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