Free Practice Questions for Percentage in Maths

$
\begin{aligned}
&A+B=40000-(i) \\
&\text { and }, \frac{15}{100} \times A=\frac{5}{100} \times B \\
&\Rightarrow \frac{A}{B}=1: 3
\end{aligned}
$
Salary of $A=\frac{40000}{(1+3)} \times 1=Rs. 10000$ 

$\%$ Loss $=\frac{100-99}{100} \times 100$
$\%$ decreases $=\frac{1}{100} \times 100=1 \%$ decreases $.$

ATQ,
Marketing $=10 \%$
On salary of employees $=65 \%$
Raw materials $=25 \%$
$
\begin{aligned}
&\Rightarrow 65 \%=65000 \\
&\Rightarrow \frac{65}{100}=65000 \\
&\Rightarrow 10,000
\end{aligned}
$
Hence,
Marketing expendicture $=10,000 \mathrm{Rs}$.

ATQ,
$36=2 \times 2 \times 3 \times 3$
$144=2 \times 2 \times 2 \times 2 \times 3 \times 3$
H.C.F $=2 \times 2 \times 3 \times 3=36$
$=$ H.C.F $=\frac{36}{2}=18$

According to question,
$
\begin{aligned}
&\Rightarrow 270 \times \frac{(x+25)}{100}=240 \times \frac{(x)}{100} \times \frac{(100+20)}{100} \\
&\Rightarrow 270 \times \frac{(x+25)}{100}=\frac{288 x}{100} \\
&\Rightarrow 270 x+6750=288 x
\end{aligned}
$
$
\Rightarrow 288 x-270 x=6750
$
$
\Rightarrow 18 x=6750
$
$
\Rightarrow x=375
$
Now,
The value of $12 \%$ of $(x+125)=(375+125) \times \frac{12}{100}=\frac{500 \times 12}{100}=60$
The value of $10 \%$ of $x=(x) \times \frac{10}{100}=375 \times \frac{10}{100}=37.5$
Therefore, required percentage $=\frac{(60-37.5)}{37.5} \times 100=\frac{22.5}{37.5} \times 100=60 \%$
Hence, option D is correct.

Amount invested by Arun $=$ ₹ 8000

Amount invested by Suresh $=8000 \times \frac{(100-40)}{100}=8000 \times \frac{60}{100}= $ ₹ 4800

Amount invested by Ramesh $=4800 \times \frac{(100+30)}{100}=48 \times 130=₹ 6240$ .

Therefore, total amount invested by all of them = 8000 + 4800 + 6240 = ₹ 19040

Hence, option C is correct.

MATHS MIRROR METHOD -

Let radius and height be 10.

Volume of a cylinder =

Previous volume $= 10 \times 10 \times 10 = 1000$

New volume = $= 6 \times 6 \times 16 = 576$

So, volume will decrease by = $\frac{(1000-576)}{1000} \times 100$

= $\frac{424}{1000} \times 100$ = 42.4%

BASIC APPROACH -

Let the radius of cylinder = r and height of the cylinder = h

We know that:

Volume of a cylinder =

Volume of a new cylinder =

= 

= 

Therefore, required percentage = 

= 

= 

= 42.4 %

Hence, option D is correct.

Let the price of sugar $=x$ per $\mathrm{kg}$
According to question,
$\Rightarrow \frac{272}{\left(x \times \frac{85}{100}\right)}-\frac{272}{x}=6$
$\Rightarrow \frac{272 \times 20}{17 x}-\frac{272}{x}=6$
$\Rightarrow \frac{320}{x}-\frac{272}{x}=6$
$\Rightarrow \frac{320-272}{x}=6$
$\Rightarrow x=\frac{48}{6}$
$\Rightarrow x=8$
Therefore, reduced price of sugar $=8 \times \frac{85}{100}= Rs. 6.80$
Hence, option C is correct.

The value of $B=C \times \frac{65}{100}=\frac{65 C}{100}$
The value of $A=B \times \frac{120}{100}=\frac{13 C}{20} \times \frac{120}{100}=\frac{78 C}{100}$
According to question,
$
\Rightarrow A+B+C=121 \frac{1}{2}
$
$
\Rightarrow \frac{78 C}{100}+\frac{65 C}{100}+C=\frac{243}{2}
$
$\Rightarrow \frac{78 C+65 C+100 C}{100}=\frac{243}{2}$
$\Rightarrow \frac{243 C}{100}=\frac{243}{2}$
$\Rightarrow C=50$
$A=\frac{78 C}{100}=\frac{78 \times 50}{100}=39$
$B=\frac{65 C}{100}=\frac{65 \times 50}{100}=32.5$
Therefore, required value of $2 C-4 B+A$
$
=2 \times 50-4 \times 32.5+39
$
$
=100-130+39
$
$
=9
$
Hence, option B is correct.

Alternate Method 

$ A : B = 6:5 , B :C = 13:20 $

Equating the value of B in both equations 

$Now ,A:B:C = 39 : 32.5 : 50$

Therefore, required value of $2 C-4 B+A$
$
=2 \times 50-4 \times 32.5+39
$
$
=100-130+39
$
$
=9
$ 

Calculation:

40% of 120% of 80% of 750

⇒ $750 \times \frac{80}{100} \times \frac{120}{100} \times \frac{40}{100}$

⇒ 288

∴ Required answer is 288.

ATQ,
P X Q = Expenditure

$\begin{array}{cc}Q_{1} & Q_{2} \\ 100 & 92\end{array}$
loss% $\Rightarrow$ $\frac{8}{100} \times 100$
$\Rightarrow$ 8%

Let total number of populations in a constituency $=100$
Number of males in constituency $=100 \times \frac{55}{100}=55$
Number of females in constituency $=100-55=45$
Number of males who are literate $=55 \times \frac{40}{100}=22$

Number of females who are illiterate $= 45 \times 33 \frac{1}{3} \% = 15$
Number of females who are literate $=45-15=30$
Total number of voters who are literate $=22+30=52$
Percentage of the voters who are literate $=\frac{52}{100} \times 100=52 \%$

ATQ,
Let alok per month salary = Rs.100
Expenditure of alok 85%
=Rs. 85 

15% = 1200
1 % = $\frac{1200}{15}$
$1 \%=80$
$100 \%=80 \times 100$
= Rs.8000 

 Kritika had 21450 rupees with him

Amount left after 16 spend= 84% 21450=18018

Amount left after 50% spent= 50%18018=9009 Amount left after spent on internet= 9009-991=8018 Hence, option 1 is the correct answer

ATQ,
Total anita's total expenditure
$=$ Rs. 50000 per month
Payment for tailoring
$=50000 \times \frac{40}{100}$
$=20,000 \mathrm{Rs} .$

Let initial salary of employee $=100 \mathrm{x}$
Salary of employee after two successive increments $=2.5 \times 100 x=250 x$
Let first increment in salary $=\mathrm{y} \%$
Second increment in salary $=4 \mathrm{y} \%$
Now, $100 x \times \frac{(100+y)}{100} \times \frac{(100+4 y)}{100}=250 x$
$
\Rightarrow(100+y) \times(100+4 y)=25000
$
$\Rightarrow 10000+400 y+100 y+4 y^{2}=25000$
$\Rightarrow 4 y^{2}+500 y-15000=0$
$\Rightarrow y^{2}+125 y-3750=0$
$\Rightarrow y^{2}+150 y-25 y-3750=0$
$\Rightarrow y(y+150)-25(y+150)=0$
$\Rightarrow(y+150)(y-25)=0$
$\Rightarrow y=-150$ (not possible),so, $y=25$

Let number of workers in the company = 100x
Number of skilled workers in the company $=\left(\frac{75}{100}\right) \times 100 x=75 x$
Number of unskilled workers in the company $=100 x-75 x=25 x$
Number of permanent skilled workers in the company $=\left(\frac{84}{100}\right) \times 75 x=63 x$
Number of permanent unskilled workers in the company $=\left(\frac{28}{100}\right) \times 25 x=7 x$
Number of permanent workers in the company $=63 x+7 x=70 x$
Number of temporary workers in the company $=100 x-70 x=30 x$
According to question
$
\begin{aligned}
&\Rightarrow 30 x=180 \\
&\Rightarrow x=6
\end{aligned}
$
Total number of workers in the company = 600

Final sale price (to the nearest rupee) of a microwave oven =$
\begin{aligned}
15485 \times \frac{(100-40)}{100} \times \frac{(100-25)}{100}=15485 \times \frac{(60)}{100} \times \frac{(75)}{100}=\text { Rs. } 6968.25 \\
& \approx \text { Rs. } 6968
\end{aligned}
$

Net salary $= 100 \times \frac{(100-50)} {100} \times \frac{(100+50)}{100}$

$= 100 \times \frac{50} {100} \times \frac{150}{100}$

= 75

Loss % = (100-75) = 25%

Let the number $=\mathrm{y}$
Then, Girls : Boys $=16 \mathrm{y}: 9 \mathrm{y}$
$
\begin{aligned}
&\Rightarrow 16 y=48 \\
&\Rightarrow y=3
\end{aligned}
$
Now, boys $= 9y = 9 \times 3=27$
Total number girls and boys $=32+27=59$
Percentage $=(48 / 59) \times 100=81.35 \%$