NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6.



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

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 2.6 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 2.6 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

Solution:


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 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 2.6 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) 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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