Exercise 1.1 - Chapter 1 Numbers 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.1$  : Chapter 1 Number Systems Term 1 Class 8th std Maths Guide Samacheer Kalvi Solutions
Question $1 .$
Fill in the blanks:
(i) $\frac{-19}{5}$ lies between the integers __and__
Answer:
$-4$ and $-3$
(ii) The decimal form of the rational number $\frac{15}{-4}$ is
Answer:
$-3.75$
(iii) The rational numbers $\frac{-8}{3}$ and $\frac{8}{3}$ are equidistant from
Answer:
0
(iv) The next rational number in the sequence $\frac{-15}{24}, \frac{20}{-32}, \frac{-25}{40}$ is
Answer:
$\frac{30}{-48}$
(v) The standard form of $\frac{58}{-78}$ is
Answer:
$\frac{-29}{39}$

Question $2 .$
Say True or False
(i) 0 is the smallest rational number.
Answer:
False
(ii) $\frac{-4}{5}$ lies to the left of $\frac{-3}{4}$.
Answer:
True
(iii) $\frac{-19}{5}$ is greater than $\frac{15}{-4}$.
Answer:
False

(iv) The average of two rational numbers lies between them.
Answer:
True
(v) There are an unlimited number of rational numbers between 10 and 11 .
Answer:
True

Question $3 .$
Find the rational numbers represented by each of the question marks marked on the following number lines.

Answer:
The number lies between $-3$ and 4 . The unit part between $-3$ and $-4$ is divided into 3 equal parts and the second part is asked.
$\therefore$ The required number is $-3 \frac{2}{3}=-\frac{11}{3}$

Answer:
The required number lies between 0 and $-1$. The unit part between 0 and $-1$ is divided into 5 equal parts, and the second part is taken.
$\therefore$ The required number is $-\frac{2}{5}$

Answer:
The required number lies between 1 and 2. The unit part between 1 and 2 is divided into 4
equal parts and the third part is taken.
$\therefore$ The required number is $1 \frac{3}{4}=\frac{7}{4}$
 

Question $4 .$
The points $\mathrm{S}, \mathrm{Y}, \mathrm{N}, \mathrm{C}, \mathrm{R}, \mathrm{A}, \mathrm{T}$, I and $\mathrm{O}$ on the number line are such that $\mathrm{CN}=\mathrm{NY}=\mathrm{YS}$ and $\mathrm{RA}=\mathrm{AT}=\mathrm{TI}=\mathrm{IO}$. Find the rational numbers represented by the letters $\mathrm{Y}, \mathrm{N}, \mathrm{A}, \mathrm{T}$ and $\mathrm{I}$.

Answer:
$\begin{aligned}
\mathrm{Y} &=-2+\frac{1}{3}=\frac{-6+1}{3}=\frac{-5}{3} \\
\mathrm{~N} &=\frac{-5}{3}+\frac{1}{3}=\frac{-5+1}{3}=\frac{-4}{3} \\
\mathrm{RA} &=\mathrm{AT}=\mathrm{TI}=\mathrm{IO}=\frac{1}{4} \\
\mathrm{~A} &=2+\frac{1}{4}=\frac{8+1}{4}=\frac{9}{4} \\
\mathrm{~T} &=\frac{9}{4}+\frac{1}{4}=\frac{9+1}{4}=\frac{10}{4} \\
\mathrm{I} &=\frac{10}{4}+\frac{1}{4}=\frac{10+1}{4}=\frac{11}{4}
\end{aligned}$

Question $5 .$
Draw a number line and represent the following rational numbers on it.

(i) $\frac{9}{4}$
(ii) $\frac{-8}{3}$
(iii) $\frac{-17}{-5}$
(iv) $\frac{15}{-4}$
Answer:
(i) $\frac{9}{4}$
$\frac{9}{4}=2 \frac{1}{4}$

(ii) $\frac{-8}{3}$
$\frac{-8}{3}=-2 \frac{2}{3}$
$-2 \frac{2}{3}$ lies between $-2$ and 3

$\begin{aligned}
&\text { (iii) } \frac{-17}{-5} \\
&\frac{-17}{-5}=3 \frac{2}{5}
\end{aligned}$
$3 \frac{2}{5}$ lies between 3 and 4 in the number line.

$\begin{aligned}
&\text { (iv) } \frac{15}{-4} \\
&\frac{15}{-4}=-3 \frac{3}{4}
\end{aligned}$
$-3 \frac{3}{4}$ lies between $-3$ and $-4$

Question $6 .$
Write the decimal form of the following rational numbers.
(i) $\frac{1}{11}$
(ii) $\frac{13}{4}$
(iii) $\frac{-18}{7}$
(iv) $1 \frac{2}{5}$
(v) $-3 \frac{1}{2}$
Answer:

Question $7 .$
List any five rational numbers between the given rational numbers.
(i) 2 and 0
(ii) $\frac{-1}{2}$ and $\frac{3}{5}$
(iii) $\frac{1}{4}$ and $\frac{7}{20}$
(iv) $\frac{-6}{4}$ and $\frac{-23}{10}$
Answer:
(i) 2 and 0
i.e., $\frac{-2}{1}$ and $\frac{0}{1}$
$\begin{aligned}
\frac{-2}{1} &=\frac{-2 \times 10}{1 \times 10}=\frac{-20}{10} \\
\frac{0}{1} &=\frac{0 \times 10}{1 \times 10}=\frac{0}{10}
\end{aligned}$
$\therefore$ Five rational number between $\frac{-20}{10}(=-2)$ and $\frac{0}{10}(=0)$ are $\frac{-20}{10}, \frac{-19}{10}, \frac{-18}{10}, \frac{-7}{10}, \frac{-6}{10}, \frac{-5}{10}, \frac{0}{10}(=0)$.
(ii) $\frac{-1}{2}$ and $\frac{3}{5}$
LCM of 2 and $5=2 \times 5=10$
$\begin{aligned}
-\frac{1}{2} &=\frac{-1 \times 5}{2 \times 5}=\frac{-5}{10} \\
\frac{3}{5} &=\frac{3 \times 2}{5 \times 2}=\frac{6}{10}
\end{aligned}$
$\therefore$ Five rational number between

$\frac{-1}{2}\left(=\frac{-5}{10}\right)$ and $\frac{3}{5}\left(=\frac{6}{10}\right)$ are $\frac{-3}{10}, \frac{-1}{10}, 0, \frac{1}{10}, \frac{2}{10}, \frac{5}{10}$
(iii) $\frac{1}{4}$ and $\frac{7}{20}$
$\frac{-1}{2}\left(=\frac{-5}{10}\right)$ and $\frac{3}{5}\left(=\frac{6}{10}\right)$ are $\frac{-3}{10}, \frac{-1}{10}, 0, \frac{1}{10}, \frac{2}{10}, \frac{5}{10}$
$\therefore$ Five rational number between
$\frac{1}{4}\left(=\frac{15}{60}\right)$ and $\frac{7}{20}\left(=\frac{21}{60}\right)$ are
$\frac{16}{60}, \frac{17}{60}, \frac{18}{60}, \frac{19}{60}, \frac{20}{60}$

$\begin{aligned}
&\text { (iv) } \frac{-6}{4} \text { and } \frac{-23}{10} \\
&\frac{-6}{4}=\frac{-6 \times 5}{4 \times 5}=\frac{-30}{20} \\
&\frac{-23}{10}=\frac{23 \times 2}{10 \times 2}=\frac{-46}{20}
\end{aligned}$
$\therefore$ Five rational number between
$\frac{-6}{4}\left(=\frac{-30}{20}\right)$ and $\frac{-23}{10}\left(=\frac{-46}{20}\right)$ are
$\frac{-31}{20}, \frac{-32}{20}, \frac{-33}{20}, \frac{-34}{20}, \frac{-35}{20}$

Question 8 .
Use the method of averages to write 2 rational numbers between $\frac{14}{5}$ and $\frac{16}{3}$

Answer:

The average of $a$ and $b$ is $\frac{1}{2}(a+b)$
The average of $\frac{14}{5}$ and $\frac{16}{3}$ is $C_{1}=\frac{1}{2}\left(\frac{14}{5}+\frac{16}{3}\right)$
$C_{1}=\frac{1}{2}\left(\frac{42+80}{15}\right)$
$\begin{aligned}
C_{1} &=\frac{122}{30} \\
C_{1} &=\frac{61}{15} \\
\frac{14}{5} &<\frac{61}{15}<\frac{16}{3}
\end{aligned}$
The average of $\frac{14}{5}$ and $\frac{61}{15}$ is $\mathrm{C}_{2}=\frac{1}{2}\left(\frac{14}{5}+\frac{61}{15}\right)$
$\begin{aligned}
C_{2} &=\frac{1}{2} \times\left(\frac{42+61}{15}\right) \\
C_{2} &=\frac{1}{2} \times \frac{103}{15}=\frac{103}{30} \\
\therefore \frac{14}{5} &<\frac{103}{30}<\frac{61}{15}
\end{aligned}$
From (1), (2) we get, $\quad \frac{14}{5}<\frac{103}{30}<\frac{61}{15}<\frac{16}{3}$

Question $9 .$
Compare the following pairs of rational numbers.
(i) $\frac{-11}{5}, \frac{-21}{8}$
(ii) $\frac{3}{-4}, \frac{-1}{2}$
(iii) $\frac{2}{3}, \frac{4}{5}$
Answer:
(i) $\frac{-11}{5}, \frac{-21}{8}$
LCM of 5,8 is 40
$\begin{aligned}
&\frac{-11}{5}=\frac{-11 \times 8}{5 \times 8}=\frac{-88}{40} \\
&\frac{-21}{8}=\frac{-21 \times 5}{8 \times 5}=\frac{-105}{40}
\end{aligned}$

$\frac{-105}{40}<\frac{-88}{40}$ $\therefore \frac{-21}{8}<\frac{-11}{5}$ (ii) $\frac{3}{-4}, \frac{-1}{2}$ LCM of 4 and $2=4$ $\frac{3}{-4}=\frac{-3}{4}$ $\frac{-1}{2}=\frac{-1 \times 2}{2 \times 2}=\frac{-2}{4}$ $\frac{3}{-4}<\frac{-2}{4}$ $\frac{3}{4}<\frac{-1}{2}$
LCM of 4 and $2=4$
$\frac{3}{-4}=\frac{-3}{4}$
$\frac{-1}{2}=\frac{-1 \times 2}{2 \times 2}=\frac{-2}{4}$
$\frac{3}{-4}<\frac{-2}{4}$
$-\frac{3}{4}<\frac{-1}{2}$
(iii) $\frac{2}{3}, \frac{4}{5}$
LCM of 3 and 5 is 15 .
$\begin{aligned}
\frac{2}{3} &=\frac{2 \times 5}{3 \times 5}=\frac{10}{15} \\
\frac{4}{5} &=\frac{4 \times 3}{5 \times 3}=\frac{12}{15} \\
\frac{10}{15} &<\frac{12}{15} \\
\therefore \quad \frac{2}{3} &<\frac{4}{5}
\end{aligned}$

Question $10 .$
Arrange the following rational numbers in ascending and descending order.
(i) $\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}$
(ii) $\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}$
Answer:
(i) $\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}$
LCM of $12,8,24,9,36$ is $4 \times 3 \times 2 \times 3=72$

$\begin{aligned}
\frac{-5}{12} &=\frac{-5 \times 6}{12 \times 6}=\frac{-30}{72} \\
\frac{-11}{8} &=\frac{-11 \times 9}{8 \times 9}=\frac{-99}{72} \\
\frac{-15}{24} &=\frac{-15 \times 3}{24 \times 3}=\frac{-45}{72} \\
\frac{-7}{-9} &=\frac{7 \times 8}{9 \times 8}=\frac{56}{72} \\
\frac{12}{36} &=\frac{12 \times 2}{36 \times 2}=\frac{24}{72}
\end{aligned}$
Now comparing the numerators $-30,-99,-45,56,24$ we get $56>24>-30>-45>-99$ i.e $\frac{56}{72}>\frac{24}{72}>\frac{-30}{72}>\frac{-45}{72}>\frac{-99}{72}$ and so $\frac{-7}{-9}>\frac{12}{36}>\frac{-5}{12}>\frac{-15}{24}>\frac{-11}{8}$ $\therefore$ Descending order $\frac{-7}{-9}>\frac{12}{36}>\frac{-5}{12}>\frac{-15}{24}>\frac{-11}{8}$ Ascending order $\frac{-11}{8}<\frac{-15}{24}<\frac{-5}{12}<\frac{12}{36}<\frac{-7}{-9}$
(ii) $\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}$
LCM of $10,5,4,20$ is $5 \times 2 \times 2=20$

$\begin{aligned}
\frac{-17}{10} &=\frac{-17 \times 2}{10 \times 2}=\frac{-34}{20} \\
\frac{-7}{5} &=\frac{-7 \times 4}{5 \times 4}=\frac{-28}{20} \\
\frac{-2}{4} &=\frac{-2 \times 5}{4 \times 5}=\frac{-10}{20} \\
\frac{-19}{20} &=\frac{-19}{20}
\end{aligned}$
Negative numbers are less than zero.
$\therefore$ Arranging the numerators we get
$-34<-28<-19<-10<0$
$\therefore \frac{-34}{20}<\frac{-28}{20}<\frac{-19}{20}<\frac{-10}{20}<0$
Ascending order $=\frac{-17}{10}<\frac{-7}{5}<\frac{-19}{20}<\frac{-2}{4}<0$
Descending order $0>\frac{-2}{4}>\frac{-19}{20}>\frac{-7}{5}>\frac{-17}{10}$
 

Objective Type Questions:
Question $11 .$
The number which is subtracted from $\frac{-6}{11}$ to get $\frac{8}{9}$ is

(A) $\frac{34}{99}$
(B) $\frac{-142}{99}$
(C) $\frac{142}{99}$
(D) $\frac{-34}{99}$
Answer:
(B) $\frac{-142}{99}$
Hint:
Let $x$ be the number to be subtracted
$\begin{aligned}
&\frac{-6}{11}-x=\frac{8}{9} \\
&\frac{-6}{11}-\frac{8}{9}=x \\
&x=\frac{(-6 \times 9)+(-8 \times 11)}{11 \times 9}=\frac{-54+(-88)}{99}=\frac{-142}{99}
\end{aligned}$

Question $12 .$
Which of the following pairs is equivalent?

(A) $\frac{-20}{12}, \frac{5}{3}$
(B) $\frac{16}{-30}, \frac{-8}{15}$
(C) $\frac{-18}{36}, \frac{-20}{44}$
(D) $\frac{7}{-5}, \frac{-5}{7}$
Answer:
(B) $\frac{16}{-30}, \frac{-8}{15}$
$\begin{aligned}
&\text { Hint: } \\
&\begin{array}{l}
\frac{-20}{12}=\frac{-20 \div 4}{12 \div 4}=\frac{-5}{3} \neq \frac{5}{3} \\
\frac{16}{-30}=\frac{-16 \div 2}{30+2}=\frac{-8}{15} \\
\frac{-18}{36}=\frac{-18 \div 9}{36 \div 9}=\frac{-2}{4}=\frac{-2 \times 11}{4 \times 11}=\frac{-22}{44} \neq \frac{-20}{44} \\
\therefore \frac{16}{-30} \text { and } \frac{-8}{15}
\end{array}
\end{aligned}$

Question $13 .$
$\frac{-5}{4}$ is a rational number which lies between
(A) 0 and $\frac{-5}{4}$
(B) $-1$ and 0
(C) $-1$ and $-2$
(D) $-4$ and $-5$
Answer:
(C) $-1$ and $-2$

Question $14 .$
Which of the following rational numbers is the greatest?
(A) $\frac{-17}{24}$
(B) $\frac{-13}{16}$
(C) $\frac{7}{-8}$
(D) $\frac{-31}{32}$
Answer:
(A) $\frac{-17}{24}$

Question $15 .$
The sum of the digits of the denominator in the simplest form of is $\frac{112}{528}$ is
(A) 4
(B) 5
(C) 6
(D) 7
Answer:
(C) 6