AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.2 Textbook Questions and Answers.

## AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.2

Question 1.

Find the product of the following.

i) 3 × \(\frac{5}{12}\)

ii) \(\frac{15}{8}\) × 12

iii) 1\(\frac{3}{4}\) × \(\frac{12}{21}\)

iv) \(\frac{4}{5}\) × \(\frac{12}{7}\)

Answer:

Question 2.

Which is greater?

i) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)

ii) \(\frac{2}{7}\) of \(\frac{3}{4}\) or \(\frac{3}{5}\) of \(\frac{5}{8}\)

Answer:

i) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)

\(\frac{3}{7}\) or \(\frac{2}{7}\)

So, \(\frac{3}{7}\) > \(\frac{2}{7}\)

∴ \(\frac{1}{2}\) of \(\frac{6}{7}\) > \(\frac{2}{3}\) of \(\frac{3}{7}\)

ii) \(\frac{2}{7}\) of \(\frac{3}{4}\) or \(\frac{3}{5}\) of \(\frac{5}{8}\)

Convert them into like fractions

LCM of denominators = 2 × 2 × 2 × 7 = 56

\(\frac{3 \times 4}{14 \times 4}\) or \(\frac{3 \times 7}{8 \times 7}\)

\(\frac{12}{56}\) or \(\frac{21}{56}\)

So, \(\frac{12}{56}\) < \(\frac{21}{56}\)

∴ \(\frac{2}{7}\) of \(\frac{3}{4}\) < \(\frac{3}{5}\) of \(\frac{5}{8}\)

Question 3.

Find:

i) \(\frac{7}{11}\) of 330

ii) \(\frac{5}{9}\) of 108

iii) \(\frac{2}{7}\) of 16

iv) \(\frac{1}{7}\) of \(\frac{3}{10}\)

Answer:

i) \(\frac{7}{11}\) of 330

= \(\frac{7}{11}\) × 330

= \(\frac{7}{11}\) × 11 × 30

= 7 × 30 = 210

ii) \(\frac{5}{9}\) of 108

= \(\frac{5}{9}\) × 108

= \(\frac{5}{9}\) × 9 × 12

= 5 × 12 = 60

iii) \(\frac{2}{7}\) of 16

= \(\frac{2}{7}\) × 16

= \(\frac{2×16}{7}\)

= \(\frac{32}{7}\) or 4\(\frac{4}{7}\)

iv) \(\frac{1}{7}\) of \(\frac{3}{10}\)

= \(\frac{1}{7}\) × \(\frac{3}{10}\)

= \(\frac{1 \times 3}{7 \times 10}\)

= \(\frac{3}{70}\)

Question 4.

If the cost of a notebook is Rs. 10\(\frac{3}{4}\). Then, find the cost of 36 books.

Answer:

Cost of each notebook = Rs. 10\(\frac{3}{4}\) (or) \(\frac{43}{4}\)

Cost of 36 notebooks = 36 × 10\(\frac{3}{4}\) = 36 × \(\frac{43}{4}\) = \(\frac{9 \times 4 \times 43}{4}\) = 9 × 43

Cost of 36 notebooks = Rs. 387

Question 5.

A motor bike runs 52\(\frac{1}{2}\) km using 1 litre of petrol. How much distance will it cover for 2\(\frac{3}{4}\) litres of petrol ?

Answer:

Distance covered by the motor bike 1 litre petrol = 52\(\frac{1}{2}\) km (or) \(\frac{105}{2}\) km

Distance covered by the motor bike for 2\(\frac{3}{4}\) litres of petrol = 2\(\frac{3}{4}\) of 52\(\frac{1}{2}\)

= \(\frac{11}{4}\) × \(\frac{105}{2}\)

= \(\frac{11 \times 105}{4 \times 2}\)

∴ Distance covered by the motor bike for 2\(\frac{3}{4}\) litres of petrol = \(\frac{1155}{8}\) = 144\(\frac{3}{8}\) km.