Free Practice Questions for Percentage in Maths

Invested by Divya = ₹ $6,000$
Reshma has invested $30 \%$ less than Divya
Invested by Reshma $=$₹ 6,000 $\times \frac{100-30}{100}$ = ₹ $4,200$
Sushma has invested $40 \%$ more than Reshma
Invested by Sushma =₹ $4,200 \times 140 \%$
=₹ $4,200 \times \frac{140}{100}$
=₹ $5,880$

Let Saurabh's monthly income $=100 \mathrm{x}$
Saurabh's Saving $=100 x \times \frac{80}{100} \times \frac{70}{100} \times \frac{50}{100} \times \frac{50}{100}$
₹ $11,200=14 \mathrm{x}$
x=₹ 800
his monthly income $=100 \mathrm{x}$
= ₹ 80,000

After 1 st year =₹ $65,00,000 \times \frac{95}{100}$
= ₹6,175,000
After 2 st year =₹ $6,175,000 \times \frac{96}{100}$
= ₹5,928,000
After 32 st year =₹ $5,928,000 \times \frac{97}{100}$
= ₹ 57,50,160

Let Rehana income is 100 unit
Sazia's : Rehana $=80: 100$
Rehana income $=\frac{20}{80} \times 100 \%$
$
=25 \%
$
Rehana's income is $25 \%$ more than Sazia's income

Let the value of $y$ is 100
According to question
$
\begin{aligned}
&\mathrm{x}=100 \times \frac{112.25}{100} \\
&=112.25 \\
&\text { Required } \%=\frac{112.25-100}{112.25} \times 100 \\
&=\frac{12.25}{112.25} \times 100=10.9 \%
\end{aligned}
$

According to question

$
\begin{aligned}
&30 \% \text { of }(\mathrm{x}+60)-25 \% \text { of } \mathrm{x}=40 \\
&\frac{3}{100} \times(x+60)-\frac{25}{100} \times x=40 \\
&30 \mathrm{x}+1800-25 \mathrm{x}=4000 \\
&5 \mathrm{x}=2200 \\
&\mathrm{x}=440
\end{aligned}
$
Now, $35 \%$ of $(\mathrm{x}-40)=\frac{7}{20} \times(440-40)$
$
=\frac{7}{20} \times(400)=140
$
Required $\%=\frac{140-120}{120} \times 100=16 \frac{2}{3} \%$

Let 100 be the number
When number is decreased by $25 \%$ and increased by $25 \%=100 \times \frac{75}{100} \times \frac{125}{100}=$ $93.75$
Net percentage decrease $=\frac{100-93.75}{100} \times 100=\frac{6.25}{100} \times 100=6 \frac{1}{4} \%$

Let the required number be $x$
According to question,
$80 \%$ of $\frac{3}{4}$ of $x=63$
$\frac{4}{5} \times \frac{3}{4} \times x=63$
$x=\frac{63 \times 5}{3}$
$x=105$

$
\begin{aligned}
&3 \% \text { of } X+2 \% \text { of } Y=\frac{2}{3} \times(2 \% \text { of } X+6 \% \text { of } Y) \\
&3 \times X+2 \times Y=\frac{2}{3} \times(2 \times X+6 \times Y) \\
&5 \times \frac{X}{3}=2 \times Y \\
&\frac{X}{Y}=\frac{6}{5}
\end{aligned}
$
Now, $\frac{3 X}{2 Y}=\frac{3 \times 6}{2 \times 5}=\frac{18}{10}=\frac{9}{5}$

According to question
$
\begin{aligned}
&x=\frac{125}{100} \times y \\
&\frac{x}{y}=\frac{5}{4}
\end{aligned}
$
Required $\%=\frac{(5-4)}{5} \times 100=20 \%$

Let the number is 100 , then
after $10 \%$ increase, the number $=\frac{(100+10)}{100} \times 100=110$
Now it is decreased by $10 \%$, So the number $=\frac{(100-10)}{100} \times 110=\frac{90 \times 11}{10}=99$
$\%$ decrease $=\frac{(100-99)}{100} \times 100=1 \%$
Hence, the number is decreased by $1 \%$.

$
\begin{aligned}
&A=\frac{130}{100} \times C \\
&B=\frac{150}{100} \times C
\end{aligned}
$
Required ratio $=\frac{130 C}{100}: \frac{150 C}{100}=13: 15$

$
\begin{aligned}
&\Rightarrow 12 \% \text { of } 4 \% \text { of } 7 \% \text { of } 2 \times 10^6 \\
&=\frac{12}{100} \times \frac{4}{100} \times \frac{7}{100} \times 2 \times 100 \times 100 \times 100 \\
&=672
\end{aligned}
$

$\frac{30-x}{x-25}=\frac{26}{74}=\frac{13}{37}$
$1110-37 x=13 x-325$
$50 x=1435$
$x=28.7 \%$

Let the income of B be 100 units.
Then, $A=100 \times \frac{170}{100}=170$ units
Income of $\mathrm{C}=\frac{80}{100} \times 270=216$ units
Required $\%=\frac{216-170}{216} \times 100$
$
=\frac{4600}{216}=21.29 \%=21.3 \%
$

According to question
$
\begin{aligned}
&90 \times \frac{(100-k)}{100}=60 \times \frac{(100+k)}{100} \\
&3(100-k)=2(100+k) \\
&300-3 k=200+2 k \\
&5 k=100 \\
&k=20
\end{aligned}
$
So, $k \%$ of $120=\frac{20}{100} \times 120=24$
And $(k+20) \%$ of $150=\frac{40}{100} \times 150=60$
Required $\%=\frac{24}{60} \times 100=40 \%$

Let the total number of votes polled is $100 x$, then
Number of votes received by the rejected candidate $=\frac{38}{100} \times 100 x=38 x$
So, the Number of votes received by the accepted candidate $=100 x-38 x=62 x$
Difference of votes $=62 x-38 x=24 x$
According to question
$
24 x=28800
$
$
x=1200
$
Hence, Total number of votes $=100 x=100 \times 1200=1,20,000$

We know,
Revenue $=$ Price $\times$ Sale
Let price $=10$ and sale $=10$
As per the question,
Revenue $=12.5 \times 8.4=105$
$\%$ increase $=\frac{105-100}{100} \times 100=5 \%$

Let monthly salary of Vignesh is ₹ $x$.
As per the question,
His total expenditure $=42 \%+16 \%+11 \%+7 \%=76 \%$
So, $x \times \frac{24}{100}=(18000-12000)$
$x$=₹ 25,000

Let the amount of $\mathrm{A}$ and $\mathrm{C}$ is $8 x$ and $5 x$ respectively, then
Amount of $\mathrm{B}=\frac{75}{100} \times 8 x=6 x$
According to question
$8 x+6 x+5 x=$₹ $9500$
$19 x=$₹ $9500$
$x=$₹ $500$
Hence, Amount of $B=6 \times 500=$₹ $3000$