Percentage: Core Ideas
Before solving complex CAT-level problems, you must have a complete understanding of the CAT percentage topic and its applications. A complete overview of the important CAT concepts is given below for the reference of the candidates. It is important for the candidates to study these topics in depth to ensure a good CAT score under the quantitative aptitude section.
Understanding "Percent"
"Percent" means “per hundred.” A 40% increase means 40 parts out of 100 added to the original.
Value-to-Percentage
To find what percentage one value is of another:
Percentage = (Part / Whole) × 100
Example: What percent is 60 of 150?
= (60 / 150) × 100 = 40%
Percentage-to-Value
To find the value represented by a percentage:
Value = (Percentage × Total) / 100
Example: 25% of 480 = (25 × 480)/100 = 120
These simple conversions are frequently used in longer CAT word problems.
Previous Year CAT Questions on Percentage
The topic percentage is one of the most important CAT quantitative aptitude topics. It is widely applied in questions under topics such as profit and loss, data interpretation, time, speed and distance, ratio and proportion and so on. It is beneficial for the candidates to look into the CAT previous year question papers and solve the questions on percentage, as this will provide them an idea of what they can expect on the CAT exam day. A few of the selected previous year CAT percentage questions are as follows.
Question 1:
The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is:
Options
1. 100
2. 400
3. 200
4. 300
Solution:
Percentage of indigo in bottles 1 and 2: 33\% and $17 \%$ respectively.
A part of the solution (say $\mathrm{y} c \mathrm{c}$ ) from the first bottle is thrown away and replaced by an equal volume ( y cc ) of the solution from the second bottle
Thus, in the first bottle, there is a mix of $33 \%$ indigo (say $x c c$ ) with $17 \%$ indigo ( $y c c$ ) $⇒$ the resultant solution has $21 \%$ indigo.
$\frac {(33 x+17 y)}{(x+y)}=21$
$⇒12 x=4 y $
$⇒ x: y=1: 3$
Since the total volume is 800 cc, we have: $y=\frac {3}{1+3} \times 800=600 \mathrm{cc}$ Thus, 600 cc of solution from bottle 2 was moved to bottle 1.
Thus, volume remaining in bottle $1=800-600=200 \mathrm{cc}$
The correct answer is 200 cc.
Hence, the correct answer is option (3)
Question 2:
If the price of a commodity is raised by 20% then by how much % does a householder reduce his consumption so that the expenditure does not change?
Options:
1. 16.67%
2. 18.66%
3. 10.66%
4. 1.66%
Solution:
Let the initial consumed quantity of the commodity be 100 and the initial price is 100.
So, total expenditure = 100 × 100 = 10000
New price = 100 + 20 = 120
Total expenditure will be the same.
So, new consumed quantity $=\frac{10000}{120}=83.33$
$\therefore$ Reduced percentage in consumption = 100 – 83.33 = 16.67%
Hence, the correct answer is 16.67%.
Question 3:
In an election between two candidates, the candidate who got 57% of valid votes won by a majority of 420 votes. Find the total valid votes.
Options:
1. 3000 votes
2. 2000 votes
3. 4000 votes
4. 1000 votes
Solution:
Let the total number of votes be x.
According to the question,
0.57x – 0.43x = 420
⇒ 0.14x = 420
⇒ x = $\frac {420}{0.14}$ = 3000
Hence, the correct answer is 3000 votes.
Question 4:
A man spends 35% of his income on food, 25% on children's education and 80% of the remaining on house rent. What percent of his income he is left with?
Options:
1. 6
2. 8
3. 10
4. 12
Solution:
Let original income $=100$
$35 \%$ on food $+25 \%$ on education $=60\%$
Remaining = $100-60=40$
$80 \%$ of $40$ on rent $=32%$
Income left $=100-(60+32)=8 \%$
Hence, the correct answer is 8%.
Question 5:
In an exam 52% of the candidates failed in science, 42% in maths and 17% in both. The number of those who passed in both subjects is:
Options:
1. 83%
2. 64%
3. 23%
4. 55.5%
Solution:
Total students $=100$
Failed in science $n(s)=52$
Failed in math $\mathrm{n}(\mathrm{m})=42$
Failed in both $n(s \& m)=17$
Failed in either math or science $\mathrm{n}(\mathrm{m}$ or $\mathrm{s})$
$\begin{aligned} & \mathrm{N}(\mathrm{m} \text { or } \mathrm{s})=\mathrm{n}(\mathrm{m})+\mathrm{n}(\mathrm{s})-\mathrm{n}(\mathrm{m} \& \mathrm{~s}) \\ & \mathrm{N}(\mathrm{m} \text { or } \mathrm{s})=52+42-17=77\end{aligned}$
Passed in both the subjects $=100-77=23\%$
Hence, the correct answer is 23%.
Question 6:
If the present population of a town is 10000 and the annual increase is 20%. What will be the population after 3 years?
Options:
1. 17280
2. 12000
3. 12325
4. 15625
Solution:
We know, $\text{Total population}=\text{initial population}×(1+\frac{\text{Rate}}{100})^{\text{Time}}$
$\therefore \text{Total population}=10000(1+\frac{20}{100})^3=17280$
Hence, the correct answer is 17280.
Question 7:
Gaurav spends 30% of his monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. If his monthly salary is Rs. 18,400, how much money does he save every month?
Options:
1. 3864
2. 4903
3. 5849
4. 6789
Solution:
So, his savings = $18400×\frac{70}{100}×\frac{60}{100}×\frac{50}{100}=3864$
Hence, the correct answer is Rs. 3864.
Question 8:
If the cost of a calculator worth Rs. 250 is increased by Rs. 100, the rate of increase is:
Options:
1. 100%
2. 40%
3. 25%
4. None
Solution:
Rate of increase $=\frac{100}{250}×100=40\%$
Hence, the correct answer is 40%.
Question 9:
What percent of $\frac{7}{8}$ is:
Options:
1. 25.5%
2. 87.5%
3. 75%
4. 12.5%
Solution:
Required percentage $=\frac{7}{8}×100=87.5\%$
Hence, the correct answer is 87.5%.
Question 10:
A student multiplied a number by $\frac{3}{5}$ instead of $\frac{5}{3}$, What is the percentage error in the calculation?
Options:
1. 54%
2. 64%
3. 74%
4. 84%
Solution:
Let the original number be $\mathrm{x}$.
Correct number after multiplication $=\frac{5 \mathrm{x}}{3}$
Incorrect number after multiplication $=\frac{3 \mathrm{x}}{5}$
So, Error $=\frac{5 \mathrm{x}}{3}-\frac{3 \mathrm{x}}{5}=\frac{16 x}{15}$
$\therefore$ Error $\%=[\frac{(\frac{16 \mathrm{x} }{15})}{(\frac{5 \mathrm{x} }{3})}] ×100=64 \%$
Hence, the correct answer is 64%.
Question 11:
The salaries of John, Sara, and Romi were in the ratio of 4 : 7 : 11 in 2012, and in the ratio of 5 : 11 : 17 in 2015. If John’s salary increased by 30% during 2012- 2015, then the average percentage increase in the total salaries of Sara and Romi during this period is closest to
Options:
1. 62
2. 31
3. 19
4. 40
Solution:
$\text{John's new salary} = 130\% \text{ of } 4x = 5.2x \\
\text{Sara's new salary} = 5.2x \times \frac{11}{5} = 11.44x \\
\text{Romi's new salary} = 5.2x \times \frac{17}{5} = 17.68x \\
\text{Sara's and Romi's new salary} = 11.44x + 17.68x = 29.12x \\
\text{Sara's and Romi's old salary} = 7x + 11x = 18x \\
\text{Percentage increase in Sara's and Romi's salary} = \frac{29.12x - 18x}{18x} \times 100 = \frac{11.12x}{18x} \times 100 = 61.77\% \\
\text{Average percentage increase} = \frac{61.77\%}{2} = 30.88\%$
Question 12:
A marathon runner embarks on a marathon consisting of a pleasant as well as a hot and humid running condition. He fills up his water bottle at the beginning of the race. He drinks 12% of his water while covering 18% of the total race in hot and humid running conditions. He knows he has to cover another 24% of the total race in similar conditions. What should be the percentage decrease in his water consumption during pleasant conditions over the hot and humid conditions, so that he just completes the entire race without a refill of the water bottle?
Options:
1. 29.70
2. 38.20
3. 45.00
4. 46.30
Solution:
For 18% of races in hot and humid conditions, water consumption = 12%
So for the total $(18 \%+24 \%)=42 \%$ race in hot and humid conditions, water
consumption $=12 \times\left(\frac{42}{18}\right)=28 \%$
Race left in pleasant condition $=100 \%-42 \%=58 \%$
Water left $=100 \%-28 \%=72 \%$
Rate of water consumption in hot and humid conditions $=\frac{18}{12}=\frac{3}{2}$
Rate of water consumption in pleasant conditions $=\frac{58}{72}=\frac{29}{36}$
Hence, the percentage decrease in water consumption
$=\left\{\left(\frac{3}{2}-\frac{29}{36}\right)\right] \times\left(\frac{100 \times 2}{3}\right) \%=\frac{1250}{27} \%=46.30 \%$
Hence, the correct answer is 46.30.
Question 13:
Of the population over 18 years in Singapore, 36% of men and 45% of women are married. What percentage of the total population aged more than 18 years is men? (Assume that no man marries more than one woman and vice versa)?
Options
1. 44.44%
2. 55.55%
3. Cannot be determined
4. None of these
Solution:
\text{Let there be } 100x \text{ and } 100y \text{ men and women respectively (aged more than 18 years)}. \\
\text{Married Men} = \text{Married Women} \Rightarrow 36x = 45y \Rightarrow x = \frac{5y}{4} \\
\text{Total Men} = 100x = 100 \times \frac{5y}{4} = 125y \\
\text{Total Population (more than 18 years)} = 125y + 100y = 225y \\
\text{Men \%} = \frac{125y}{225y} \times 100 = 55.55\% \\
\text{Hence, the correct answer is } 55.55\%.
Question 14:
Fresh fish contain 59% water by weight, while sun-dried fish contains 5% water by weight. A fisherman caught fresh fish, added salt in the ratio of 4: 1, and prepared the sun-dried salted fish weighing 150 kg. How many kg of fish had the fisherman caught?
Options:
1. 432.70
2. 316.25
3. 237.50
4. 170.50
Solution:
\text{Let the amount of caught fresh fish be } x \text{ kg}. \\
\text{So the amount of salt added} = \frac{x}{5} \text{ kg}. \\
\text{Now since water in fresh fish is } 59\% \text{ of weight, actual flesh in fresh fish is } 41\% \text{ of weight} \\
\Rightarrow \text{Actual flesh in fresh fish} = \frac{41x}{100} \text{ kg} \\
\text{In sun-dried salted fish, the amount of salt present is also } \frac{x}{5} \text{ kg} \\
\text{Now since water in sun-dried fish is } 5\% \text{ of weight, actual flesh in sun-dried fish is } 95\% \text{ of weight} \\
\Rightarrow \text{Actual flesh in sun-dried fish} = \left(150 - \frac{x}{5} \right) \times \frac{95}{100} \text{ kg} \\
= \left(142.5 - \frac{19x}{100} \right) \text{ kg} \\
\text{But the actual flesh must be the same in both cases, so:} \\
142.5 - \frac{19x}{100} = \frac{41x}{100} \Rightarrow x = 237.50 \text{ kg}
Question 15:
Agro giant Keventers, packages and sells frozen peas in two different packages. The smaller package has an MRP (maximum retail price) of 33.33% of the MRP of the larger one. However, it was found that the MRP per unit of peas in the larger packet is 5% less than the same unit of peas in the smaller one. What percent of the larger package is the weight of the smaller package of peas?
Options:
1. 37.30
2. 33.33
3. 31.66
4. 35.08
Solution:
Let the MRP of the larger pack be x.
So, MRP of smaller pack = 0.33x
Let the weight of the larger pack be l and the smaller pack be s.
According to the question,
$\frac xl=\frac{95}{100}×\frac{0.33x}{s}$
$⇒s=\frac{0.3333×95}{100}×l$
$\therefore s = 0.3166l$
$\therefore$ Required percentage = $0.3166\times 100 = 31.66\%$
Hence, the correct answer is option (3).
Question 16:
\text{X is } P\% \text{ less than Y and Z is } P\% \text{ more than Y. If X is } 2.5P\% \text{ more than Z, then find } P. \ (\text{Note: } P \ne 0)
Options:
1. 20
2. 40
3. 60
4. 50
Solution:
$X =(\frac{100-p}{100})Y$
$Z =(\frac{100+p}{100})Y$
$X =(\frac{100+2.5p}{100})Z$
From the 1st and 2nd relations, we get
$X=\frac{Z(100-p)}{(100+p)}$
⇒ $\frac{Z(100-p)}{(100+p)}=(\frac{100+2.5p}{100})Z$
⇒ $(100+p)(100+2.5p)=100(100-p)$
⇒ $2.5p^2=50p$
⇒ $p=20$
Hence, the correct answer is 20.
Question 17:
You have a container with a 25% saltwater solution and another container with a 10% saltwater solution. You want to create a 15-litre mixture that contains 15% salt. How many litres of each solution should we mix to achieve the desired concentration?
Options:
1. 15
2. 20
3. 10
4. 5
Solution:
Let x be the number of litres of the 25% solution, and (15 - x) be the number of litres of the 10% solution.
The equation is as follows:
(0.25x) + (0.10(15 – x)) = 0.15 × 15
⇒ 0.25x + 1.5 – 0.10x = 2.25
⇒ 0.15x + 1.5 = 2.25
⇒ 0.15x = 2.25 – 1.5
⇒ 0.15x = 0.75
$\therefore$ x = 5
So, you should mix 5 litres of the 25% salt water solution with (15 - 5) = 10 litres of the 10% salt water solution to achieve the desired 15% salt concentration in a 15-litre mixture.
Hence, the correct answer is 5.
Question 18:
\text{Manoj scored } 30\% \text{ in an examination and failed by } m \text{ marks. He got his marks reviewed, and even though his marks increased by } 50\%, \text{ he failed by } n \text{ marks. In the same exam, Sunil had also appeared. Sunil got } 20\% \text{ more marks than post-review marks of Manoj. He got just passing marks. } \left( \frac{m - n}{x} \right) \text{ is what percent of the maximum marks?}
Options:
1. 9
2. 5
3. 15
4.20
Solution:
\text{Let the maximum marks in the exam be } 100x. \\
\textbf{1st Condition:} \\
\text{Passing marks } = 30x + m \\
\textbf{2nd Condition:} \\
\text{His marks increased by } 50\%, \text{ i.e., } 30x + 15x = 45x \\
\text{Passing marks } = 45x + n \\
\Rightarrow 30x + m = 45x + n \Rightarrow 15x = m - n \\
\textbf{3rd Condition:} \\
\text{Sunil gets } 20\% \text{ more marks than the post-review marks of Manoj, i.e., } 45x + \frac{20}{100} \cdot 45x = 45x + 9x = 54x \\
\text{So, } 30x + m = 54x \Rightarrow m = 24x \\
\text{And, } 45x + n = 54x \Rightarrow n = 9x \\
\Rightarrow m - n = 15x \Rightarrow \left( \frac{m - n}{x} \right) = 15\% \text{ of the maximum marks.}
Hence, the correct answer is 15.
Question 19:
After receiving two successive raises, Harish’s salary became equal to $\frac{21}{7}$ times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as high (in percent) as the first?
Options:
1. 15%
2. 20%
3. 25%
4. 50%
Solution:
\text{Let the first raise in salary be } x\%. \\
\text{Then, the second raise is } 2x\%. \\
\text{Net change } = [x + 2x + \frac{x \cdot 2x}{100}]\% \\
\text{Also, let the initial salary be } 100. \\
\text{After change, salary becomes } 100 \times \frac{21}{7} = 300 \\
\text{So, net percentage change } = 300 - 100 = 200\% \\
\text{Therefore,} \\
x + 2x + \frac{x(2x)}{100} = 200 \Rightarrow 3x + \frac{2x^2}{100} = 200 \\
\Rightarrow 300x + 2x^2 = 20000 \Rightarrow x^2 + 150x - 10000 = 0 \\
\text{On solving, } x = 50\% \\
\text{Hence, the correct answer is } \boxed{50\%}.
Question 20:
\text{After receiving two successive raises, Harish's salary became equal to } \left( \frac{21}{10} \right) \text{ times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as high (in percent) as the first?}
Options:
1. -5/8
2. -5/4
3. 5/8
4. 5/5
Solution:
\text{Let the first rise in salary be } x\%. \\
\text{Then, the second rise is } 2x\%. \\
\text{Net change } = \left[x + 2x + \frac{x \cdot 2x}{100}\right]\% \\
\text{Also, let the initial salary be } 100. \\
\text{After change, salary becomes } 100 + \left( \frac{21}{10} \cdot 100 - 100 \right) = 210 - 100 = 110 \text{ increase} \\
\text{So, net percentage change } = 110\% \\
\text{Therefore:} \\
x + 2x + \frac{2x^2}{100} = 110 \Rightarrow 3x + \frac{2x^2}{100} = 110 \Rightarrow 300x + 2x^2 = 11000 \Rightarrow x^2 + 150x - 10000 = 0
Question 21:
The number of girls appearing for CAT are half of that of boys. If 
of the girls and 
of the boys cleared the CAT cut off. If only 
of students who cleared the cutoff got admission in IIMs, candidates who cleared the cut off and got admission in IIMs is what percent of who did not clear the cutoff?Options:
1. 9.59
2. 10.71
3. 2.25
4. 12.17
Solution:
| Girls | Boys | Total |
Appeared | 50 | 100(Let) | 150 |
Cleared the cut-off | 10 | 25 | 35 |
Not cleared the cut-off | 40 | 75 | 115 |
$40 \%$ of cleared candidates got admitted in IIMs, i.e. 14 Not cleared $=115$
Required answer $=\left(\frac{14}{115}\right) \times 100=12.17 \%$
Hence, the correct answer is 12.17.
Question 22:
Three gift hampers contain four items in each as follows:
| Gift Hamper A | Gift Hamper B | Gift Hamper C |
Fairness Cream | 5 | 8 | 8 |
Body Lotion | 5 | 4 | 4 |
Eyeliner | 8 | 8 | 5 |
Lipstick | 3 | 2 | 4 |
The price of gift hampers A, B, and C are equal. Also, the cost of 1 lipstick is 50% more than 1 eyeliner. If another gift hamper consists of 15 lipsticks only, it costs Rs 230 more than any of the gift hamper above. Find the cost of 1 lipstick. (in Rs.)
Options:
1. 240
2. 90
3. 110
4. 160
Solution:
Let F = cost of 1 fairness cream, B = cost of 1 body lotion, E = cost of 1 eyeliner, and L = cost of 1 lipstick.
According to the question,
5F + 5B + 8E + 3L = 8F + 4B + 8E + 2L = 8F + 4B + 5E + 4L ------------------ (1)
Also, L = (B + 50% of B) = $\frac{3B}{2}$ --------------------------(2)
Solivng, 5F + 5B + 8E + 3L = 8F + 4B + 8E + 2L
⇒ B + L = 3F
⇒ $\frac{2L}{3}$ + L = 3F
⇒ F = $\frac{5L}{9}$ -------------------(3)
Solivng, 8F + 4B + 8E + 2L = 8F + 4B + 5E + 4L
⇒ 3E = 2L
⇒ E = $\frac{2L}{3}$ -------------------(4)
One more condition is given,
15L = (5F + 5B + 8E + 3L) + 230 ---------------------(5)
On solving the above equations,we get L = 90
Hence, the correct answer is option (2).
Question 24:
A dosage of 24 cubic centimetres of a certain drug is prescribed to a patient whose body weight is 80 pounds per day. If the typical dosage is 4 cubic centimetres per 20 pounds of body weight per day, but he had taken 6 cubic centimetres already in the morning, what percent of the prescribed dosage did he have to take according to the typical dosage?
Options:
1. 33.33
2. 66.66
3. 41.66
4. 58.33
Solution:
The typical dosage for a patient of 80 pounds 
The remaining amount of dosage 
which is 
Hence, the correct answer is 
.
Question 25.
In the entrance exam of JEE, $\mathrm{m}$ students appeared. 
of the students are girls and the rest are boys. There are 7200 more girls than boys. If 
of the students, including 1600 boys, cleared the entrance exam, the percentage of the girls who failed to clear the entrance is:Options:
1. 6
2. 84
3. 94
4. 16
Solution:
\text{According to the question,} \\
56\% \text{ of } m - 44\% \text{ of } m = 7200 \Rightarrow 12\% \text{ of } m = 7200 \Rightarrow m = 60000 \\
\therefore \text{Number of girls} = 56\% \text{ of } 60000 = 33600 \\
\text{Number of students who cleared the exam} = 6\% \text{ of } 60000 = 3600 \\
\text{Number of girls who cleared the exam} = 3600 - 1600 = 2000 \\
\text{Percentage of girls who did not clear the exam} = \left( \frac{33600 - 2000}{33600} \right) \times 100 = 94.04\% \approx 94\% \\
\text{Hence, the correct answer is } \boxed{94}.
Hence, the correct answer is 94.
Must-Know Percentage Applications for CAT 2025
In the CAT examination, the questions related to the concept of percentages will be closely related to the direct application of the equations. Hence, it is important for the candidates to know the important applications of the concept of percentages.
Successive Percentage Changes
In real-world contexts, values often increase or decrease multiple times. CAT tests your ability to calculate compound effects.
Formula:
Net Change = A + B + (A × B)/100
For example, a price first increases by 20%, and then drops by 10%. The net change is:
20 – 10 + (20 × –10)/100 = 10 – 2 = 8% increase
Reverse Percentage Logic
CAT questions often give the final value and ask you to reverse-calculate the original using percentage.
If the salary becomes ₹13,200 after a 10% increase, what was it before?
Let the original be x. Then x + 10% of x = 13200 → 1.1x = 13200 → x = 12000
This is a standard CAT-style setup.
Comparison Questions Using Percentages
These involve relative percentage differences and frequently trip up students.
If A is 25% more than B, then B is (25/125) × 100 = 20% less than A.
Not 25%. That’s a classic CAT trap.
CAT rewards students who understand that percentage increases and decreases are not symmetric.
