Exercise 4.1 - Chapter 4 Life Mathematics 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 4.1 : Chapter 4  Life Mathematics Class 8th std Maths Guide Samacheer Kalvi Solutions
Question 1.
Fill in the blanks:
(i) If $30 \%$ of $x$ is 150 , then $x$ is
Answer:
500
Hint:
Given $30 \%$ of $x$ is 150
i.e $\frac{30}{100} \times x=150$
$\therefore x=\frac{180 \times 100}{30}$
$\therefore \mathrm{x}=500$
(ii) 2 minutes is $\%$ to an hour.
Answer:
$3 \frac{1}{3} \%$
Hint:
Let $2 \min$ be $x \%$ of an hour
and $1 \mathrm{hr}=60 \mathrm{~mm}$
$x \%=\frac{2}{60} \times 100=\frac{200}{60}=\frac{10}{3}=3 \frac{1}{3}$
$x=3 \frac{1}{3} \%$
(iii) If $x \%$ of $x=25$, then $x=$
Answer:
50
Hint:
Given that $x \%$ of $x$ is 25
$\therefore \frac{z}{100} \times \mathrm{x}=25$
$\therefore x^{2}=25 \times 100=2500$
$\therefore x=\sqrt{2500}=50$

(iv) In a school of 1400 students, there are 420 girls. The percentage of boys in the school is
Answer:
70
Hint:
Given total number of students in school $=1400$
Number of girls in school $=420$
$\therefore$ Number of boys in school $=1400-420=980$
$\%$ of boys in school $=\frac{\text { Number of boys }}{\text { Total number of students }} \times 100=\frac{980}{14,00} \times 100$
$=\frac{980}{14}=70$
$\%$ of boys $=70 \%$
(v) $0.5252$ is $\%$.
Answer:
$52.52 \%$
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
$\therefore$ to express $0.5252$ as percentage, we should multiply by 100

$\therefore 0.5252 \times 100=52.52 \%$

Question $2 .$
Rewrite each underlined part using percentage language.
(i) One half of the cake is distributed to the children.
Answer:
$50 \%$ of the cake is distributed to the children
Hint:
One half is nothing but $\frac{1}{2}$
as percentage, we need to multiply by 100
$\therefore \frac{1}{2} \times 100=50 \%$
(ii) Aparna scored $7.5$ points out of 10 in a competition.
Answer:
Aparna scored $75 \%$ in a competition
Hint:
$7.5$ points out of 10 is $\frac{7.5}{10}=0.75$
For percentage, we need to multiply by 100
We get $0.75 \times 100=75 \%$
(iii) The statue was made of pure silver.
Answer:
The statue was made of $100 \%$ pure silver
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver $=\frac{100}{100}$
$\therefore$ to express as percentage, $\frac{100}{100} \times 100 \%=100 \%$

(iv) 48 out of 50 students participated in sports.
Answer:
$96 \%$ students participated in sports.
Hint:
48 out of 50 students in fraction form is $\frac{48}{50}$
As a percentage, we need to multiply by 100
$\therefore \frac{48}{50} \times 100=96 \%$
(v) Only 2 persons out of 3 will be selected in the interview.
Answer:
Only $66 \frac{2}{3} \%$ will be selected in the interview.
Hint:
2 out of 3 in fraction form is $\frac{2}{3}$
to express as percentage, we need to multiply by 100
$\frac{2}{3} \times 100=\frac{200}{3}=66 \frac{2}{3} \%$
 

Question $3 .$
48 is $32 \%$ of which number?
Answer:
Let the number required to be found be ' $\mathrm{x}$ '
Given that $32 \%$ of $x$ is 48
i.e., $\frac{32}{100} \times \mathrm{x}=48$
$\therefore x=\frac{48^{6^{3}} \times 100}{32 \pi}=\frac{300}{2}=150$
$\therefore \mathrm{x}=150$

Question $4 .$
What is $25 \%$ of $30 \%$ of 400 ?
Answer:

Required to find $25 \%$ of $30 \%$ of 400
First $30 \%$ of $400=\frac{30}{100} \times 400=120$
Next $25 \%$ of the above is $=\frac{25^{1}}{100^{4}} \times 120=30$
 

Question $5 .$
If a car is sold for ₹ $2,00,000$ from its original price of $₹ 3,00,000$, then find the percentage of decrease in the value of the car.
Answer:
original price of car $=₹ 3,00,000$
actual selling price of car $=₹ 2,00,000$
Decrease in amount from original $=3,00,000-2,00,000=1,00,000$
Percentage decrease $=\frac{\text { Decrease }}{\text { Original value }} \times 100$

$=\frac{1,00,000}{3,00,000} \times 100=\frac{100}{3}=33 \frac{1}{3} \%$

Question $6 .$
If the difference between $75 \%$ of a number and $60 \%$ of the same number is $82.5$, then find $20 \%$ of that number.
Answer:
Given that $75 \%$ of number less $60 \%$ of number is $82.5$
Let the number be ' $x$ '
$\begin{aligned}
&\therefore \frac{75}{100} \mathrm{xx}-\frac{60}{100} \mathrm{xx}=82.5 \\
&\therefore 0.75 \mathrm{x}-0.60 \mathrm{x}=82.5 \\
&\therefore 0.15 \mathrm{x}=82.5 \\
&\therefore \mathrm{x}=\frac{82.5}{0.15}=\frac{8250}{15}=550
\end{aligned}$
Required to find $20 \%$ of number ie $20 \%$ of $x$.
$\frac{20}{100} \times x=\frac{2 \emptyset}{1 \emptyset \emptyset} \times 55 \emptyset=110$

Question 7.
A number when increased by $18 \%$ gives 236 . Find the number.
Answer:
Let the number be x. Given that when it is increased by $18 \%$, we get 236 .
$x+\frac{18}{100} x=236$
$\frac{100 x+18 x}{100}=236$
$\therefore \frac{118}{100} x=236$
$\therefore$ the number $=x=\frac{236 \times 100}{118}=200$

Question $8 .$
A number when decreased by $20 \%$ gives 80 . Find the number.
Answer:
Let the number be $x$. Given that when it is increased by $20 \%$ we get 80 .
$\begin{aligned}
x-\frac{20}{100} \times x &=80 \\
\frac{100 x-20 x}{100} &=80 \\
\frac{80 x}{100} &=80 \Rightarrow x=\frac{80 \times 100}{80} \\
x=100
\end{aligned}$

Question $9 .$
A number is increased by $25 \%$ and then decreased by $20 \%$. Find the percentage change in that number.
Answer:
Method $1 .$
Let the number be $x$.
First it is increased by $25 \%$
$\therefore$ It becomes $\mathrm{x}+\frac{25}{100} \times \mathrm{x}=\frac{125}{100}$
Secondary it is decreased by $20 \%$
$\frac{125 x}{100}-\frac{20}{100} \times \frac{125}{100} x=\frac{125}{100} x \times \frac{80}{100}=x$
Now we get back $x$, therefore there is no change.
Hence percentage change in that number is $0 \%$
Method $2 .$
[to understand, let us assume that number is 100]
So, first when we increase by $25 \%$, we get
$100+\frac{25}{100} \times 100=100+25=125$
Now this 125 is decreased by $20 \%$, we get
$125-\frac{25}{100} \times 125=125-25=100$
$\therefore$ We get back $100 \Rightarrow$ No change

Hence percentage change in that number is $0 \%$
 

Question 10 .
The ratio of boys and girls in a class is $5: 3$. If $16 \%$ of boys and $8 \%$ of girls failed in an examination, then find the percentage of passed students.
Answer:
Let number of boys be ' $b$ ' and number of girls be ' $g$ '
Ratio of boys and girls is given as $5: 3$
$\mathrm{b}: \mathrm{g}=5: 3 \Rightarrow \frac{b}{g}=\frac{5}{3} \ldots . . \text { (A) }$
Failure in boys $=16 \%=\frac{16}{100} \times \vec{b}=\frac{16 b}{100}$
Failure in girls $=8 \%=\frac{8}{100} \times \mathrm{g}=\frac{8 g}{100}$
Pass in boys $=100-16 \%=84 \%=\frac{84}{100} b \ldots . .$ (1)
Pass in girls $=100-8 \%=92 \%=\frac{92}{100} g \ldots \ldots$ (2)
From A, we have $\frac{b}{g}=\frac{5}{3}$, adding I on both sides, we get
$\frac{b}{g}+1=\frac{5}{3}+1$ $\frac{b+g}{g}=\frac{5+3}{3}=\frac{8}{3}$ $\therefore \mathrm{g}=\frac{3}{8}(\mathrm{~b}+\mathrm{g}) \ldots \ldots(3)$ Similarly $\mathrm{b}=\frac{5}{8}(\mathrm{~b}+\mathrm{g}) \ldots \ldots(4)$ Total pass $=$ pass in girls + pass in boys $=(1)+(2)=\frac{84}{100} b+\frac{92}{100} g$ $=(1)+(2)=\frac{84}{100} b+\frac{92}{100} g$

Total pass percentage $=\frac{\text { total pass }}{\text { total students }} \times 100$
Total pass $=$ boys passed $+$ girls passed $=\left(\frac{\frac{84}{100} b+\frac{92}{100} g}{b+g}\right) \times 100$
Substituting (3) \& (4) in the above, we get
$\begin{aligned}
&=\left(\frac{\frac{84}{100} b+\frac{92}{100} g}{b+g}\right) \times 100 \\
&=\left[\frac{\frac{84}{100} \times \frac{5}{8}(b+g)}{(b+g)}+\frac{\frac{92}{100} \times \frac{3}{8}(b+g)}{(b+g)}\right] \times 100 \\
&=\frac{84}{100} \times \frac{5}{8}+\frac{92}{100} \times \frac{3}{8}=\left[\frac{420}{800}+\frac{276}{800}\right] \times 100 \\
&=\frac{696}{800} \times 100=87 \%
\end{aligned}$

Objective Type Questions
Question $11 .$
$12 \%$ of 250 litre is the same as of 150 litre.
(A) $10 \%$
(B) $15 \%$
(C) $20 \%$

(D) $30 \%$
Answer:
(C) 20\%
Hint:
$12 \%$ of $250=\frac{12}{100} \times 250=30$ lit.
Percentage: $\frac{30}{150} \times 100=20 \%$
 

Question $12 .$
If three candidates $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in a school election got 153,245 and 102 votes respectively, then the percentage of votes got by the winner is
(A) $48 \%$
(B) $49 \%$
(C) $50 \%$
(D) $45 \%$
Answer:
(B) $49 \%$
Hint:
Candidate 1: 153
Candidate 2: 245 - winner [as maximum votes)
Candidate 3: 102
Total votes $=1+2+3=153+245+102=500$
$\%$ of votes for winner $=\frac{\text { no. of votes that winner got }}{\text { total votes }} \times 100$
$=\frac{245}{500} \times 100=49 \%$

Question $13 .$
$15 \%$ of $25 \%$ of $10000=$
(A) 375
(B) 400
(C) 425
(D) 475
Answer:
(A) 375

Question $14 .$
When 60 is subtracted from $60 \%$ of a number to give 60 , the number is
(A) 60
(B) 100
(C) 150
(D) 200
Answer:
(D) 200
Hint:
Let the number be ' $X$ '
$60 \%$ of the number is $\frac{60}{100} \times x=\frac{60 z}{100}$
Given that when 60 is subtracted from $60 \%$, we get 60
$\begin{aligned}
&\text { i.e } \frac{60}{100} \times-60=60 \\
&\therefore \frac{60}{100} \times 60+60=120 \\
&\therefore x=\frac{120 \times 100}{60}=200
\end{aligned}$

Question $15 .$
If $48 \%$ of $48=64 \%$ of $x$, then $x=$
(A) 64
(B) 56
(C) 42
(D) 36
Answer:
(D) 36