The DAV Class 7 Maths Book Solutions Pdf and **DAV Class 7 Maths Chapter 2 Worksheet 6** Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 2 WS 6 Solutions

Question 1.

Correct the following statements:

(i) Between two rational numbers, we can find only one rational number.

Answer:

Between two fractions, we can find infinitely many fractions.

(ii) Between two rational numbers, we can find as many integers as we like.

Answer:

Between two rational numbers, we can find as many rational numbers as we like.

(iii) Between two integers, we can find as many integers as we like.

Answer:

Between two integers, we can find as many rational numbers as we like.

Question 2.

Find a rational number between:

(i) 2 and 4

Answer:

Rational number between

2 and 4 = \(\frac{1}{2}\)(2 + 4) = \(\frac{1}{2}\) x 6 = 3

(ii) – 2 and – 6

Answer:

Rational number between

– 2 and – 6 = \(\frac{1}{2}\)[-2 + (-6)]

= \(\frac{1}{2}\)[-2 – 6]

= \(\frac{1}{2}\)[-8]

= -4

(iii) \(\frac{1}{4}\) and \(\frac{-3}{4}\)

Answer:

Rational number between

\(\frac{1}{4}\) and \(\frac{-3}{4}\)

= \(\frac{1}{2}\left[\frac{1}{4}+\left(\frac{-3}{4}\right)\right]\)

= \(\frac{1}{2}\left[\frac{1}{4}-\frac{3}{4}\right]\)

= \(\frac{1}{2}\left[\frac{1-3}{4}\right]\)

= \(\frac{1}{2}\left[\frac{-2}{4}\right]=\frac{-1}{4}\)

(iv) \(\frac{-2}{3}\) and \(\frac{-7}{3}\)

Answer:

Rational number between

\(\frac{-2}{3}\) and \(\frac{-7}{3}\)

= \(\frac{1}{2}\left[\frac{-2}{3}+\left(\frac{-7}{3}\right)\right]\)

= \(\frac{1}{2}\left[\frac{-2}{3}-\frac{7}{3}\right]\)

= \(\frac{1}{2}\left[\frac{-2-7}{3}\right]\)

= \(\frac{1}{2}\left[\frac{-9}{3}\right]\)

= \(\frac{-3}{2}\)

Question 3.

Insert three rational numbers between:

(i) \(\frac{4}{13}\) and \(\frac{1}{13}\)

Answer:

First rational number between \(\frac{4}{13}\) and \(\frac{1}{13}\)

= \(\frac{1}{2}\left(\frac{4}{13}+\frac{1}{13}\right)\)

= \(\frac{1}{2}\left(\frac{4+1}{13}\right)\)

= \(\frac{1}{2} \times \frac{5}{13}=\frac{5}{26}\)

Second rational number = \(\frac{1}{2}\left(\frac{4}{13}+\frac{5}{26}\right)\)

= \(\frac{1}{2}\left(\frac{4 \times 2+5 \times 1}{26}\right)\)

= \(\frac{1}{2}\left(\frac{8+5}{26}\right)\)

= \(\frac{1}{2} \times \frac{13}{26}=\frac{1}{4}\)

Third rational number = \(\frac{1}{2}\left(\frac{5}{26}+\frac{1}{13}\right)\)

= \(\frac{1}{2}\left(\frac{5 \times 1+1 \times 2}{26}\right)\)

= \(\frac{1}{2}\left(\frac{5+2}{26}\right)\)

= \(\frac{1}{2} \times \frac{7}{26}=\frac{7}{52}\)

Hence the required rational numbers are \(\frac{5}{26}, \frac{1}{4}, \frac{7}{52}\)

(ii) \(\frac{-7}{10}\) and \(\frac{11}{10}\)

Answer:

First rational number between \(\frac{-7}{10}\) and \(\frac{11}{10}\)

= \(\frac{1}{2}\left(\frac{-7}{10}+\frac{11}{10}\right)\)

= \(\frac{1}{2}\left(\frac{-7+11}{10}\right)\)

Second rational number = \(\frac{1}{2}\left(\frac{-7}{10}+\frac{1}{5}\right)\)

= \(\frac{1}{2}\left(\frac{-7+2}{10}\right)\)

Third rational number = \(\frac{1}{2}\left(\frac{1}{5}+\frac{11}{10}\right)\)

= \(\frac{1}{2}\left(\frac{-7+2}{10}\right)\)

= \(\frac{1}{2} \times \frac{13}{10}=\frac{13}{20}\)

Hence the required rational numbers are

(iii) \(\frac{-4}{3}\) and \(\frac{-19}{3}\)

Answer:

First rational number between \(\frac{-4}{3}\) and \(\frac{-19}{3}\)

= \(\frac{1}{2}\left(\frac{-4}{3}-\frac{19}{3}\right)\)

= \(\frac{1}{2} \times\left(\frac{-4-19}{3}\right)\)

= \(\frac{1}{2} \times\left(\frac{-23}{3}\right)=\frac{-23}{6}\)

Second rational number = \(\frac{1}{2}\left(\frac{-4}{3}-\frac{23}{6}\right)\)

= \(\frac{1}{2}\left(\frac{-8-23}{6}\right)\)

= \(\frac{1}{2} \times\left(-\frac{31}{6}\right)\)

= \(\frac{1}{2} \times \frac{-31}{6}=\frac{-31}{12}\)

Third rational number = \(\frac{1}{2}\left(\frac{-23}{6}-\frac{19}{3}\right)\)

= \(\frac{1}{2}\left(\frac{-23}{6}-\frac{19}{3}\right)\)

= \(\frac{1}{2} \times\left(\frac{-61}{6}\right)=\frac{-61}{12}\)

Hence the required rational number are \(\frac{-23}{6}, \frac{-31}{12}\) and \(\frac{-61}{12}\)

(iv) \(\frac{1}{8}\) and \(\frac{-15}{8}\)

Answer:

First rational number between \(\frac{1}{8}\) and \(\frac{-15}{8}\)

= \(\frac{1}{2}\left(\frac{1}{8}-\frac{15}{8}\right)\)

= \(\frac{1}{2} \times\left(\frac{1-15}{8}\right)\)

= \(\frac{1}{2} \times\left(\frac{-14}{8}\right)\)

Second rational number = \(\frac{1}{2}\left(\frac{1}{8}-\frac{7}{8}\right)\)

= \(\frac{1}{2} \times\left(\frac{1-7}{8}\right)\)

= \(\frac{1}{2} \times\left(\frac{-6}{8}\right)\)

Third rational number = \(\frac{1}{2}\left(\frac{-7}{8}-\frac{15}{8}\right)\)

= \(\frac{1}{2}\left(\frac{-7-15}{8}\right)\)

= \(\frac{1}{2} \times\left(\frac{-22}{8}\right)\)

Hence the required rational number are \(\frac{-7}{8}, \frac{-3}{8}\) and \(\frac{-11}{8}\)

Question 4.

Find five rational numbers between:

(i) \(\frac{-4}{7}\) and \(\left|\frac{-4}{7}\right|\)

Answer:

The given rational numbers are \(\frac{-4}{7}\) and \(\left|\frac{-4}{7}\right|\)

First rational number = \(\frac{1}{2}\left(\frac{-4}{7}+\frac{4}{7}\right)\)

= \(\frac{1}{2}\) × 0 = 0

Second rational number = \(\frac{1}{2}\) (\(\frac{-4}{7}\) + 0)

= \(\frac{1}{2} \times \frac{-4}{7}=\frac{-2}{7}\)

Third rational number = \(\frac{1}{2}\)(0 + \(\frac{4}{7}\))

= \(\frac{1}{2} \times \frac{4}{7}=\frac{2}{7}\)

Fourth rational number = \(\frac{1}{2}\)(0 + \(\frac{-2}{7}\))

= \(\frac{1}{2}\)(0 – \(\frac{2}{7}\))

= \(\frac{1}{2} \times \frac{-2}{7}=\frac{-1}{7}\)

Fifth rational number = \(\frac{1}{2}\)(0 + \(\frac{2}{7}\))

= \(\frac{1}{2} \times \frac{2}{7}=\frac{1}{7}\)

Hence the required rational numbers are 0, \(\frac{-2}{7}, \frac{2}{7}, \frac{-1}{7}\) and \(\frac{1}{7}\)

(ii) \(\frac{-8}{3}\) and \(\left|\frac{-8}{3}\right|\)

Answer:

= \(\frac{-8}{3}\) and \(\frac{8}{3}\)

First rational number = \(\frac{1}{2}\left(\frac{-8}{3}+\frac{8}{3}\right)\)

= \(\frac{1}{2}\) × 0 = 0

Second rational number = \(\frac{1}{2}\)(\(\frac{-8}{3}\) + 0)

= \(\frac{1}{2} \times \frac{-8}{3}=\frac{-4}{3}\)

Third rational number = \(\frac{1}{2}\)(0 + \(\frac{8}{3}\))

= \(\frac{8}{3}\)

Fourth rational number = \(\frac{1}{2}\)(0 + \(\frac{-4}{3}\))

= \(\frac{1}{2}\)(0 – \(\frac{4}{3}\))

= \(\frac{1}{2} \times \frac{-4}{3}=\frac{-2}{3}\)

Fifth rational number = \(\frac{1}{2}\)(0 + \(\frac{4}{3}\))

= \(\frac{1}{2} \times \frac{4}{3}=\frac{2}{3}\)

Hence the required rational numbers are 0, \(\frac{-4}{3}, \frac{4}{3}, \frac{-2}{3}\) and \(\frac{2}{3}\)

### DAV Class 7 Maths Chapter 2 Value Based Questions

Question 1.

Rohit donated \(\frac{1}{5}\) of his monthly income to an Non-Government Organisation working for the education of the girl child, spend \(\frac{1}{4}\) of his salary on food, \(\frac{1}{3}\) on rent and \(\frac{1}{15}\) on other expenses. He is left with ₹ 9000.

(i) Find Rohit’s monthly salary.

(ii) What values of Rohit are depicted here?

(iii) Why is the education, specially for girls, important?

Answer:

(i) Let Rohit’s monthly salary be ₹ x.

Amount donated to NGO = \(\frac{1}{5}\) × x = \(\frac{x}{5}\)

Amount spent on food = \(\frac{1}{4}\) × x = \(\frac{x}{4}\)

Amount spent on rent = \(\frac{1}{3}\) × x = \(\frac{x}{3}\)

Amount spent on other expenses = \(\frac{1}{15}\) × x = \(\frac{x}{15}\)

According to the question, we have

x – \(\left(\frac{x}{5}+\frac{x}{4}+\frac{x}{3}+\frac{x}{15}\right)\) = 9000

⇒ x – \(\left(\frac{60 x+75 x+100 x+20 x}{300}\right)\) = 9000

⇒ x – \(\left(\frac{255 x}{300}\right)\) = 9000

⇒ \(\frac{300 x-255 x}{300}\) = 9000

⇒ \(\frac{45 x}{300}\) = 9000

⇒ x = \(\frac{300 \times 9000}{45}\) = 60,000

Therefore, Rohit’s monthly salary is ₹ 60,000.

(ii) Values: Sensitivity towards girls education, kindness, etc.

(iii) Education helps in overall development of girls themselves and society.