CAT Questions On Percentages With Detailed Video Solutions PDF

In order to score above the 99 percentile in the CAT exam, the Quantitative Ability section plays a vital role as this is the scoring section of the CAT exam. If you can score above the 99 percentile in the CAT quant section, there will be more chances to get a 99 overall percentile in the CAT exam, even if you got a decent score in the remaining sections. There are some key concepts in this section that the aspirants must take a special focus on preparing. Among them, arithmetic is the most vital topic, and we can expect at least 30 to 40 percent of the weightage in the CAT quants section. Arithmetic is the integration of several concepts, such as percentages, ratios and proportions, profit and loss, Time speed distance, Time & Work etc. In this article, we will provide you with the 20 most important previous year CAT Questions on percentages with video solutions PDF.

Aspirants are advised to be well-versed in the basic concepts and formulas. One can download this CAT quant formula cheat sheet which consists of every formula from all the concepts from the Quantitative ability of CAT.

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Percentage Questions For CAT With Video Solutions

Question 1: Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be
[CAT 2003 leaked]

1. 85.5
2. 92.5
3. 90.5
4. 87.5

1) Answer (D)

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Solution:

Surface area of sphere A (of radius a) is $4\pi*a^2$
Surface area of sphere B (of radius b) is $4\pi*b^2$
=> $4\pi*a^2$/$4\pi*b^2$ = 1/4 => a:b = 1:2
Their volumes would be in the ratio 1:8
Therefore, k = 7/8 * 100% = 87.5%

Question 2: At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > 0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?
[CAT 2003 leaked]

1. p = q
2. p < q
3. p > q
4. p = q/2

2) Answer (C)

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Solution:

By the end of the year 2002, Shepard bought 4 times and sold 4 times. He is left with the initial number of goats that he had in 1998. If the percentage of goats bought is equal to or lesser than the percentage of goats sold, then there would be a net decrease in the total number of goats. For the number of goats to remain the same, p has to be greater than q, because p% is being calculated in a lesser number and q% is being calculated on a greater number. Hence, p > q.

Question 3: Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight and the remaining proportion being pulp. What is the weight of dry grapes available from 20 kg of fresh grapes?

1. 2 kg
2. 2.4 kg
3. 2.5 kg
4. None of these

3) Answer (C)

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Solution:

Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg.

In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

Suppose the weight of the dry grapes be D.

80% of the weight of dry grapes = weight of the pulp = 2 kg

(80/100) x D =2 kg.

D = 2.5 kg

Question 4: Flights A and B are scheduled from an airport within the next one hour. All the booked passengers of the two flights are waiting in the boarding hall after check-in. The hall has a seating capacity of 200, out of which 10% remained vacant. 40% of the waiting passengers are ladies. When boarding announcement came, passengers of flight A left the hall and boarded the flight. Seating capacity of each flight is two-third of the passengers who waited in the waiting hall for both the flights put together. Half the passengers who boarded flight A are women. After boarding for flight A, 60% of the waiting hall seats became empty. For every twenty of those who are still waiting in the hall for flight B, there is one air hostess in flight A. What is the ratio of empty seats in flight B to the number of air hostesses in flight A?

1. 10 : 1
2. 5 : 1
3. 20 : 1
4. 1 : 1

4) Answer (A)

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Solution:

Out of 200 of the seating capacity, 180 seats are filled out of which 108 are males and 72 are females. Remaining 20 seats are vacant. According to given condition seating capacity in both the planes is 120 . Considering flight A – we can find that 100 passenger in waiting hall will be taking fight A . So 80 people remain in in the waiting hall who will be taking flight B . Now for every 20 people taking flight B we have a air hostess in flight A . So in total there are 4 air hostess in flight A. Flight B having 120 as seating capacity, 40 remain vacant. So required ratio 40:4 = 10:1 .

Question 5: Fresh grapes contain 90% water while dry grapes contain 20% water. What is the weight of dry grapes obtained from 20 kg fresh grapes?

1. 2 kg
2. 2.5 kg
3. 2.4 kg
4. 10 kg

5) Answer (B)

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Solution:

Let the total weight of fresh grapes be 100 gm.

=> Fresh grapes have 90 gm of water and 10 gm of fruit.

When these grapes are dried, the amount of fruit does not change.

=> 10 grams will become 80% of the content in dry grapes

=> Weight of dry grapes = $\frac{10}{0.8}$ = 12.5 gm

So, the weight of fresh grapes reduces to 1/8th of its original weight.

=> 20 kg of fresh grapes give 2.5 kg of dry grapes.

Question 6: One bacterium splits into eight bacteria of the next generation. But due to environmental condition only 50% survives and remaining 50% dies. If the seventh generation number is 4,096 million, what is the number in first generation?

1. 1 million
2. 2 million
3. 4 million
4. 8 million

6) Answer (A)

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Solution:

let’s say x is the initial number of bacterias :
So in 2nd generation no. of bacterias = $\frac{8x}{2} = 4x$
In 3rd generation, it will be = 16x
4th gen. = 64x
5th gen. = 256x
6th gen. = 1024x
7th gen. = 4096x
Hence x = 1 million

Question 7: Arun’s present age in years is 40% of Barun’s. In another few years, Arun’s age will be half of Barun’s. By what percentage will Barun’s age increase during this period?

7) Answer: 20

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Solution:

Let Arun’s current age be A. Hence, Barun’s current age is 2.5A
Let Arun’s age be half of Barun’s age after X years.
Therefore, 2*(X+A) = 2.5A + X
Or, X = 0.5A
Hence, Barun’s age increased by 0.5A/2.5A = 20%

Question 8: The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is

1. 35
2. 50
3. 60
4. 65

8) Answer (D)

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Solution:

Let the number of girls be 2x and number of boys be x.

Girls getting admission = 0.6x

Boys getting admission = 0.45x

Number of students not getting admission = 3x – 0.6x -0.45x = 1.95x

Percentage = (1.95x/3x) * 100 = 65%

Question 9: Out of the shirts produced in a factory, 15% are defective, while 20% of the rest are sold in the domestic market. If the remaining 8840 shirts are left for export, then the number of shirts produced in the factory is

1. 13600
2. 13000
3. 13400
4. 14000

9) Answer (B)

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Solution:

Let the total number of shirts be x.
Hence number of non defective shirts = x – 15% of x = 0.85x
Number
of shirts left for export = No of non defective shirts – number of
shirts sold in domestic market
= No of non defective shirts – 20% of No of non defective shirts
= 80% of No of non defective shirts

Hence 8840 = 0.8 * (0.85x) .
Solving for x we get, x = 13000

Question 10: In a village, the production of food grains increased by 40% and the per capita production of food grains increased by 27% during a certain period. The percentage by which the population of the village increased during the same period is nearest to

1. 16
2. 13
3. 10
4. 7

10) Answer (C)

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Solution:

Let initial population and production be x,y and final population be z
Final production = 1.4y, final percapita = 1.27 times initial percapita
=> $\frac{1.4y}{z} $ = $ 1.27 \times \frac{y}{x}$
=> $\frac{z}{x} = \frac{1.4}{1.27} \approx 1.10$
Hence the percentage increase in population = 10%

Question 11: Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

1. 17 : 25
2. 18 : 25
3. 19 : 24
4. 21 : 25

11) Answer (C)

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Solution:

The selling price of the mixture is Rs.40/kg.
Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x.
It has been given that 1.1x = 40
x = 40/1.1
Price per kg of the mixture in ratio 3:2 = $\frac{3a+2b}{5} $
$\frac{3a+2b}{5} = \frac{40}{1.1}$
$3.3a+2.2b=200$ ——–(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
Price per kg of the mixture in ratio 2:3 = $\frac{2a+3b}{5}$
$\frac{2a+3b}{5} = \frac{40}{1.05}$
$2.1a+3.15b=200$ ——(2)

Equating (1) and (2), we get,
$3.3a+2.2b = 2.1a+3.15b$
$1.2a=0.95b$
$\frac{a}{b} = \frac{0.95}{1.2}$
$\frac{a}{b} = \frac{19}{24}$

Therefore, option C is the right answer.

Question 12: A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

1. 96
2. 98
3. 86
4. 84

12) Answer (A)

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Solution:

Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80
On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer

Question 13: In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

1. $N \leq 200$.
2. $243 \leq N \leq 252$.
3. $201 \leq N \leq 242$.
4. $N \geq 253$.

13) Answer (B)

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Solution:

Total marks = N
Pass marks = 45% of N = 0.45N
Marks obtained = 36
It is given that, obtained marks is 68% less than that pass marks
=>the obtained marks is 32% of the pass marks.
So, 0.32 * 0.45N = 36
On solving, we get N = 250
Hence, option B is the correct answer.

Question 14: A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

1. 30.3
2. 35.2
3. 25.4
4. 20.5

14) Answer (B)

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Solution:

Final quantity of alcohol in the mixture = $\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$ = 567 ml

Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $\dfrac{308}{875}\times 100$ = 35.2 %

Question 15: The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is

1. 15
2. 13
3. 12
4. 14

15) Answer (D)

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Solution:

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.

Percentage of salt in vessel A =$\ \frac{\ 70}{500}\times\ 100$

=14%

Question 16: In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was

16) Answer: 80

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Solution:

Let the score of D = 100d

The score of C = 20% less than that of D = 80d

The score of B = 25% more than C = 100d

The score of A = 10% less than B =90d

90d=72

100d= 72*100/90

= 80

Question 17: In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

17) Answer: 20

View Video Solution

Solution:

Assuming the number of students =100x

Hence, the number of girls = 60x and the number of boys = 40x

We have, 60x-40x=30  => x=1.5

The number of girls = 60*1.5=90

Number of girls that pass = 68x-30=68*1.5-30 = 102-30=72

The number of girls who do not pass = 90-72=18

Hence the percentage of girls who do not pass = 1800/90=20

Question 18: A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

1. 80
2. 70
3. 85
4. 75

18) Answer (A)

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Solution:

The weight/volume(g/L) for liquid 1 = 1000

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1)=80

Question 19: Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

1. 90%
2. 94%
3. 92%
4. 89%

19) Answer (C)

View Video Solution

Solution:

Initially let’s consider A and B as one component

The volume of the mixture is doubled by adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be ‘x’.

Using allegations , $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

Percentage of alcohol in A = 60% => Let’s percentage of alcohol in B = x%

The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations , $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$

=> x= 92%

Question 20: Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to

1. 144
2. 127
3. 135
4. 124

20) Answer (B)

View Video Solution

Solution:

Let the selling price of the Large and Small boxes of chocolates be Rs.200 and Rs.100 respectively. Let us consider that the Large box has $L$ grams of chocolate while the Small box has $S$ grams of chocolate.

The relation between the selling price per gram of chocolate can be represented as: $\frac{200}{L}=0.88\times\ \frac{100}{S}$

On solving we obtain the ratio of the amount of chocolate in each box as: $\frac{L}{S}=\frac{25}{11}$

The percentage by which the weight of chocolate in the large box exceeds that in the small box = $\left(\frac{25}{11}-1\right)\times\ 100\approx\ 127\%$

CAT percentage Questions – FAQs

How do you calculate percentage questions for CAT?

• There are a few important formulas to calculate percentage questions. You can find the easy tricks to calculate the percentages questions here – Percentage Formulas For CAT PDF.

Do percentage questions repeat in the CAT exam?

• Yes, the questions based on the percentages appear every year in the CAT exam. The exact questions will not repeat in the CAT, but similar questions will be asked.

How do you solve percentage-based questions?

• Percentage questions in the CAT exam are straightforward and can be solved within a few minutes. These questions are entirely formula-based. So, being well-versed in the formulas is very important to solve the percentage questions.

What is the trick to percentages?

• There are some best tricks to solve the percentage questions in CAT exams. One of the best tricks in percentages is remembering the Percent Fraction conversions and vice versa. For example, 1/5th is 20%, 1/6 is 16.66% etc. These tricks are very handy and extremely helpful in solving problems quickly. You can find all such important tricks on our formula PDFs provided below.

What is the easiest way to understand percentages?

• Here is a simple trick. If you want to calculate the percentage of a number, divide the number by the whole and multiply by 100.

What are all the formulas of percentages?

• If a quantity x is increased by y%, new quantity = (100+y/100)*x
• If a quantity x is decreased by y%, new quantity = (100-y/100)*x
• Percentage change = New quantity-old quantity/old * 100
• By what per cent A is greater than B, required percentage = A-B/B*100

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