NCERT Solution class 7 chapter 2 exercise 2.6

Exercise 2.6

Question 1

Find:

(i) 0.2 \times 6

Sol :

0.2 \times 6 =\dfrac{2}{10} \times 6 =\dfrac{12}{10} = 1.2


(ii) 8 \times 4.6

Sol :

8 \times 4.6 =8\times \dfrac{46}{10} =\dfrac{368}{10} = 36.8


(iii) 2.71 \times 5

Sol :

2.71 \times 5 =\dfrac{271}{100}\times 5 =\dfrac{1355}{100} = 13.55


(iv) 20.1 \times 4

Sol :

20.1 \times 4 =\dfrac{201}{10} \times 4 =\dfrac{804}{10} = 80.4


(v) 0.05 \times 7

Sol :

0.05 \times 7 =\dfrac{5}{100} \times 7 =\dfrac{35}{100} = 0.35


(vi) 211.02 \times 4

Sol :

211.02 \times 4 =\dfrac{21102}{100} \times 4 =\dfrac{84408}{100} = 844.08


(vii) 2 \times 0.86

Sol :

2 \times 0.864 =2 \times \dfrac{86}{100} =\dfrac{172}{100} = 1.72


 

Question 2

Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.

Sol :

Length = 5.7 cm
Breadth = 3 cm
Area = Length x Breadth

Area = 5.7\times 3 =17.1 cm^2


 

Question 3

Find:

(i) 1.3 \times 10

Sol :

= 13


(ii) 36.8 \times 10

Sol :

= 368


(iii) 153.7 \times 10

Sol :

= 1537


(iv) 168.07 \times 10

Sol :

= 1680.7


(v) 31.1 \times 100

Sol :

= 3110


(vi) 156.1 \times 100

Sol :

= 15610


(vii) 3.62 \times 100

Sol :

= 362


(viii) 43.07 \times 100

Sol :

= 4307


(ix) 0.5 \times 10

Sol :

= 5


(x) 0.08 \times 10

Sol :

= 0.8


(xi) 0.9 \times 100

Sol :

= 90


(xii) 0.03 \times 1000

Sol :

= 30


Notes : We know that when a decimal number is multiplied by 10, 100, 1000, the decimal point in the product is shifted to the right by as many places as there are zeroes.


 

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?

Sol :

Distance covered in 1 litre of petrol = 55.3 km

Distance covered in 10 litre of petrol = 1 0 \times 55.3 = 553 km

Therefore, it will cover 553 km distance in 10 litre petrol.


 

Question 5

Find:

(i) 2.5 \times 0.3

Sol :

2.5 \times 0.3 =\dfrac{25}{10}\times \dfrac{3}{10} =\dfrac{75}{100} = 0.75


(ii) 0.1 \times 51.7

Sol :

0.1 \times 51.7 =\dfrac{}{} \times \dfrac{}{} =\dfrac{517}{100} = 5.17


(iii) 0.2 \times 316.8

Sol :

0.2 \times 316.8 =\dfrac{2}{10} \times \dfrac{3168}{10} =\dfrac{6336}{100} = 63.36


(iv) 1.3 \times 3.1

Sol :

1.3 \times 3.1 =\dfrac{13}{10} \times \dfrac{31}{10} =\dfrac{403}{100} = 4.03


(v) 0.5 \times 0.05

Sol :

0.5 \times 0.05 =\dfrac{5}{10} \times \dfrac{5}{100} =\dfrac{25}{100} = 0.025


(vi) 11.2 \times 0.15

Sol :

11.2 \times 0.15 =\dfrac{112}{10} \times \dfrac{15}{100} =\dfrac{1680}{1000} = 1.680 or 1.68


(vii) 1.07 \times 0.02

Sol :

1.07 \times 0.02 =\dfrac{107}{100} \times \dfrac{2}{100} =\dfrac{214}{1000} = 0.0214


(viii) 10.05 \times 1.05

Sol :

10.05 \times 1.05 =\dfrac{1005}{100} \times \dfrac{105}{100} =\dfrac{105525}{10000} = 10.5525


(ix) 101.01 \times 0.01

Sol :

101.01 \times 0.01 =\dfrac{10101}{100} \times \dfrac{1}{100} =\dfrac{10101}{10000} = 1.0101


(x) 100.01 \times 1.1

Sol :

100.01 \times 1.1 =\dfrac{10001}{100} \times \dfrac{11}{10} =\dfrac{110011}{1000} = 110.011

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