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 = 24.8cm^{2}

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