The DAV Class 7 Maths Book Solutions Pdf and **DAV Class 7 Maths Chapter 2 Worksheet 4** Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 2 WS 4 Solutions

Question 1.

Find the reciprocals of:

(i) \(\frac{-1}{5}\)

Answer:

Reciprocal of \(\frac{-1}{5}\) = -5

(ii) 4

Answer:

Reciprocal of 4 = \(\frac{1}{4}\)

∴ 4 × \(\frac{1}{4}\) = 1

(iii) \(\frac{11}{-12}\)

Answer:

Reciprocal of \(\frac{11}{-12}=\frac{-12}{11}\)

∴ \(\frac{11}{-12} \times \frac{-12}{11}\) = 1

(iv) \(\frac{-2}{2}\)

Answer:

Reciprocal of \(\frac{-2}{-19}=\frac{-19}{-2}\)

∴ \(\frac{-2}{-19} \times \frac{-19}{-2}\) = 1

Question 2.

Check, if the reciprocal of \(\frac{-2}{3}\) is \(\frac{3}{2}\)?

Answer:

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

\(\frac{3}{2}\) is not the reciprocal of \(\frac{-2}{3}\)

Question 3.

Check, if the reciprocal of \(\frac{-1}{5}\) is – 5?

Answer:

\(\frac{-1}{5}\) × -5 = 1

∴ -5 is the reciprocal of \(\frac{-1}{5}\)

Question 4.

Verify that (x – y)^{-1} ≠ x^{-1} – y^{-1} by taking x = \(\frac{-2}{7}\), y = \(\frac{4}{7}\)

Answer:

(x – y)^{-1} = \(\left(\frac{-2}{7}-\frac{4}{7}\right)^{-1}\)

= \(\left(\frac{-2-4}{7}\right)^{-1}\)

= \(\left(\frac{-6}{7}\right)^{-1}\)

= \(\frac{7}{-6}\)

x^{-1} – y^{-1} = \(\left(\frac{-2}{7}\right)^{-1}-\left(\frac{4}{7}\right)^{-1}\)

= \(-\frac{7}{2}-\frac{7}{4}=\frac{-14-7}{4}\)

= \(=\frac{-21}{4}\)

So, \(\frac{7}{-6} \neq \frac{-21}{4}\)

Hence verified.

Question 5.

Verify that (x + y)^{-1} ≠ x^{-1} + y^{-1} by taking x = \(\frac{5}{9}\), y = \(\frac{- 4}{3}\)

Answer:

(x + y)^{-1} = \(\left(\frac{5}{9}-\frac{4}{3}\right)^{-1}\)

= \(\left(\frac{5-12}{9}\right)^{-1}=\left(\frac{-7}{9}\right)^{-1}\)

= \(=\frac{9}{-7}\)

x^{-1} + y^{-1} = \(\left(\frac{5}{9}\right)^{-1}+\left(\frac{-4}{3}\right)^{-1}\)

= \(\frac{9}{5}-\frac{3}{4}=\frac{9 \times 4-3 \times 5}{20}\)

= \(\frac{36-15}{20}\)

= \(\frac{21}{20}\)

∴ \(\frac{9}{-7} \neq \frac{21}{20}\)

Hence Verified.

Question 6.

Verify that (x × y)^{-1} = x^{-1} × y^{-1} by taking x = \(\), y = \(\)

Answer:

(x × y)^{-1} = \(\left[\left(\frac{-2}{3}\right) \times\left(\frac{-3}{4}\right)\right]^-1\)

= \(\left[\frac{-2 \times-3}{3 \times 4}\right]^{\prime}=\left[\frac{6}{12}\right]^{\prime}\)

= \(\frac{12}{6}\)

= 2

x^{-1} × y^{-1} = \(\left(\frac{-2}{3}\right)^{-1} \times\left(\frac{-3}{4}\right)^{-1}\)

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

= \(\frac{12}{6}\)

= 2

Hence (x × y)^{-1} = x^{-1} × y^{-1} = 2

Question 7.

Fill in the blanks:

(i) The number _________ has no reciprocal.

Answer:

zero

(ii) _________ and _________ are their own reciprocals.

Answer:

1 and -1

(iii) If a is the reciprocal ofb then b is the reciprocal of _________.

Answer:

a

(iv) (11 × 5)^{-1} = (11)^{-1} × _________.

Answer:

(5)^{-1}

(v) \(\frac{-1}{8}\) × _________ = 1.

Answer:

-8

(vi) _________ × (-5\(\frac{1}{3}\)) = 1.

Answer:

\(\frac{-3}{16}\)