# DAV Class 7 Maths Chapter 2 Worksheet 2 Solutions

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

## DAV Class 7 Maths Ch 2 WS 2 Solutions

Question 1.
Find the value of:
(i) $$\frac{6}{7}-\frac{-5}{7}$$
$$\frac{6}{7}-\frac{-5}{7}$$ = $$\frac{6-(-5)}{7}$$
= $$\frac{6+5}{7}$$
= $$\frac{11}{7}$$

(ii) $$\frac{5}{24}-\frac{7}{36}$$
$$\frac{5}{24}-\frac{7}{36}$$
= $$[\frac{5 \times 3-7 \times 2}{72}/latex] = [latex]\frac{15-14}{72}$$
= $$\frac{1}{72}$$

(iii) $$\frac{9}{10}-\frac{7}{-15}$$
$$\frac{9}{10}-\frac{7}{-15}$$
= $$\frac{9}{10}-\frac{-7}{15}$$
= $$\frac{9 \times 3-(-7) \times 2}{30}$$
= $$\frac{27+14}{30}$$
= $$\frac{31}{30}$$

(iv) $$\frac{-3}{8}-\frac{(-6)}{20}$$
$$\frac{-3}{8}-\frac{(-6)}{20}$$
= $$\frac{-3}{8}+\frac{6}{20}$$
= $$\frac{-3 \times 5+6 \times 2}{40}$$
= $$\frac{-15+12}{40}$$
= $$\frac{-3}{40}$$

Question 2.
Subtract:
(i) $$\frac{5}{9}$$ from $$\frac{-7}{9}$$
$$\frac{-7}{9}-\frac{5}{9}$$
= $$\frac{-7-5}{9}$$
= $$\frac{-12}{9}=\frac{-4}{3}$$

(ii) $$\frac{-5}{7}$$ from 0
0 – $$\left(-\frac{5}{7}\right)$$
= 0 + $$\frac{5}{7}$$
= $$\frac{5}{7}$$

(iii) $$\frac{5}{11}$$ from $$\frac{-8}{23}$$
$$\frac{-8}{23}-\frac{5}{11}$$
= $$\frac{-8 \times 11-5 \times 23}{253}$$
= $$\frac{-88-115}{253}$$
= $$\frac{-203}{253}$$

(iv) $$\frac{-2}{9}$$ from $$\frac{7}{6}$$
$$\frac{7}{6}-\left(-\frac{2}{9}\right)$$
= $$\frac{7}{6}+\frac{2}{9}$$
= $$\frac{3 \times 7+2 \times 2}{18}$$
= $$\frac{21+4}{18}$$
= $$\frac{25}{18}$$

Question 3.
The sum of two rational numbers is – 5. If one of the number is $$\frac{2}{3}$$, find the other.
The required number = Sum of the numbers – the given number
= 5 – $$\frac{2}{3}$$
= $$\frac{-5}{1}-\frac{2}{3}$$
= $$\frac{-5 \times 3-2 \times 1}{3}$$
= $$\frac{-15-2}{3}$$
= $$\frac{-17}{3}$$
Hence the other number = $$\frac{-17}{3}$$.

Question 4.
What number should be added to $$\frac{-3}{7}$$ to get 1?
Here sum of the two numbers is 1 and one of them is $$\frac{-3}{7}$$
∴ The other number = 1 – ($$\frac{-3}{7}$$ )
= 1 + $$\frac{3}{7}$$
= $$\frac{1}{1}+\frac{3}{7}$$
= $$\frac{1 \times 7+3 \times 1}{7}$$
= $$\frac{7+3}{7}$$
= $$\frac{10}{7}$$
Hence the required number is $$\frac{10}{7}$$.

Question 5.
What number should be subtracted from – 1 so as to get $$\frac{5}{3}$$?
Here sum of the two numbers = – 1
and one of them is $$\frac{5}{3}$$
∴The required number = – 1 – $$\frac{5}{3}$$
= $$\frac{-1}{1}-\frac{5}{3}$$
= $$\frac{-1 \times 3-5 \times 1}{3}$$
= $$\frac{-3-5}{3}$$
= $$\frac{-8}{3}$$
Hence the required number = $$\frac{-8}{3}$$

Question 6.
Simplify:
(i) $$\frac{-4}{5}-\frac{3}{15}+\frac{7}{20}$$
$$\frac{-4}{5}-\frac{3}{15}+\frac{7}{20}$$
= $$\frac{-4 \times 12-3 \times 4+7 \times 3}{60}$$
= $$\frac{-48-12+21}{60}$$
= $$\frac{-60+21}{60}$$
= $$-\frac{39}{60}=-\frac{13}{20}$$

(ii) $$\frac{-5}{13}-\frac{-3}{26}-\frac{9}{-52}$$
$$\frac{-5}{13}-\frac{-3}{26}-\frac{9}{-52}$$
= $$\frac{-5}{13}+\frac{3}{26}+\frac{9}{52}$$
= $$\frac{-5 \times 4+3 \times 2+9 \times 1}{52}$$
= $$\frac{-20+6+9}{52}$$
= $$\frac{-20+15}{52}$$
= $$\frac{-5}{52}$$

(iii) $$\frac{7}{24}+\frac{5}{12}-\frac{11}{18}$$
$$\frac{7}{24}+\frac{5}{12}-\frac{11}{18}$$
= $$\frac{7 \times 3+5 \times 6-11 \times 4}{72}$$
= $$\frac{21+30-44}{72}$$
= $$\frac{51-44}{72}$$
= $$\frac{7}{72}$$

(iv) $$\frac{-11}{30}-\frac{8}{15}+\frac{7}{6}+\frac{-2}{5}$$
$$\frac{-11}{30}-\frac{8}{15}+\frac{7}{6}+\frac{-2}{5}$$
= $$\frac{-11 \times 1-8 \times 2+7 \times 5-2 \times 6}{30}$$
= $$\frac{-11-16+35-12}{30}$$
= $$\frac{-39+35}{30}$$
= $$\frac{-4}{30}= \frac{-2}{15}$$

Question 7.
Find the value of x – y and y – x for x = $$\frac{2}{3}$$, y = $$\frac{5}{9}$$. Are they equal?
Here x = $$\frac{2}{3}$$, y = $$\frac{5}{9}$$
x – y = $$\frac{2}{3}$$ – $$\frac{5}{9}$$
= $$\frac{2 \times 3-5 \times 1}{9}$$
= $$\frac{6-5}{9}=\frac{1}{9}$$

and y – x = $$\frac{5}{9}-\frac{2}{3}$$
= $$\frac{5 \times 1-2 \times 3}{9}$$
= $$\frac{5-6}{9}=\frac{-1}{9}$$
∴ $$\frac{1}{9} \neq-\frac{1}{9}$$
So they are not equal.

Question 8.
For x = $$\frac{1}{10}$$, y = $$\frac{-3}{5}$$, z = $$\frac{7}{20}$$, find the values of expressions (x – y) – z and x – (y – z). Are they equal?
Here, x = $$\frac{1}{10}$$,
y = $$\frac{-3}{5}$$,
z = $$\frac{7}{20}$$

(x – y) – z = $$\left(\frac{1}{10}-\frac{-3}{5}\right)-\frac{7}{20}$$
= $$\left(\frac{1}{10}+\frac{3}{5}\right)-\frac{7}{20}$$
= $$\left(\frac{1 \times 1+3 \times 2}{10}\right)-\frac{7}{20}$$
= $$\left(\frac{1+6}{10}\right)-\frac{7}{20}=\frac{7}{10}-\frac{7}{10}$$
= 0

and x – (y – z) = $$\frac{1}{10}-\left(\frac{-3}{5}-\frac{7}{20}\right)$$
= $$\frac{1}{10}-\left(\frac{-3 \times 4-7 \times 1}{20}\right)$$
= $$\frac{1}{10}-\left(\frac{-12-7}{20}\right)$$
= $$\frac{1}{10}-\left(\frac{-19}{20}\right)$$
= $$\frac{1}{10}+\frac{19}{20}$$
= $$\frac{1 \times 2+19 \times 1}{20}$$
= $$\frac{2+19}{20}$$
= $$\frac{21}{20}$$

∴ 0 ≠ $$\frac{21}{20}$$
Hence they are not equal.