Free Practice Questions for Percentage in Maths

Population on 1 January $2021=45000 \times \frac{110}{ 100} \times  \frac{110}{ 100}=54450$
$\therefore$ Population on 1 January 2023 is 54450.

Population of town $=2500$
Number of males $=2500 \times \frac{55}{100}=1375$
Number of females $=2500-1375=1125$
Number of literate males $=1375 \times \frac{92}{100}=1265$
Literate population in town $=2500 \times \frac{85}{100}=2125$
Number of literate females $=2125-1265=860$
Literacy rate of females in the town $=\frac{860}{2500} \times 100=34.4$%

Area of the shaded region $=80 \times 50=4000 \mathrm{~m}^{2}$
Area of the unshaded region is $35 \%$ of the area of the shaded region.
Area of the unshaded region $=\frac{35}{100} \times 4000=1400 \mathrm{~m}^{2}$
Total area of shaded region and unshaded region $=4000+1400=5400 \mathrm{~m}^{2}$
From figure:
$
\begin{aligned}
&\Rightarrow(80+x)(50+x)=5400 \\
&\Rightarrow 4000+80 x+50 x+x^{2}=5400 \\
&\Rightarrow x^{2}+130 x-1400=0 \\
&\Rightarrow x^{2}+140 x-10 x-1400=0 \\
&\Rightarrow x(x+140)-10(x+140)=0 \\
&\Rightarrow(x+140)(x-10)=0 \\
&\Rightarrow \mathrm{x}=-140 \text { or } \mathrm{x}=10
\end{aligned}
$
As $x$ is length therefore $x$ cannot be negative. Hence, $x=10$

Let the monthly income of the person be 100x.

the amount he spends on rent

= $\frac{22}{100}$ × 100x = 22x

The amount that one spends on household expenses and groceries

= $\frac{40}{100}$ × 100x = 40x$

∴ balance is= 100x - 22x - 40x = 38x

50% of the remaining on children's education and other heads

= $\frac{50}{100}$ × 38x = 19x

Person saves 19x his income

⇒ 19x = 3800

∴ x = Rs 200

Amount that spends on children's education and other heads

⇒ 19 × 200 = Rs 3800

The person spends Rs 3800 on children's education and other heads.

Alternate Method:

Let the total income be 100%.

He spends = 22% + 40% = 62%

⇒ Education = $\frac{50}{100}$ × 38 = 19%

∴ 19% is the savings of the person

⇒ Savings = Education = 19% = 3800

The person spends Rs 3800 on children's education and other heads.

Given:

Profit = 120% of the cost price

The cost price is increased by 25%.

Used Concept:

Profit = $ \frac{SP–CP}{CP}$× 100

SP = selling price

CP = cost price

Calculation:

Let the cost price be $5x$ Rs.

Profit = $5x$ × $ \frac{6}{5} = 6x$

Selling price = $5x + 6x = 11x$

New cost price $= 5x × \frac{5}{4} = \frac{25x}{4} $

New profit $= 11x - \frac{25x}{4}$

$= \frac{19x}{4}$

New profit % $= [\frac{19x}{4} 11x] × 100$

$= \frac{19x}{44x} × 100$

= 43.18 ≈ 43%

∴ New profit % is 43%.

Let the marked price = Rs $x$

According to the question:

Selling price = [$x$ × $\frac{100–15}{100}$ × $\frac{100–20}{100}$ × $\frac{100–18}{100}$]

⇒ 2230.40 = $x$ × $\frac{85}{100}$ × $\frac{80}{100}$ × $\frac{82}{100}$

⇒ 2230.40 = $\frac{697x}{1250}$

⇒ $x$ =$\frac{2230.40 × 1250 }{697}$ = 4000

∴ The marked price (in Rs) of the article is Rs 4000.

One-third of a journey is covered at a rate of $40 \mathrm{~km} / \mathrm{h}$, one-fourth at the $35 \mathrm{km} / \mathrm{hr}$ and rest at the rate of $45 \mathrm{~km} / \mathrm{h}$.
Let total distance $=\operatorname{LCM}(40,35,45)=2520 \mathrm{km}$ 

Distance covered @ $40 \mathrm{~km} / \mathrm{hr}=2520 \times \frac{1}{3}=840 \mathrm{~km}$ 

Time taken in covering this distance $=\frac{840}{40}=21$ hours 

Distance covered @35 km/hr $=2520 \times \frac{1}{4}=630 \mathrm{~km}$ 

Time taken in covering this distance $=\frac{630}{35}=18$ hours
Remaining distance $=2520-840-630=1050 \mathrm{~km}$
Time taken in covering remaining distance $=\frac{1050}{45}$ hour $=\frac{70}{3}$ hours
Average speed for the whole journey (in km/h) $=\frac{2520}{21+18+\frac{70}{3}}=\frac{7560}{63+54+70}=\frac{7560}{187} \mathrm{~km} / \mathrm{hr}$

Profit (Revenue - Expenditure) in year $2014=120-75=45$ crores
Profit (Revenue - Expenditure) as a percentage of revenue in year $2014=$ $\frac{45}{120} \times 100=37.5 \%$
Profit (Revenue - Expenditure) in year $2015=140-100=40$ crores
Profit (Revenue - Expenditure) as a percentage of revenue in year $2015=$ $\frac{40}{140} \times 100=28.57 \%$
Profit (Revenue - Expenditure) in year $2016=150-120=30$ crores
Profit (Revenue - Expenditure) as a percentage of revenue in year $2016=$ $\frac{30}{150} \times 100=20 \%$
Profit (Revenue - Expenditure) in year $2017=200-150=50$ crores
Profit (Revenue - Expenditure) as a percentage of revenue in year $2017=$ $\frac{50}{200} \times 100=25 \%$
Profit (Revenue - Expenditure) in year $2018=250-175=75$ crores
Profit (Revenue - Expenditure) as a percentage of revenue in year $2018=$ $\frac{75}{250} \times 100=30 \%$
Clearly, there are 3 years in which the profit (Revenue - Expenditure) as a percentage of revenue is more than $28 \%$ i.e., 2014, 2015 & 2018.

SMART APPROACH -

Let the number be N.

25% of N = 75

40% of N = $ \frac {75}{25} \times 40$ = 120

BASIC APPROACH -

From the question,

When 75 is added to $75 \%$ of a number, the answer obtained is the number itself.

Calculation,

Let the number be x.

According to the question, $75 \% $ of x + 75 = x 

x × $ \frac {75}{100}$ + 75 = x

x - $ \frac {3x}{4}$ = 75

x = 75 × 4 = 300 (number)

$40 \%$ of x = 300 × $ \frac {40}{100}$ = 120

So, $40 \%$ of the number = 120 

Given,

Marked Price = Rs.12000

Selling price = Rs.11000

Formula used,

Difference = Marked Price - Selling Price

Discount % = $ \frac{Difference}{Marked Price} \times 100$

Calculation,

Difference = Marked Price - Selling Price =12000 - 11000 = Rs.1000

Discount % = $ \frac{Difference}{Marked Price} \times 100$

Discount % = $ \frac{1000}{12000} \times 100$

  = $ \frac{25}{3}$ = 8$ \frac{1}{3}$%

At a recruitment test, the candidates were tested for General Awareness $(\mathrm{GA})$ and Quantitative techniques (QT).
$54 \%$ and $45 \%$ failed in GA and QT, respectively, while $16 \%$ failed in both.
Total failed candidate in atleast one subject = Candidates failed in GA + Candidates failed in QT - Candidates failed in GA and QT both 

Total failed candidate in atleast one subject = $=54 \%+45 \%-16 \%=83 \%$
Total passed candidate in both subjects $=100 \%-83 \%=17 \%$
According to question:
$
\begin{aligned}
&\Rightarrow 17 \%=136 \\
&\Rightarrow 1 \%=8 \\
&\Rightarrow 100 \%=800
\end{aligned}
$
Total number of candidates $=800$

In 1997, the cost price of new oven = ₹2500

In 2000, the cost price of that type of oven= ₹4800

Increased price(From 1997 to 2000) = 4800-2500=2300

∴ increased %$ =\frac{2300}{2500} \times 100 $ = $92 \%$

∴ Percentage increase between 1997 and 2000 = $92 \%$

Let the actual price be 100%

After giving 20% discount, the cost price =Rs 100

Therefore,

$80 \% \rightarrow \quad 100$

$100 \% \rightarrow \frac{100}{80} \times 100=125$

Hence, the actual price is Rs 125

Let the initial price be $100 x$.
Then, price after first discount of $10 \%$
$
=(100 x)-(100 x) \times 10 \%=100 x-10 x=90 x
$
Price after second discount of $15 \%$
$
=(90 x)-(90 x) \times 15 \%=90 x-13.5 x=76.5 x
$
And final price after discount of $18 \%$
$
=(76.5 x)-(76.5 x) \times 18 \%=76.5 x-13.77 x=62.73 x
$
Therefore, effective discount equivalent to three successive discounts of $10 \%$, $15 \%$ and $18 \%=$$\left(\frac{100 x-62.73 x}{100 x}\right) \times 100=\left(\frac{37.27 x}{100 x}\right) \times 100$
$
=37.27 \%
$

MATH MIRROR SOLUTION -

Required Decrease (in $\%$) $= \frac{20 -18 x}{20 } = \frac{2 }{20 }\times 100=10 \%$

BASIC SOLUTION -

Let the initial income of the girl be Rs. 100x.
Then, her expenditure $=(100 \mathrm{x}) \times 80 \%=$ Rs. $80 \mathrm{x}$
And her savings $=100 x-80 x=$ Rs. $20 x$
Now, increased income $=(100 x)+(100 x) \times 18 \%=100 x+18 x=$ Rs. $118 x$
And increased expenditure $=(80 x)+(80 x) \times 25 \%=80 x+20 x=$ Rs. $100 x$
New savings $=118 x-100 x=$ Rs. $18 x$
Decrease (in $\%$) in savings $=\left(\frac{20 x-18 x}{20 x}\right) \times 100=\left(\frac{2 x}{20 x}\right) \times 100=10 \%$

MATHS MIRRO METHOD -

New Price = $1200 \times \frac{25}{100}$ = 300

New price per unit = 300/10 = 30

Original price = $30 \times \frac{100}{75}$ = 40

BASIC SOLUTION -

Given information,

$\Rightarrow$ Percentage reduction in the price of sugar = 25% 

$\Rightarrow$ Price of sugar = 1200

$\Rightarrow$ Change in the quantity of sugar = 10 kg

Method used,

To maintain the expenditure,

Price of Commodity is inversely proportional to the Consumption.

$\Rightarrow$ Let the original price of sugar be A

$\Rightarrow$ Then the quantity of sugar = $\frac{1200}{A}$

$\Rightarrow$ Reduced price = $\frac{100 -25}{100}$ $\times$ A =   $\frac{3}{4}$$\times$ A

$\Rightarrow$ Increased quantity of sugar = $\frac{1200}{(3A/4)}$  = $\frac{1600}{A}$

$\Rightarrow$ According to the question,

$\Rightarrow$ $\frac{1600}{A}$ - $\frac{1200}{A}$ = 10

$\Rightarrow$ 400 = 10A

$\Rightarrow$ A = 40

$\therefore$ The original price of sugar = 40 Rs. per kg.

Hence, option(4) is the correct answer.

Let B's salary is 100

According to the question,

A's salary is 25% less than B

A's salary = 100 - $\frac{25}{100}$ $\times$ 100

A's salary = 100 - 25

A's salary = 75

Now we can see that B's salary is 25 more than A's salary 

Then % by which B's salary is more than A = $\frac{25}{75}$ $\times$ 100

$\Rightarrow$ 33.33% or $33 \frac{1}{3} \%$

Hence, option(3) is the correct answer.

Let the price of the sugar be 100a

According to the question,

Price of the sugar after decreased = 100a $\times \frac{90}{100} = 90$a

Price ratio of the sugar before and after the price decreased is 100a and 90a respectively.

We know that

If there is no change in total expenditure, then the consumption ratio is inversely proportional to the price.

Then

 the consumption ratio of the sugar before and after the price decreased is 90a and 100a respectively

$\therefore$ Consumption increased by = 100a - 90a = 10a

Required percentage = $\frac{10a}{90a}$ $\times$ 100 =  $\frac{100}{9}$ %

$\Rightarrow$ $11 \frac{1}{9} \%$

Hence, option(4) is the correct answer.

We have,

Ratio of girls and boys $=16$ : 9

$ \therefore \text { Girls percentage }(\%)=\frac{\text { Number of girls }}{\text { Total number of students }} \times 100 \\$

$=\frac{16}{25} \times 100 \\$

$=64 \%$

Let the first number be A and the second number be B.

According to the question,

20% of A + B = B + 50% of B

20% of A = 50% of B

$\frac{20}{100}$ $\times$ A = $\frac{50}{100}$ $\times$ B

 $\frac{A}{B}$ =$\frac{50}{20}$ = $\frac{5}{2}$

So, the ratio of the first to the second number = 5:2

Hence, option(3) is the correct answer.