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Aptitude::Percentage

Home > Quantitative Aptitude > Percentage > Subjective Solved Examples

Example 1 / 28

What is 10% of 20% of 25% of 100?
It can be solved as,
= 0.1 * 0.2 * 0.25 * 100
= 0.5%
So, the answer is a. 0.5 Ans.
Formula Used:
x% of y
= \(\frac{x}{100}\times y\)

NA
We can write directly as,
(i) 10% of 100 = 10
(ii) 1% of 100 = 1
e.g;
25% of 100 = (20+5)% of 100
= 2*10% of 100 + 5* 1% of 100
= 20+ 5 = 25
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Example 2 / 28

If 30% of a number is 300, then 50% of that number is:
Let the number be 'x'.
So, 0.3*x = 300
=> x = 1000
Thus, 50% of 1000 = 5*(100) = 500 Ans.
Formula Used:
x% of y
= \(\frac{x}{100}\times y\)
NA
We can directly have a formula like:
Let the no. to be obtained be x
x = \(\frac{(The Obtained Percentage)}{\%Age Given}\)
e.g;
=> 30% of x = 300
=> x = \(\frac{300}{30%}\)
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Example 3 / 28

30% of a% of b is 25% of b% of c. Which of the following is c?
Given,
0.3* \(\frac{ab}{100}\) = 0.25*\(\frac{bc}{100}\)
0.3a = 0.25c
So, c = \(\frac {0.3}{0.25}\)
c = 1.20 a Ans.
Formula Used:
x% of y% of z
\(\frac{xyz}{10000}\)
NA
When 'n' number of % is given of some number is given then,
\(\frac{'n' numbers}{10^{(n-1)}}\)
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Example 4 / 28

When 40% of a number A is added to another number B, B becomes 125% of its previous value. Then which of the following is true regarding the values of A and B?
0.4A + B = 1.25B
0.4 A = 0.25 B
A = 0.625 B
Since we don't know the values of A and B,
so, it can be depending upon the values of A and B.

Formula Used:
X% of y
= \(\frac{x}{100} \times y\)

NA
To calculate p% of y, it is, (p% of y) = (y% of p)
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Example 5 / 28

Two numbers A and B are such that the sum of 5% of A and 10% of B is \(\frac{1}{2}\) of the sum of 20% of A and 10% of B. Find the ratio of A:B?
0.05A + 0.1B = \(\frac{1}{2}\)*( 0.2A + 0.1B)
0.05A + 0.1B = 0.5*(0.2A + 0.1B)
0.05A + 0.1B = 0.1A + 0.05B
0.05A = 0.05B
So, A:B = 1:1 Ans.

Formula Used:
X% of y
= \(\frac{x}{100} \times y\)


NA
We can directly have a formula like:
To calculate p% of y, it is, (p% of y) = (y% of p)

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Example 6 / 28

In an election between 2 candidates, Chaman gets 80% of the total valid votes. If the total votes were 12000, what is the number of valid votes that the other candidate Dhande gets if 15% of the total votes were declared invalid?
Total votes = 12000
Invalid votes = 15% of 12000 = 0.15* 12000 = 1800
So, valid votes = 12000-1800 = 10200
No. of votes Chaman gets = 80% of 10200 = 0.80 * 10200 = 8160
No. of valid votes other candidates get = (10200 - 8160) = 2040 Ans.

Formula Used:

X% of y
= \(\frac{x}{100}\times y\)

NA
First take out total invalid vote and subtract it from the total votes, then find the no. of votes by taking out the percentage of the votes.
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Example 7 / 28

Raunak generally wears his father’s coat. Unfortunately, his cousin Vikas told him one day that he was wearing a coat of length more than his height by 15%. If the length of Raunak’s father’s coat is 345 cm then find the actual length(in cm) of his coat.
Let the actual length of the coat be 'x'.
So, x + 15% of x = 345
=> x + 0.15 x = 345
=> 1.15 x= 345
=> x = 300cm Ans.

y+ X% of y
=> \(y + \frac{x}{100} \times y\)

NA
To calculate x, we can directly do as
X = \(\frac{Given Length}{1 + Given%}\)

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Example 8 / 28

Sachin wanted to subtract 10 from a number. Unfortunately, he added 10 instead of subtracting. Find the percentage change in the result due to this mistake.
% change = \(\frac{(10+10)}{10}\) * 100 = 200%

Formula Used:

= \(\frac{Initial - Final}{Initial}\times 100\)

NA
To calculate the percent change = \(\frac{Initial - Final}{Initial}\)
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Example 9 / 28

The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid.
Let the length, breadth, and height be L, B, H respectively.
So, initial volume = L*B*H = LBH
So, now,
Increased length = L + 10% of L = 1.1 L
Increased breadth = B + 20% of B = 1.2 B
Increased height = H + 50% of H = 1.5 H
New volume = 1.1L * 1.2B * 1.5H = 1.98 LBH
% change in volume = \(\frac{new volume – initial volume}{initial volume}\)
= \(\frac{1.98 LBH - LBH}{LBH}\) = 98% Ans.

Formula Used:

= \(\frac{New Volume – Initial Volume}{Initial Volume}\)
NA
To increase X by Y%, we have formula,
\(X \left(1 + \frac{Y}{100}\right)\)

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Example 10 / 28

A student multiplied a number by 3/5 instead of 5/3, What is the percentage error in the calculation ?
Let the number be x.
Then, ideally, he should have multiplied by x by 5/3.
Hence Correct result was x * \(\frac{5x}{3}\) = \(\frac{5x}{3}\)
By mistake he multiplied x by \(\frac{3}{4}\).
Hence the result with error = \(\frac{3x}{5}\)
Then, error = \(\left(\frac{5x}{3} - \frac{3x}{5}\right)\) = \(\frac{16}{15}\)
Error % = (error/True value) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64% Ans.

Formula Used:

Error% = \(\frac{error}{True Value} \times 100\)
NA
To calculate error%:
\(\frac{\left( true - incorrect \right)}{true}\)*100

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Example 11 / 28

The price of salt is reduced by 50% but, inspite of the decrease, Aayush ends up increasing his expenditure on salt by 50%. What is the percentage change in his monthly consumption of salt?
Let the initial price of salt be Rs. 100 per kg.
Let the total consumption be 100 kg.
So, total expenditure= 100*100= Rs. 10,000.
Now, new price= Rs. 50.
Also, let new consumption be x kg.
So, new expenditure= Rs. 50x

Total expenditure increased by 50%.
So, 10,000* \(\frac{150}{100}\)= 50x
X = 300
His consumption increased by (300 - 100)kg = 200 kg.
So, 200% increase on monthly consumption. Ans.

Formula Used:

= \(\frac{New Volume – Initial Volume}{Initial Volume}\)
NA
To increase X by Y%, we have formula,
\(X \left(1 + \frac{Y}{100}\right)\)

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Example 12 / 28

At an election, the candidate who got 60% of the votes cast won by 200 votes. Find the total number of voters on the voting list if 66.67% people cast their vote and were no invalid votes.
At an election, the candidate who got 60% of the votes cast won by 200 votes. Find the total number of voters on the voting list if 66.67% people cast their vote and there were no invalid votes.
Let the total votes be 3V
People who casted Votes = \(\frac{66.67}{100}\)*3V = 2V
Candidate who won got = \(\frac{60}{100}\)*2V = 1.2V
Other Candidate got = 2V - 1.2V = 0.8V
Difference = 1.2V - 0.8V = 0.4V
0.4V = 200
=> V = 500
3V = 1500
Total Number of Voters on Voting List = 1500 Ans.

Formula Used:

= \(\frac{New Volume – Initial Volume}{Initial Volume}\)
NA
To increase X by Y%, we have formula,
\(X\left(1 + \frac{Y}{100}\right)\)

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Example 13 / 28

The population of the city Gotham is 50,000 at this moment. It increases by 20% in the first year. However, in the second year, due to immigration, the population drops by 10%. Find the population at the end of the third year if in the third year the population increases by 30%.
Population = 50,000
In first year = (50000 + 2% of 50000) = 60,000
In second year = (60,000 - 10% of 60000) = 54000
In the third year = (54000 + 30% of 54000) = 70200

Formula Used:
To calculate increase of x by y%
New x = (x + y% of x)
To calculate decrease of x by y%
New x = (x - y% of x)

NA
If the current population is P and it increases at the rate of R% per annum, then,
=>population after n years = \(P \left(1 + \frac{R}{100}\right)^{n}\)
=> population n years ago = \(\frac{P}{\left(1 + \frac{R}{100}\right)^{n}}\)

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Example 14 / 28

Sharad spents 20% of his monthly income on his household expenditure, 30% of the rest on food, 10% of the rest on clothes and saves the rest. On counting, he comes to know that he has finally saved rs. 10080. Find his monthly income (in Rs.).
Let the monthly income be Rs. X.
Now, for household expenditure = 20%
For food = 30%
For clothes = 10%
So, remaining = 100% -(20% + 30% + 10%) = 40%
\(\frac{40x}{100}\) = 10080
=> X = 25200

Formula Used:
= x% of y
NA
If amounts spent are given, then, remaining will be Total - (sum of all the amount spent).
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Example 15 / 28

The population of the town is 6000. If the number of males increases by 10% and the number of females increases by 20%, then the population becomes 6800. Find the population of females in the town.
Total population is 6000
Let no. of males = x and females = y
so (x + y) = 6000 ------(1)
x increases by 10% the new x = \(\frac{110x}{100}\)
y increases by 20% then new y = \(\frac{120y}{100}\)
Now, \(\frac{110x}{100}\) + \(\frac{120y}{100}\) = 6800 -------(2)
By solving (1) and (2)
y = 2000. Ans.

Formula Used:

=> if x increases by y%, then % increase= x+(y% of x)
NA
To increase X by Y%, we have shortcut formula,
\(X \left(1 + \frac{Y}{100}\right) \)

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Example 16 / 28

In an election contested by two parties, Party SJP secured 12% points of the total votes more than Party SJD. If party SJD got 132,000 votes and there are no invalid votes, by how many votes did it lose the elections?
Let Assume total votes = 100V
Party SJP got = 12V + 132000
Party SJD Got = 132000
132000 + 12V + 132000 = 100V
88 V = 264000
V = 3000
Party SJD lost by 12V
12*3000 = 36000 Ans.

Formula Used:

= votes of SJP + votes of SJD + 0 invalid votes = Total Votes
NA
To find total votes, there is no need of taking out %. Directly we can add all the given votes and equate by total votes.
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Example 17 / 28

Vinay’s salary is 75% more than Ashok’s. Vinay got a raise of 40% on his salary while Ashok got a raise of 25% on his salary. By what percentage is Vinay’s salary more than Ashok’s?
If the salary of Ashok is Rs. 100 then, Vinay’s salary is Rs. 175
again,
If Ashok's salary is Rs. 125 then, Vinay's salary is 245
now,
Ashok's salary is less than Vicky by 120

less% = \(\frac{120}{125}\times 100\) = 96%

Formula Used:

if there is y% increase in x, then, new x = (x + y)% of x
NA
To increase X by Y%, we have formula,
\(X\left(1 + \frac{Y}{100}\right)\)
Also, we must remember, suppose we have considered a number 1000
Then,10% of 1000 = 100
1% of 1000 = 10
5% of 1000 = 5*(1% of 1000) = 50
75% of 1000 = 7*(10% of 1000) + 5*(1% of 1000) = 750

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Example 18 / 28

An ore contains 20% of an alloy that has 50% copper. Other than this, in the remaining 80% of the ore, there is no copper. How many kilograms of the ore are needed to obtain 10kg of pure copper?
Let there is x kg of ore.
So, \(\frac{20x}{100}\)= alloy
So, 50% of \(\frac{20x}{100}\) = copper
For, 10kg of copper \(\frac{10x}{100}\) copper
So, 80% of ore has no copper, so, \(\frac{80x}{100}\)
So, we can say that,
\(\frac{10x}{100}\) = 10kg
=> X = 100 kg. Ans.
Formula Used:
to take x% of y, we use
= x%*y

NA
x% of y% of z = xy% of z
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Example 19 / 28

The population of a village is 4,00,000. Increase rate per annum is 20%. Find the population at the starting of the 4th year?
P = 4,00,000, t = 3 years, r = 20%
So, population at starting of 4th year = \(P\left(1 + \frac{r}{100}\right)^{3}\)
= \(4,00,000\left(1 + \frac{20}{100}\right)^{3}\) = 691200

Formula Used:

To calculate increase of x by y%
New x = (x + y% of x)
To calculate decrease of x by y%
New x = (x - y% of x)

NA
If the current population is P and it increases at the rate of R% per annum, then,
=>population after n years = \(P\left(1 + \frac{R}{100}\right)^{n}\)
=> population n years ago = \(\frac{P}{\left(1 + \frac{R}{100}\right)^{n}}\)

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Example 20 / 28

A table and a chair are priced at Rs. 3000 and Rs. 1000 respectively. If the price of the table and that of the chair is increased by 10% and 20% respectively, then the price of 10 tables and 20 chairs is:
Let T be the price of tables and C be the price of the chairs.
So, New price of table = (3000 + 300) = 3300
New price of chair = (1000 + 200) = 1200
So, the price of 10 tables and 20 chairs = (10 * 3300 + 20 * 1200)
= 33000 + 24000 = 57,000 Ans.

Formula Used: = increase of x by y%= new x= x+y%of x
NA
To increase X by Y%, we have formula,
\(X\left(1 + \frac{Y}{100}\right)\)

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Example 21 / 28

Ramesh buys a house for Rs. 2,00,000. The annual repair cost comes to 6.0% of the price of purchase. Besides, he has to pay an annual tax of Rs. 12000. At what monthly rent must he rent out the house to get a return of 20% on his net investment(in Rs.) of the first year?
CP of house = Rs. 2,00,000
Repair cost = 6% of 2,00,000 = 12000
Annual tax = 12000
So, total = Rs. 2,24,000
Now, to get 20% on his net investment = 20% of 224000 = 44800
So, monthly amount = \(\frac{44800}{12}\) = Rs.3733.33 Ans.

Formula Used: monthly rent = \(\frac{Total Expense * given%}{12}\)
NA
For monthly rent = \(\frac{return\% \times Total Expense}{12}\)
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Example 22 / 28

\(\frac{4}{5}\)th of the voters in Kanpur promised to vote for Modi and the rest promised to vote for Advani. Of these voters, 10% of the voters who had promised to vote for Modi, did not vote on election day, while 20% of the voters who had promised to vote for Advani did not vote on the election day. What is the total number of votes polled if Modi got 216000 votes?
Given that \(\frac{4}{5}\) of the voters in Kanpur promised to vote for Modi and the rest promised to vote for Advani
Now, let the voters in Kanpur promised 4x votes for Modi and x votes for Advani.
The voters who actully voted to Modi = (4x - 10% of 4x) = (4x - 0.4x) = 3.6x
The number of votes Modi got = 216000
So,3.6x = 216000
=> x = 216000/3.6 = 60000
And
The voters who actully voted to Advani = (x - 20% of x) = (x- 0.2x) = 0.8x
Now,
Total polled votes = (3.6x + 0.8x) = 4.4x
So, Total polled votes = 4.4*60000 = 264000

Formula Used:
Decrease in x by y% = (x - y% of x)
NA
Tips: first, take our respective vots by given formula, then add and equate by total votes, and multiply by total polled
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Example 23 / 28

Sudhir spends 25% of his salary on house rent, 20% of the rest he spends on his children’s education and 10% of the total salary he spends on clothes. After his expenditure, he is left with Rs. 20,000. What is Sudhir’s salary?
Let the salary be x.
So, house rent = 25% of x.
Remaining= 75% of x.
So, for children's education = 20% of 75% of x = 15% of x
For clothes = 10% of x.
Money left with him = Rs.20,000
So, x - (25% of x) + (15% of x) + (10% of x) = 20000
so, x = 40,000 Ans.

Formula Used:

Remaining salary = Total - (sum of all expenses)
NA
To find the total salary, first find the total expense, add all and subtract from the salary and equate it to the remaining salary given.
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Example 24 / 28

Raman’s monthly salary is A rupees. Of this, he spends X rupees. The next month he has an increase of C% in his salary and D% in his expenditure. The new amount saved is:
Monthly salary = A
Expenditure = X.
So, Remaining = (A-X).
Now, for next month, increase of C%
So, (A) + C% of (A) = \(A\left[1 + \frac{C}{100}\right]\)
New Expenditure = x + D% of x
= \(X\left(1 + \frac{D}{100}\right)\)
Remaining = (New Salary - New Expenditure)
= \(A\left[1+ \frac{C}{100}\right] - X\left[1+\frac{D}{100}\right]\) Ans.

Formula Used:

Savings = (Salary - Expenditure)
Increase in salary by x% = salary + x% of salary
Increase in expenditure by y% = expenditure + y% of expenditure
New savings = (increased salary - increased expenditure)

NA
To increase X by Y%, we have formula,
X(1+\(\frac{Y}{100}\)
Also, Savings = (salary - expenditure)

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Example 25 / 28

Ranjan buys goods worth Rs. 10,000. He gets a rebate of 20% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
Rebate = 20% of 10000
i.e price after rebate is 80% of 10000 = 8000.
sales tax is 10% of 8000 = 800
Therefore,Final amount paid = 8000 + 800 = Rs. 8800 Ans.

Formula Used:

Sale price(SP) = (P - R*P)
Where, P = original price, R = rebate(discount %)

Then Sales = S * SP, where S = sales%, SP = sales price
So, final amount paid = Sales + SP.

NA
Rebate is nothing but simple discount on the given original price.
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Example 26 / 28

The salary of Anuj is 20% lower than Bhuwan’s salary and the salary of Chauhan is 56.25% greater than Anuj’s salary. By how much percent the salary of Bhuwan is less than the salary of Chauhan?
Let Bhuwan's salary be Rs.100.
So, Anuj's salary = Rs.80.
Salary of Chauhan = Rs.80 + 56.25% of 80 = Rs.125
So, Bhuwan's salary is less than Chahuhan's salary by Rs.(125 - 100) = Rs.25
% Decrease = \(\frac{25}{125} \times 100\) = 20% Ans.

Formula Used:% Increase in x by y% = x + y% of x
NA
To increase X by Y%, we have formula,
\(X\left(1 + \frac{Y}{100}\right)\)

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Example 27 / 28

A batsman scored 100 runs which included 4 boundaries and 6 sixes. What percent of his total score did he make by running between the wickets?
Total Runs = 100
Total boundaries = 4, Total sixes = 6
So, Total score made by running = 100-(4*4+6*6) = 100-52 = 48 runs
So, % Score = \(\frac{48}{100} \times 100\) = 48% Ans.

Formula Used:

Total score by running = (Total Score - (sixes + boundaries)
% score = \(\frac{Total Score By Running}{Total Runs}\times 100\)

NA
NA
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Example 28 / 28

A landowner increased the length and breadth of a rectangular plot by 20% and 30% respectively. Find the percentage change in the cost of the plot assuming land prices are uniform throughout his plot.
Let the length of the plot 'x' m and breadth be 'y' m respectively.
So, initial area= 'xy' sqm.

Now,
Increased length= x + 20% of x = \(\frac{120x}{100}\)
Increased breadth= y + 30%of y = \(\frac{130y}{100}\)
So, New Area = \(\frac{120x}{100}\)*\(\frac{130y}{100}\)

So, % change = \(\frac{ (New Area - Initial Area)}{Initial Area}\)*100

= \(\frac{\left(\frac{120x}{100} \times \frac{130y}{100}-xy \right)}{xy} \times 100\)
= 56% Ans.

Formula Used:

Increase in x by y% = x+y% of x.
= \(\frac{\left(New Area - Initial Area\right)}{Initial Area} \times 100\)

NA
Take out the new length and new breadth by the formula of % increase, i.e;
To increase X by Y%, we have formula,
\(X\left(1 + \frac{Y}{100}\right)\)
So, New Area = (New length * New breadth).
So, % change = \(\frac{\left(New Area - Initial Area\right)}{Initial Area} \times 100\)

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