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 \(\frac{-3}{8}\)
Answer:
\(\frac{11}{7} \times \frac{-3}{8}=\frac{11 \times-3}{7 \times 8}\)
= \(\frac{-33}{56}\) (Standard form)
(ii) \(\frac{-7}{4}\) by \(\frac{2}{3}\)
Answer:
\(\frac{-7}{4} \times \frac{2}{3}=\frac{-7 \times 2}{4 \times 3}\)
= \(\frac{-14}{12}=\frac{-7}{6}\) (Standard form)
(iii) \(\frac{-3}{12}\) by -48
Answer:
\(\frac{-3}{12} \times \frac{-48}{1}\)
= \(\frac{-3 \times-48}{12 \times 1}=\frac{144}{12}\) (Standard form)
(iv) \(\frac{-14}{9}\) by \(\frac{-3}{7}\)
Answer:
\(\frac{-14}{9} \times \frac{-3}{7}=\frac{-14 \times-3}{9 \times 7}\)
= \(\frac{42}{63}=\frac{2}{3}\) (Standard form)
(v) \(\frac{23}{5}\) by \(\frac{-25}{11}\)
Answer:
\(\frac{23}{5} \times \frac{-25}{11}=\frac{23 \times-25}{5 \times 11}\)
= \(\frac{-575}{55}=\frac{-115}{11}\) (Standard form)
(vi) 7 by \(\frac{-15}{63}\)
Answer:
\(\frac{7}{1} \times \frac{-15}{63}\)
= \(\frac{7}{1} \times \frac{-15}{63}\)
= \(\frac{-5}{3}\) (Standard form)
Question 2.
For the following values of x and y, verify x × y = y × x
(i) x = \(\frac{7}{9}\), y = \(\frac{3}{2}\)
Answer:
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 = \(\frac{5}{8}\)
Answer:
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 = \(\frac{-5}{11}\)
Answer:
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 = \(\frac{-96}{51}\)
Answer:
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 = \(\frac{8}{11}\)
Answer:
(x × y) × z = x × (y × z)
(ii) x = \(\frac{-7}{11}\), y = \(\frac{4}{5}\), z = \(\frac{3}{8}\)
Answer:
(x × y) × z = x × (y × z)
(iii) x = \(\frac{-4}{7}\), y = \(\frac{-3}{8}\), z = \(\frac{16}{5}\)
Answer:
(x × y) × z = x × (y × z)
(iv) x = -3, y = \(\frac{-4}{9}\), z = \(\frac{-7}{3}\)
Answer:
(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 = \(\frac{1}{7}\)
Answer:
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 = \(\frac{1}{2}\)
Answer:
L.H.S
L.H.S = R.H.S
Hence verified
Question 5.
Show that:
\(\frac{-4}{3} \times\left(\frac{2}{5}+\frac{-7}{10}\right)=\left(\frac{-4}{3} \times \frac{2}{5}\right)+\left(\frac{-4}{3} \times \frac{-7}{10}\right)\)
Answer:
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:
\(\frac{3}{5} \times\left(\frac{-1}{7}-\frac{5}{14}\right)=\left(\frac{3}{5} \times \frac{-1}{7}\right)-\left(\frac{3}{5} \times \frac{5}{14}\right)\)
Answer:
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 × \(\left(\frac{7}{3}-\frac{9}{10}\right)\)
Answer:
(ii) \(\frac{7}{3} \times\left(\frac{9}{8}+3\right)\)
Answer:
(iii) \(\left(\frac{-4}{3}+\frac{5}{7}\right) \times \frac{7}{9}\)
Answer:
(iv) \(\left(\frac{5}{4}-\frac{6}{20}\right) \times \frac{8}{11}\)
Answer:
Question 8.
Fill in the blanks:
(i) \(\frac{-4}{7}\) × _______ = \(\frac{-4}{7}\)
Answer:
1
(ii) \(\frac{3}{8}\) × _______ = \(\frac{-3}{8}\)
Answer:
-1
(iii) \(\left(\frac{-1}{3} \times \frac{4}{5}\right) \times \frac{6}{7}\) = _______ × \(\left(\frac{4}{5} \times \frac{6}{7}\right)\)
Answer:
\(\frac{-1}{3}\)
(iv) \(\frac{5}{3} \times\left(\frac{-7}{8} \times \frac{11}{3}\right)\) = (\(\frac{5}{3}\) × _______) × \(\frac{11}{3}\)
Answer:
\(\frac{-7}{8}\)
(v) \(\frac{3}{7} \times \frac{-6}{11}=\frac{-6}{11}\) × _______
Answer:
\(\frac{3}{7}\)
(vi) \(\frac{4}{3}\) × _______ = 0
Answer:
0
(vii) For any rational number x, x × 5 = x + x + = _______ times
Answer:
x + x + x
(viii) \(\frac{2}{3} \times\left(\frac{7}{5}-\frac{2}{9}\right)=\frac{2}{3} \times \frac{7}{5}\) – _______
Answer:
\(\frac{2}{3} \times \frac{2}{9}\)
(ix) \(\frac{-5}{7} \times \frac{1}{3}+\frac{-5}{7} \times \frac{1}{6}=\frac{-5}{7}\) × (\(\frac{1}{3}\) + _______)
Answer:
\(\frac{1}{6}\)
(x) \(\frac{4}{7} \times \frac{2}{3}-\frac{4}{7} \times \frac{5}{6}=\frac{4}{7}\) × (_______ – _______)
Answer:
\(\frac{2}{3}, \frac{5}{6}\)