Free Practice Questions for Percentage in Maths

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

Let Saurabh's monthly income $=100\mathrm{x}$
Saurabh's Saving $=100x\times \frac{80}{100}\times \frac{70}{100}\times \frac{50}{100}\times \frac{50}{100}$
₹ $11,200=14\mathrm{x}$
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 $=\frac{20}{80}\times 100\mathrm{\%}$
$=25\mathrm{\%}$
Rehana's income is $25\mathrm{\%}$ more than Sazia's income

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

According to question

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

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

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

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

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

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

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

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

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

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 $\mathrm{\%}=\frac{216-170}{216}\times 100$
$=\frac{4600}{216}=21.29\mathrm{\%}=21.3\mathrm{\%}$

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

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

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

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

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