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.