NCERT Answers for Class 7 Maths
Chapter 2 Fractions and Decimals Exercise 2.6
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Chapter 2 Exercise 2.1 |
Page 52
Ex 2.6 Class 7 Maths Question 1. Find:
(i) 0.2 × 6 (ii) 8 × 4.6 (iii) 2.71 × 5 (iv) 20.1 × 4
(v) 0.05 × 7 (vi) 211.02 × 4 (vii) 2 × 0.86
(i) 0.2 × 6
∵ 2 × 6 = 12 and we have 1 digit right to the decimal point in 0.2.
Thus 0.2 × 6 = 1.2
(ii) 8 × 4.6
∵ 8 × 46 = 368 and there is one digit right to the decimal point in 4.6.
Thus 8 × 4.6 = 36.8
(iii) 2.71 × 5
∵ 271 × 5 = 1355 and there are two digits right to the decimal point in 2.71.
Thus 2.71 × 5 = 13.55
(iv) 20.1 × 4
∵ 201 × 4 = 804 and there is one digit right to the decimal point in 20.1.
∵ 20.1 × 4 = 80.4
(v) 0.05 × 7
∵ 5 × 7 = 35 and there are 2 digits right to the decimal point in 0.05.
Thus 0.05 × 7 = 0.35
(vi) 211.02 × 4
∵ 21102 × 4 = 84408 and there are 2 digits right to the decimal point in 211.02.
Thus 211.02 × 4 = 844.08
(vii) 2 × 0.86
∵ 2 × 86 = 172 and there are 2 digits right to the decimal point in 0.86.
Thus 2 × 0.86 = 1.72
Ex 2.6 Class 7 Maths Question 2. Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
From the question, it is given that,
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
Then,
Area of the rectangle = length × Breadth
= 5.7 × 3
= 17.1 cm2
Ex 2.6 Class 7 Maths Question 3. Find:
(i) 1.3 × 10 (ii) 36.8 × 10 (iii) 153.7 × 10 (iv) 168.07 × 10
(v) 31.1 × 100 (vi) 156.1 × 100 (vii) 3.62 × 100 (viii) 43.07 × 100
(ix) 0.5 × 10 (x) 0.08 × 10 (xi) 0.9 × 100 (xii) 0.03 × 1000
(i) 1.3 × 10
Answer
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 1.3 × 10 = 13
(ii) 36.8 × 10
Answer
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 36.8 × 10 = 368
(iii) 153.7 × 10
Answer
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 153.7 × 10 = 1537
(iv) 168.07 × 10
Answer
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 168.07 × 10 = 1680.7
(v) 31.1 × 100
Answer
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 31.1 × 100 = 3110
(vi) 156.1 × 100
Answer
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 156.1 × 100 = 15610
(vii) 3.62 × 100
Answer
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 3.62 × 100 = 362
(viii) 43.07 × 100
Answer
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 43.07 × 100 = 4307
(ix) 0.5 × 10
Answer
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 0.5 × 10 = 5
(x) 0.08 × 10
Answer
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 0.08 × 10 = 0.8
(xi) 0.9 × 100
Answer
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 0.9 × 100 = 90
(xii) 0.03 × 1000
Answer
On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.
= 0.03 × 1000 = 30
Ex 2.6 Class 7 Maths Question 4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Answer
From the question, it is given that,
Distance covered by two-wheeler in 1 litre of petrol = 55.3 km
Then, distance covered by two wheeler in 10 litres of petrol = (10 × 55.3)
= 553 km
∴ Two-wheeler covers a distance in 10 litres of petrol is 553 km.
Ex 2.6 Class 7 Maths Question 5. Find:
(i) 2.5 × 0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1
(v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02
(viii) 10.05 × 1.05 (ix) 101.01 × 0.01 (x) 100.01 × 1.1
Solution:
(i) 0 2.5 × 0.3
∵ 25 × 3 = 75 and there are 2 digits (1 + 1) right to the decimal points in 2.5 and 0.3.
Thus 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7
∵ 1 × 517 = 517 and there are two digits (1 + 1) right to the decimal places in 0.1 and 51.7.
(iii) 0.2 × 316.8
∵ 2 × 3168 = 6336 and there are 2 digits (1 + 1) right to the decimal points in 0.2 and 316.8.
Thus 0.2 × 316.8 = 63.36.
(iv) 1.3 × 3.1
∵ 13 × 31 = 403 and there are 2 digits (1 + 1) right to the decimal points in 1.3 and 3.1.
Thus 1.3 × 3.1 – 4.03
(v) 0.5 × 0.05
∵ 5 × 5 = 25 and there are 3 digits (1 + 2) right to the decimal points in 0.5 and 0.05.
Thus 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15
∵ 112 × 15 = 1680 and there are 3 digits (1 + 2) right to the decimal points in 11.2 and 0.15.
Thus 11.2 × 0.15 = 1.680
(vii) 1.07 × 0.02
∵ 107 × 2 = 214 and there are 4-digits (2 + 2) right to the decimal places is 1.07 × 0.02.
Thus 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05
∵ 1005 × 105 = 105525 and there are 4 digits (2 + 2) right to the decimal places in 10.05 × 1.05.
Thus 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01
∵ 10101 × 1 = 10101 and there are 4 digits (2 + 2) right to the decimal places in 101.01 and 0.01.
Thus 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1
∵ 10001 × 11 = 110011 and there are 3 digits (2 + 1) right to the decimal points in 100.01 and 1.1.
Thus 100.01 × 1.1 = 110.011.