AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 6

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 6

Question 1.
Solve the following.
(i) 0.3 × 6
(ii) 7 × 2.7
(iii) 2.71 × 5
(iv) 19.7 × 4
(v) 0.05 × 7
(vi) 210.01 × 5
(vii) 2 × 0.86
Solution:
(i) 0.3 × 6 = 1.8
(ii) 7 × 2.7 = 18.9
(iii) 2.71 × 5 = 13.55
(iv) 19.7 × 4 = 78.8
(v) 0.05 × 7 = 0.35
(vi) 210.01 × 5 = 1050.05
(vii) 2 × 0.86 = 1.72

Question 2.
Find the area of a rectangle whose length is 6.2 cm and breadth is 4 cm.
Solution:
Length of the rectangle = 6.2 cm
Breadth of the rectangle = 4 cm
Area of the rectang’e = Length × Breadth
= 6.2 × 4 = 24.8cm2

Question 3.
Solve the following.
(i) 21.3 × 10
(ii) 36.8 × 10
(ii) 53.7 × 10
(iv) 168.07 × 10
(v) 131.1 × 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:
(i) 21.3 × 10 = 213
(ii) 36.8 × 10 = 368
(ii) 53.7 × 10 = 537
(iv) 168.07 × 10 = 1680.7
(v) 131.1 × 100 = 13110
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

Question 4.
A motor bike covers a distance of 62.5 km.consuming one litre of petrol. How much distance does it cover for 10 litres of petrol?
Solution:
Distance covered for 1 lit, of petrol = 62.5 km
∴ Distance covered for 10 lit, of petrol = 62.5 × 10 = 625 km

Question 5.
Solve the following.
(i) 1.5 × 0.3
(ii) 0.1 × 47.5
(iii) 0.2 × 210.8
(iv) 4.3 × 3.4
(v) 0.5 × 0.05
(vi) 11.2 × 0.10
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 70.01 × 1.1
Solution:
(i) 1.5 × 0.3 = 0.45
(ii) 0.1 × 47.5 = 4.75
(iii) 0.2 × 210.8 = 42.16
(iv) 4.3 × 3.4 = 14.62
(v) 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.10 = 1.12
(vii) 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01 = 1.0101
(x) 70.01 × 1.1 = 77.011

Question 6.
Solve the following.
(i) 2.3 ÷ 100
(ii) 0.45 ÷ 5
(iii) 44.3 ÷ 10
(iv) 127.1 ÷ 1000
(v) 7 ÷ 35
(vi) 88.5 ÷ 0.15
(vii) 0.4 ÷ 20
Solution:
(i) 2.3 ÷ 100 = $$\frac{23}{10}$$ ÷ 100 = $$\frac{23}{10} \times \frac{1}{100}=\frac{23}{1000}$$ = 0.023
(ii) 0.45 ÷ 5 = $$\frac{45}{100}$$ ÷ 5 = $$\frac{45}{100} \times \frac{1}{5}=\frac{9}{100}$$ = 0.09
(iii) 44.3 ÷ 10 = 44.3 × $$\frac{1}{10}$$ = 4.43
(iv) 127.1 ÷ 1000 = 127.1 × $$\frac{1}{1000}$$ = 0.1271
(v) 7 ÷ 35 = 7 × $$\frac{1}{3.5}=\frac{7 \times 10}{3.5 \times 10}=\frac{70}{35}$$ = 2
(vi) 88.5 ÷ 0.15 = $$\frac{885}{10} \div \frac{15}{100}$$ = $$\frac{885}{10} \times \frac{100}{15}$$ = 590
(vii) 0.4 ÷ 20 = $$\frac{4}{10}$$ ÷ 20 = $$\frac{4}{10} \times \frac{1}{20}=\frac{1}{10 \times 5}=\frac{1}{50}$$ = 0.02

Question 7.
A side of a regular polygon is 3.5 cm in length. The perimeter of the polygon is 17.5 cm.
How many sides does the polygon have?
Solution:
Side of each length of the polygon . = 3.5 cm
Total length of all sides = perimeter = 17.5 cm
Number of sides of the polygon = 17.5 + 3.5
= $$\frac{175}{10} \div \frac{35}{10}$$ = $$\frac{175}{10} \times \frac{10}{35}$$ = 5

Question 8.
A rain fall of 0.896 cm. was recorded in 7 hours, what was the average amount of rain per
hour?
Solution:
Total rainfall recorded in 7 hours = 0896 cm
∴ Average rainfall (for 1 hour) = 0.896 ÷ 7
= $$\frac{896}{1000} \div 7=\frac{896}{1000} \times \frac{1}{7}=\frac{128}{1000}$$ = 0.128