Exercise 1.2 - Chapter 1 Numbers 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Fill in the blanks:
(i) The value of $\frac{-5}{12}+\frac{7}{15}=$
Answer:
$\frac{1}{20}$
(ii) The value of $\left(\frac{-3}{6}\right) \times\left(\frac{18}{-9}\right)$ is $=$
Answer:
1
(iii) The value of $\left(\frac{-15}{23}\right) \div\left(\frac{30}{-46}\right)$ is
Answer:
1
(iv) The rational number does not have a reciprocal.
Answer:
0

(v) The multiplicative inverse of $-1$ is
Answer:
$-1$

Question $2 .$
Say True or False
(i) All rational numbers have an additive inverse.
Answer:
True

(ii) The rational numbers that are equal to their additive inverses are 0 and $-1$.
Answer:
False
(iii) The additive inverse of $\frac{-11}{-17}$ is $\frac{11}{17}$
Answer:
False
(iv) The rational number which is its own reciprocal is $-1$.
Answer:
True
(v) The multiplicative inverse exists for all rational numbers.
Answer:
False
 

Question $3 .$
Find the sum
(i) $\frac{7}{5}+\frac{3}{5}$
(ii) $\frac{7}{5}+\frac{5}{7}$
(iii) $\frac{6}{5}+\left(\frac{-14}{15}\right)$
(iv) $-4 \frac{2}{3}+7 \frac{5}{12}$
Answer:
(i) $\frac{7}{5}+\frac{3}{5}$
$\frac{7}{5}+\frac{3}{5}=\frac{7+3}{5}=\frac{10}{5}=2$

(ii) $\frac{7}{5}+\frac{5}{7}$
$\frac{7}{5}+\frac{5}{7}=\frac{7 \times 7+5 \times 5}{35}=\frac{49+25}{35}=\frac{74}{35}$
(iii) $\frac{6}{5}+\left(\frac{-14}{15}\right)$
$\frac{6}{5}+\left(\frac{-14}{15}\right)=\frac{6 \times 3+(14)}{15}=\frac{18+(-14)}{5}=\frac{4}{5}$
(iv) $-4 \frac{2}{3}+7 \frac{5}{12}$
$-4 \frac{2}{3}+7 \frac{5}{12}=\frac{14}{3}+\frac{18}{12}=\frac{-14 \times 4+89}{12}=\frac{-56+89}{12}=\frac{-33}{\underset{42}{4}}=\frac{-11}{4}$

Question $4 .$
Subtract $\frac{-8}{44}$ from $\frac{-17}{11}$
Answer:

$\frac{-17}{11}-\left(\frac{-8}{44}\right)=\frac{-17}{11}+\frac{8}{44}=\frac{-17 \times 4+8}{44}=\frac{-68+8}{44}=\frac{60}{44}=\frac{-15}{11}$

Question 5 .
Evaluate
(i) $\frac{9}{132} \times \frac{-11}{3}$
(ii) $\frac{-7}{27} \times \frac{24}{-35}$
Answer:
(i) $\frac{9}{132} \times \frac{-11}{3}$
$\frac{\not \beta}{132} \times \frac{-111}{\not 2}=\frac{-1}{4}$

Question $6 .$
Divide
(i) $\frac{-21}{5}$ by $\frac{-7}{-10}$
(ii) $\frac{-3}{13}$ by $-3$
(iii) $-2$ by $\frac{-6}{15}$
Answer:

Question $7 .$
Find $(a+b) \div(a-b)$ if
(i) $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{2}{3}$
(ii) $\mathrm{a}=\frac{-3}{5}, \mathrm{~b}=\frac{2}{15}$
Answer:
(i) $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{2}{3}$
$\begin{aligned}
&a+b=\frac{1}{2}+\frac{2}{3}=\frac{1 \times 3+2 \times 2}{6}=\frac{3+4}{6}=\frac{7}{6} \\
&a-b \quad=\frac{1}{2}-\frac{2}{3}=\frac{1 \times 3-2 \times 2}{6}=\frac{3-4}{6}=\frac{-1}{6}
\end{aligned}$

$(a+b) \div(a-b)=\frac{7}{6} \div \frac{-1}{6}=\frac{7}{6} \times \frac{6}{-1}=-7$
(ii) $\mathrm{a}=\frac{-3}{5}, \mathrm{~b}=\frac{2}{15}$
$\begin{aligned}
a+b &=\frac{-3}{5}+\frac{2}{15}=\frac{-3 \times 3+2}{15}=\frac{-9+2}{15}=\frac{-7}{15} \\
a-b &=\frac{-3}{5}-\frac{2}{15}=\frac{-3 \times 3-2}{15}=\frac{-9-2}{15}=\frac{-11}{15} \\
(a+b) \div(a-b) &=\frac{-7}{15} \div \frac{-11}{15}=\frac{-7}{15} \times \frac{15}{-11}=\frac{7}{11}
\end{aligned}$

Question $8 .$
Simplify $\frac{1}{2}+\left(\frac{3}{2}-\frac{2}{5}\right) \div \frac{3}{10} \times 3$ and show that it is a rational number between 11 and 12 .
Answer:
$\begin{aligned}
\frac{1}{2}+\left(\frac{3}{7}-\frac{2}{5}\right) \div \frac{3}{10} \times 3 &=\frac{1}{2}+\left(\frac{15-4}{10}\right) \div \frac{3}{10} \times 3=\frac{1}{2}+\frac{11}{10} \times \frac{10}{\not} \times \not \beta \\
&=\frac{1}{2}+11=11 \frac{1}{2}=\frac{23}{2}
\end{aligned}$

Question $9 .$
Simplify
(i) $\left[\frac{11}{8} \times\left(\frac{-6}{33}\right)\right]+\left[\frac{1}{3}+\left(\frac{3}{5} \div \frac{9}{20}\right)\right]-\left[\frac{4}{7} \times \frac{-7}{5}\right]$
(ii) $\left[\frac{4}{3} \div\left(\frac{8}{-7}\right)\right]-\left[\frac{3}{4} \times \frac{4}{3}\right]+\left[\frac{4}{3} \times\left(\frac{-1}{4}\right)\right]$
Answer:
(i) $\left[\frac{11}{8} \times\left(\frac{-6}{33}\right)\right]+\left[\frac{1}{3}+\left(\frac{3}{5} \div \frac{9}{20}\right)\right]-\left[\frac{4}{7} \times \frac{-7}{5}\right]$

$=-\frac{1}{4}+\frac{5}{3}+\frac{4}{5}=\frac{-15+100+48}{60}=\frac{133}{60}$
(ii) $\left[\frac{4}{3} \div\left(\frac{8}{-7}\right)\right]-\left[\frac{3}{4} \times \frac{4}{3}\right]+\left[\frac{4}{3} \times\left(\frac{-1}{4}\right)\right]$
$\left[\frac{4}{3} \div\left(\frac{8}{-7}\right)\right]-\left[\frac{3}{4} \times \frac{4}{3}\right]+\left[\frac{4}{3} \times\left(\frac{-1}{4}\right)\right]=$

$\begin{aligned}
&=\left(\frac{-7}{6}\right)-1+\left(\frac{-1}{3}\right) \\
&=\frac{-7-6+(-2)}{6}=\frac{-15}{6}=\frac{-5}{2}
\end{aligned}$

Question $10 .$
A student had multiplied a number by $\frac{4}{3}$ instead of dividing it by $\frac{4}{3}$ and got 70 more than the correct answer. Find the number.
Answer:

Let the number $=a$
$\begin{aligned}
a \times \frac{4}{3}-a \div \frac{4}{3} &=70 \\
a \times \frac{4}{3}-a \times \frac{3}{4} &=70 \\
a\left[\frac{4}{3}-\frac{3}{4}\right] &=70 \\
a\left[\frac{16-9}{12}\right] &=70 \\
a\left[\frac{7}{12}\right] &=70 \\
a &=70 \\
a &=70 \times \frac{12}{7}=120 \\
a[&=120
\end{aligned}$

Objective Type Questions

Question 11.
The standard form of the sum is
(A) 1
(B) $\frac{-1}{2}$
(C) $\frac{1}{12}$
(D) $\frac{1}{22}$
Answer:
1

Question $12 .$
$\left(\frac{3}{4}-\frac{5}{8}\right)+\frac{1}{2}=$
(A) $\frac{15}{64}$
(B) 1
(C) $\frac{5}{8}$
(D) $\frac{1}{16}$
Answer:
(C) $\frac{5}{8}$

Question $13 .$
$\frac{3}{4}+\left(\frac{5}{8}+\frac{1}{2}\right)=$
(A) $\frac{13}{10}$
(B) $\frac{2}{3}$
(C) $\frac{3}{2}$
(D) $\frac{5}{8}$
Answer:
(B) $\frac{2}{3}$

Question $15 .$
Which of these rational number which have additive inverse?
(A) 7
(B) $\frac{-5}{7}$
(C) 0
(D) all of these
Answer:
(D) all of these