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