# PSEB 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers Ex 1.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1

1. Using appropriate properties find.

Question (i).
$$-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}$$
Solution:
$$-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}$$
= $$-\frac{2}{3} \times \frac{3}{5}-\frac{3}{5} \times \frac{1}{6}+\frac{5}{2}$$ (Commutative)
= $$\frac{3}{5} \times\left[-\frac{2}{3}-\frac{1}{6}\right]+\frac{5}{2}$$ (Distributive)
= $$\frac{3}{5}\left[\frac{-4-1}{6}\right]+\frac{5}{2}$$
= $$\frac{3}{5}\left[\frac{-5}{6}\right]+\frac{5}{2}$$
= $$\frac{3}{5} \times \frac{-5}{6}+\frac{5}{2}$$
= $$-\frac{1}{2}+\frac{5}{2}$$
= $$\frac{-1+5}{2}$$
= $$\frac {4}{2}$$
= 2

Question (ii).
$$\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}$$
Solution:
$$\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}$$
= $$\frac{2}{5} \times\left(\frac{-3}{7}\right)+\frac{1}{14} \times \frac{2}{5}-\frac{1}{6} \times \frac{3}{2}$$ (Commutative)
= $$\frac{2}{5} \times\left(\frac{-3}{7}+\frac{1}{14}\right)-\frac{1}{6} \times \frac{3}{2}$$ (Distributive)
= $$\frac{2}{5} \times\left[\frac{-6+1}{14}\right]-\frac{1}{4}$$
= $$\frac{2}{5} \times \frac{-5}{14}-\frac{1}{4}$$
= $$-\frac{1}{7}-\frac{1}{4}=\frac{-4-7}{28}$$
= $$\frac{-11}{28}$$

2. Write the additive inverse of each of the following:

Question (i).
$$\frac{2}{8}$$
Solution:
Additive inverse of $$\frac{2}{8}$$ = $$\frac{-2}{8}$$

Question (ii).
$$\frac{-5}{9}$$
Solution:
Additive inverse of $$\frac{-5}{9}$$ = $$\frac{5}{9}$$

Question (iii).
$$\frac{-6}{-5}$$
Solution:
Additive inverse of $$\frac{-6}{-5}$$ means $$\frac{6}{5}$$ = $$\frac{-6}{5}$$

Question (iv).
$$\frac{2}{-9}$$
Solution:
Additive inverse of $$\frac{2}{-9}$$ = $$\frac{2}{9}$$

Question (v).
$$\frac{19}{-6}$$
Solution:
Additive inverse of $$\frac{19}{-6}$$ = $$\frac{19}{6}$$

3. Verify that – (- x) = x for

(i) x = $$\frac {11}{15}$$
Solution:
x = $$\frac {11}{15}$$
∴ (-x) = $$\left(\frac{-11}{15}\right)$$
-(-x) = –$$\left(\frac{-11}{15}\right)$$
= $$\frac {11}{15}$$ = x
∴ -(-x) = x

(ii) x = $$\frac {-13}{17}$$
Solution:
x = $$\frac {-13}{17}$$
∴ (-x) = $$\left(\frac{-13}{17}\right)$$
= $$\frac {13}{17}$$
-(-x) = –$$\left(\frac{-13}{17}\right)$$
= $$\frac {-13}{17}$$ = x
∴ -(-x) = x

4. Find the multiplicative inverse of the following:

Question (i).
-13
Solution:
Multiplicative inverse of -13 = $$\frac {-1}{13}$$

Question (ii).
$$\frac {-13}{19}$$
Solution:
Multiplicative inverse of $$\frac {-13}{19}$$ $$\frac {-19}{13}$$

Question (iii).
$$\frac {1}{5}$$
Solution:
Multiplicative inverse of $$\frac {1}{5}$$ = 5

Question (iv).
$$\frac{-5}{8} \times \frac{-3}{7}$$
Solution:
$$\left(\frac{-5}{8}\right) \times\left(\frac{-3}{7}\right)$$
= $$\frac{(-5 \times-3)}{8 \times 7}$$
= $$\frac {15}{56}$$
Multiplicative inverse of $$\frac {15}{56}$$ = $$\frac {56}{15}$$

Question (v) .
1 × $$\frac {-2}{5}$$
Solution:
-1 × $$\frac {-2}{5}$$ = $$\frac{(-1 \times-2)}{5}$$
= $$\frac {2}{5}$$
Multiplicative inverse of $$\frac {2}{5}$$ = $$\frac {5}{2}$$

Question (vi).
-1
Solution:
Multiplicative inverse of -1 = (-1)
(∵ $$\frac{1}{(-1)}$$ = (-1))

5. Name the property under multiplication used in each of the following:

Question (i).
$$\frac{-4}{5} \times 1=1 \times \frac{-4}{5}=-\frac{4}{5}$$
Solution:
1 is the multiplicative identity.

Question (ii).
$$-\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}$$
Solution:
Commutative property of multiplication.

Question (iii).
$$\frac{-19}{29} \times \frac{29}{-19}=1$$
Solution:
Existence of multiplicative inverse.

6. Multiply $$\frac {6}{13}$$ by the reciprocal of $$\frac {-7}{16}$$.
Solution:
Reciprocal of $$\frac{-7}{16}=\frac{-16}{7}$$
∴ $$\frac{6}{13} \times \frac{-16}{7}$$
= $$\frac{6 \times(-16)}{13 \times 7}$$
= $$\frac {-96}{91}$$

7. Tell what property allows you to compute.
$$\frac{1}{3} \times\left(6 \times \frac{4}{3}\right)$$ as $$\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}$$
Solution:
In computing
$$\frac{1}{3} \times\left(6 \times \frac{4}{3}\right)$$ as $$\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}$$
we use the associativity.

8. Is $$\frac {8}{9}$$ the multiplicative inverse of – 1 $$\frac {1}{8}$$ ? Why or why not ?
Solution:
$$-1 \frac{1}{8}=\frac{-9}{8}$$
$$\frac{8}{9} \times \frac{-9}{8}$$ = (-1)
∴ $$\frac {8}{9}$$ is is not the multiplicative inverse of -1 $$\frac {1}{8}$$ as product of two multiplicative inverse is always 1.

9. Is 0.3 the multiplicative inverse of 3 $$\frac {1}{3}$$ ? Why or why not?
Solution:
0.3 = $$\frac {3}{10}$$ and 3 $$\frac {1}{3}$$ = $$\frac {10}{3}$$
$$\frac{3}{10} \times \frac{10}{3}$$ = 1
∴ the multiplicative inverse of 3 $$\frac {1}{3}$$ is 0.3.

10. Write:

Question (i).
The rational number that does not have a reciprocal.
Solution:
The rational number that does not have a reciprocal is 0.

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

Question (iii).
The rational number that is equal to its negative.
Solution:
The rational number that is equal to its negative is zero (0).

11. Fill in the blanks:

Question (i).
Zero has ……………. reciprocal.
Solution:
Zero has no reciprocal.

Question (ii).
The numbers ……………. and ……………. are their own reciprocals.
Solution:
The numbers 1 and -1 are their own reciprocals.

Question (iii).
The reciprocal of – 5 is …………….
Solution:
The reciprocal of – 5 is $$\frac {-1}{5}$$

Question (iv).
Reciprocal of $$\frac{1}{x}$$, where x ≠ 0 is …………….
Solution:
Reciprocal of $$\frac{1}{x}$$, where x ≠ 0 is x

Question (v) .
The product of two rational numbers is always a …………….
Solution:
The product of two rational numbers is always a rational number.

Question (vi).
The reciprocal of a positive rational number is …………….
Solution:
The reciprocal of a positive rational number is positive.