# DAV Class 7 Maths Chapter 2 Worksheet 3 Solutions

The DAV Class 7 Maths Book Solutions Pdf and DAV Class 7 Maths Chapter 2 Worksheet 3 Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 2 WS 3 Solutions

Question 1.
Multiply and express the result as a rational number in standard form.
(i) $$\frac{11}{7}$$ by $$\frac{-3}{8}$$
$$\frac{11}{7} \times \frac{-3}{8}=\frac{11 \times-3}{7 \times 8}$$
= $$\frac{-33}{56}$$ (Standard form)

(ii) $$\frac{-7}{4}$$ by $$\frac{2}{3}$$
$$\frac{-7}{4} \times \frac{2}{3}=\frac{-7 \times 2}{4 \times 3}$$
= $$\frac{-14}{12}=\frac{-7}{6}$$ (Standard form)

(iii) $$\frac{-3}{12}$$ by -48
$$\frac{-3}{12} \times \frac{-48}{1}$$
= $$\frac{-3 \times-48}{12 \times 1}=\frac{144}{12}$$ (Standard form)

(iv) $$\frac{-14}{9}$$ by $$\frac{-3}{7}$$
$$\frac{-14}{9} \times \frac{-3}{7}=\frac{-14 \times-3}{9 \times 7}$$
= $$\frac{42}{63}=\frac{2}{3}$$ (Standard form)

(v) $$\frac{23}{5}$$ by $$\frac{-25}{11}$$
$$\frac{23}{5} \times \frac{-25}{11}=\frac{23 \times-25}{5 \times 11}$$
= $$\frac{-575}{55}=\frac{-115}{11}$$ (Standard form)

(vi) 7 by $$\frac{-15}{63}$$
$$\frac{7}{1} \times \frac{-15}{63}$$
= $$\frac{7}{1} \times \frac{-15}{63}$$
= $$\frac{-5}{3}$$ (Standard form)

Question 2.
For the following values of x and y, verify x × y = y × x
(i) x = $$\frac{7}{9}$$, y = $$\frac{3}{2}$$
x × y = y × x
= $$\frac{7}{9} \times \frac{3}{2}=\frac{3}{2} \times \frac{7}{9}$$
= $$\frac{7 \times 3}{9 \times 2}=\frac{3 \times 7}{2 \times 9}$$
= $$\frac{7 \times 3}{9 \times 2}=\frac{3 \times 7}{2 \times 9}$$
Hence verified.

(ii) x = $$\frac{-2}{7}$$, y = $$\frac{5}{8}$$
x × y = y × x
= $$\frac{-2}{7} \times \frac{5}{8}=\frac{5}{8} \times \frac{-2}{7}$$
= $$\frac{-2 \times 5}{7 \times 8}=\frac{5 \times-2}{8 \times 7}$$
= $$\frac{-10}{56}=\frac{-10}{56}$$
Hence verified.

(iii) x = $$\frac{4}{9}$$, y = $$\frac{-5}{11}$$
x × y = y × x
= $$\frac{4}{9} \times \frac{-5}{11}=\frac{-5}{11} \times \frac{4}{9}$$
= $$\frac{4 \times-5}{9 \times 11}=\frac{-5 \times 4}{11 \times 9}$$
= $$\frac{-20}{99}=\frac{-20}{99}$$
Hence verified.

(iv) x = $$\frac{-17}{48}$$, y = $$\frac{-96}{51}$$
x × y = y × x
= $$\frac{-17}{48} \times \frac{-96}{51}=\frac{-96}{51} \times \frac{-17}{48}$$
= $$\frac{-17 \times-96}{48 \times 51}=\frac{-96 \times-17}{51 \times 48}$$
= $$\frac{1632}{2248}=\frac{1632}{2248}$$
Hence verified.

Question 3.
For the following values of x, y and z, find the products (x × y) × z and x × (y × z) and observe the result (x × y) × z = x × (y × z).
(i) x = $$\frac{3}{5}$$, y = $$\frac{-7}{3}$$, z = $$\frac{8}{11}$$
(x × y) × z = x × (y × z)

(ii) x = $$\frac{-7}{11}$$, y = $$\frac{4}{5}$$, z = $$\frac{3}{8}$$
(x × y) × z = x × (y × z)

(iii) x = $$\frac{-4}{7}$$, y = $$\frac{-3}{8}$$, z = $$\frac{16}{5}$$
(x × y) × z = x × (y × z)

(iv) x = -3, y = $$\frac{-4}{9}$$, z = $$\frac{-7}{3}$$
(x × y) × z = x × (y × z)

Question 4.
Verify the property x × (y + z) = x × y + x × z by taking
(i) x = $$\frac{1}{3}$$, y = $$\frac{1}{5}$$, z = $$\frac{1}{7}$$
L.H.S = x × (y + z)
= $$\frac{1}{3} \times\left(\frac{1}{5}+\frac{1}{7}\right)$$
= $$\frac{1}{3} \times\left(\frac{1 \times 7+1 \times 5}{35}\right)$$
= $$\frac{1}{3} \times\left(\frac{7+5}{35}\right)$$
= $$\frac{1}{3} \times \frac{12}{35}=\frac{12}{105}$$

R.H.S = x × y + x × z
= $$\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{1}{7}$$
= $$\frac{1 \times 1}{3 \times 5}+\frac{1 \times 1}{3 \times 7}=\frac{1}{15}+\frac{1}{21}$$
= $$\frac{1 \times 7+1 \times 5}{105}=\frac{7+5}{105}$$
= $$\frac{12}{105}$$

L.H.S = R.H.S
Hence Verified

(ii) x = $$\frac{-3}{7}$$, y = $$\frac{2}{5}$$, z = $$\frac{1}{2}$$
L.H.S

L.H.S = R.H.S
Hence verified

Question 5.
Show that:
$$\frac{-4}{3} \times\left(\frac{2}{5}+\frac{-7}{10}\right)=\left(\frac{-4}{3} \times \frac{2}{5}\right)+\left(\frac{-4}{3} \times \frac{-7}{10}\right)$$
L.H.S = $$\frac{-4}{3} \times\left(\frac{2}{5}+\frac{-7}{10}\right)$$
= $$\frac{-4}{3} \times\left(\frac{2 \times 2+1 \times-7}{10}\right)$$
= $$\frac{-4}{3} \times\left(\frac{4-7}{10}\right)$$
= $$\frac{-4}{3} \times \frac{-3}{10}=\frac{12}{30}$$

R.H.S = $$\left(\frac{-4}{3} \times \frac{2}{5}\right)+\left(\frac{-4}{3} \times \frac{-7}{10}\right)$$
= $$\left(\frac{-4 \times 2}{3 \times 5}\right)+\left(\frac{-4 \times-7}{3 \times 10}\right)$$
= $$\frac{-8}{15}+\frac{28}{30}$$
= $$\frac{-16+28}{30}=\frac{12}{30}$$

L.H.S = R.H.S
Hence verified

Question 6.
Show that:
$$\frac{3}{5} \times\left(\frac{-1}{7}-\frac{5}{14}\right)=\left(\frac{3}{5} \times \frac{-1}{7}\right)-\left(\frac{3}{5} \times \frac{5}{14}\right)$$
L.H.S = $$\frac{3}{5} \times\left(\frac{-1}{7}-\frac{5}{14}\right)$$
= $$\frac{3}{5} \times\left(\frac{-1 \times 2-5 \times 1}{14}\right)$$
= $$\frac{3}{5} \times\left(\frac{-2-5}{14}\right)$$
= $$\frac{3}{5} \times \frac{-7}{14}=\frac{3 \times-7}{5 \times 14}=\frac{-21}{70}$$

R.H.S = $$\left(\frac{3}{5} \times \frac{-1}{7}\right)-\left(\frac{3}{5} \times \frac{5}{14}\right)$$
= $$\frac{3 \times-1}{5 \times 7}-\frac{3 \times 5}{5 \times 14}$$
= $$\frac{-3}{35}-\frac{15}{70}$$
= $$\frac{-3 \times 2-15 \times 1}{70}=\frac{-6-15}{70}$$
= $$\frac{-21}{70}$$

L.H.S = R.H.S
Hence verified

Question 7.
Simplify and express the result in standard form.
(i) -4 × $$\left(\frac{7}{3}-\frac{9}{10}\right)$$

(ii) $$\frac{7}{3} \times\left(\frac{9}{8}+3\right)$$

(iii) $$\left(\frac{-4}{3}+\frac{5}{7}\right) \times \frac{7}{9}$$

(iv) $$\left(\frac{5}{4}-\frac{6}{20}\right) \times \frac{8}{11}$$

Question 8.
Fill in the blanks:
(i) $$\frac{-4}{7}$$ × _______ = $$\frac{-4}{7}$$
1

(ii) $$\frac{3}{8}$$ × _______ = $$\frac{-3}{8}$$
-1

(iii) $$\left(\frac{-1}{3} \times \frac{4}{5}\right) \times \frac{6}{7}$$ = _______ × $$\left(\frac{4}{5} \times \frac{6}{7}\right)$$
$$\frac{-1}{3}$$

(iv) $$\frac{5}{3} \times\left(\frac{-7}{8} \times \frac{11}{3}\right)$$ = ($$\frac{5}{3}$$ × _______) × $$\frac{11}{3}$$
$$\frac{-7}{8}$$

(v) $$\frac{3}{7} \times \frac{-6}{11}=\frac{-6}{11}$$ × _______
$$\frac{3}{7}$$

(vi) $$\frac{4}{3}$$ × _______ = 0
0

(vii) For any rational number x, x × 5 = x + x + = _______ times
(viii) $$\frac{2}{3} \times\left(\frac{7}{5}-\frac{2}{9}\right)=\frac{2}{3} \times \frac{7}{5}$$ – _______
$$\frac{2}{3} \times \frac{2}{9}$$
(ix) $$\frac{-5}{7} \times \frac{1}{3}+\frac{-5}{7} \times \frac{1}{6}=\frac{-5}{7}$$ × ($$\frac{1}{3}$$ + _______)
$$\frac{1}{6}$$
(x) $$\frac{4}{7} \times \frac{2}{3}-\frac{4}{7} \times \frac{5}{6}=\frac{4}{7}$$ × (_______ – _______)
$$\frac{2}{3}, \frac{5}{6}$$