = 0 and U = 9.
So, we have U = 9, T = 1 and A = 0.

InText Questions

Question 1.
Kishor has a set of number cards and 5 boxes. If each box must contain exactly one number card, suggest an arrangement to help him distribute the cards, so that 5 cards add to 30? Is it possible?

Solution:
No, it is not possible, as the sum of 5 odd numbers is always odd and 30 is an even number.

Question 2.
In a 3 × 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3 × 4 grid, there are 12 small squares, which is an even number.
Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

Solution:
Yes, we can determine the parity of the number of small squares in a grid without directly calculating the full product, simply by observing the parity of the dimensions.
Rule: The product of two numbers is:

• Even if at least one of the numbers is even.
• Odd if both numbers are odd.

Question 3.
We can describe how the numbers within the magic square are related to each other, i.e. the structure of the magic square.
Choose any magic square that you have made so far using consecutive numbers. If m is the letter- number of the number in the centre, express how other numbers are related to m, how much more or less than m.
Solution:
Consider the magic square

We can express it using the letter- number m for the number in the centre as:

Question 4.
Write the next 3 numbers in the sequence:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, , , , …
If you have to write one more number in the Sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)?
Solution:
The next 3 terms in the sequence are:
55 + 89 = 144; 89 + 144 = 233; 144 + 233 = 377
Yes, we can determine the parity without adding the two previous number.
Since odd + odd = even
Hence, parity of next number in sequence is even.

Question 5.
Find out what each letter stands for.

Solution:
(i) YY is a two-digit number where both digits are the same. So it can be 99, 88, …
But Z is a 1-digit number and ZOO is a 3-digit number.
So, Y = 9, Z = 1 and O = 0.

\(\begin{array}{r}
99 \\
+\quad 1 \\
\hline 1 \quad 00 \\
\hline
\end{array}\)

(ii)

(iii) Here, KP is a 2-digit number and PRR is a 3-digit number. Basically, 2 × (KP) = PRR.
If P = 1, then R = 2.
Hence, K = 6, P = 1 and R = 2.

Here, C + 1 is a two-digit number i.e. 10.

Number Play Class 7 Extra Questions

Number Play Class 7 Very Short Question Answer

Question 1.
A bag holds 100 balls, each marked with an odd number. If two balls are picked at random and their numbers are added, what will be the parity of the result?
Solution:
We know that odd number + odd number = even number.
Since each ball is marked with an odd number. If two balls are picked at random and their numbers are added, the parity of the sum will be even.

Question 2.
A number has odd parity. What will be the parity of:
(i) Number + 7
(ii) Number – 2
(iii) Number + Number
Solution:
Given, a number has odd parity.
(i) Since the parity of the sum of two odd numbers is even, the parity of (Number + 7) will be even.
(ii) Since the parity of the difference of an even and an odd number is odd, parity of (N umber – 2) will be odd.
(iii) Since the parity of the sum of two odd numbers is even, the parity of (Number + Number) will be even.

Question 3.
Neha wants to climb a staircase with 7 steps. She has a simple rule: She can take either 1 step or 2 steps at a time. For example, one possible way to climb is: 2, 1, 2, 2.
In how many different ways can Neha climb to the top of the 7-step staircase?
Solution:
The number of different ways in which Neha can climb to the top of the 7-step staircase taking either 1 step or 2 steps at a time, is the 7^th element of Virahanka sequence.
Virahanka sequence: 1, 2, 3, 5, 8, 13, 21, …
7^th element of Virahanka sequence = 21
∴ Neha can climb to the top of the 7-step staircase in 21 different ways.

Number Play Class 7 Short Question Answer

Question 1.
During a maths quiz, a student says, “Two consecutive numbers add up to 98.” Is this claim correct? Justify your answer with reasoning.
Solution:
(i) The claim is not correct.
(ii) The counting numbers 1, 2, 3, 4, 5, … alternate between even and odd numbers.
In any two consecutive numbers, one will always be even and the other will always be odd.
We know that the parity of the sum of an even number and an odd number is odd.
Since 98 is an even number, his claim is incorrect.

Question 2.
Form two expressions, one that always gives odd parity and another that always gives even parity.
Solution:
We know that the parity of the product is even if at least one of them is even.
Also, in any two consecutive numbers, one is alwavs even and the other is always odd.
∴ For any number n, the expression n(n + 1) = n² + n always has even parity.
Since even number + odd number = odd number.
∴ The expression n² + n + 1 always has odd parity.

Question 3.
Solve the following cryptarithms:
(i)
\(\begin{array}{r}
\mathrm{B} 2 \\
+6 \mathrm{C} \\
\hline \mathrm{EC} 6
\end{array}\)
(ii)
\(\begin{array}{r}
\mathrm{PQ} \\
+\mathrm{QR} \\
\hline \mathrm{QRQ}
\end{array}\)
Solution:
We know that in cryptarithms:

• Each letter stands for unique digit and different letters represent different digits.
• The same letter always means the same digit.

(i) Here, 2 + C = 6 ⇒ C = 4
Now, B + 6 = EC = E4 [Since C = 4]
⇒ B = 8 and E = 1

(ii) Here, Q + R = Q ⇒ R = 0
Now, P + Q = QR
⇒ P + Q = Q0
⇒ P = 9 and Q = 1

Number Play Class 7 Long Question Answer

Question 1.
Aarav stores his stamp collection in small boxes. He has:

• an odd number of boxes with 5 stamps each,
• an odd number of boxes with 7 stamps each, and
• an even number of boxes with 10 stamps each. He counts all his stamps and says the total is 175.

Did Aarav make a mistake? Explain your reasoning.
Solution:
We know that the parity of the product of two odd numbers is odd and the parity of the product of two even numbers is even. Thus,

• An odd number of boxes with 5(odd) stamps each gives an odd number of stamps.
• An odd number of boxes with 7(odd) stamps each also gives an odd number of stamps.
• An even number of boxes with 10(even) stamps each gives an even number of stamps.

Since, odd number + odd number = even number And even number + even number = even number, the final total must be even number. But Aarav claims the total is 175, which is odd. Therefore, he must have made a mistake in his counting.

Question 2.
Two consecutive numbers in the Virahanka sequence are 377 and 610. What are the previous 2 numbers in the sequence?
Solution:
We know that in Virahanka sequence, a number is the sum of previous two numbers.
Let the previous two numbers be x and y respectively.
The sequence will be as: …, x,y, 377, 610, …
Then, y + 377 = 610
⇒ v = 610-377 = 233
Now, x + v = 377 ,
⇒ x + 233 = 377
⇒ x = 377 – 233 = 144
Thus, the previous two numbers in the Virahanka sequence are 144 and 233.

Question 3.
Solve the following cryptarithms:
(i)
\(\begin{array}{r}
\mathrm{Tl} \\
+1 \mathrm{~T} \\
\hline 66
\end{array}\)
(ii)
\(\begin{array}{r}
\mathrm{D} 3 \\
+3 \mathrm{D} \\
\hline \mathrm{PPQ}
\end{array}\)
Solution:
(i) T + 1 = 6 ⇒ T = 5
Thus,
\(\begin{array}{r}
51 \\
+\quad 15 \\
\hline 66
\end{array}\)

(ii) A three-digit number abc can be written as 100a + 10b + c.
D3 = 10D + 3, 3D = 3 × 10 + D = 30 + D and PPQ = 100P + 10P + Q = 110P + Q
Now, D3 + 3D = (10D + 3) + (30 + D)
= 11 D + 33
When D ranges from 1 to 6, the expression 11 D + 33, gives a two-digit number.
At D = 7, 11 × 7 + 33 = 77 + 33 = 110
Putting P = 1 and Q = 0, we get
110P + Q = 110 × 1 + 0 = 110
∴ P = 1, Q = 0 and D = 7

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 7 A Tale of Three Intersecting Lines Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines Solutions

Ganita Prakash Class 7 Chapter 7 Solutions

Class 7 Maths Ganita Prakash Chapter 7 Solutions A Tale of Three Intersecting Lines

Question 1.
Use the points on the circle and/or the centre to form isosceles triangles.

Solution:
Select any two points on the circle and connect them with the centre of the circle.
Also, join these points to each other. This will form an isosceles triangle as the two radii are equal in length.

Question 2.
Can we say anything about the existence of a triangle for each of the following sets of lengths?
(i) 10 km, 10 km and 25 km
(ii) 5 mm. 10 mm and 20 mm
(iii) 12 cm. 20 cm and 40 cm
Solution:
(i) When we take direct path = 25 km.
Then roundabout path =10 km + 10 km = 20 km.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

(ii) When we take direct path = 20 mm.
Then roundabout path = 10 mm + 5 mm = 15 mm.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

(iii) When we take direct path = 40 cm.
Then roundabout path =12 cm + 20 cm = 32 cm.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

Question 3.
Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.
Solution:
Yes, an equilateral triangle with sides 50, 50, 50 exists because the sum of two sides is greater than the third side. For any positive number say x > 0, x + x > x. So, an equilateral triangle with all side lengths ‘x’ exists.

Question 4.
For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values can also be chosen): ;
(i) 1, 100
(ii) 5, 5
(iii) 3, 7
Solution:
(i) 5 possible values for the third length would be 99.5. 99.8, 100, 100.5, 100.9
Since, 100 < 1 + 99.5, 100 < 1 + 99.8, 100 < 1 + 100, 100.5 < 1 + 100, and 100.9 < 1 + 100

(ii) 5 possible values for the third length would be 1, 3.5, 5, 7.5, 8.9
Since, 5 < 1 + 5, 5 < 5 + 3.5, 5 < 5 + 5, 7.5 < 5 + 5, and 8.9 < 5 + 5

(iii) 5 possible values for the third length would be 4.5, 5, 6.9, 8, 9.8
Since, 7 < 3 + 4.5, 7 < 5 + 3, 7 < 3 + 6.9, 8 < 3 + 7, 9.8 < 3 + 7

Question 5.
Construct triangles for the measurement, 3 cm, 120°, 8 cm, where the angle is included between the sides.
Solution:

Steps of construction are given below:
Step 1: Construct a side AB of length 8 cm.
Step 2: Construct ∠d = 120° by drawing the other arm of-the angle.
Step 3: Mark the point C on the other arm such that AC, = 3 cm.
Step 4: Join BC to get the required triangle.

Question 6.
Construct triangles for the measurements, 25°, 3 cm, 60° where the side is included between the angles.
Solution:

Steps of construction are given below:
Step 1: Draw the base AB of length 3 cm.
Step 2: Draw ∠d and ∠B of measure 25° and 60° respectively.
Step 3: Make the point of intersection of the two new arms of ∠d and ∠B as point C to get the required triangle.

Question 7.
Determine which of the following pairs can be the angles of a triangle and which cannot:
(i) 35°, 150°
(ii) 70°, 30°
Solution:
(i) The sum of the given angles 35° + 150° = 185°.
This is not possible because sum of the angles of the triangle exceeds 180°.

(ii) The sum of the given angles 70° + 30° = 100°.
Possible third angle 180°- 100° = 80°.
Since the possible third angle comes out positive (80°), the given angles can be angles of a triangle.

Question 8.
Find the third angle of a triangle (using a parallel line) when two of the angles are:
(i) 36°, 72°
(ii) 150°, 15°
Solution:
(i) Here ∠B = 36° and ∠C = 72°.

Since the line BC is parallel to AT.
So, ∠XAB = ∠B = 36° [Alternate angles] ………. (i)
and ∠YAC = ∠C = 72 [Alternate angles] ………. (i)
Also, ∠XAB + ∠B AC + ∠YAC = 180° [∠XAY is a straight angle]
⇒ 36° + ∠BAC + 72° = 180° [Using (i) and (ii)]
⇒ ∠BAC = 180°- 108° = 72°.

(ii) Here ∠B = 150° and ∠C = 15°.

Since the line BC is parallel to AY.
So, ∠XAB = ∠B = 150° [Alternate angles] ………. (i)
and ∠YAC = ∠C = 15° [Alternate angles] ………. (ii)
Also, ∠XAB + ∠BAC + ∠YAC = 180° [∠XAY is a straight angle]
⇒ 150° + ∠BAC + 15° = 180° [Using (i) and (ii)]
⇒ ∠BAC = 180° – 165° = 15°.

Question 9.
Here is a triangle in which we know ∠B = ∠C and ∠A = 50°, Can you find ∠B and ∠C?

Solution:
Given, ∠A = 50° and ∠B – ∠C.

Draw a line Xy which is parallel to BC.
Now, ∠XA B = ∠B and ∠YAC = ∠C [Alternate angles] ………… (i)
Also, ∠XAB + ∠BAC + ∠YAC = 180°
⇒ ∠B + 50° + ∠C = 180° [Using (i)]
⇒ ∠B + ∠C = 180° – 50° = 130°
⇒ 2 ∠B = 130° [∵ ∠B = ∠C]
⇒ ∠B = 65° = ∠C.

Question 10.
Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.
Solution:
Steps of construction are given below:

Step 1: Construct a side TR of length 7 cm.
Step 2: Construct ∠R = 140° by drawing the other arm of the angle.
Step 3: Mark the point 1 on the other arm such that RY = 4 cm.
Step 4: Join TY to get the required triangle.
Step 5: Keep the ruler aligned to RY. Place the set square along the ruler such that one of the edges of the right angle touches the ruler.
Step 6: Slide the set square along the ruler till the perpendicular edge of the set square touches the vertex T.
Step 7: Extend the line YR and then draw the altitude through T on extended YR using the perpendicular edge of the set square.

InText Questions

Question 1.
What happens when the three vertices lie on a straight line? Will these points form a triangle?
Solution:
When the three vertices lie on a straight line, they become collinear. This means they no longer form a triangle because the three points do not enclose any area.

Question 2.
Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle?
Solution:
Let the triangle be ABC, where AB = 8 cm.

Since, the arcs from the points A and B do not meet. So, we are not able to construct the triangle with side lengths 3 cm, 4 cm, and 8 cm.

A Tale of Three Intersecting Lines Class 7 Extra Questions

A Tale of Three Intersecting Lines Class 7 Very Short Question Answer

Question 1.
In the adjoining figure:

(i) Name the vertex opposite to side PQ.
(ii) Name the side opposite to vertex Q.
(iii) Name the angle opposite to side QR.
(iv) Name the side opposite to ∠R.
Solution:
In the given figure,
(i) The vertex opposite to side PQ is R.
(ii) The side opposite to vertex Q is PR.
(iii) The angle opposite to side QR is ∠P.
(iv) The side opposite to ∠R is PQ.

Question 2.
In the given triangle ABC, find the value of x.

Solution:
Given, ∠T = 7x – 3°, ∠B = 130°, ∠C = 6x + 1°
We know that sum of all the angles in a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ 7x – 3° + 130° + 6x + 1° = 180°
⇒ 13x + 128° = 180° ⇒ 13x = 180°- 128° = 52°
⇒ x = \(\frac{52^{\circ}}{13},\) = 4°
Thus, the value of x is 4°.

Question 3.
Can a triangle be formed for the following set of angles?
(i) 80°, 70° and 50°
(ii) 56°, 64° and 60°
Solution:
We know that the sum of all angles in a triangle is 180°.
(i) Given set of angles are 80°, 70° and 50°.
Now, sum of the angles = 80° + 70° + 50° = 200° ≠ 180°
Thus, the given set of angles cannot form a triangle.

(ii) Given set of angles are 56°, 64° and 60°.
Now, sum of the angles = 56° + 64° + 60° = 180°
Thus, the given set of angles can form a triangle.

Question 4.
In the following figure, find the value of p.

Solution:
In the given figure, ∠SPR is an exterior angle, and ∠PRQ and ∠PQR are two interior opposite angles to ∠SPR.
∴ ∠SPR = ∠PRQ + ∠PQR [Exterior angle property]
⇒ p = 105° + 45° = 150°
Thus, the value of p is 150°.

Question 5.
In ∆XYZ, YX is extended to O. If ∠Y = 57° and XY = XZ, then find ∠ZXO.

Solution:
Given, ∠Y = 57° and XY = XZ
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠Z = ∠Y = 57° ……… (i)
In the given figure, ∠ZXO is an exterior angle.
And, ∠Y and ∠Z are two interior opposite angle to ∠ZXO.
We know, Exterior angle = Sum of two interior opposite angles
∴ ∠ZXO = ∠Y + ∠Z
⇒ x – 57° + 57° = 114° [Using (i)]
Thus, the value of ∠ZXO is 114°.

A Tale of Three Intersecting Lines Class 7 Short Question Answer

Question 1.
Construct a triangle PQR such that PQ =3.5 cm, QR = 6.5 cm and PR = 4 cm.
Solution:
Given, PQ = 3.5 cm, QR = 6.5 cm and PR = 4 cm

To construct the triangle PQR of given sides, we follow the steps laid down below:
Step 1: Using ruler, draw a line segment PQ, 3.5 cm long.
Step 2: With P as centre, draw an arc with radius equal to 4 cm.
Step 3: With Q as centre, draw an arc with radius equal to 6.5 cm to cut the previous arc at R.
Step 4: Join R to P and Q. Thus, ∆PQR is required triangle.

Question 2.
Construct a triangle PQR such that PQ = 7 cm, QR = 8.5 cm and PR = 9 cm.
Solution:
Given, PQ = 7 cm, QR = 8.5 cm and PR = 9 cm.

To construct the triangle PQR of given side lengths, follow the steps laid down below:
Step 1: Using ruler, draw a line segment PQ, 7 cm long.
Step 2: With P as centre, draw an arc with radius equal to 9 cm.
Step 3: With Q as centre, draw an arc with radius equal to 8.5 cm to cut the previous arc at R.
Step 4: Join R to P and Q. Thus, ∆PQR is required triangle.

Question 3.
In the given figure, PQ is parallel to RS. Find the value of y.

Solution:
In ∆PQR, we have
∠QPR = 82°, ∠PRQ = 42° and ∠PQR = x
We know that the sum of all angles in a triangle is 180°. ‘
∴ ∠QPR + ∠PRQ + ∠PQR = 180°
⇒ 82° + 42° + x = 180°
⇒ 124° + x = 180°
⇒ x = 180°-124° = 56°
Since PQ is parallel to RS and QR is a transversal, ∠PQR and ∠QRS are alternate interior angles, they are equal.
⇒ ∠QRS = ∠PQR
⇒ Y = ,x = 56°
Thus, the value of y is 56°.

Question 4.
In the given figure, find the value of x.

Solution:
In ∆DBE, ∠DEF = 150° is an exterior angle.
We know that the exterior angle of a triangle is equal to the sum of opposite interior angles.
∴ ∠DEF = ∠ERD + ∠EDB
⇒ 150° = 50° + ∠EDB
⇒ ∠EDB = 150° – 50° = 100°
Again, in ADAG, ∠EDB is an exterior angle.
∴ ∠EDB = ∠DAG + ∠AGD
⇒ 100° = ∠DAG + 70°
[∵ ∠EDB = 100°, ∠AGD = 70°]
⇒ ∠DAG = 100° – 70° = 30°
Also, ∠DAG and ∠GAH form a linear pair of angles.
∴ ∠DAG + ∠GAH = 180°
[∵ ∠DAG = 100°, ∠GAH = x]
⇒ 30° + x = 180°
⇒ x = 180°-30° = 150°

Question 5.
In ∆IJK, if IJ = IK, ∠IKJ = 50°, ∠JIK = m, find the value of m.

Solution:
Given, IJ = IK, ∠IKJ = 50°, ∠JIK = m
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠IKJ = ∠IJK = 50°
Now, sum of all the angles = 180°
⇒ ∠JIK + ∠IKJ + ∠IJK = 180°
⇒ m + 50° + 50° = 180°
⇒ m + 100° = 180°
⇒ m = 180°- 100° = 80°
Thus, the value of m is 80°.

Question 6.
In the given figure, if ∠NMO = 84° and LM = MN, then find the values of ∠N and ∠L.

Solution:
Given, ∠NMO = 84° and LM = MN
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠N = ∠L ………. (i)
In the given figure, ∠NMO = 84° is an exterior angle.
And, ∠N and ∠L are two interior opposite angles to ∠NMO.
We know,
Exterior angle = Sum of two interior opposite angles
∴ ∠NMO = ∠N + ∠L
⇒ 84° = ∠N + ∠N [Using (i)]
⇒ 2∠N = 84° ⇒ ∠N = \(\frac{84^{\circ}}{2}\) = 42°
Thus, ∠N = ∠L = 42°

Question 7.
In the given figure, find the value of y.

Solution:
In the given figure, ∠BCD is an exterior angle.
And, ∠BAC and ∠ABC are two interior opposite angles to ∠BCD.
We know, Exterior angle = Sum of two interior opposite angles
∴ ∠BCD = ∠BAC + ∠ABC
⇒ 7y + 6° = 50° + 96°
⇒ 7v + 6° = 146°
⇒ 7y = 146° – 6° = 140°
⇒ y = \(\frac{140^{\circ}}{7}\) = 20°
Thus, the value of y is 20°.

Question 8.
In ∆PQR, ZQ is thrice of ∠P and ∠R is twice of sum of ∠P and ∠Q. Find the angles.
Solution:
Let ∠P be x. Then,
∠Q = 3 x, ∠P = 3x
and ∠R = 2(∠P + ∠Q) = 2(x + 3x) = 2 × 4x = 8x
We know that sum of all the angles in a triangle is 180°.
∴ ∠P + ∠Q + ∠R = 180°
⇒ x + 3x + 8x = 180°
⇒ 12x = 180°
⇒ x = \(\frac{180^{\circ}}{12}\) = 15°
Thus, ∠P = x = 15°, ∠Q = 3x = 3 × 15° = 45° and ∠R = 8x = 8 × 15° = 120°

A Tale of Three Intersecting Lines Class 7 Long Question Answer

Question 1.
Mrs. Roy gave straws of length 21 cm to her students and asked them to cut the straw to get three pieces, which can be used to form each of the three types of triangles. The lengths of the three pieces were supposed to be whole numbers (1, 2, 3, …) only. Based on these criteria, write any one possible combination of straw lengths in Column 3 of the below table.

Solution:
We know that if a given set of three lengths satisfies the triangle inequality (each length < sum of the other two lengths), then a triangle exists having those as side lengths.
(a) If all three sides of a triangle are different in length, then it is called a scalene triangle.
Thus, the possible sets of side lengths are:
(i) 2 cm, 9 cm and 10 cm
(ii) 3 cm, 8 cm and 10 cm
(iii) 4 cm, 7 cm and 10 cm
(iv) 4 cm, 8 cm and 9 cm
(v) 5 cm, 6 cm and 10 cm
(vi) 5 cm, 7 cm and 9 cm
(vii) 6 cm, 7 cm and 8 cm

(b) If any two sides of a triangle are equal in length, then it is called an isosceles triangle.
Thus, the possible sets of side lengths are:
(i) 6 cm, 6 cm and 9 cm
(ii) 8 cm, 8 cm and 5 cm
(iii) 9 cm, 9 cm and 3 cm
(iv) 10 cm, 10 cm and 1 cm

(c) If all three sides of a triangle are equal in length, then it is called an equilateral triangle. Thus, the possible set of side lengths is 7 cm, 7 cm, 7 cm.

Question 2.
In the given figure, find ∠EAB + ∠ABC + ∠BCD + ∠CDE + ∠DEA.

Solution:
We know that the sum of angles of a triangle is 180°.
∴ In ∆ABC, we have
∠CAB + ∠ABC + ∠BCA = 180° ………… (i)
In ∆ACD, we have
∠ACD + ∠CDA + ∠DAC = 180° ………. (ii)
And in AADE, we have
∠ADE + ∠DEA + ∠EAD = 180° ……….. (iii)
Adding (i), (ii) and (iii), we get
∠CAB + ∠ABC + ∠BCA + ∠ACD + ∠CDA + ∠DAC + ∠ADE + ∠DEA + ∠EAD = 180° + 180° + 180°
⇒ (∠EAD + ∠DAC + ∠CAB) + ∠ABC + (∠BCA + ∠ACD) + (∠CDA + ∠ADE) + ∠DEA = 540°
⇒ ∠EAB + ∠ABC + ∠BCD + ∠CDE + ∠DEA = 540°
[∵ ∠EAD + ∠DAC + ∠CAB = ∠EAB, ∠BCA + ∠ACD = ∠BCD and ∠CDA + ∠ADE = ∠CDE]

Question 3.
Two line segments PS and QR intersect at O. Joining PQ and SR, we get two triangles, ∆POQ and ∆ROS as shown in the figure. Find the values of ∠P and ∠Q.

Solution:
In ∆ROS, ∠R = 35° and ∠S = 70°
We know that sum of all the angles in a triangle is 180°.
∴ ∠R + ∠ROS + ∠S = 180°
⇒ 35° + ∠ROS + 70° = 180°
⇒ ∠ROS + 105° = 180°
⇒ ∠ROS = 180° – 105° = 75°
Since ∠ROS and ∠POQ are vertically opposite angles, ∠ROS = ∠POQ = 75°.
Now, in ∆POQ, ∠P = 5x, ∠POQ =75° and ∠Q = 4x + 6°
We know that sum of all the angles in a triangle is 180°.
∴ ∠P + ∠POQ + ∠Q = 180°
⇒ 5i + 75° + 4x + 6° = 180°
⇒ 9x + 81° = 180°
⇒ 9x = 180° – 81° = 99°
⇒ x = \(\frac{99^{\circ}}{9}\) = 11°
Thus, ∠P = 5x = 5 × 11° = 55° and ∠Q = 4x + 6° = 4 × 11° + 6° = 44° + 6° = 50°

Question 4.
Construct a triangle ∆XYZ with XY = 4 cm, YZ = 7 cm and ∠Y = 108°.
Solution:
Given, in ∆XYZ, XY = 4 cm, YZ = 7 cm and ∠Y = 108°
Steps of construction of ∆XYZ are as follows:
Step 1: Using a ruler, draw a line segment XY, 4 cm long.
Step 2: Using protractor, draw the given vertex angle, ∠Y = 108°.
Step 3: On the new arm created from point F, mark a point Z such that YZ = 7 cm using a ruler and a compass.
Step 4: Using a ruler, connect points X and Z to complete the required triangle XYZ.

Question 5.
Construct a triangle IJK if IJ = 5 cm, JK = 6 cm and included angle between sides IJ and JK is 37°.
Solution:
Given, in ∆IJK, IJ = 5 cm, JK = 6 cm and included angle between sides IJ and JK is 37°.
Steps of construction of ∆IJK are as follows:
Step 1: Using a ruler, draw a line segment IJ of 5 cm length.
Step 2: Using protractor, draw the given vertex angle, ∠J = 37°.

Step 3: On the new arm created from point K mark a point K such that JK = 6 cm using a ruler anti a compass.
Step 4: Using a ruler, connect points I and K to complete the triangle IJK.

Question 6.
Construct a triangle XYZ with ∠Z = 30°, ∠Y = 115° and YZ = 5 cm.
Solution:
Given, in ∆XYZ, ∠Z = 30°, ∠Y = 115° and YZ = 5 cm
Steps of construction of ∆XYZ are as follows:
Step 1: Using a ruler, draw a line segment YZ of 5 cm length.
Step 2: Using protractor, draw the given vertex angle, ∠Y = 115°.

Step 3: Using protractor, draw the given vertex angle, ∠Z = 30°.
Step 4: Mark the intersecting point of two new lines (arms of angles) as X. Thus, the required triangle, ∆XFZ is constructed.

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 8 Working with Fractions Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 8 Working with Fractions Solutions

Ganita Prakash Class 7 Chapter 8 Solutions

Class 7 Maths Ganita Prakash Chapter 8 Solutions Working with Fractions

Question 1.
A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make _______ km of the water canal. If they work 5 days a week, they can make _______ km of the water canal in a week.
Solution:
Water canal made by team of workers in 8 days = 1 km
So, water canal made by team of workers in 1 dav = \(\frac{1}{8}\) km
The length of the water canal made by the team of workers in 5 days = 5 × \(\frac{1}{8}\) km = \(\frac{5}{8}\) km
Hence, the team can make \(\frac{5}{8}\) km of the water canal in one week.

Question 2.
Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets \(\frac{5}{6}\) hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?
Solution:
There are 3 days from Monday to Thursday. Since the Moon sets \(\frac{5}{6}\) hours later than previous day, the number of hours the Moon will set later on Thursday than Monday = 3 × \(\frac{5}{6}\) hours = \(\frac{15}{6}\) hours = \(\frac{5}{2}\) hours.
We know that 1 hour = 60 minutes
⇒ \(\frac{5}{2}\) hours = \(\frac{5}{2}\) × 60 minutes = \(\frac{300}{2}\) minutes = 150 minutes
Now, 150 minutes = 120 minutes + 30 minutes = 2 hours 30 minutes
Hence, on Thursday the Moon will set 2 hours 30 minutes after 10 pm.

Question 3.
Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.
(i) \(\frac{2}{3}\) × \(\frac{4}{5}\)
(ii) \(\frac{1}{4}\) × \(\frac{2}{3}\)
Solution:
(i) Each row represents \(\frac{1}{5}\) and each column represents \(\frac{1}{3}\). The whole is divided into 5 rows and 3 columns creating 5 x 3 = 15 equal parts and 2 × 4 = 8 of the parts is double-shaded, that is \(\frac{8}{15}\) of the whole is double-shaded.
Therefore, \(\frac{2}{3}\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

(ii) Each row represents \(\frac{1}{3}\) and each column represents \(\frac{1}{4}\). The whole is divided into 3 rows and 4 columns, creating 3×4=12 equal parts and 2 of the parts are double-shaded, that is \(\frac{2}{12}\) of the whole is double-shaded.
Therefore, \(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Question 4.
A water tank is filled from a tap. If the tap is open for 1 hour, \(\frac{7}{10}\) of the tank gets filled. How much of the tank is filled if the tap is open for
(i) \(\frac{1}{3}\) hours
(ii) \(\frac{7}{10}\) hours
Solution:
In 1 hour, part of the tank gets filled = \(\frac{7}{10}\)
(i) In \(\frac{1}{3}\) hours, part oLthe tank gels filled = \(\frac{1}{3}\) × \(\frac{7}{10}\) = \(\frac{1 \times 7}{3 \times 10}\) = \(\frac{7}{30}\)
Therefore, in \(\frac{1}{3}\) hours, \(\frac{7}{30}\) of the tank gets filled.

(ii) In \(\frac{7}{10}\) hours, part of the tank gets filled = \(\frac{7}{10}\) × \(\frac{7}{10}\) = \(\frac{7 \times 7}{10 \times 10}\) = \(\frac{49}{100}\)
Therefore, in \(\frac{7}{10}\) hours, \(\frac{49}{100}\) of the tank gets filled.

Question 5.
Tsewang plants four saplings in a row in his garden. The distance between two saplings is \(\frac{3}{4}\) m. Find the distance between the first and last sapling.
Solution:

The distance between the first and the last sapling = \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = 3 × \(\frac{3}{4}\) = \(\frac{9}{4}\) m

Question 6.
Which is heavier: \(\frac{12}{15}\) of 500 grams or \(\frac{3}{20}\) of 4 kg?
Solution:
We have \(\frac{12}{15}\) of 500 g = \(\frac{12}{15}\) × 500 g = \(\frac{12 \times 500}{15}\) = 400 g
And \(\frac{3}{20}\) × 4000 g = \(\frac{3 \times 4000}{20}\) g = 600 g [∵ 1 kg = 1000 g]
∵ 600 g is heavier than 400 g.
∵ \(\frac{3}{20}\) of 4 kg is heavier than \(\frac{12}{15}\) of 500 grams.

Question 7.
(i) 3 ÷ \(\frac{7}{9}\)
(ii) \(\frac{14}{6}\) \(\frac{7}{3}\)
(iii) \(\frac{1}{6}\) ÷ \(\frac{11}{12}\)
(iv) 3\(\frac{2}{3}\) ÷ 1\(\frac{3}{8}\)
Solution:
(i) 3 ÷ \(\frac{7}{9}\) = 3 × \(\frac{9}{7}\) = \(\frac{9}{7}\) = \(\frac{3 \times 9}{7}\)
= 3\(\frac{6}{7}\)

(ii) \(\frac{14}{6}\) \(\frac{7}{3}\) = \(\frac{14}{6}\) × \(\frac{3}{7}\) = \(\frac{14 \times 3}{6 \times 7}\)
= \(\frac{42}{42}\) = 1

(iii) \(\frac{1}{6}\) ÷ \(\frac{11}{12}\) = \(\frac{1}{6}\) × \(\frac{12}{11}\) = \(\frac{1 \times 12}{6 \times 11}\)
= \(\frac{2}{11}\)

(iv) 3\(\frac{2}{3}\) ÷ 1\(\frac{3}{8}\) = \(\frac{11}{3}\) ÷ \(\frac{11}{8}\) = \(\frac{11}{3}\) × \(\frac{11}{8}\)
= \(\frac{11 \times 8}{3 \times 11}\)
= \(\frac{8}{3}\) = 2\(\frac{2}{3}\)

Question 8.
Patiganita a book written by Sridharacharya in the 9^th century CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together 1÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) and 1 ÷ \(\frac{1}{2}\). What should the friend say?
Solution:
1÷ \(\frac{1}{6}\) = 1 × 6 = 6, 1 ÷ \(\frac{1}{10}\) = 1 × 10 = 10, 1 ÷ \(\frac{1}{13}\) = 1 × 13 = 13, 1 ÷ \(\frac{1}{9}\) = 1 × 9 = 9 and 1 ÷ \(\frac{1}{2}\) = 1 × 2 = 2
Therefore, the sum obtained by adding together
1÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) and 1 ÷ \(\frac{1}{2}\)
= 6 + 10 + 13 + 9 + 2 = 40
Thus, the friend should say the ‘sum’ is 40.

Question 9.
Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5\(\frac{1}{6}\) hours to get there. If he takes a plane, it will take him \(\frac{1}{2}\) hour. How many hours does the plane save?
Solution:
The difference between the two durations = 5\(\frac{1}{6}\) – \(\frac{1}{2}\) = \(\frac{31}{6}\) – \(\frac{1}{2}\) [∵ 5\(\frac{1}{6}\) = \(\frac{31}{6}\)]
= \(\frac{31}{6}\) – \(\frac{3}{6}\) = \(\frac{31-3}{6}\) = \(\frac{28}{6}\)
= \(\frac{14}{3}\) = 4\(\frac{2}{3}\) hours
Hence, the plane saves 4\(\frac{2}{3}\) hours.

Question 10.
What fraction of the whole square is shaded?

Solution:
In the given figure, the big square is divided into 4 identical squares. So, one small square occupies \(\frac{1}{4}\) of the area of the big square. Now, consider the smaller square.
The small square in the figure is divided into 8 identical triangles in which 3 are shaded.

So, the shaded part is \(\frac{3}{8}\) of the small square.
But the small square is \(\frac{3}{8}\) of the big square.
∴ The shaded part is \(\frac{1}{4}\) × \(\frac{3}{8}\) = \(\frac{3}{32}\) of the big square.
Hence, \(\frac{3}{32}\) of the whole square is shaded.

Question 11.
A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the Fig. 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Solution:
At first point ants split in two ways. So, fraction of ants at each way is 1 ÷ \(\frac{1}{2}\) = \(\frac{1}{2}\).
At the second point. ants split in two ways.
So the fraction of ants at each war \(\frac{1}{2}\) ÷ 2 = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

At the third point, ants split in four ways.
So, fraction of ants at each way
= \(\frac{1}{4}\) ÷ 4 = \(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)
At the fourth point, ants split in 2 wars.
So fraction of ants at each way
= \(\frac{1}{16}\) ÷ 2 = \(\frac{1}{16}\) × \(\frac{1}{2}\) = \(\frac{1}{32}\)
Hence, fraction of ants at mango tree = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{16}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\)
= \(\frac{16}{32}\) + \(\frac{8}{32}\) + \(\frac{2}{32}\) + \(\frac{2}{32}\) + \(\frac{1}{32}\) = \(\frac{16+8+2+2+1}{32}\)
= \(\frac{29}{32}\)
Fraction of ants near sugarcane field = \(\frac{1}{32}\) + \(\frac{1}{16}\) = \(\frac{1}{32}\) + \(\frac{2}{32}\)
= \(\frac{1+2}{32}\) = \(\frac{3}{32}\)

Question 12.
What is (1 – \(\frac{1}{2}\))?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\) ?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) ?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\)) ?
Make a general statement and explain.
Solution:
1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\) = \(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\);
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) = \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{1}{5}\);
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\))
= \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) × \(\frac{4}{5}\) × \(\frac{5}{6}\) × \(\frac{6}{7}\) × \(\frac{7}{8}\) × \(\frac{8}{9}\) × \(\frac{9}{10}\) = \(\frac{1}{10}\)
Here, we observe that in this pattern, denominator of each term cancels the numerator of the next term, and found that the final product has the numerator of the first term and the denominator of the last term.
In general, (1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × ………… × (1 – \(\frac{1}{n}\)) = \(\frac{1}{n}\)

InText Questions

Question 1.
In each of the figures given below, find the fraction of the big square that the shaded region occupies.

Solution:
(i) In the figure. the smaller square in the up- right corner is divided into 4 smaller squares and each square within it is further divided into 2 triangles.

∴ Total number of triangles in whole square
= (4 × 2) × 4 = 32
Out of these 32 triangles. 12 triangular regions are shaded.
∴ Fraction of shaded region = \(\frac{12}{32}\) = \(\frac{3}{8}\)
Thus, the shaded region occupies \(\frac{3}{8}\) of the area of the whole square.

(ii) In the figure, the smaller square is divided into 4 triangles and a square, which can be divided into 4 triangles having the same area. Therefore, total triangle in 1 smaller square is 8.

∴ Total number of triangles in whole square = 8 × 4 = 32
Out of these 32 triangles, 2 triangular regions are shaded.
∴ Fraction of shaded region = \(\frac{2}{32}\) = \(\frac{1}{16}\)
Thus, the shaded region occupies \(\frac{1}{16}\) of the area of the whole square.

Working with Fractions Class 7 Extra Questions

Working with Fractions Class 7 Very Short Question Answer

Question 1.
A library has 2400 books, and \(\frac{5}{8}\) of them are placed on the ground floor. The rest are kept on the first floor. How many books are on the first floor?
Solution:
Given, total number of books in library = 2400
Number of books on ground floor
= \(\frac{5}{8}\) of total books = \(\frac{5}{8}\) × 2400 = 5 × 300 = 1500
Hence, number of books on the first floor
= 2400 – 1500 = 900.

Question 2.
A water tank can hold 1250 litres of water. If 2\(\frac{1}{5}\) of such tanks are filled, how much water is used in total?
Solution:
Given, capacity of one tank = 1250 litres
And, 2\(\frac{1}{5}\) of such tanks are filled.
∴ Total water used = 2\(\frac{1}{5}\) × capacitv of one tank
= \(\frac{11}{5}\) × capacity of one tank
= \(\frac{11}{5}\) × 125o = 11 × 250
= 2750 litres

Question 3.
Divide:
(i) 25 by \(\frac{1}{3}\)
(ii) 48 by 3\(\frac{3}{4}\)
Solution:
(i) 25 ÷ \(\frac{1}{3}\) = 25 × \(\frac{3}{1}\)
= \(\frac{25 \times 3}{1}\) = 75

(ii) 48 ÷ 3\(\frac{3}{4}\) = 48 ÷ \(\frac{15}{4}\)
= 48 × \(\frac{4}{15}\) [∵ 3\(\frac{3}{4}\) = \(\frac{15}{4}\)]
= \(\frac{16 \times 4}{5}\) = \(\frac{64}{5}\) = 12\(\frac{4}{5}\)

Question 4.
Solve the following:
(i) \(\frac{3}{4}\) ÷ \(\frac{2}{5}\)
(ii) 2\(\frac{1}{2}\) ÷ \(\frac{5}{8}\)
Solution:
(i) \(\frac{3}{4}\) ÷ \(\frac{2}{5}\) = \(\frac{3}{4}\) ÷ \(\frac{5}{2}\)
= \(\frac{3 \times 5}{4 \times 2}\) = \(\frac{15}{8}\)

(ii) 2\(\frac{1}{2}\) ÷ \(\frac{5}{8}\) = \(\frac{5}{2}\) ÷ \(\frac{5}{8}\)
= \(\frac{5}{2} \times \frac{8}{5}\) = \(\frac{5 \times 8}{2 \times 5}\) = 4

Question 5.
Solve:
(i) \(\frac{5}{6}\) ÷ 3
(ii) 1\(\frac{2}{3}\) ÷ 4
Solution:
(i) \(\frac{5}{6}\) ÷ 3 = \(\frac{5}{6}\) × \(\frac{1}{3}\) = \(\frac{5 \times 1}{6 \times 3}\) = \(\frac{5}{18}\)

(ii) 1\(\frac{2}{3}\) ÷ 4 = \(\frac{5}{3}\) ÷ 4 = \(\frac{5}{3}\) × \(\frac{1}{4}\) = \(\frac{5 \times 1}{3 \times 4}\) = \(\frac{5}{12}\) [∵ 1\(\frac{2}{3}\) = \(\frac{5}{3}\)]

Question 6.
During a community health program, each participant was provided \(\frac{2}{5}\) litres of clean water per day. If 40 litres were distributed on a particular day, calculate the number of participants who received the water.
Solution:
Quantity of water for I participant = \(\frac{2}{5}\) litres
Total quantity oÍ waler = 40 litres
Now, number of participants
= \(\frac{\text { Total quantity of water }}{\text { Quantity of water for } 1 \text { participant }}\)
= 40 ÷ \(\frac{2}{5}\) = 40 × \(\frac{5}{2}\) = 100

Question 7.
What number should be multiplied by 5\(\frac{3}{4}\) to get 3\(\frac{3}{5}\)?
Solution:
Let the required number be x. Then,
x × 5\(\frac{3}{4}\) = 3\(\frac{3}{5}\)
⇒ x = \(\frac{3 \frac{3}{5}}{5 \frac{3}{4}}\)
⇒ x = \(\frac{\frac{18}{5}}{\frac{23}{4}}\) = \(\frac{15}{5}\) × \(\frac{4}{23}\) = \(\frac{18 \times 4}{5 \times 23}\)
= \(\frac{72}{115}\)

Question 8.
Each guest is served 1\(\frac{3}{5}\) litres of juice. If 40 litres of juice is available, how many guests can be served?
Solution:
Given, juice served to each guest.
= 1\(\frac{3}{5}\) litres = \(\frac{8}{5}\)litres
Total quantity of juice available = 40 litres
Now, number of guests that can be served
= \(\frac{\text { Total quantity of juice available }}{\text { Juice served to each guest }}\)
= \(\frac{40}{\frac{8}{5}}\) = 40 × \(\frac{5}{8}\) = 5 × 5 = 25

Working with Fractions Class 7 Short Question Answer

Question 1.
Multiply the following fractions and express as mixed fraction:
(i) 1\(\frac{3}{4}\) × 2\(\frac{2}{5}\)
(ii) 4\(\frac{1}{3}\) × \(\frac{3}{7}\)
(iii) 5\(\frac{2}{9}\) × 1\(\frac{1}{6}\)
(iv) 3\(\frac{4}{5}\) × 2\(\frac{3}{8}\)
Solution:
(i) 1\(\frac{3}{4}\) × 2\(\frac{2}{5}\) = \(\frac{7}{4}\) × \(\frac{12}{5}\)
= \(\frac{7 \times 3}{1 \times 5}\) = \(\frac{21}{5}\)
= 4\(\frac{1}{5}\)

(ii) 4\(\frac{1}{3}\) × \(\frac{3}{7}\) = \(\frac{13}{3}\) × \(\frac{3}{7}\)
= \(\frac{13 \times 1}{1 \times 7}\) = \(\frac{13}{7}\)
= 1\(\frac{6}{7}\)

(iii) 5\(\frac{2}{9}\) × 1\(\frac{1}{6}\) = \(\frac{47}{9}\) × \(\frac{7}{6}\)
= \(\frac{47 \times 7}{9 \times 6}\) = \(\frac{329}{54}\)
= 6\(\frac{5}{54}\)

(iv) 3\(\frac{4}{5}\) × 2\(\frac{3}{8}\) = \(\frac{19}{5}\) × \(\frac{19}{8}\)
= \(\frac{19 \times 19}{5 \times 8}\) = \(\frac{361}{40}\)
= 9\(\frac{1}{4}\)

Question 2.
Meena invited \(\frac{2}{7}\) of her students to a weekend workshop. If \(\frac{3}{5}\) of the invited students were girls, how many boys were present at the workshop if her class has 105 students?
Solution:
Given, total students in the class = 105
∴ Nunber of students invited
= \(\frac{2}{7}\) × Total students =
= \(\frac{2}{7}\) × 105 = 2 × 15 = 30
Now, girls among invited students
= \(\frac{3}{5}\) of the invited students
= \(\frac{3}{5}\) × 30 = 3 × 6 = 18
Hence, number of boys present at the workshop
= 30 – 18 = 12.

Question 3.
A library donated \(\frac{3}{8}\) of its books to a village school. If \(\frac{2}{3}\) of the donated books were story books, how many non-story books were donated to the school if the library originally had 1280 books?
Solution:
Given, total number of books in library = 1280
Number of books donated = \(\frac{3}{8}\) of total books
= \(\frac{3}{8}\) × 1280 = 3 × 160 = 480
Story books among donated books = \(\frac{2}{3}\) × of donated books = \(\frac{2}{3}\) × 480 = 2 × 160 = 320
Hence, number of non-story books among donated books = 480 – 320 = 160

Question 4.
The area of a rectangular field is 63\(\frac{3}{5}\) m². If its length is 7\(\frac{1}{2}\)m, find its breadth?
Solution:
Given area of rectangular field = 63\(\frac{3}{5}\) = \(\frac{318}{5}\) m²
length of rectangular field = 7\(\frac{1}{2}\) = \(\frac{15}{2}\) m
We know.
Area of rectangle length × Breadth
⇒ Breadth = \(\frac{\text { Area of rectangle }}{\text { Length }}\) = \(\frac{318}{5}\) ÷ \(\frac{15}{2}\)
= \(\frac{318}{5}\) × \(\frac{2}{15}\) = \(\frac{106 \times 2}{5 \times 5}\)
= \(\frac{212}{25}\) = 8\(\frac{12}{25}\) m

Question 5.
Determine the number by which 5\(\frac{4}{7}\) must be multiplied to obtain 4\(\frac{3}{5}\).
Solution:
We have. 5\(\frac{4}{7}\) = \(\frac{39}{7}\) and 4\(\frac{3}{5}\) = \(\frac{23}{5}\)
Let the number to be found b x. Then,
\(\frac{39}{7}\) × x = \(\frac{23}{5}\)
Multiplying by the reciprocal of \(\frac{39}{7}\) on both sides, we get
\(\frac{7}{39}\) × \(\frac{39}{7}\) × x = \(\frac{7}{39}\) × \(\frac{23}{5}\) [∵ Reciprocal of \(\frac{39}{7}\) is \(\frac{7}{39}\)]
⇒ x = \(\frac{7 \times 23}{39 \times 5}\) = \(\frac{161}{195}\)

Question 6.
A bookstore sells journals at ₹ 6\(\frac{1}{2}\) per copy. If the shop’s revenue from journal sales totalled ₹ 975, how many dozens of journals were sold?
Solution:
Given, total revenue from journal sales = ₹ 975
Price of 1 journal = ₹ 6\(\frac{1}{2}\) = ₹ \(\frac{13}{2}\)
Now, number of journals sold
= \(\frac{\text { Total revenue from journal sales }}{\text { Price of } 1 \text { journal }}\)
⇒ Number of journals sold = \(\frac{975}{\frac{13}{2}}\) = 975 × \(\frac{2}{13}\)
= \(\frac{75 \times 2}{1}\) = 150
Now, the number of journals sold (in dozens)
= \(\frac{150}{12}\) = \(\frac{25}{2}\) = 12\(\frac{1}{2}\) [∵ 1 dozen = 12]
Thus, 12\(\frac{1}{2}\) dozens of journals were sold.

Question 7.
Divide:
(i) 36 by \(\frac{3}{8}\)
(ii) 54 by 2\(\frac{2}{5}\)
(iii) 42 by \(\frac{7}{6}\)
Solution:
(i) 36 ÷ \(\frac{3}{8}\) = 36 × \(\frac{8}{3}\) = 12 × 8 = 96

(ii) 54 ÷ 2\(\frac{2}{5}\)
= 54 ÷ \(\frac{12}{5}\) = 54 × \(\frac{5}{12}\)
= \(\frac{9 \times 5}{2}\) = \(\frac{45}{2}\) = 22\(\frac{1}{2}\) [∵ 2\(\frac{2}{5}\) = \(\frac{12}{5}\)]

(iii) 42 ÷ \(\frac{2}{5}\) = 42 × \(\frac{6}{7}\)
= 6 × 6 = 36

Question 8.
How many square tiles with a side length of \(\frac{1}{2}\) m are required to cover an area of 12\(\frac{1}{4}\) m²?
Solution:
Given, the side length of a square tile is \(\frac{1}{2}\) m.
Therefore, area of one tile = Side × Side
= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\) m²
Area to be covered by the tiles = 12\(\frac{1}{4}\) m² = \(\frac{49}{4}\) m²
Now, number of required tiles = Total area that need to be covered ÷ Area of one tile
= \(\frac{49}{4}\) ÷ \(\frac{1}{4}\) = \(\frac{49}{4}\) × \(\frac{4}{1}\) = 49
Thus, the number of required tiles is 49.

Question 9.
A fruit seller earns ₹ 4\(\frac{1}{2}\) per apple. He earns ₹864. How many dozens of apples did he sell?
Solution:
Given, total earnings = ₹864
Amount earned per apple = ₹4\(\frac{1}{2}\) = ₹\(\frac{9}{2}\)
Now, number of apples sold = \(=\frac{\text { Total earnings }}{\text { Amount earned per apple}}\)
= \(\frac{864}{\frac{9}{2}}\) = 864 × \(\frac{2}{9}\) = 96 × 2 = 192
So, number of apples sold (in dozens)
= \(\frac{192}{12}\) = 16 [∵ 1 dozen = 12]

Working with Fractions Class 7 Long Question Answer

Question 1.
Fill in the boxes using <, > or = without actually finding the product:
(i) \(\frac{3}{8}\) × \(\frac{2}{7}\) ☐ \(\frac{2}{7}\)
(ii) \(\frac{5}{9}\) × \(\frac{4}{11}\) ☐ \(\frac{5}{9}\)
(iii) 4\(\frac{1}{2}\) × 3\(\frac{3}{4}\) ☐ 4\(\frac{1}{2}\)
(iv) \(\frac{6}{13}\) × \(\frac{15}{4}\) ☐ \(\frac{15}{4}\)
Solution:
(i) We know, the product of two proper fractions is less than each of them.
Here, \(\frac{3}{8}\) and \(\frac{2}{7}\) are proper tractions.
∴ \(\frac{3}{8}\) × \(\frac{2}{7}\) < \(\frac{2}{7}\)

(ii) We know, the product of two proper fractions is less than each of them.
Here. \(\frac{5}{9}\) and \(\frac{4}{11}\) are proper fractions.
∴ \(\frac{5}{9}\) × \(\frac{4}{11}\) < \(\frac{5}{9}\)

(iii) We know, the product of two improper fractions is greater than both the fractions or equal to either of them.
Here, 4\(\frac{1}{2}\) and 3\(\frac{3}{4}\) are mixed tractions and mixed tractions can he converted into improper fractions.
∴ 4\(\frac{1}{2}\) × 3\(\frac{3}{4}\) > 4\(\frac{1}{2}\)

(iv) We know, the product of a proper fraction and an improper fraction (greater than 1) lies between both the fractions.
Here. \(\frac{6}{13}\) is a proper traction and \(\frac{15}{4}\) is an improper traction (greater than 1).
∴ \(\frac{6}{13}\) × \(\frac{15}{4}\) < \(\frac{15}{4}\)

Question 2.
Solve the following and write the result in simplest form:
(i) 8 × \(\frac{5}{6}\)
(ii) 4 × \(\frac{2}{5}\)
(iii) 7 × \(\frac{3}{8}\)
(iv) 5 × \(\frac{4}{3}\)
(v) \(\frac{11}{4}\) × 3
(vi) 10 × \(\frac{1}{5}\)
Solution:
(i) 8 × \(\frac{5}{6}\) = \(\frac{48 \times 5}{6}\) = \(\frac{4 \times 5}{3}\)
= \(\frac{20}{3}\)

(ii) 4 × \(\frac{2}{5}\) = \(\frac{4 \times 2}{5}\)
= \(\frac{8}{5}\)

(iii) 7 × \(\frac{3}{8}\) = \(\frac{7 \times 3}{8}\)
= \(\frac{21}{8}\)

(iv) 5 × \(\frac{4}{3}\) = \(\frac{5 \times 4}{3}\)
= \(\frac{20}{3}\)

(v) \(\frac{11}{4}\) × 3 = \(\frac{11 \times 3}{4}\)
= \(\frac{33}{4}\)

(vi) 10 × \(\frac{1}{5}\) = 2 × 1 = 2

Question 3.
Solve the following and express as a mixed fraction:
(i) 3\(\frac{5}{8}\) × 2\(\frac{1}{4}\)
(ii) 7\(\frac{2}{3}\) × 4\(\frac{3}{5}\)
(iii) 5\(\frac{3}{10}\) × 3\(\frac{4}{7}\)
(iv) \(\frac{4}{9}\) × 6\(\frac{2}{5}\)
(v) \(\frac{7}{10}\) × 9\(\frac{1}{6}\)
(vi) \(\frac{3}{8}\) × 8\(\frac{5}{9}\)
Solution:
(i) 3\(\frac{5}{8}\) × 2\(\frac{1}{4}\) = \(\frac{29}{8}\) × \(\frac{9}{4}\) = \(\frac{29 \times 9}{8 \times 4}\)
= \(\frac{261}{32}\) = 8\(\frac{5}{32}\)

(ii) 7\(\frac{2}{3}\) × 4\(\frac{3}{5}\) = \(\frac{23}{3}\) × \(\frac{23}{5}\) = \(\frac{23 \times 23}{3 \times 5}\)
= \(\frac{529}{15}\) = 35\(\frac{4}{15}\)

(iii) 5\(\frac{3}{10}\) × 3\(\frac{4}{7}\) = \(\frac{53}{10}\) × \(\frac{25}{7}\)
= \(\frac{265}{14}\) = 18\(\frac{13}{14}\)

(iv) \(\frac{4}{9}\) × 6\(\frac{2}{5}\) = \(\frac{4}{9}\) × \(\frac{32}{5}\) = \(\frac{4 \times 32}{9 \times 5}\)
= \(\frac{128}{45}\) = 2\(\frac{38}{45}\)

(v) \(\frac{7}{10}\) × 9\(\frac{1}{6}\) = \(\frac{7}{10}\) × \(\frac{55}{6}\)
= \(\frac{77}{12}\) = 6\(\frac{5}{12}\)

(vi) \(\frac{3}{8}\) × 8\(\frac{5}{9}\) = \(\frac{3}{8}\) × \(\frac{77}{9}\) = \(\frac{1 \times 77}{8 \times 3}\)
= \(\frac{77}{24}\) = 3\(\frac{5}{24}\)

Question 4.
A car travels 5\(\frac{1}{4}\) km north, then \(\frac{1}{2}\) km west and 4\(\frac{3}{8}\) km north. Find total distance travelled (in km)?
What fraction of the journey was travelled in the north direction?
Solution:
Given: Distance travelled in north direction
d₁ = 5\(\frac{1}{4}\) km = \(\frac{21}{4}\) km
Distance travelled in west direction, d₂ = \(\frac{1}{2}\) km
Distance travelled in north direction again.
d₃ = 4\(\frac{3}{8}\)km = \(\frac{35}{8}\) km
∴ Total distance travelled = d₁ + d₂ + d₃
= \(\frac{21}{4}\) + \(\frac{1}{2}\) + \(\frac{35}{8}\) = \(\frac{42}{8}\) + \(\frac{4}{8}\) + \(\frac{35}{8}\)
= \(\frac{42+4+35}{8}\) = \(\frac{81}{8}\) = 10\(\frac{1}{8}\) km
Total distance travelled in north direction
= d₁ + d₃ = \(\frac{21}{4}\) + \(\frac{35}{8}\) = \(\frac{42+35}{8}\) = \(\frac{77}{8}\) km
∴ Fraction of the journey travelled in the north direction
= \(\frac{d_1+d_3}{d_1+d_2+d_3}\) = \(\frac{\frac{77}{8}}{\frac{81}{8}}\)
= \(\frac{77}{8}\) × \(\frac{8}{81}\) = \(\frac{77}{81}\)

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 2 Arithmetic Expressions Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 2 Arithmetic Expressions Solutions

Ganita Prakash Class 7 Chapter 2 Solutions

Class 7 Maths Ganita Prakash Chapter 2 Solutions Arithmetic Expressions

Question 1.
Fill in the blanks to make the expressions equal on both sides of the ‘=’ sign:
(i) 13 + 4 = _____ + 6
(ii) 22 + = 6 × 5
(iii) 8 × = 64 4 2
(iv) 34 – _____ = 25
Solution:
(i) 13 + 4 = 11 + 6 [∵ 13 + 4 = 17 and 11 + 6 = 17]
(ii) 22 + 8 = 6 × 5 [∵ 6 × 5 = 30 and 22 + 8 = 30]
(iii) 8 × 4 = 64 ÷ 2 [∵ 64 ÷ 2 = 32 and 8 × 4 = 32]
(iv) 34 – 9 = 25

Question 2.
Arrange the following expressions in ascending (increasing) order of their values.
(i) 67 – 19
(ii) 67 – 20
(iii) 35 × 25
(iv) 5 × 11
(v) 120 ÷ 3
Solution:
(i) 67 – 19 = 48
(ii) 67 – 20 = 47
(iii) 35 + 25 = 60
(iv) 5 × 11 = 55
(v) 120 ÷ 3 = 40
Clearly, 40 < 47 < 48 < 55 < 60
∴ 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 4 25
Hence, (v) < (ii) < (i) < (iv) < (iii).

Question 3.
For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(i) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
(ii) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(a) for four adults and three children?
(b) for two groups having three adults each?
(iii) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.

Solution:
(i) Number of gold coins Princess Elsa got = 100
Number of gold coins Princess Anna got = 100
Princess Elsa used the coins to start the business and double her coins.
So, the number of coins Princess Elsa has = 2 × 100
Princess Anna bought jewellery and has only half of the coins left.
So, the total number of coins Princess Anna has = \(\frac{100}{2}\)
Therefore, the total number of gold coins Princess Elsa and Princess Anna have together
= 2 × 100 + \(\frac{100}{2}\) = 200 + 50 = 250
Thus, the expression describing the above situation is, 2 × 100 + \(\frac{100}{2}\).
Terms: 2 × 100, \(\frac{100}{2}\)

(ii) (a) Fare of metro train ticket for an adult = ₹40
So, the fare of metro train ticket for four adults (in ₹) = 4 × 40
Fare of metro train ticket for a child = ₹20
So, the fare of metro train ticket for three children (in ₹) = 3 × 20
Therefore, the expression describing the total cost of tickets (in ₹) for four adults and three children is 4 × 40 + 3 × 20.
Total fare = 4 × ₹40 + 3 × ₹20 = ₹ 160 + ₹60 = ₹220
Terms: 4 × 40, 3 × 20

(b) Fare of metro train ticket for an adult = ₹40
So, the fare of metro train ticket for a group of three adults (in ₹) = 3 × 40
Therefore, the expression describing the total cost of tickets (in ₹) for the two groups having three adults each is 2 × (3 × 40).
Total fare = 2 × (3 × ₹ 40) = 2 × ₹120 = ₹240
Terms: 2 × (3 × 40)

(iii) By observing the given picture, the total height of the window
= Number of gaps × 5 cm + Number of grills × 2 cm + Number of borders × 3 cm
Here, total number of gaps = 7; Total number of grills = 6; Total number of borders = 2
∴ Total height of window (in cm) = 7 × 5 + 6 × 2 + 2 × 3 = 35 + 12 + 6 = 47 + 6 = 53
Terms: 7 × 5, 6 × 2, 2 ×3

Question 4.
Remove the brackets and write the expression having the same value.
(i) 14 + (12 + 10)
(ii) 14 – (12 + 10)
(iii) 14 + (12 – 10)
(iv) 14 – (12 – 10)
(v) -14 + 12 – 10
(vi) 14 – (-12 – 10)
Solution:
(i) 14 + (12 + 10) = 14 + 12 + 10 = 14 + 22 = 36
(ii) 14 – (12 + 10) = 14 – 12 – 10 = 14 – 22 = -8
(iii) 14 + (12 – 10) = 14 + 12 – 10 = 14 + 2 = 16
(iv) 14 – (12 – 10) = 14 – 12 + 10 = 14 – 2 = 12
(v) – 14 + 12 – 10 = – 14 + 2 = – 12
(vi) 14 – (-12 – 10) = 14 + 12 + 10 = 14 + 22 = 36
Here, expressions given in (i) and (vi) have same value.

Question 5.
Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(i) (6 + 10) – 2 and 6 + (10 – 2)
(ii) 16 – (8 – 3) and (16 – 8) – 3
(iii) 27 – (18 + 4) and 27 + (-18 – 4)
Solution:
(i) (6 + 10) – 2 = 16 – 2 = 14 and 6 + (10 – 2) = 6 + 8 = 14
Clearly, (6 + 10) – 2 = 6 + (10 – 2)
Hence, both the expressions have the same value.

(ii) 16 – (8 – 3) = 16 – 5 = 11 and (16 – 8) – 3 = 8 – 3 = 5
Clearly, 16 – (8 – 3) ≠ (16 – 8) – 3
Hence, both the expressions do not have the same value.

(iii) 27 – (18 + 4) = 27 – 22 = 5 and 27 + (-18 – 4) = 27 + (- 22) = 27 – 22 = 5
Clearly, 27 – (18 + 4) = 27 + (-18 – 4)
Hence, both the expressions have the same value.

Question 6.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(i) 34 – 9 + 12 = 13
(ii) 56 – 14 – 8 = 34
(iii) – 22 – 12 + 10 + 22 = – 22
Solution:
(i) 34 – (9 + 12) = 34 – 21 = 13
(ii) (56 – 14) – 8 = 42 – 8 = 34
(iii) – 22 – (12 + 10) + 22 = – 22 – 22 + 22 = – 22

Question 7.
Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

Solution:
For I : 5 × 4 + 4 × 8 = 20 + 32 = 52 or 4 × (4 + 8) + 4 = 4 × 12 + 4 = 52
For II : 8 × (5 + 6) = 8 × 11 = 88 or 8 × 5 + 8 × 6 = 40 + 48 = 88

Question 8.
Read the situations given below. Write appropriate expressions for each of them and find their values.
(i) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
(ii) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
(iii) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?
Solution:
(i) Amount of mangoes supplied by Rahim each day = 9 kg
Amount of mangoes supplied by Shyam each day = 11 kg
Total supplies of mangoes in the market on each day = (9 + 11) kg
∴ Total supplies of mangoes in the market in a week (7 days) = 7 × (9 + 11) = 7 × 20 = 140 kg

(ii) Binu’s per month earning = ₹20,000
Binu’s total monthly expenditures
= ₹5,000 on rent + ₹5,000 on food + ₹ 2,000 on other expenses
= ₹(5,000 + 5,000 + 2,000)
Therefore, Binu’s monthly savings = ₹20,000 – ₹(5,000 + 5,000 + 2,000) = ₹20,000 – ₹12,000 = ₹8,000
Thus, Binu’s total yearly savings = 12 × 8000 = ₹96000
Hence, Binu will save ₹96000 by the end of the year.

(iii) Since the snail climbs 3 cm up the post in daytime and slips down by 2 cm at night.
The distance climbed by the snail in a day = 3 – 2 = 1 cm
∴ The distance climbed in 7 days = 7 cm
The height of the post is 10 cm.
The distance climbed on the 8th day before slipping = 7 + 3=10 cm
So, the snail will take 8 days to reach the top of the post and get the delicious treat.

Question 9.
Find different ways of evaluating the following expressions:
(i) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(ii) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Solution:
(i) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 + 3 + 5 + 7 + 9) + (-2 – 4 – 6 – 8 – 10) = 25 + (- 30) = – 5
OR
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)
= (-1) + (-1) + (-1) + (-1) + (-1) = -5

(ii) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)
= 0 + 0 + 0 + 0 + 0 = 0
OR
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1= (1 + 1 + 1 + 1 + 1) + (- 1 – 1 – 1 – 1 – 1)
= 5 + (- 5) = 0

Question 10.
Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
(i) 83 – 37 – 12
(a) 84 – 38 – 12
(b) 84 – (37 + 12)
(c) 83 – 38 – 13
(d) -37 + 83 – 12
(ii) 93 + 37 × 44 + 76
(a) 37 + 93 × 44 + 76
(b) 93 + 37 x× 76 + 44
(c) (93 + 37) × (44 + 76)
(d) 37 × 44 + 93 + 76
Solution:
(i) 83 – 37 – 12 = 83 – 37 – 12 + (1 – 1) = (83 + 1) – 37 – 1 – 12 = 84 – 38 – 12
Also, 83 – 37 – 12 = – 37 + 83 – 12
Hence, (a) and (d) are equal to the given expression 83 – 37 – 12.

(ii) 93 + 37 × 44 + 76
Rearranging the terms, we get 37 × 44 + 93 + 76, which is equal to the given expression in option (d). Hence, (d) is equal to the given expression 93 + 37 × 44 + 76.

InText Questions

Question 1.
Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.

Solution:
(i)
∴ 245 + 289 > 246 + 285

(ii)
∴ 273 – 145 = 272 – 144

(iii)
∴ 364 + 587 < 363 + 589

(iv)
∴ 142 + 245 < 129 + 245

(v)
∴ 213 – 77 < 214 – 76

Question 2.
Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Solution:
In the token model:
Subtracting a positive number (e.g. subtracting 3) means removing 3 positive tokens.
Adding a negative number (e.g. adding- 3) meaning adding 3 negative tokens. These 3 negative tokens cancel out 3 existing positive tokens (by forming zero pairs), which is equivalent to removing 3 positive tokens.
Since both actions result in removing the same number of positive tokens, the final value is the same.

Question 3.
Does adding the terms of an expression in any order give the same value? Take some expressions and check. Consider expressions with more than 3 terms also.
Solution:
Yes
(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (- 5) + (-4) + (-5)
= (-9) + (-5) = -14

(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (-2) + (-7) + (-5)
= (-2) + (-12) = -14

(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-9)
= (-5) + (- 9) = -14
Note: Students can do on their own by taking different numbers.

Question 4.
5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?
Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?
Solution:
Expression 5 × 4 + 3 means 3 more than 5 × 4, which is equal to 23, but 5 × (4 + 3) means 5 times the sum of 3 and 4 which is equal to 35.
Hence, 5 × 4 + 3 + 5 ×(4 + 3)
Now, 5 × (4 + 3), 5 × (3 + 4), and (3 + 4) × 5 have the same meaning, which is 5 times the sum of 3 and 4 and give the same value.
Hence, 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5

Question 5.
Use distributive property to find the following products:
(i) 95 × 8
(ii) 104 × 15
(iii) 49 × 50
Is this quicker than the multiplication procedure you use generally?
Solution:
(i) 95 × 8
= (100 – 5) × 8
= (100 × 8) – (5 × 8)
= 800 – 40 = 760

(ii) 104 × 15
= (100 + 4) × 15
= (100 × 15) + (4 × 15)
= 1500 + 60 = 1560

(iii) 49 × 50
= (50 – 1) × 50
= (50 × 50) – (50 × 1)
= 2500 – 50 = 2450
Yes, this procedure is quicker than the general multiplication procedure.

Arithmetic Expressions Class 7 Extra Questions

Arithmetic Expressions Class 7 Very Short Question Answer

Question 1.
Riya buys 5 notebooks per day for 4 days and 7 notebooks per day for the remaining 3 days of a week. Form an expression to represent the total number of notebooks she buys in that week.
Solution:
For 4 days: Riya buys 5 notebooks per day
For 3 days: Riya buys 7 notebooks per day
Thus, expression for total number of notebooks
= 4 × 5 + 3 × 7

Question 2.
Compare which is greater: 79 – 96 or 117 – 130.
Solution:
The value of 79 – 96 is – 17.
The value of 117 – 130 is – 13.
Clearly, – 17 < -13
Hence, 117 – 130 is greater than 79 – 96.

Question 3.
Anaya is preparing for a temple festival. She decorates 5 pillars, placing 7 marigold garlands on each. However, 2 garlands fall and cannot be used. Find the total number of garlands finally used.
Solution:
Total garlands before any fall = 5 × 7 = 35
Garlands that cannot be used = 2
Now, total garlands finally used = 35 – 2 = 33

Question 4.
Priya buys 2 packs of biscuits for ₹15 each and a juice for ₹20. What is the total cost that Priya needs to pay?
Solution:
Given, Priya buys 2 packs of biscuits for ₹15 each and a juice for ₹20.
Cost of biscuits = 2 × ₹15
Cost of juice = ₹ 20
Now, total cost = 2 × ₹ 15 + ₹20 = ₹30 + ₹20
= ₹50

Question 5.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(i) 25 – 7 + 12 = 6
(ii) 42 – 15 – 7-9 = 29
(iii) – 32 – 18 + 14 + 32 = – 32
(iv) 35 – 18 – 5 + 10 = 32
Solution:
(i) 25 – (7 + 12) = 6
(ii) 42 – 15 – (7 – 9) = 29
(iii) – 32 – (18 + 14) + 32 = – 32
(iv) 35 – (18 – 5) + 10 = 32

Arithmetic Expressions Class 7 Short Question Answer

Question 1.
Ankit is planning a birthday party. He buys:

• 5 party hats, each costing ₹ 60
• 2 big balloons, each costing ₹90
• If the total cost exceeds ₹400, he receives a discount of ₹50.

Write the expression that shows the amount (in ₹) Ankit has to pay.
Solution:
Given, cost of five party hats = 5 × ₹ 60
Cost of two big balloons = 2 × ₹90
Discount = ₹50 [If total cost > ₹400]
Now, total cost (in ₹) = 5 × 60 + 2 × 90 = 300 + 180 = 480
As 480 > 400, Ankit gets a discount of ₹50.
So, the expression that shows the amount (in ₹) Ankit has to pay (after applying the discount) is
5 × 60 + 2 × 90 – 50.

Question 2.
Simplify:
(i) (-40) × (-1) + 28 ÷ 7
(ii) 7 – [13 – 2{4 × (- 4)}]
(iii) 81 × [59 – {7 × 8 + (13 – 2 × 5)}]
Solution:
(i) (-40) × (- 1) + 28 ÷ 7
= (40 × 1) + 28 ÷ 7
= 40 + 4 = 44 [∵ (-) × (-) = ( + )]

(ii) 7 – [13 – 2{4 × (- 4)}]
= 7 – [13 – 2 × {-16}] [∵ (+) × (-) = (-)]
= 7 – [13 + 32] = 7 – 45 = -38

(iii) 81 × [59 – {7 × 8 + (13 – 2 × 5)}]
= 81 × [59 – {7 × 8 + (13 – 10)}]
= 81 × [59 – {56 + 3}]
= 81 × [59 – 59] = 81 × 0 = 0

Question 3.
Find the following products using the distributive property.
(i) 107 × 12
(ii) 98 × 14
Solution:
(i) Let’s break 107 as 100 + 7.
Now, 107 × 12 = (100 + 7) × 12
= 100 × 12 + 7 × 12 [Using distributive property]
= 1200 + 84 = 1284
Hence, 107 × 12 = 1284

(ii) Let’s break 98 as 100 – 2.
Now, 98 × 14 = (100 – 2) × 14
= 100 × 14 – 2 × 14
= 1400 – 28 = 1372
Hence, 98 × 14 = 1372

Question 4.
Simplify: 659 – [219 – (750 — 255 ÷ 5 × 9)]
Solution:
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
Thus, 659 – [219 – (750 – 255 ÷ 5 × 9)]
= 659 – 219 + (750 – 255 ÷ 5 × 9)
= 659 – 219 + (750 – 51 × 9) [As 255 ÷ 5 = 51]
= 659 – 219 + (750 – 459) [As 51 × 9 = 459]
= 1409-678
[As 659 + 750 = 1409 and – 219 – 459 = – 678]
= 731
Hence, 659 – [219 – (750 – 255 ÷ 5 × 9)] = 731

Question 5.
Simplify:
63 – (- 3){- 2 – 8 – 3} ÷ {5 + (- 2)(- 1)}
Solution:
63 – (-3){-2 – 8 – 3} ÷ {5 + (-2)(-1)}
= 63 – (- 3){-2 – 5} ÷ {5 + (- 2)(- 1)}
[Removal of bar]
= 63 + 3{-2 – 5} ÷ {5 + 2}
[As – (-3) = 3 and (-2) (-1) = 2]
= 63 + 3{-7} ÷ 7
= 63 – 21 ÷ 7[As 3(-7) = -21]
= 63 – 3 = 60[As 21 ÷ 7 = 3]

Question 6.
Ravi took part in a painting competition and got scores 34, 37 and 31 from three judges. Later, it was found that second judge had mistakenly given 37 instead of the correct score 35. What are Ravi’s initial and updated total scores?
Solution:
Given, Ravi got scores 34, 37 and 31 from three judges.
So, the initial total score = 34 + 37 + 31 = 102
Later, the second judge’s score was corrected from 37 to 35, which is 2 less than the initial score.
Since only the second score changed, the updated total score = 102 – 2 = 100
Hence, Ravi’s updated total score is 100.

Question 7.
Remove the brackets and write the expression having the same value.
(i) 18 + (12 + 6)
(ii) 18 + (12 – 6)
(iii) 18 – (12 + 6)
(iv) 18 – (12 – 6)
(v) 18 – (-12 – 6)
(vi) 18 – (-12 + 6)
Solution:
On removing the brackets preceded by a plus sign, the signs of all the terms inside the brackets remain same.
(i) 18 + (12 + 6) = 18 + 12 + 6
(ii) 18 + (12 – 6) = 18 + 12 – 6
On removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
(iii) 18 – (12 + 6) = 18 – 12 – 6
(iv) 18 – (12 – 6) = 18 – 12 + 6
(v) 18 – (-12 – 6) = 18 + 12 + 6
(vi) 18 – (-12 + 6) = 18 + 12 – 6

Question 8.
Leela is organizing candles for a festival. She places 36 candles in one box and 27 in another. She gives away 8 candles from the second box to her neighbour.
Write an expression for the number of candles Leela is left with.
Solution:
Candles in the first box = 36
Candles in the second box = 27
Candles given to neighbour = 8
Now, the number of candles left with Leela
= Candles in the first box + Candles in the second box – Candles given to neighbour
= 36 + 27 – 8, which is the required expression.

Question 9.
In three sections of a school library, 64, 93 and 81 books were recorded respectively.
Later, the librarian found that there were mistakes in two sections:

• The first section actually had 84 books, not 64.
• The second section actually had 73 books, not 93.

What is the difference between total number of books initially and after correcting the record?
Solution:
The error in the number of books in the first section was + 20 (i.e., 84 – 64 = +20).
The error in the second section was -20
(i.e., 73 – 93 = – 20).
Since these two errors cancel each other out, the overall total remains unchanged.
Therefore, there is no difference in the total number of books.

Arithmetic Expressions Class 7 Long Question Answer

Question 1.
Find the following products using the distributive property.
(i) 125 × 16
(ii) 87 × 13
(iii) 106 × 104
(iv) 107 × 91
Solution:
(i) Let’s break 125 as 100 + 25.
Now, 125 × 16 = (100 + 25) × 16
= 100 × 16 + 25 × 16 = 1600 + 400 = 2000
Hence, 125 × 16 = 2000

(ii) Let’s break 87 as 100 -13.
Now, 87 × 13 = (100 – 13) × 13
= 100 × 13 – 13 × 13 = 1300 – 169 = 1131
Hence, 87 × 13 = 1131

(iii) Let’s break 106 as 100 + 6.
Now, 106 × 104 = (100 + 6) × 104
= 100 × 104 + 6 × 104 = 10400 + 6 × (100 + 4)
[As 104 = 100 + 4]
= 10400 + 6 × 100 + 6 × 4
= 10400 + 600 + 24 = 11024
Hence, 106 × 104 = 11024

(iv) Let’s break 107 as 100 + 7.
Now, 107 × 91 = (100 + 7) × 91
= 100 × 91 + 7 × 91 = 9100 + 7 × (100 – 9)
[As 91 = 100 – 9]
= 9100 + 7 × 100 – 7 × 9
= 9100 + 700 – 63
= 9800 – 63 = 9737
Hence, 107 × 91 = 9737.

Question 2.
Simplify:
(i) 121 ÷ [17 – {15 – 3(7 – 4)}]
(ii) 32 ÷ [32 + {32 – (32 + 32 – 32 )}]
(iii) 15 – (-3) × [{4- 7 – 3} + 3 × {5 + (-3) × (-6)}]
Solution:
(i) 121 ÷ [17 – {15 – 3(7 – 4)}]
= 121 ÷ [17 – {15 – 3 × 3}]
= 121 ÷ [17 – {15 – 9}]
= 121 ÷ [17 – 6]
= 121 ÷ 11 = 11

(ii) 32 ÷ [32 + {32 – (32 + 32 – 32 )}]
= 32 ÷ [32 + {32 – (32 + 0)}]
= 32 ÷ [32 + {32 – 32}]
= 32 ÷ [32 + 0] = 32 + 32 = 1

(iii) 15 – (-3) × [{4 – 7 – 3 } ÷ 3 × {5 + (-3) × (-6)}]
= 15 + 3 × [{4 – 4} ÷ 3 × {5 + 18}]
= 15 + 3 × [0 ÷ 3 × 23]
= 15 + 3 × [0 × 23] [As 0 ÷ 3 = 0]
= 15 + 3 × 0 = 15

Arithmetic Expressions Class 7 Case Based Questions

Question 1.
At Surya Vidya Mandir, the annual Art Festival is being celebrated. Ishan is participating in the event and earning performance points based on his art scenes.
Here’s how the points are awarded:

• In the 1^st scene, he puts in a solid effort and earns 22 points.
• In the 2^nd scene, he gets more confident and scores 8 more points than in the 1^st scene.
• In the 3^rd scene, he is slightly exhausted and earns half the points he got in the 2^nd scene.

Based on the above information, answer the following questions:
(i) Write arithmetic expressions to represent the number of points earned in the 2nd and 3rd scenes respectively.
(ii) What is the total number of points earned in all three scenes?
(iii) If each point is worth ₹50, how much money can Ishan claim?
Solution:
Points earned in 1^st scene = 22
Points earned in 2^nd scene
= 8 more points than in the 1^st scene
= 22 + 8 = 30
Points earned in 3^rd scene
= \(\frac{1}{2}\) × (Points earned in 2^nd scene)
= \(\frac{1}{2}\) × 30 = 15

(i) Arithmetic expression to represent the number of points earned in 2^nd scene = 22 + 8
Arithmetic expression to represent the number of points earned in 3^rd scene = \(\frac{1}{2}\) × 30

(ii) Total number of points earned in all three scenes = 22 + 30 + 15 = 67

(iii) Given, 1 point = ₹50
Total money that Ishan can claim
= ₹50 × 67 = ₹3,350

Question 2.
Ria buys 5 packs of pencils for her art class.
Each pack has 6 coloured pencils and 4 graphite pencils. But she loses 3 coloured pencils on the way.
Based on the above information, answer the following questions

(i) How many coloured pencils does Ria have?
(ii) How many graphite pencils does she have?
(iii) What is the total number of pencils Ria finally has?
Solution:
(i) In 1 pack, there are 6 coloured pencils.
So, number of coloured pencils in 5 packs
= 5 × 6 = 30
Given, Ria lost 3 coloured pencils on the way.
Now, number of remaining coloured pencils
= 30 – 3 = 27

(ii) In 1 pack, there are 4 graphite pencils.
So, number of graphite pencils in 5 packs
= 4 × 5 = 20

(iii) Total number of pencils with Ria
= Number of coloured pencils + Number of graphite pencils
= 27 + 20 = 47