Free Practice Questions for Percentage in Maths

Let the $3 \mathrm{rd}$ number is $100$ , then
$1\mathrm{st}$ number $=\frac{12}{100} \times 100=12$
$2\mathrm{nd}$ number $=\frac{48}{100} \times 100=48$
Required $\%=\frac{12}{48} \times 100=25 \%$

Runs scored with the help of sixes = 6 × 15 = 90 runs

 Runs made with the help of fours = 4 × 18 = 72 runs

Total number of runs = 200 runs

Runs scored by running between wickets $=200-162=38$  runs

% of runs scored by running between the wickets

=$\frac{38}{200} \times 100=19 \%$

As per the question,
$
\begin{aligned}
&\Rightarrow 200 \% \text { of } \mathrm{k}-(\mathrm{k}+2) \% \text { of } 50=\mathrm{k} \\
&\Rightarrow 2 \mathrm{k}-(\mathrm{k}+2) \frac{1}{2}=\mathrm{k} \\
&\Rightarrow 4 \mathrm{k}-\mathrm{k}-2=2 \mathrm{k} \\
&\Rightarrow 4 \mathrm{k}-3 \mathrm{k}=2 \\
&\Rightarrow k=2
\end{aligned}
$

As per the question,

$\mathrm{p} + \mathrm{p} \times \frac{99}{100}=99 + \frac{99}{100} \times \mathrm{p}$

$\mathrm{p}=99$

So the value of 

$\mathrm({p}+51) \%\\\ \text {of} \\\ 928+72$ 

$=(99+51) \% \text { of } 928+72$

$=\frac{150}{100} \times 928+72 \\$

$=1392+72=1464$

Let saving initial saving of renu $=100$
initial expenditure $=100+100 \times \frac{150}{100}=250$
So, income $=100+250=350$
New savings when increased by $50 \%=100+100 \times \frac{50}{100}=150$
New when income increased by $20 \%=350+350 \times \frac{20}{100}=420$
New expenditure $=420-150=270$
Required percentage $=\frac{20}{250} \times 100=8 \%$

The original price of 15 gram of gold $=71820 \times \frac{100}{112}=$₹ 64125
The original price of 1 gram of gold $=\frac{64125}{15}=$₹ 4275

Rahul paid as tax $=100 \%-60 \%=40 \%$
According to question
$
40 \% \Rightarrow 12800
$
$
\begin{aligned}
&1 \% \Rightarrow 320 \\
&100 \% \Rightarrow 32000
\end{aligned}
$
So, the total tax to be paid by Rahul is $=32000$

According to Question ,

7 umbrellas are sold $=$ Rs. 800
9 umbrellas are sold $=\frac{800}{7} \times 9=1028.52$
Profit $\%=\frac{1028.52-800}{800} \times 100=28.57 \%$

Let the total number of votes = 100%

Let $\mathrm{R}$ got ${x}$ votes in election
$\mathrm{D}$ got votes in election $={x}+12 \%$
We know, ${x}+{x}+12 \%=100 \%$
$
\begin{aligned}
&2 {x}+12 \%=100 \% \\
&2 {x}=100 \%-12 \% \\
&{x}(\mathrm{R})=44 \% \\
&\text { Given, } 44 \%(\mathrm{R}) \rightarrow 176000 \\
&1 \%=4000
\end{aligned}
$
D got votes in election $=44 \%+12 \%=56 \%$
He lose the election by the votes $=(56 \%-44 \%) \times 4000$
$=12 \% \times 4000$
$=48000$

Net change $=5 \%-1 \%=4 \%$
Population of city after 3 year $=5,00,000 \times \frac{104}{100} \times \frac{104}{100} \times \frac{104}{100}$
$
=5,00,000 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}
$
$=562432$

Difference of price $=\frac{150}{7.5}-\frac{150}{8}=150 \times \frac{0.5}{60}=1.25$

Let $x$ be the number.
$3 \%$ of $x=4 \%$ of $12+6 \%$ of 80
$
\begin{aligned}
&\frac{3 x}{100}=0.48+4.8 \\
&3 \mathrm{x}=528 \\
&\mathrm{X}=176
\end{aligned}
$

Let value of the machine at the end of first year be Rs.

x Value of the machine at the end of second year $=x \times \frac{75}{100}$

Value of the machine at the end of third year $=x \times\left(\frac{75}{100}\right)^2$
Given,
$
\begin{aligned}
&\mathrm{x} \times\left(\frac{75}{100}\right)^2-\mathrm{x} \times \frac{75}{100}=24000 \\
&\mathrm{x} \times \frac{3}{4} \times \frac{3}{4}-\mathrm{x} \times \frac{3}{4}=24000 \\
&\frac{9 x}{16}-\frac{3 x}{4}=24000 \\
&\mathrm{x}=\mathrm{Rs} .128000
\end{aligned}
$

Total expense of salary $=25 \%+15 \%+18 \%=58 \%$
Saving $=100 \%-58 \%=42 \%$
Here, $42 \% \rightarrow 7560$
$
\begin{aligned}
&1 \% \rightarrow 180 \\
&100 \% \rightarrow 18000
\end{aligned}
$

A candidate got votes in an election $=41250$
He Defeated his opponent by a margin of 7500 votes
Defeated candidate got votes in an election $=41250-7500=33750$
Total votes $=41250+33750=75000$
Percentage of votes obtained by the winning candidate $=\frac{41250}{75000} \times 100=$ $55 \%$

$78 \%$ of $\mathrm{A}=32.5 \%$ of $\mathrm{B}$
$\Rightarrow \frac{78}{100} \times \mathrm{A}=\frac{32.5}{100} \times B$
$\Rightarrow \frac{A}{B}=\frac{32.5}{78}$
$\Rightarrow \frac{A}{B}=\frac{12}{5}$

Let x be the money spent by Sushant
Share of wife $=\mathrm{x} \times \frac{30}{100}=0.30 \mathrm{x}$
Share of two sons $=x \times \frac{70}{100} \times \frac{18}{100}=0.252 \mathrm{x}$
Miscellaneous items $=(\mathrm{x}-0.30 \mathrm{x}-0.252 \mathrm{x}) \times \frac{3}{4}$
$
=0.448 \mathrm{x} \times \frac{3}{4}=0.366 \mathrm{x}
$
According to question
$
\begin{aligned}
&\Rightarrow \mathrm{x}=0.30 \mathrm{x}+0.252 \mathrm{x}+0.336 \mathrm{x}+19600 \\
&\Rightarrow \mathrm{x}=0.888 \mathrm{x}+19600 \\
&\Rightarrow \mathrm{x}-0.888 \mathrm{x}=19600 \\
&\Rightarrow 0.112 \mathrm{x}=19600 \\
&\Rightarrow \mathrm{x}=175000
\end{aligned}
$

$15 \%$ of 2011 's attendance at a cricket stadium $=2160$
Hence, 2011's attendance at a cricket stadium $=\frac{2160 \times 100}{15}=14400$

$10 \%=\frac{1}{10}$
Let the price of the article $=1000$, then
Price at the end of 1st year $=1000 \times \frac{11}{10}=1100$
Price at the end of 3rd year $=1000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}=1331$
Price at the end of 4th year $=1000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}=1464.1$
Difference between the price at the end of 3 rd year to that at the end of 1st year $=1331-1100=231$
According to question
$231 \rightarrow $₹ $69.3$
$1 \rightarrow $₹ $0.3$
So, Price at the end of the 4th year
$1464.1 \rightarrow $₹ $0.3$ $\times 1464.1=$₹ $439.23$

Let the amount of gravel, sand and lime be $9{x}, 2 {x}$ and $6{x}$ respectively.
According to question
$
\begin{aligned}
&9{x}+2{x}+6{x}=136 \\
&\Rightarrow 17{x}=136 \\
&\Rightarrow {x}=\frac{136}{17} \\
&\Rightarrow {x}=8
\end{aligned}
$
Difference between sand and gravel $=9 {x}-2 {x}$
$
=7 {x}=7 \times 8=56 {~kg}
$