In Text Questions (Text Book Page No. 3,4,6,7,15,16,20,22,25,26) - Chapter 1 Numbers 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
The simplest form of $\frac{125}{200}$ is
Answer:
$\begin{aligned}
&\frac{125}{200}=\frac{125 \div 25}{200 \div 25}=\frac{5}{8} \\
&=\frac{5}{8}
\end{aligned}$
 

Question $2 .$
Which of the following is not an equivalent fraction of $\frac{8}{12}$ ?
(A) $\frac{2}{3}$
(B) $\frac{16}{24}$
(C) $\frac{32}{60}$
(D) $\frac{24}{36}$
Answer:
(C) $\frac{32}{60}$
$\frac{8}{12}=\frac{8+4}{12 \div 4}=\frac{2}{3}$ $\frac{s}{12}=\frac{8 \times 2}{12 \times 2}=\frac{16}{24}$ $\frac{s}{12}=\frac{8 \times 3}{12 \times 3}=\frac{24}{36}$
But $\frac{32}{60}=\frac{32 \div 5}{60 \div 5}=\frac{6.4}{12}$
$\therefore \frac{32}{60}$ is not equivalent fraction of $\frac{8}{12}$

Question $3 .$
Which is bigger $\frac{8}{9}$ or $\frac{4}{5}$ ?
Answer:
LCM of 5 and $9=45$
$\begin{aligned}
\frac{4}{5} &=\frac{4 \times 9}{5 \times 9}=\frac{36}{45} \\
\frac{8}{9} &=\frac{8 \times 5}{9 \times 5}=\frac{40}{45} \\
\therefore \quad \frac{40}{45} &>\frac{36}{45} \\
\frac{8}{9} &>\frac{4}{5}
\end{aligned}$
$\frac{8}{9}$ is bigger than $\frac{4}{5}$

Question $4 .$
Add the fractions : $\frac{3}{5}+\frac{5}{8}+\frac{7}{10}$.
Answer:
$\underset{=40}{\operatorname{LCM}}$ of $5,8,10=5 \times 2 \times 4$

$\begin{aligned}
\frac{3}{5}+\frac{5}{8}+\frac{7}{10} &=\frac{(3 \times 8)+(5 \times 5)+(7 \times 4)}{40} \\
&=\frac{24+25+28}{40} \\
&=\frac{77}{40}=1 \frac{37}{40}
\end{aligned}$

Question $5 .$
Simplify: $\frac{1}{8}-\left(\frac{1}{6}-\frac{1}{4}\right)$
Answer:
$\begin{aligned}
\frac{1}{8}-\left(\frac{1}{6}-\frac{1}{4}\right) &=\frac{1}{8}-\left[\frac{(1 \times 2)-(1 \times 3)}{12}\right] \\
&=\frac{1}{8}-\left(\frac{2-3}{12}\right) \\
&=\frac{1}{8}-\left(-\frac{1}{12}\right) \\
&=\frac{1}{8}+\frac{1}{12}=\frac{(1 \times 3)+(1 \times 2)}{24} \\
&=\frac{3+2}{24}=\frac{5}{24}
\end{aligned}$

Question $6 .$
Multiply: $2 \frac{3}{5}$ and $1 \frac{4}{7}$.
Answer:
$2 \frac{3}{5} \times 1 \frac{4}{7}=\frac{13}{5} \times \frac{11}{7}=\frac{143}{35}=4 \frac{3}{35}$

Question $7 .$
Divide $\frac{7}{36}$ by $\frac{35}{81}$.
Answer:
$\frac{7}{36}+\frac{35}{81}=\frac{7}{36} \times \frac{81}{35}=\frac{9}{20}$

Question $8 .$

Fill in the boxes : $\frac{\square}{66}=\frac{70}{\square}=\frac{28}{44}=\frac{\square}{121}=\frac{7}{\square}$.
Answer:
$\begin{aligned}
&\frac{28}{44}=\frac{28+4}{44+4}=\frac{7}{11} \\
&\frac{7}{11}=\frac{28}{44}=\frac{42}{66}=\frac{70}{110}=\frac{77}{121} \\
&\frac{42}{66}=\frac{70}{110}=\frac{28}{44}=\frac{77}{121}=\frac{7}{11}
\end{aligned}$

Question $9 .$
In a city $\frac{7}{20}$ of the population are women and $\frac{1}{4}$ are children. Find the fraction of the population of men.
Answer:
Let the total population $=1$
Population of men = Total population $-$ Women $-$ Children
$\begin{aligned}
&=1-\frac{7}{20}-\frac{1}{4}=\frac{20}{20}-\frac{7}{20}-\frac{5}{20} \\
&=\frac{20-7-5}{20}=\frac{8}{20}=\frac{2}{5}
\end{aligned}$
$\therefore$ Population of men $=\frac{2}{5}$

Question $10 .$
Répresenent $\left(\frac{1}{2}+\frac{1}{4}\right)$ by a điâgram.
Answer:

Try These (Text Book Page No. 3)
Question 1.
Is the number $-7$ a rational number? Why?
Answer:
A rational number, Because $-7=\frac{-14}{2}=\frac{p}{q}$
 

Question $2 .$
Write any 6 rational numbers between 0 and 1 .
Answer:
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$
 

Try These (Text Book Page No. 4)
Write the decimal forms of the following rational numbers:
Question $1 .$
$\frac{4}{5}$
Answer:
$\frac{4}{5}=\frac{4 \times 20}{5 \times 20}=\frac{80}{100}=0.80$

Question $2 .$
$\frac{6}{25}$
Answer:
$\frac{6}{25}=\frac{6 \times 4}{25 \times 4}=\frac{24}{100}=0.24$

Question $3 .$
$\frac{486}{1000}$
Answer:
$\frac{486}{1000}=0.486$

Question $4 .$
$\frac{1}{9}$
Answer:

$\frac{1}{9}=0.11 \ldots$

Question $5 .$
$3 \frac{1}{4}$
Answer:

Question 6
$-2 \frac{3}{5}$
Answer:

Try These (Text Book Page No. 6)
Question 1 .
$\frac{7}{3}=\frac{7}{9}=\frac{49}{?}=\frac{-21}{?}$
Answer:
$\begin{aligned}
\frac{7}{3} &=\frac{7 \times 3}{3 \times 3}=\frac{21}{9} \\
\frac{7}{3} &=\frac{7 \times 7}{3 \times 7}=\frac{49}{21} \\
\frac{7}{3} &=\frac{7 \times(-3)}{3 \times(-3)}=\frac{-21}{-9} \\
\therefore \frac{7}{3} &=\frac{21}{9}=\frac{49}{21}=\frac{-21}{-9}
\end{aligned}$

Question $2 .$
$\frac{-2}{5}=\frac{?}{10}=\frac{6}{?}=\frac{-8}{?}$
Answer:
$\begin{aligned}
\frac{-2}{5} &=\frac{-2 \times 2}{5 \times 2}=\frac{-4}{10} \\
\frac{-2}{5} &=\frac{-2 \times-3}{5 \times-3}=\frac{6}{-15} \\
\frac{-2}{5} &=\frac{-2 \times 4}{5 \times 4}=\frac{-8}{20} \\
\therefore \frac{-2}{5} &=\frac{-4}{10}=\frac{6}{-15}=\frac{-8}{20}
\end{aligned}$
 

Try These (Text Book Page No. 7)
Question 1.
Which of the following pairs represents equivalent rational numbers?
(i) $\frac{-6}{4}, \frac{18}{-12}$
(ii) $\frac{-4}{-20}, \frac{1}{-5}$
(iii) $\frac{-12}{-17}, \frac{60}{85}$
Answer:
(i) $\frac{-6}{4}, \frac{18}{-12}$
$\frac{-6}{4}=\frac{-6 \times 3}{4 \times 3}=\frac{-18}{12}$
$\therefore \frac{-6}{4}$ equivalent to $\frac{-18}{12}$
(ii) $\frac{-4}{-20}, \frac{1}{-6}$
$\frac{-4}{-20}=\frac{-4 \div(-4)}{-20 \div(-4)}=\frac{1}{5} \neq-\frac{1}{5}$
$\therefore \frac{-4}{-20}$ equivalent to $\frac{1}{-5}$
(iii) $\frac{-12}{-17}, \frac{60}{85}$
$\frac{-12}{-17}=\frac{-12 z-5}{-17 z-5}=\frac{60}{85}$
$\therefore \frac{-12}{-17}$ equivalent to $\frac{60}{85}$

Question $2 .$
Find the standard form of
(i) $\frac{36}{-96}$
(ii) $\frac{-56}{-72}$
(iii) $\frac{27}{18}$
Answer:
(i) $\frac{36}{-96}$ $=\frac{-36 \div 12}{96 \div 12}=\frac{-3}{8}$
(ii) $\frac{-56}{-72}$
$=\frac{-56 \div(-8)}{-72 \div(-8)}=\frac{7}{9}$
(iii) $\frac{27}{18}$
$=1 \frac{9}{18}=1 \frac{1}{2}$

Question $3 .$
Mark the following rational numbers on a number line.
(i) $\frac{-2}{3}$
Answer:
$\frac{-2}{3}$ lies betveen 0 and $-1$.
T?ie unit part between $\mathrm{O}$ and - lis divided into 3 equal parts and second part is taken.

(ii) $\frac{-8}{-5}$
Answer:
$\frac{-8}{-5}=1 \frac{3}{5}$
$1 \frac{3}{5}$ lies between I and 2, The unit part between I and 2 is divided into 5 equal parts and the third part is taken.

(iii) $\frac{5}{-4}$
Answer:
$\frac{5}{-4}=-\frac{5}{4}=-1 \frac{1}{4}$
$-1 \frac{1}{4}$ lies between $-1$ and $-2$. The unit part between $-1$ and $-2$ is divided into four equal parts and the first part is taken.

Think (Text Book Page No. 15)
Is zero a rational number? If so, what is its additive inverse

Answer: Yes zero a rational number Additive inverse of zero is zero.

Think (Text Book Page No. 16)
What is the multiplicative inverse of 1 and $-1$ ?
Answer:
Multiplicative inverse of 1 is 1 and $-1$ is $-1$.

Try These (Text Book Page No. 16)
Divide
(i) $\frac{-7}{3}$ by 5
(ii) 5 by $\frac{-7}{3}$
(iii) $\frac{-7}{3}$ by $\frac{35}{6}$
Answer:
(i) $\frac{-7}{3}$ by 5
$\frac{-7}{3} \div 5=\frac{-7}{3} \div \frac{5}{1}=\frac{-7}{3} \times \frac{1}{5}=\frac{-7}{15}$

(ii) 5 by $\frac{-7}{3}$
$5 \div\left(\frac{-7}{3}\right)=\frac{5}{1} \times \frac{3}{-7}=\frac{15}{-7}=-2 \frac{1}{7}$
(iii) $\frac{-7}{3}$ by $\frac{35}{6}$
$\frac{-7}{3} \div \frac{35}{62}=\frac{-7}{\not} \times \frac{\not 6}{35}=-\frac{2}{5}$

Try These (Text Book Page No. 20)
The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Answer:
Let 0 and $\frac{1}{2}$ te two rational numbers $0-\frac{1}{2}=-\frac{1}{2}$ is a rational numter
$\therefore$ Closure property for subtraction holds for rational numbers.
But consider the two rational number $\frac{5}{2}$ and 0 .
$\frac{5}{2}+0=\frac{5}{2 \times 0}=\frac{5}{0}$
Here denominator $=0$ and it is not a rational number.
$\therefore$ Closure property is not true for division of rational numbers.

Try These (Text Book Page No. 22)
(i) Is $\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}$ ?
Answer:
$\mathrm{LHS}=\frac{3}{5} \div \frac{7}{8}=\frac{(3 \times 8)-(7 \times 5)}{40}=\frac{24-35}{40}=\frac{-11}{40}$
$\mathrm{RHS}=\frac{7}{8}-\frac{3}{5}=\frac{(7 \times 5)-(3 \times 8)}{40}=\frac{35-24}{40}=\frac{11}{40}$
LHS $\neq$ RHS
$\therefore \frac{3}{5} \div \frac{7}{8} \neq \frac{7}{8}-\frac{3}{5}$
$\therefore$ Subtraction of rational numbers is not commutative.
(ii) $\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}$ ? So, what do you conclude?
Answer:
$\begin{aligned}
&\mathrm{LHS}=\frac{3}{5} \div \frac{7}{8}=\frac{3}{5} \times \frac{8}{7}=\frac{24}{35} \\
&\mathrm{RHS}=\frac{7}{8} \div \frac{5}{3}=\frac{7}{8} \times \frac{3}{5}=\frac{21}{40} \\
&\text { LHS } \neq \text { RHS }
\end{aligned}$

$\therefore \frac{3}{5} \div \frac{7}{8} \neq \frac{7}{8} \div \frac{5}{3}$
$\therefore$ Commutative property not hold good br division of rational numbers.

Try This (Text Book Page No. 22)
Check whether associative property holds for subtraction and division.
Answer:
Consider for rational numbers $\frac{2}{3}, \frac{1}{2}$ and $\frac{3}{4}$
To verify $\left(\frac{2}{3}-\frac{1}{2}\right)-\frac{3}{4}=\frac{2}{3}-\left(\frac{1}{2}-\frac{3}{4}\right)$
$\begin{aligned}
\mathrm{LHS} &=\left(\frac{2}{3}-\frac{1}{2}\right)-\frac{3}{4}=\left(\frac{(2 \times 2)-(1 \times 3)}{6}\right)-\frac{3}{4} \\
&=\left(\frac{4-3}{6}\right)-\frac{3}{4}=\frac{1}{6}-\frac{3}{4}=\frac{(1 \times 2)-(3 \times 3)}{12}=\frac{2-9}{12}=\frac{-7}{12} \\
\mathrm{RHS} &=\frac{2}{3}-\left(\frac{1}{2}-\frac{3}{4}\right)=\frac{2}{3}-\left(\frac{2-3}{4}\right)=\left(\frac{2}{3}-\left(\frac{-1}{4}\right)\right) \\
&=\frac{2}{3}+\frac{1}{4}=\frac{(2 \times 4)+(1 \times 3)}{12}=\frac{8+3}{12}=\frac{11}{12}
\end{aligned}$

$\begin{aligned}
\text { LHS } & \neq \text { RHS } \\
\therefore\left(\frac{2}{3}-\frac{1}{2}\right)-\frac{3}{4} & \neq \frac{2}{3}-\left(\frac{1}{2}-\frac{3}{4}\right)
\end{aligned}$
$\therefore$ Associative property not holds for subraction of rational numbers
Also to verify $\left(\frac{2}{3}+\frac{1}{2}\right)+\frac{3}{4}=\frac{2}{3} \div\left(\frac{1}{2} \div \frac{3}{4}\right)$
$\begin{aligned}
\text { LHS } &=\left(\frac{2}{3} \div \frac{1}{2}\right) \div \frac{3}{4}=\left(\frac{2}{3} \times \frac{2}{1}\right) \div \frac{3}{4} \\
&=\frac{4}{3} \div \frac{3}{4}=\frac{4}{3} \times \frac{4}{3}=\frac{16}{9} \\
\text { RHS } &=\frac{2}{3}+\left(\frac{1}{2} \div \frac{3}{4}\right)=\frac{2}{3} \div\left(\frac{1}{2} \times \frac{4}{3}\right)=\frac{2}{3} \div\left(\frac{2}{3}\right) \\
&=\frac{2}{3} \times \frac{3}{2}=1 \\
\text { LHS } & \neq \text { RHS } \\
\div \frac{3}{4} & \neq \frac{2}{3} \div\left(\frac{1}{2} \div \frac{3}{4}\right)
\end{aligned}$
i.e. $\left(\frac{2}{3} \div \frac{1}{2}\right) \div \frac{3}{4} \neq \frac{2}{3} \div\left(\frac{1}{2} \div \frac{3}{4}\right)$
$\therefore$ Associative property not holds for division of rational numbers

Try This (Text Book Page No. 25)
Question 1.
Observe that,
$\begin{aligned}
&\frac{1}{1.2}+\frac{1}{2.3}=\frac{2}{3} \\
&\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=\frac{3}{4} \\
&\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}=\frac{4}{5}
\end{aligned}$

Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.
Answer:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{7}{8}$
 

Think (Text Book Page No. 26)
Question 1 .
Is the square of a prime number, prime?
Answer:
No, the square of a prime number ' $\mathrm{P}$ ' has at Least 3 divisors $1, \mathrm{P}$ and $\mathrm{P}^{2}$. But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.
 

Question 2.
Will the sum of two perfect squares always be a perfect square? What about their difference and their product?
Answer:
The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. Bu the product of two perfect square is a perfect square.