Free Practice Questions for Percentage in Maths

Required income in 2016 was$=70,98,000 \times \frac{100}{(100+30)} \times \frac{100}{(100+30)}$
$=70,98,000 \times \frac{100}{(130)} \times \frac{100}{(130)}$
$=70,98,000 \times \frac{10}{(13)} \times \frac{10}{(13)}$
$=$₹ 42,00,000

Required pure acid $=\frac{30}{100} \times 15=4.5$ litre

Let his capital is $x$,
According to question
$
\begin{aligned}
&\frac{x}{2} \times 6 \%+\frac{x}{4} \times 8 \%+\left(x-\frac{x}{2}-\frac{x}{4}\right) \times 12 \%=528 \\
&x \times 3 \%+x \times 2 \%+\left(\frac{4 x-2 x-x}{4}\right) \times 12 \%=528 \\
&x \times 3 \%+x \times 2 \%+\frac{x}{4} \times 12 \%=528 \\
&x(3 \%+2 \%+3 \%)=528 \\
&x \times \frac{8}{100}=528 \\
\end{aligned}
$
$x$=₹ $6600$

Let the number be $x$.
According to question
$x-\frac{35 x}{100}=26$
$
\begin{aligned}
&\frac{100 x-35 x}{100}=26 \\
&\frac{65 x}{100}=26 \\
&x=\frac{2600}{65}=40
\end{aligned}
$
Hence, the number is $40$ .

Expenditure : Savings = $100 : 20$

Let the savings increases by $x \%$

$
\begin{aligned}
&\frac{38-35}{35-x}=\frac{20}{100} \\
&\frac{3}{35-x}=\frac{1}{5} \\
&35-x=15 \\
&x=20 \%
\end{aligned}
$
Hence, Savings increases by $20 \%$.

Let the income of Reshma $=100$ units
So, her expenditure $=84$ units and savings $=16$ units
Now, as per the question
Her income $=130$ units
And savings $=16 \times \frac{140}{100}=22.4$ units
Her expenditure $=130-22.4=107.6$ units
$\%$ increase in expenditure $=\frac{107.6-84}{84} \times 100=28.1 \%$

$\Rightarrow 45 \%$ of $360+30 \%$ of $400=25 \%$ of $\mathrm{x}$
$\Rightarrow \frac{45}{100} \times 360+\frac{30}{100} \times 400=\frac{25}{100} \times x$
$\Rightarrow 162+120=\frac{x}{4}$
$\Rightarrow x=282 \times 4$
$\Rightarrow x=1128$

$18 \mathrm{gm}$ in $\mathrm{kg}=\frac{18}{1000}=0.018 \mathrm{~kg}$
Required $\%=\frac{0.018}{4.5} \times 100=0.4 \%$

Let the income be 100 units.
Then, Expenditure $=75$ units
Savings $=25$ units
Now, income becomes $=100 \times \frac{150}{100}=150$ units
And Expenditure becomes $=75 \times \frac{180}{100}=135$ units
Total increase in expenditure $=60$ units but increase in income $=50$ units
So decrease in Savings $=10$ units or Savings $=150-135=15$ units
$\%$ decrease $=\frac{10}{25} \times 100$

Let the income of $y$ is 100 , then
Income of $x=\frac{(100+80)}{100} \times 100=180$
Income of $z=\frac{60}{100} \times(100+180)=168$
Required $\%=\frac{180-168}{180} \times 100=\frac{1200}{180}=6.67 \% \approx 6.7 \%$

Let her monthly income is $100 x$, then
Earmarked donation $=\frac{8}{100} \times 100 x=8 x$
She donated $20 \%$ less than earmarked amount, So
$\frac{80}{100} \times 8 x=$₹ 2880
$x=$₹ 450
Hence, Monthly income
$ 100 x=$₹ 45000

Let the marks of $\mathrm{B}=x$
and marks of $\mathrm{A}=x+9$
As per the question,
$
\begin{aligned}
&x+9=\frac{3}{5}(2 x+9) \\
&5 x+45=6 x+27 \\
&x=18
\end{aligned}
$
So, marks of $\mathrm{A}=x+9$ $=27$

$\Rightarrow 280 \%$ of $160-25 \%$ of $720=\mathrm{y} \%$ of 200
$\Rightarrow \frac{280}{100} \times 160-\frac{25}{100} \times 720=\frac{y}{100} \times 200$
$\Rightarrow 28 \times 16-180=2 y$
$\Rightarrow 2 y=268$
$\Rightarrow y=134$

Let the number of seats are $70, 80$ and $90$

then the seats after increment , $70\times\frac{160}{100}=112$, $80\times\frac{150}{100}=120$ and $90\times\frac{170}{100}=153$

Required Ratio $=112:120:153$

Let the number of total students $=10x$

then the number of students under $10$ year's age $=\frac{20}{100}\times10x = 2x$

Ratio of students above $10$ year's age to those of $10$ year's age $=1:3$

$4\text{ Unit} = 10x-2x$

$1\text{ Unit} = 2x$

Number of students of $10$ year's age $=3\times2x = 6x$

Given that, 

$6x=24$

$x=4$ 

So, $10x = 40$

So the total number of students in the class is $40$

Let the value of C is $100$ , then 

B $=\frac{80}{100}\times100 = 80$

A $=\frac{130}{100}\times80=104$

D $=\frac{80}{100}\times104=83.2$

D is $4\%$ more than B

Let the total number of students be $100$

then the number of boys $=60$

and the number of girls $= 40$

Number of boys passed $= \frac{45}{100}\times60 = 27$

Number of total students passed $=\frac{58}{100}\times100 = 58$

So number of girls passed $=58-27 = 31$

$\%$ of passed girls $=\frac{31\times100}{40} = 77.5\%$

Total number of girls in college $=2800 \times \frac{35}{100}=28 \times 35=980$
Total number of boys in college $=2800-980=1820$
Number of boys who passed the final examination $=1820 \times \frac{75}{100}=1365$
Number of girls who passed the final examination $=980 \times \frac{85}{100}=833$
Total number of students who passed the final examination $=1365+833=2198$
Therefore, required percentage of total students who passed the final examination $=\frac{2198}{2800} \times 100=78.5 \%$

Let the number = $10$ (We have assumed such number which is LCM of 5 & 2 so that we can save our time in calculation)
If number is multiplied by $\frac{7}{5}=(10) \times \frac{7}{5}=14$
If number is multiplied by $\frac{3}{2}=(10) \times \frac{3}{2}=15$
Therefore, Required percentage change= $\frac{(15-14)}{15} \times 100 =6 \frac{2}{3} \%$

$\Rightarrow$ Valid votes $=5000 \times \frac{92}{100}=4600$

Now, loser got $(100-60) \%$ of the valid votes
$\Rightarrow$ Required number of votes $=\frac{40}{100} \times 4600$ $=1840$