NCERT Solution class 7 chapter 2 exercise 2.6

Exercise 2.6

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Q 1, Ex 2.6 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT

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Question 1
Find:
(i) $0.2 \times 6$
Sol :
$0.2 \times 6$ $=\dfrac{2}{10} \times 6$ $=\dfrac{12}{10}$ = 1.2

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(ii) $8 \times 4.6$
Sol :
$8 \times 4.6$ $=8\times \dfrac{46}{10}$ $=\dfrac{368}{10}$ = 36.8

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(iii) $2.71 \times 5$
Sol :
$2.71 \times 5$ $=\dfrac{271}{100}\times 5$ $=\dfrac{1355}{100}$ = 13.55

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(iv) $20.1 \times 4$
Sol :
$20.1 \times 4$ $=\dfrac{201}{10} \times 4$ $=\dfrac{804}{10}$ = 80.4

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(v) $0.05 \times 7$
Sol :
$0.05 \times 7$ $=\dfrac{5}{100} \times 7$ $=\dfrac{35}{100}$ = 0.35

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(vi) $211.02 \times 4$
Sol :
$211.02 \times 4$ $=\dfrac{21102}{100} \times 4$ $=\dfrac{84408}{100}$ = 844.08

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(vii) $2 \times 0.86$
Sol :
$2 \times 0.864$ $=2 \times \dfrac{86}{100}$ $=\dfrac{172}{100}$ = 1.72

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Q 2, Ex 2.6 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT

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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 $= 5.7\times 3$ $=17.1 cm^2$

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Q 3, Ex 2.6 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT

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Question 3
Find:
(i) $1.3 \times 10$
Sol :
= 13

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(ii) $36.8 \times 10$
Sol :
= 368

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(iii) $153.7 \times 10$
Sol :
= 1537

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(iv) $168.07 \times 10$
Sol :
= 1680.7

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(v) $31.1 \times 100$
Sol :
= 3110

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(vi) $156.1 \times 100$
Sol :
= 15610

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(vii) $3.62 \times 100$
Sol :
= 362

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(viii) $43.07 \times 100$
Sol :
= 4307

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(ix) $0.5 \times 10$
Sol :
= 5

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(x) $0.08 \times 10$
Sol :
= 0.8

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(xi) $0.9 \times 100$
Sol :
= 90

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(xii) $0.03 \times 1000$
Sol :
= 30

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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.

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Q 4, Ex 2.6 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT

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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.

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Q 5, Ex 2.6 - Fractions and Decimals - Chapter 2 - Maths Class 7th - NCERT

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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

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(ii) $0.1 \times 51.7$
Sol :
$0.1 \times 51.7$ $=\dfrac{}{} \times \dfrac{}{}$ $=\dfrac{517}{100}$ = 5.17

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(iii) $0.2 \times 316.8$
Sol :
$0.2 \times 316.8$ $=\dfrac{2}{10} \times \dfrac{3168}{10}$ $=\dfrac{6336}{100}$ = 63.36

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(iv) $1.3 \times 3.1$
Sol :
$1.3 \times 3.1$ $=\dfrac{13}{10} \times \dfrac{31}{10}$ $=\dfrac{403}{100}$ = 4.03

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(v) $0.5 \times 0.05$
Sol :
$0.5 \times 0.05$ $=\dfrac{5}{10} \times \dfrac{5}{100}$ $=\dfrac{25}{100}$ = 0.025

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(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

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(vii) $1.07 \times 0.02$
Sol :
$1.07 \times 0.02$ $=\dfrac{107}{100} \times \dfrac{2}{100}$ $=\dfrac{214}{1000}$ = 0.0214

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(viii) $10.05 \times 1.05$
Sol :
$10.05 \times 1.05$ $=\dfrac{1005}{100} \times \dfrac{105}{100}$ $=\dfrac{105525}{10000}$ = 10.5525

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(ix) $101.01 \times 0.01$
Sol :
$101.01 \times 0.01$ $=\dfrac{10101}{100} \times \dfrac{1}{100}$ $=\dfrac{10101}{10000}$ = 1.0101

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(x) $100.01 \times 1.1$
Sol :
$100.01 \times 1.1$ $=\dfrac{10001}{100} \times \dfrac{11}{10}$ $=\dfrac{110011}{1000}$ = 110.011