Free Practice Questions for Percentage in Maths

Given:
Present value of an old car $=$ Rs. 540000
Depreciation rate $=25 \%$ p.a.
Now, value one year ago $=\frac{540000}{(100-25)} \times 100=\frac{540000}{75} \times 100$ $=$ Rs. 720000
Value, two years ago $=\frac{720000}{(100-25)} \times 100=\frac{720000}{75} \times 100$ $=$ Rs. 960000

The total number of girls who appeared in 2012, 2013 and $2015=325+350+325=1000$

The total number of boys who appeared in all the five years $=225+250+300+280+375=$ 1430
Therefore, required percentage $=\frac{(1430-1000)}{1430} \times 100=\frac{430}{1430} \times 100=30.1 \%$

Let the total voters be $x$.
$
\begin{aligned}
&\text { Number of male voters }= (x) \times \frac{60}{100}=\frac{3 x}{5}\\
&\text { Number of female voters }=x-\frac{3 x}{5}=\frac{2 x}{5}\\
&\text { Number of literate males }=\frac{60}{100}=\frac{3 x}{5} \\
&\text { Number of literate males }=\frac{3 x}{5} \times \frac{(100-40)}{100}=\frac{3 x}{5} \times \frac{60}{100}=\frac{9 x}{25} \\
&\text { Number of illiterate females }=\frac{2 x}{5} \times \frac{(100-25)}{100}=\frac{2 x}{5} \times \frac{75}{100}=\frac{3 x}{10} \\
&\text { Therefore, required percentage }=\frac{\left(\frac{9 x}{25}-\frac{3 x}{10}\right)}{\left(\frac{3 x}{10}\right)} \times 100
\end{aligned}
$
= $\begin{aligned} & \frac{\left(\frac{18 x-15 x}{50}\right)}{\left(\frac{3 x}{10}\right)} \times 100 \\=& \frac{3 x}{50} \times \frac{10}{3 x} \times 100 \\=& 20 \% \end{aligned}$
Hence, option A is correct.

ALTERNATE METHOD

Increase in population = 16300 - 15000 = 1300

Increase in population (in %) = $\frac{1300}{15000}\times100$ = 26/3

Now, by mixture & alligation method -

Female population (in ratio) = 26/3 - 8% = 2/3

Male population (in ratio) = 10% - 26/3 = 4/3

Ratio of Male & Female = 4/3 : 2/3 = 2 : 1

Female population = 15000/3 = 5000

BASIC SOLUTION -

According to the given information,

Let the number of male of the town be A

Then, Number of female = 15000 - A

ATQ,

If the number of males increases by 8% and number of female by 10%  then,

 New number of male = A + 8% of A

 $\Rightarrow$ 1.08A

 New number of Female = (15000 - A) + 10% of (15000 - A)

 New population  = New number of male  + New number of female

16300 = 1.08A + (15000 - A) + 10% of  (15000 - A)

16300 = 1.08A +  (15000 - A) + 1500 - 0.1A

A = $\frac{16500 - 16300}{0.02}$

A = 10000

Hence, number of males = 10,000

And number of females = 15000 - 10000 = 5000

Hence, option(4) is the correct answer.

Let the length of side be $x$ then new length $=1.4 x$
Old area $=x^{2}$
New area $=(1.4 x)^{2}=1.96 x^{2}$
Percentage increase in area $=\frac{1.96 x^{2}-x^{2}}{x^{2}} \times 100$ = $96 \%$
Hence, option B is the correct answer. 

Let the price of an item be $x$ and quantity be $y$ So,
$
x y=280 \ldots \text {...(1) }
$
Also,
$
\begin{aligned}
&\left(1-\frac{12.5}{100}\right) \times(y+2.5)=280 \\
&\Rightarrow \frac{7}{8} x(y+2.5)=280 \quad \ldots . \text { eq.(2) }
\end{aligned}
$
Eq.2 divided by eq.1
$
\begin{aligned}
&\frac{7 x(y+2.5)}{8 x y}=\frac{280}{280} \\
&\Rightarrow \frac{7(y+2.5)}{8 y}=1
\end{aligned}
$
$
\begin{aligned}
&\Rightarrow 7 y+17.5=8 y \\
&\Rightarrow y=17.5
\end{aligned}
$
Put this value in eq.2
$
\begin{aligned}
&\frac{7}{8} \times(17.5+2.5)=280 \\
&\Rightarrow \frac{7}{8} x=14
\end{aligned}
$
So, reduced price of item $=14$
Hence, option D is the correct answer. 

Selling price of both the items together $=$ Rs. 51750
Now, cost price of both the items together $=42000+3000=$ Rs. 45000
Profit $($ in $\%)=\frac{S P-C P}{C P} \times 100$
$=\frac{51750-45000}{45000} \times 100=\frac{6750}{450}$
$=15 \%$

MATHS MIRROR METHOD -

Let total number of votes = 100

The successful candidate got 60% of valid votes and won by a margin of 567 votes.  

Difference = 20% of 75.6

20% of 75.6 = 567 (given)

100 = 3750

The number of voters enrolled in the voter list is 3750.

BASIC SOLUTION -

Let the number of voters enrolled in the voter list be $x$.
Then, number of actual vote cast $=(x)-(x) \times 16 \%=x-0.16 x=0.84 x$
Number of valid votes $=(0.84 x)-(0.84 x) \times 10 \%=0.84 x-0.084 x$ $=0.756 x$
Number of votes the successful candidate got $=(0.756 x) \times 60 \%=0.4536 x$
Number of votes the other candidate got $=0.756 x-0.4536 x=0.3024 x$
Difference between votes of both the candidates (winning margin) $=567$
$
\begin{aligned}
&\Rightarrow 0.4536 x-0.3024 x=567 \\
&\Rightarrow 0.1512 x=567 \\
&\Rightarrow x=567 / 0.1512=3750
\end{aligned}
$
Therefore, the number of voters enrolled in the voter list is 3750 .

$\frac{23.5}{100} \times$ Number $=11.75$
$\Rightarrow$ Number $=\frac{11.75}{23.5} \times 100=50$

Let the fraction be $\frac{x}{y}$
According to the question,
$
\Rightarrow y \times \frac{(x) \times \frac{(100+25)}{100}}{y}=\frac{5}{14}
$
$\Rightarrow \quad \frac{(x) \times \frac{125}{100}}{y \times \frac{525}{400}}=\frac{5}{14}$
$\Rightarrow y \times \frac{(x) \times \frac{5}{4}}{16}=\frac{5}{14}$
$\Rightarrow \frac{x}{y}=\frac{5 \times 4 \times 21}{14 \times 5 \times 16}$
$\Rightarrow \frac{x}{y}=\frac{3}{8}$

MATHS MIRROR SOLUTION

$\begin{aligned} \text { Required percentage } &=\frac{345}{600} \times 100 \\ &=57.5 \% \end{aligned}$

No. of vowels in word CONSONANT $=3(O, O, A)$
No. of consonants in word CONSONANT $=6(\mathrm{C}, \mathrm{N}, \mathrm{S}, \mathrm{N}, \mathrm{N}, \mathrm{T})$
Percentage of the number of vowels in the word CONSONANT
$=\frac{3}{9} \times 100=33.33 \%$
Percentage of the number of consonants in the word CONSONANT
$=\frac{6}{9} \times 100=66.66 \%$
Required percentage $=\frac{66.66-33.33}{66.66} \times 100=\frac{33.33}{66.66} \times 100=50 \%$

Let the fraction is $y$ and total number $=1$
Let the certain sum be a then profit $=3 a$
Selling price $=a+3 a=4 a$
According to question
Selling price of fraction $=\mathrm{a}$
$
\begin{aligned}
&\Rightarrow y(4 a)=a \\
&\Rightarrow y=\frac{1}{4}
\end{aligned}
$
Hence, option A is the correct answer.

Let the total number of students in the examination be $100 \mathrm{x}$
Number of girls the examination $=(100 x) \times \frac{40}{100}=40 x$
Number of boys in the examination $=100 x-40 x=60 x$
Number of boys who passed in the examination $=(60 x) \times \frac{(100-35)}{100}$
$
\Rightarrow 585=(60 x) \times \frac{65}{100}
$
$
\Rightarrow x=\frac{585}{39}
$
$
\Rightarrow x=15
$
Number of girls who passed in the examination $=(40 x) \times \frac{(100-20)}{100}$
$
=40 \times 15 \times \frac{80}{100}
$
$
=480
$
Therefore, the number of girls who passed were 480.

Let the height of $D$ be $x \mathrm{~cm}$
The height of $C=(x) \times \frac{(100+30)}{100}=\frac{130 x}{100}=\frac{13 x}{10}$
The height of $B=\left(\frac{13 x}{10}\right) \times \frac{(100-20)}{100}=\left(\frac{13 x}{10}\right) \times \frac{80}{100}=\frac{26 x}{25}$
The height of $A=\left(\frac{26 x}{25}\right) \times \frac{(100+20)}{100}=\left(\frac{26 x}{25}\right) \times \frac{120}{100}=\frac{156 x}{125}$
Given,
$
\begin{aligned}
&\Rightarrow \frac{13 x}{10}-\frac{156 x}{125}=6.8 \\
&\Rightarrow \frac{325 x-312 x}{250}=6.8 \\
&\Rightarrow \frac{13 x}{250}=6.8 \\
&\Rightarrow x=\frac{250 \times 6.8}{13} \\
&\Rightarrow x=\frac{1700}{13}
\end{aligned}
$
Therefore, the height of $B=\left(\frac{26 x}{25}\right) \times \frac{1700}{13}=136 \mathrm{~cm}$

Let the initial income of Riya be Rs. $300 \mathrm{x}$.

Then, initial expenditure of Riya $=(300 x) \times 66 \frac{2}{3} \%=$ Rs. $200 x$

Initial savings of Riya $=300 x-200 x=$ Rs. $100 x$

New income of Riya $=300 x+(300 x) \times 16 \%$ $ =300 x+48 x=\text { Rs. } 348 x $

New savings of Riya $=(100 x)+(100 x) \times 17 \%=100 x+17 x=$ Rs. $117 x$

Now, new expenditure of Riya $=348 x-117 x=$ Rs. $231 x$

Increase (in \%) in expenditure of Riya $=\left(\frac{231 x-200 x}{200 x}\right) \times 100=\frac{31 x}{2 x}$ $ =15.5 \% $

$\Rightarrow$ Let total income of $B=100 \mathrm{x}$ Rs.
$\Rightarrow$ So, total income of $\mathrm{A}=100 \mathrm{x} \times\left(1+\frac{20}{100}\right)=120 x$ Rs.
According to the question,
$\Rightarrow$ $(100 x+120 x) \times \frac{30}{100}=26400$
$\Rightarrow$ $66 \mathrm{x}=26400$
$\Rightarrow$ $x=Rs. 400$
$\Rightarrow$ So, income of $B=400 \times 100=Rs.40000 $

Hence, option (3) is the correct answer.