NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 (Rationalised Contents)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 are the part of NCERT Solutions for Class 7 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4.
- NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
- NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2
- NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3
- NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4
- NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5
Ex 2.4 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
Solution:
(i) 0.2 × 6
∵ 2 × 6 = 12
We have 1 decimal place in 0.2.
Thus, 0.2 × 6 = 1.2
(ii) 8 × 4.6
∵ 8 × 46 = 368
We have one decimal place in 4.6.
Thus, 8 × 4.6 = 36.8
(iii) 2.71 × 5
∵ 271 × 5 = 1355
We have two decimal places in 2.71.
Thus, 2.71 × 5 = 13.55
(iv) 20.1 × 4
∵ 201 × 4 = 804
We have one decimal place in 20.1.
Thus, 20.1 × 4 = 80.4
(v) 0.05 × 7
∵ 5 × 7 = 35
We have 2 decimal places in 0.05.
Thus, 0.05 × 7 = 0.35
(vi) 211.02 × 4
∵ 21102 × 4 = 84408
We have 2 decimal places in 211.02.
Thus, 211.02 × 4 = 844.08
(vii) 2 × 0.86
∵ 2 × 86 = 172
We have 2 decimal places in 0.86.
Thus, 2 × 0.86 = 1.72
Ex 2.4 Class 7 Maths Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
Area of a rectangle = length × breadth
= 5.7 × 3 = 17.1 cm2
Hence, the area of the rectangle is 17.1 cm2.
Ex 2.4 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
Solution:
Ex 2.4 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 is 10 litres of petrol?
Solution:
Distance covered in 1 litre of petrol = 55.3 km
Distance covered in 10 litres of petrol = 55.3 × 10 km
Hence, the distance covered by the two-wheeler is 553 km.
Ex 2.4 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) 2.5 × 0.3
∵ 25 × 3 = 75
There is 1 decimal place in 2.5 and 1 decimal place in 0.3. Hence, the product will have two (1 + 1 = 2) decimal places.
Thus, 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7
∵ 1 × 517 = 517
There is 1 decimal place in 0.1 and 1 decimal place in 51.7. Hence, the product will have two (1 + 1 = 2) decimal places.
Thus, 0.1 × 51.7 = 5.17
(iii) 0.2 × 316.8
∵ 2 × 3168 = 6336
There is 1 decimal place in 0.2 and 1 decimal place in 316.8. Hence, the product will have two (1 + 1 = 2) decimal places.
Thus, 0.2 × 316.8 = 63.36
(iv) 1.3 × 3.1
∵ 13 × 31 = 403
There is 1 decimal place in 1.3 and 1 decimal place in 3.1. Hence, the product will have two (1 + 1 = 2) decimal places.
Thus, 1.3 × 3.1 = 4.03
(v) 0.5 × 0.05
∵ 5 × 5 = 25
There is 1 decimal place in 0.5 and 2 decimal places in 0.05. Hence, the product will have three (1 + 2 = 3) decimal places.
Thus, 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15
∵ 112 × 15 = 1680
There is 1 decimal place in 11.2 and 2 decimal places in 0.15. Hence, the product will have three (1 + 2 = 3) decimal places.
Thus, 11.2 × 0.15 = 1.680
(vii) 1.07 × 0.02
∵ 107 × 2 = 214
There are 2 decimal places in 1.07 and 2 decimal places in 0.02. Hence, the product will have four (2 + 2 = 4) decimal places.
Thus, 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05
∵ 1005 × 105 = 105525
There are 2 decimal places in 10.05 and 2 decimal places in 1.05. Hence, the product will have four (2 + 2 = 4) decimal places.
Thus, 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01
∵ 10101 × 1 = 10101
There are 2 decimal places in 101.01 and 2 decimal places in 0.01. Hence, the product will have four (2 + 2 = 4) decimal places.
Thus, 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1
∵ 10001 × 11 = 110011
There are 2 decimal places in 100.01 and 1 decimal place in 1.1. Hence, the product will have three (2 + 1 = 3) decimal places.
Thus, 100.01 × 1.1 = 110.011
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