# DAV Class 7 Maths Chapter 2 Worksheet 4 Solutions

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

(ii) 4
Reciprocal of 4 = $$\frac{1}{4}$$
∴ 4 × $$\frac{1}{4}$$ = 1

(iii) $$\frac{11}{-12}$$
Reciprocal of $$\frac{11}{-12}=\frac{-12}{11}$$
∴ $$\frac{11}{-12} \times \frac{-12}{11}$$ = 1

(iv) $$\frac{-2}{2}$$
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 $$\frac{3}{2}$$?
$$\frac{-2}{3} \times \frac{3}{2}$$ = -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?
$$\frac{-1}{5}$$ × -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 = $$\frac{4}{7}$$
(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 = $$\frac{- 4}{3}$$
(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 = 
(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.
zero

(ii) _________ and _________ are their own reciprocals.
1 and -1

(iii) If a is the reciprocal ofb then b is the reciprocal of _________.
(v) $$\frac{-1}{8}$$ × _________ = 1.
(vi) _________ × (-5$$\frac{1}{3}$$) = 1.
$$\frac{-3}{16}$$