GSEB Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find:
(i) 0.2 x 6
(ii) 8 x 4.6
(iii) 2.71 x 5
(iv) 20.1 x 4
(v) 0.05 x 7
(vi) 211.02 x 4
(vii) 2 x 0.86
Solution:
(i) 0.2 x 6
Since 2 x 6 = 12 and there is one digit to the right of decimal point in 0.2.
ā“ 0.2 x 6 = 1.2

(ii) 8 x 4.6
āµ 8 x 46 = 368 and there is one digit to the right of decimal point in 4.6.
ā“ 8 x 4.6 = 36.8

(iii) 2.71 x 5
āµ 271 x 5 = 1355 and there are two digits to the right of decimal point in 2.71.
ā“ 2.71 x 5 = 13.55

(iv) 20.1 x 4
āµ 201 x 4 = 804 and there is 1 digit to the right of decimal point in 20.1.
ā“ 20.1 x 4 = 80.4

(v) 0.05 x 7
āµ 5 x 7 = 35 and there are 2 digits to the right of decimal point in 0.05.
ā“ 0. 05 x 7 = 0.35

(vi) 211.02 x 4
āµ 21102 x 4 = 84408 and there are two digits – to the right of the decimal point in 211.02.
ā“ 211.02 x 4 = 844.08

(vii) 2 x 0.86
āµ 2 x 86 = 172 and there are two digits to the right of the decimal point in 0.86.
ā“ 2 x 0.86 = 1.72

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth 3 cm.

Solution:
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
Area of the rectangle = Length X Breadth = 5.7 cm x 3 cm
= $$\frac { 57 }{ 10 }$$ x 3 cmĀ²
= 17.1 cmĀ²

Question 3.
Find:
(i) 1.3 x 10
(ii) 36.8 x 10
(iii) 153.7 x 10
(iv) 168.07 x 10
(v) 31.1 x 100
(vi) 156.1 x 100
(vii) 3.62 x 100
(viii) 43.07 x 100
(ix) 0.5 x 10
(x) 0.08 x 10
(xi) 0.9 x 100
(xii) 0.03 x 1000
Solution:
(i) 1.3 x 10
āµ There is one zero in 10.
ā“ The decimal point is shifted to the right by one place.
Thus, 1.3 x 10 = 13

(ii) 36.8 x 10
āµ There is one zero in 10.
ā“ The decimal point is shifted to the right by one place.
ā“ 36.8 x 10 = 368

(iii) 153.7 x 10
āµ There is one zero in 10.
ā“ The decimal point is shifted to the right by one place.
ā“ 153.7 x 10 = 1537

(iv) 168.07 x 10
āµ There is one zero in 10.
ā“ The decimal point is shifted to the right by one place.
ā“ 168.7 X 10 = 1680.7

(v) 31.1 x 100
āµ There are two zeros in 100
ā“ The decimal point is shifted to the right by two places.
ā“ 31.1 x 100 = 3110.0

(vi) 156.1 x 100
āµ There are two zeros in 100.
ā“ The decimal point is shifted to the right by two places.
ā“ 156.1 x 100 = 15610

(vii) 3.62 x 100
āµ There are two zeros in 100.
ā“ The decimal point is shifted to the right by two places.
ā“ 3.62 x 100 = 362

(viii) 43.07 x 100
āµ There are two zeros in 100.
ā“ The decimal point is shifted to the right by two places.
ā“ 43.07 x 100 = 4307

(ix) 0.5 x 10
āµ There is one zero in 10.
ā“The decimal point is shifted to the right by one place.
ā“ 0.5 x 10 = 5

(x) 0.08 x 10
āµ There is one zero in 10.
ā“ The decimal point is shifted to the right by one place.
ā“ 0.08 x 10 = 0.8

(xi) 0.9 x 100
āµ There are two zeros in 100.
ā“ The decimal point is shifted to the right by two places.
ā“ 0.9 x 100 = 90

(xii) 0.03 x 1000
āµ There are three zeros in 1000.
ā“ The decimal point is shifted to the right by three places.
ā“ 0.03 x 1000 = 30

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?
Solution:
Distance covered by the two-wheeler in 1 litre of petrol = 55.3 km
ā“ Distance covered by it in 10 litres of petrol = 55.3 x 10 km = 553 km

Question 5.
Find:
(i) 2.5 x 0.3
(ii) 0.1 x 51.7
(iii) 0.2 x 316.8
(iv) 1.3 x 3.1
(v) 0.5 x 0.05
(vi) 11.2 x 0.15
(vii) 1.07 x 0.02
(viii) 10.05 x 1.05
(ix) 101.01 x 0.01
(x) 100.01 x 1.1
Solution:
(i) 2.5 x 0.3
āµ 25 x 3 = 75 and total number of digits in the decimal parts of the given decimal numbers = 1 + 1 = 2
ā“ Decimal point is placed in the product after 2 places from the right most digit.
Thus, 2.5 x 0.3 = 0.75

(ii) 0.1 x 51.7
āµ 1 x 517 = 517 and total number of digits in the decimal parts of the given decimal numbers = 1 + 1 = 2
ā“ Decimal point is placed in the product after 2 places from the right most digit.
Thus, 0.1 x 51.7 = 5.17

(iii) 0.2 x 316.8
āµ 2 x 3168 = 6336 and total number of digits in the decimal parts of the given decimal numbers =1 + 1 = 2
ā“ Decimal point is placed in the product after 2 places from the right most digit.
ā“ 0.2 x 316.8 = 63.36

(iv) 1.3 x 3.1
āµ 13 x 31 = 403 and total number of digits in the decimal parts of the given decimal numbers = 1 + 1 = 2
ā“ Decimal point is placed in the product after 2 places from the right most digit.
ā“ 1.3 x 3.1 = 4.03

(v) 0.5 x 0.05
āµ 5 x 5 = 25 and total number of digits in the decimal parts of the given decimal numbers =1+2=3
ā“ Decimal point is placed in the product after 3 places from the right most digit.
ā“ 0.5 x 0.05 = 0.025

(vi) 11.2 x 0.15
āµ 112 x 15 = 1680 and there are 1 + 2 = 3 digits in the decimal part of the given decimal numbers.
ā“ The decimal point is placed in the product after 3 places from the right most digit.
ā“ 11.2 x 0.15 = 1.680

(vii) 1.07 x 0.02
āµ 107 x 2 = 214 and the total number of digits in the decimal part of the given decimal number is 2 + 2 = 4.
ā“ The decimal point is placed in the product after 4 places from right most digit.
ā“ 1.07 x 0.02 = 0.0214

(viii) 10.05 x 1.05
āµ 1005 x 105 = 105525 and the total number of digits in the decimal parts of the given decimal numbers is 2 + 2 = 4.
ā“ The decimal point is placed in the product after 4 places from right most digit.
ā“ 10.05 x 1.05 = 10.5525

(ix) 101.01 x 0.01
āµ 10101 x 1 = 10101 and the total number of digits in the decimal parts of the given decimal numbers is 2 + 2 = 4.
ā“ The decimal point is placed after 4 places from the right most digit in the product.
ā“ 101.01 x 0.01 = 1.0101

(x) 100.01 x 1.1
āµ 10001 x 11 = 110011 and the total number of digits in the decimal parts of the given decimal number is 2 + 1 = 3.
ā“ The decimal point is placed after 3 places from the right most digit in the product.
ā“ 100.01 x 1.1 = 110.011