# AP 8th Class Maths 1st Chapter Rational Numbers Exercise 1.1 Solutions

Well-designed AP Board Solutions Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Rational Numbers Class 8 Exercise 1.1 Solutions – 8th Class Maths 1.1 Exercise Solutions

Question 1.
Using appropriate properties find.
(i) $$-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}$$
Now using the commutativity of Rational numbers
We can re-write the given as
$$\left(\frac{-2}{3} \times \frac{3}{5}\right)-\left(\frac{3}{5} \times \frac{1}{6}\right)+\frac{5}{2}$$
= $$\frac{3}{5}\left[\frac{-2}{3}-\frac{1}{6}\right]+\frac{5}{2}=\frac{3}{5}\left[\frac{-4}{6}-\frac{1}{6}\right]+\frac{5}{2}$$ (distributivity)
= $$\left(\frac{3}{5} \times \frac{-5}{6}\right)+\frac{5}{2}=\frac{-1}{2}+\frac{5}{2}=\frac{-1+5}{2}=\frac{4}{2}$$
= 2

(ii) $$\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}$$
Using Commutativity
$$\left(\frac{2}{5} \times \frac{-3}{7}\right)+\left(\frac{2}{5} \times \frac{1}{14}\right)-\frac{1}{6} \times \frac{3}{2}$$
= $$\frac{2}{5}\left[\frac{-3}{7}+\frac{1}{14}\right]-\frac{1}{6} \times \frac{3}{2}$$ (distributivity)
= $$\frac{2}{5}\left[\frac{-6}{14}+\frac{1}{14}\right]-\frac{1}{6} \times \frac{3}{2}$$

= $$\frac{-1}{7}-\frac{1}{4}=\frac{-4-7}{28}=\frac{-11}{28}$$

Question 2.
Write the additive inverse of each of the following.
(i) $$\frac{2}{8}$$
Additive inverse of $$\frac{2}{8}$$ is $$\frac{-2}{8}$$
∴ $$\frac{2}{8}+\frac{-2}{8}$$ = 0

(ii) $$\frac{-5}{9}$$
Additive inverse of $$\frac{-5}{9}$$ is $$\frac{5}{9}$$
∴ $$\frac{-5}{9}+\frac{5}{9}$$ = 0

(iii) $$\frac{-6}{-5}$$
Additive inverse of $$\frac{-6}{-5}$$ is $$\frac{-6}{5}$$
∴ $$\frac{-6}{5}=\frac{6}{5} \Rightarrow \frac{6}{5}+\left(\frac{-6}{5}\right)$$ = 0

(iv) $$\frac{2}{-9}$$
Additive inverse of $$\frac{2}{-9}$$ is $$\frac{2}{9}$$
∴ $$\frac{2}{-9}+\frac{2}{9}$$ = 0

(v) $$\frac{19}{-6}$$
Additive inverse of $$\frac{19}{-6}$$ is $$\frac{19}{6}$$
∴ $$\frac{-19}{6}+\frac{19}{6}$$ = 0

Question 3.
Verify that – (- x) = x for.
(i) x = $$\frac{11}{15}$$
We know additive inverse of a = – a,
now let x = $$\frac{11}{15}$$, then its additive inverse of ‘x’ is – x = $$\frac{-11}{15}$$ …….(1)
Now additive inverse of – x = – (- x) …..(2)
then additive inverse of $$\frac{-11}{15}=\frac{11}{15}$$ = x ………..(3)
So from the above it is clear that
x = – (- x) (verified)

(ii) x = $$-\frac{13}{17}$$
x = $$-\frac{13}{17}$$
Now writing additive inverses of above LHS, RHS.
i.e., additive inverse of x= -x
∴ -x = $$\frac{13}{17}$$ ….(2)
Again writing additive inverses of above eq. (2)
i.e., additive inverse of – x = – (- x)
and additive inverse of  which is equal to ‘x’ ………..(3)
From above eq (2) and (3) we can say -(-x) = x.

Question 4.
Find the multiplicative inverse of the following.
(i) -13
We know a × $$\frac{1}{a}$$= 1 in which (a) and a
$$\left(\frac{1}{\mathrm{a}}\right)$$ are multiplicative inverses of each other.
Let given -13 = a, then its multiplicative inverse of a = $$\frac{1}{\mathrm{a}}=\frac{1}{-13}=\frac{-1}{13}$$

(ii) $$\frac{-13}{19}$$
Let given = $$\frac{-13}{19}$$ = a, then it’s multiplicative inverse of a = $$\frac{1}{a}=\frac{1}{\left(\frac{-13}{19}\right)}=\frac{-19}{13}$$

(iii) $$\frac{1}{5}$$
It is in the form of $$\frac{1}{a}$$
So, multiplicative inverse of $$\left(\frac{1}{a}\right)$$ is (a).
∴ Multiplicative inverse of $$\left(\frac{1}{5}\right)$$ = 5

(iv) $$\frac{-5}{8} \times \frac{-3}{7}$$
Now multiplicative inverse of $$\left(\frac{-5}{8}\right)=\frac{-8}{5}$$
Now multiplicative inverse of $$\left(\frac{-3}{7}\right)=\frac{-7}{3}$$
then multiplicative inverse of $$\left(\frac{-5}{8} \times \frac{-3}{7}\right)=\frac{-8}{5} \times \frac{-7}{3}=\frac{56}{15}$$

(v) -1 × $$\frac{-2}{5}$$
-1 × $$\frac{-2}{5}=\frac{2}{5}$$ then multiplicative inverse of $$\frac{2}{5}=\frac{5}{2}$$

(vi) -1
Multiplicative inverse of (-1) = $$\frac{1}{(-1)}$$ = -1

Question 5.
Name the property under multiplication used in each of the following.
(i) $$\frac{-4}{5}$$ × 1 = 1 × $$\frac{-4}{5}=-\frac{4}{5}$$
Multiplicative identity (a × 1) = (1 × a) = a

(ii) $$-\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}$$
Commutativity property (a × b) = (b × a)

(iii) $$\frac{-19}{29} \times \frac{29}{-19}-\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}$$ = 1
Multiplicative inverse identity a × $$\frac{1}{a}$$ = 1

Question 6.
Multiply $$\frac{6}{13}$$ by the reciprocal of $$\frac{-7}{16}$$.
Reciprocal of given $$\frac{-7}{16}=\left(\frac{-16}{7}\right)$$
Now, multipliying $$\left(\frac{6}{13}\right)$$ by the reciprocal of $$\left(\frac{-7}{16}\right)=\frac{6}{13} \times \frac{-16}{7}=\frac{-96}{91}$$

Question 7.
Tell what property allows you to compute $$\frac{1}{3} \times\left(6 \times \frac{4}{3}\right)$$ as $$\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}$$.
Sol. From the property of Associativity of rational numbers under multiplication a x (b x c) = (a x b) x c 1 4

Question 8.
Is $$\frac{8}{9}$$ the multiplicative inverse of -1$$\frac{1}{8}$$ ? Why or why not ?
Given = -1$$\frac{1}{8}=\frac{-8+1}{8}=\frac{-7}{8}$$
Multiplicative inverse of (-1$$\frac{1}{8}$$) = $$\frac{-8}{7}$$
So, multiplication inverse of -1$$\frac{1}{8}$$ = $$\frac{-8}{7}$$ but not $$\frac{8}{9}$$.

Question 9.
Is 0.3 the multiplicative inverse of 3$$\frac{1}{3}$$? Why or why not?
Given 3$$\frac{1}{3}=\frac{3(3)+1}{3}=\frac{9+1}{3}=\frac{10}{3}$$ then multiplication inverse of $$\left(\frac{10}{3}\right)=\frac{3}{10}$$ = 0.3
So, 0.3 is the multiplicative inverse of given 3$$\frac{1}{3}$$

Question 10.
Write
(i) The rational number that does not have a reciprocal.
0

(ii) The rational numbers that are equal to their reciprocals.
1 and (- 1)

(iii) The rational number that is equal to its negative.
0

Question 11.
Fill in the blanks.
(i) Zero has ___________ reciprocal.
No

(ii) The numbers ___________ and ___________ are their own reciprocals.
(1) and (- 1)

(iii) The reciprocal of – 5 is ___________.
$$\frac{-1}{5}$$
(iv) Reciprocal of $$\frac{1}{x}$$, where x ≠ 0 is ___________.