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) \(-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}\)

Answer:

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) \(\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}\)

Answer:

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) \(\frac{2}{8}\)

Answer:

Additive inverse of \(\frac{2}{8}\) is \(\frac{-2}{8}\)

∴ \(\frac{2}{8}+\frac{-2}{8}\) = 0

(ii) \(\frac{-5}{9}\)

Answer:

Additive inverse of \(\frac{-5}{9}\) is \(\frac{5}{9}\)

∴ \(\frac{-5}{9}+\frac{5}{9}\) = 0

(iii) \(\frac{-6}{-5}\)

Answer:

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) \(\frac{2}{-9}\)

Answer:

Additive inverse of \(\frac{2}{-9}\) is \(\frac{2}{9}\)

∴ \(\frac{2}{-9}+\frac{2}{9}\) = 0

(v) \(\frac{19}{-6}\)

Answer:

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 = \(\frac{11}{15}\)

Answer:

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 = \(-\frac{13}{17}\)

Answer:

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

Answer:

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) \(\frac{-13}{19}\)

Answer:

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) \(\frac{1}{5}\)

Answer:

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) \(\frac{-5}{8} \times \frac{-3}{7}\)

Answer:

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 × \(\frac{-2}{5}\)

Answer:

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

(vi) -1

Answer:

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 × \(\frac{-4}{5}=-\frac{4}{5}\)

Answer:

Multiplicative identity (a × 1) = (1 × a) = a

(ii) \(-\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\)

Answer:

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

Answer:

Multiplicative inverse identity a × \(\frac{1}{a}\) = 1

Question 6.

Multiply \(\frac{6}{13}\) by the reciprocal of \(\frac{-7}{16}\).

Answer:

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 ?

Answer:

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?

Answer:

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.

Answer:

0

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

Answer:

1 and (- 1)

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

Answer:

0

Question 11.

Fill in the blanks.

(i) Zero has ___________ reciprocal.

Answer:

No

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

Answer:

(1) and (- 1)

(iii) The reciprocal of – 5 is ___________.

Answer:

\(\frac{-1}{5}\)

(iv) Reciprocal of \(\frac{1}{x}\), where x ≠ 0 is ___________.

Answer:

x

(v) The product of two rational numbers is always a ___________.

Answer:

A rational number

(vi) The reciprocal of a positive rational number is ___________.

Answer:

Positive rational