Free Practice Questions for Percentage in Maths

Let the selling price be $x$ Rs.
Cost price $=\frac{7}{6} \times(\mathrm{x})=\frac{7 x}{6}$
We know that:
Loss percentage $=\frac{(\text { Cost price }-\text { Selling price })}{\text { Cost price }} \times 100$
Therefore, the loss percentage $=\frac{\left(\frac{7 x}{6}-x\right)}{\left(\frac{7 x}{6}\right)} \times 100=\frac{\left(\frac{7 x-6 x}{6}\right)}{\left(\frac{7 x}{6}\right)} \times 100=\left(\frac{x}{6}\right) \times\left(\frac{6}{7 x}\right) \times 100=14.29 \%$

MATHS MIRROR SOLUTION

Let CP $=\frac{7}{6}$SP

$\frac{CP}{SP}=\frac{7}{6}$

Required loss percentage  $=\frac{7-6}{7}\times100=\frac{1}{7}\times100=14.29 \%$

Let the cost price of one copy of a book be x Rs.

Cost price of 6 copies of a book = 6x

Cost price of 7 copies of a book = 7x

Simple interest = 7x – 6x = x

We know that:
Simple interest $=\frac{P \times R \times T}{100}$
$
\begin{aligned}
&\Rightarrow \mathrm{x}=\frac{(6 x) \times R \times 1}{100} \\
&\Rightarrow \mathrm{R}=\frac{100 x}{6 x} \\
&\Rightarrow \mathrm{R}=16.7 \%
\end{aligned}
$
Therefore, the rate per annum of simple interest is $16.7 \%$

The percentage of workers whose daily wages are less than 500 == = 22.5%

The percentage of workers whose daily wages are ₹600 and above == = 50%

Therefore, the required difference between the percentage of workers whose daily wages are less than 500 and the percentage of workers whose daily wages are ₹600 and above = 50% - 22.5 = 27.5%

Let the income of company in 2015 be x Rs.

According to question,

(x) ×× = 3125000

(x) ×× = 3125000

 $x =\frac{3125000 \times 4 \times 4}{5 \times 5}$

 x = Rs. 20,00,000 

Therefore, the income of company in 2015 was Rs. 20,00,000 

Total number of students in class A and B = 92

 The number of students in A is 30% more than that in B.

 ⇒

 Now,

 ⇒

⇒

⇒

 Hence, A = 92 – B = 92 – 40 = 52

 Average weight of all the students in A and B is 56 kg

 Sum of weight of all the students in A and B =

 Let average weight (in kg) of students in A = x kg

Let average weight (in kg) of students in B = y kg

 The average weight (in kg) of students in B is 50% more than that of students in A.

 Now,

 ⇒

⇒

⇒

 Now,

 Hence, average weight (in kg) of students in B = 69 kg

Given:

Amount invested by Arun = 8000 Rs.

Amount invested by Suresh = 8000 × = 8000 × = 4800 Rs.

Amount invested by Ramesh = 4800 × = 48 × 130 = 6240 Rs.

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

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 $=8 \times \frac{85}{100}=6.80$ Rs.

The value of $B=C \times \frac{65}{100}=\frac{65 C}{100}$
The value of $A=B \times \frac{120}{100}=\frac{65 C}{100} \times \frac{120}{100}=\frac{78 C}{100}$
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
$

According to the question,

x =

 x = 250

Now, the value of 18% of (x + 50) == = 54

The value of 12% of x = = 30

Therefore, required percentage == = 80%

Hence, option D is correct.

Clearly, quantity of petrol required to travel
$3045 \mathrm{~km}=\frac{3045}{15}$$=203$ litres
Now, increase in price of petrol per litre = 82 $\times \frac{5}{100}$
= 4.1 rupees
So, required increase in monthly expenses on petrol
= 203 × 4.1 = 832.3
$\approx 832$ rupees

$\left(x \times \frac{45}{100}\right)+\left(90 \times \frac{30}{100}\right)=\left(210 \times \frac{30}{100}\right)+\left(x \times \frac{20}{100}\right)$
$\Rightarrow \quad \frac{1}{100}[45 x]+[2700]=[(6300)+(20 x)] \times \frac{1}{100}$
$\Rightarrow \quad 45 x-20 x=6300-2700$
$\Rightarrow \quad 25 x=3600$
$\Rightarrow \quad x=\frac{3600}{25}$
$=144$

$x-15=x-x \times \frac{80}{100}$
$\Rightarrow \quad x-15=x\left(1-\frac{80}{100}\right)=x\left(\frac{20}{100}\right)=\frac{x}{5}$
$\Rightarrow \quad 5 x-75=x$
$\Rightarrow \quad 5 x-x=75$
$\Rightarrow \quad 4 x=75$
$\Rightarrow \quad x=75 / 4$
$\therefore$ अभीष्ट उत्तर $\quad \frac{75}{4} \times \frac{40}{100}=7.5$

अंग्रेजी में अनुत्तीर्ण परीक्षार्थी $=100-70=30 \%$
गणित में अनुत्तीर्ण परीक्षार्थी $=100-80=20 \%$
दोनों विषयों में अनुत्तीर्ण परीक्षार्थी $=10 \%$
दोनों विषयों में कुल उत्तीर्ण परीक्षार्थी
$
\begin{aligned}
&=[100-(30+20-10)] \% \\
&=(100-40) \%=60 \%
\end{aligned}
$
माना कुल परीक्षार्थियों की संख्या $x$ है।
$\therefore x$ का $60 \%$
\begin{array}{ll}
\Rightarrow & \frac{x \times 60}{100}=144 \\
\Rightarrow & x=\frac{144 \times 100}{60} \\
\Rightarrow & x=240
\end{array}

माना कुल मतदाता $x$ थे। $8 \%$ मतदाताओं ने अपने मत नहीं डाले।
जीतने वाले उम्मीदवार को प्राप्त मत $=48 \%$
हारने वाले उम्मीदवार को प्राप्त मत
$\ =[100-(48+8)] \%=44 \% $
मतों का अन्तर =1100
कुल मतों की संख्या =
$ =\frac{1100}{48-44} \times 100=\frac{1100 \times 100}{4}=27500 $

यहाँ $r_{1}=20 \%, \quad x=30$
तथा $r_{2}=32 \%, \quad y=42$
$\therefore$ परीक्षा का पूर्णांक $=\frac{x+y}{r_{2}-r_{1}} \times 100$
$=\frac{30+42}{32-20} \times 100=\frac{72}{12} \times 100=600$
$\therefore$ परीक्षा के न्यूनतम अंक $=\frac{32 \times 600}{100}-42$
$=192-42=150$
$\therefore$ न्यूनतम अंकों का प्रतिशत $=\frac{150 \times 100}{600}=25 \%$