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Every year, the CAT asks about 4-5% of the questions in Quantitative Aptitude from the percentages chapter. Too many questions from this chapter are asked in other entrance/competitive tests like the MAT, XAT, and UPMCAT, among others. There are a lot of questions about how to use percentages in all chapters of commercial arithmetic, which shows how important percentages are (especially Profit and Loss, Ratio and Proportion, Time and Work, Time, Speed and Distance).
- Percentage Questions for CAT are asked under the Arithmetic section.
- Generally, the Percentage Questions for CAT in the exam are moderately difficult.
Percentage Questions for CAT 2023 are of different types such as Percentage increase /decrease, change of base, or applications based on percentage.
Concepts
The basic percentage is the hygiene factor for all types of calculations. One should know the fractional percentage for ease of calculation in Quants and DI.
Learnings from the Conversion Table
All the values explained are in decimal. Multiply 100 in the values to get the percentage.
- The denominator of all the percentages whose whole numbers are 50, 100, 150, 200, etc., is 2.
- The denominator of all the percentages whose whole number values are 33, 66, or 100 is 3.
- If the percentage value is a multiple of 20, then the denominator of the fraction is 5.
- If the whole number is 16, 33, 50, 66, 83, or 100, then there is a 6 in the fraction's denominator.
- Similarly, if the numbers are 14, 28, 42, 57, 71, or 85, then the denominator is 7.
- If a number is a multiple of 11, like 11, 22, 33, 44, 99, etc., then the fraction's denominator is 9.
- If there are multiples of 9 like 9, 18, or 27, the denominator of the fraction is 11.
Percentage Change
There are many ways to use percentage change, such as for profit and loss, simple interest, compound interest, etc. All of these are exactly based on the original (or actual) value's percentage increase or decrease.
Percentage Increase in a value = \(\frac{Increased Value - Original Value }{Original Value}\) x 100 %
Percentage Decrease in a value = \(\frac{Original Value - Decreased Value }{Original Value}\) x 100 %
Percentage Point Change
The CAT Percentage Questions given below provide an overview of the type of questions asked on this topic. A solved example for a percentage point change problem for CAT is as follows –
| Example: The previous year, Deepak's salary was $10,000 and Sanak's was $8,000. Deepak's salary is $12,000 this year, while Sanak's is $10,000.
Solution:
Percentage increase in Deepak’s salary = 20 Required percentage = (25 – 20)/20 x 100= 25%
|
Also Check:
Special Cases of Percentage Change
CAT Percentage Questions are asked in different variations. Some of the special cases are –
Getting back the Original value
- If a value p is increased by x%, then we have to decrease the resultant value by (x/(x+100))x100% to get back to the original value p.
| Example: Age of Ravi is 20 years. If Shyam’s age is 25% greater than that of Ravi then how much percent Ravi’s age is less than Shyam’s age? Ans: As Shyam’s age is 25% more than Ravi. Ravi’s age will be (25/(25+100))x100 = 20% that of Shyam’s age |
- If a value p is decreased by x% then to get back to the original value p, we have to increase the decreased value by (x/(100-x))x100%.
| Example: Due to Ramesh's irregular work practises, his salary was reduced by 20%; however, after a few months, it was restored to its original level. What is Ramesh's compensation increase by percentage? Ans: Let the initial salary be 100. Percentage increase in salary of Ramesh = (20/(100 – 20))x100 = 25%. |
NOTE:
- If a value X is increased by p% to Y and then dropped by q% to X, p is always greater than q. (for positive values).
- If a value X is reduced by p% to Y and subsequently increased by q% to X, p is always smaller than q.
- If a value A is increased by p%, then by q%, and then by r%, the final value will be the same irrespective of the order of p, q and r.
- Rule 3 also applies to numbers that are declining. When a value 'A' is first reduced by p%, then by q%, then by r%, and so on, the resultant value is the same.
- When a value 'A' is first raised by p%, then by q%, and then by r%, the effects are the same even after changing the order of increment.
Reversing the Change by the Same Percentage
If a number's value is reduced by x percent and subsequently that number's value is increased by x percent, the net outcome is always a loss of the original number's value. To put it another way, that's a loss percentage equal to (x/10)2%.
Difference Between ‘by’ and ‘to’
Please keep in mind that "by" and "to" are not the same. "By" stands for the difference and "to" stands for the final value.
For example, if the income is cut by 40%, it means that the new income is 60% of what it was before. If the income is cut by 40%, it means that the new income is 40% of what it was before.
For example, if Deepika's income goes up by 20%, it means that its new income is 120% of what it was before. Deepika's income has gone up by 120%, which means that her new income is 120% of what it was before.
CAT Percentage Practice Questions
Percentage Questions for CAT are given below to help aspiring candidates in their preparation.
Ques 1: There are 50% more people in Class B than in Class A. There are as many girls in Class A as there are boys in Class B. The number of girls in each class is the same. How many of the students in the group are boys?
- 33%
- 40%
- 25%
- 60%
Click Here for the Answer
Ans: B. 50% more than x is 1.5x. Simple, but very useful idea that might help you in solving these kinds of problems.
Let number of girls in class A = x
Let number of boys in class A = y
Total number of students = x + y
Proportion of girls = x/(x+y)
Number of boys in class B = x
Total number of students in class B = 1.5(x + y)
Proportion of girls = 1 /(1.5(x+y))
Percentage of boys in the overall student community = (x+y)/2.5* (x + y) * 100 = 40%
Ques 2: Traders A and B each pay Rs. 1000 and Rs.2000 for two goods. Trader A puts a markup of x% on his goods, while trader B puts a markup of 2x% on his goods and gives a discount of x%. If both make the same amount of money that isn't zero, find x.
- 25%
- 5%
- 5%
- 40%
Click Here for the Answer
Ans: A. SP of trader A = 1000 (1 + x).
Profit of trader A = 1000 (1 + x) – 1000.
MP of trader B = 2000 (1 + 2x).
SP of trader B = 2000 (1 + 2x) (1 – x).
Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000
1000x = 2000 – 4000x2 + 4000x – 2000x – 2000
4000x2 -1000x = 0
1000x (4x – 1) = 0
=> x = 25%
Ques 3: The number ab is 60% of the number x. When you flip the digits of ab, you get a two-digit number that is 60% bigger than x. Find x.
- 45
- 54
- 63
- 72
Click Here for the Answer
Ans: D. 10a + b = 0.6x
10b + a = 1.6x
Subtracting one from the other, we have
9b - 9a = x or x = 9 (b - a)
x should be a multiple of 9.
10a + b = 3x/5 this implies that x should also be a multiple of 5.Or, x should be a multiple of 45
x should be equal to 45, ab = 27, ba = 72
Ques 4: A business owner buys 80 things that cost Rs. 40 each. He sells n of them at a n% profit and the rest at a (100-n)% profit. How much money could the merchant have made at the very least from this trade?
- 2160
- 1420
- 1580
- 2210
Click Here for the Answer
Ans: C. CP = 80 × 40
Profit from the n objects = n% × 40 × n.
Profit from the remaining objects = (100 – n)% × 40 × (80 – n).
We need to find the minimum possible value of n% × 40 × n + (100 – n)% × 40 × (80 – n).
Or, we need to find the minimum possible value of n2 + (100 – n) (80 – n).
Minimum of n2+ n2 – 180n + 8000
Minimum of n2 – 90n + 4000
Minimum of n2 – 90n + 2025 – 2025 + 4000
We add and subtract 2025 to this expression in order to crate an expression that can be expressed as a perfect square.
This approach is termed as the "Completion of Squares" approach. We keep revisiting this in multiple chapters.
Minimum of n2 – 90n + 2025 + 1975 = (n – 45)2 + 1975
This reaches minimum when n = 45.
When n = 45, the minimum profit made
45% × 40 × 45 + 55% × 40 × 35
18 × 45 + 22 × 35 = 810 + 770 = 1580
Ques 5: B makes 25% less money than A. C is 25% better off than A. D makes 20% less money than A. A makes 20% less than E does. A, B, C, D, and E all earn whole numbers that are less than Rs. What is the total amount that all five of them made?
- 300
- 245
- 305
- 480
Click Here for the Answer
Ans: C. Compare the income of A, B, C and D.
A=(5/4)B
C=(5/4)A
D=(5/6)A
E=(6/5)A
Finding the LCM of 4,5 & 6,which is 60.
From the above relationships,
Let A= Rs.60 consequently B= Rs.48, C= Rs.75, D= Rs.50 and E= Rs.72.
Adding the income of A,B,C,D and E equals=
Rs. 60 + Rs. 48 + Rs. 75 + Rs. 50 + Rs. 72 => Rs. 305
Ques 6: A, B, C, and D all get something. A gets a percent of the whole. B gets b% of what's left (after A has taken his share). C gets c% of what's left, and D gets everything else. D gets a percent less than A, and B and C get the same amount. b = 2a.
- What share did C get of what A got?
- What is the difference between what A got and what D got if the total amount is Rs.1000?
- 160% , A got Rs.40 more than D
- 80% , A got Rs.20 more than D
- 175% , A got Rs.50 more than D
- 150% , A got Rs.35 more than D
Click Here for the Answer
Ans: Let total amount = T
A would have got T * a. After A takes his share, there would be T (1 - a) left
B would have got T (1 - a) * b
C would have got T (1 - a)(1 - b) * c
D would have got T (1 - a) (1 - b) (1 - c)
D gets a % less than what A gets, or, T *a (1 - a) = T(1 - a) (1 - b) (1 - c), or
a = (1 - b)(1 - c) ------(1)
B and C get the same amounts, or (1 - a) * b = (1 - a) * (1 - b) * c, or b = (1 - b) c, or, b = c - cb, or
b = c/c+ 1 ------(2)
b = 2a -------(3)
Substituting, Equations (2) in and (3) in (1), we get
c/(2(c+1)) = (1 − c)/(c + 1)
, or c = 2 - 2c, or c = 2/3
b = c/(c + 1)= 2/5
, a = 1/5
A gets 20% of the loot, B gets 40% of the remaining 80%, or 32% of the loot. C gets 66.6% of 48%, or 32% of the overall loot. And D gets the final 16% of the overall amount (20% lesser than A's share).
Ques 7: Fresh grapes contain 90% water while dry grapes contain 20% water. What is the weight of dry grapes ...
- 2 kg
- 2.5 kg
- 2.4 kg
- 10 kg…
Click Here for the Answer
Ans: B. 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= 10/0.8 = 12.5gm
So, the weight of fresh grapes reduces to 1/8th of its original weight.
- 20kg of fresh grapes give 2.5kg of dry grapes.
Ques 8: Eight new bacteria are made when one bacterium divides. But because of the environment, only 50% survive and the other 50% die. If the number in the seventh generation is 4,096 million, how many are in the first generation?
- 1 million
- 2 million
- 4 million
- 8 million.
Click Here for the Answer
Ans: A. let’s say x is the initial number of bacteria count :
So in 2nd generation bacteria = 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.
Ques 9: P is x% more than Q. Q is (x - 10)% less than R. If P > R, what is the range of values x can take?
- 10% to 28%
- 10% to 25%
- 10% to 37%
- 10% to 43%
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Ans: C. P = Q ( 1 + x/100)
Q = R ( 1 – (x−10)/100)
R = Q / ( 1 – (x−10)/100)
P > R
Q ( 1 + x/100) > Q / ( 1 – (x−10)/100)
1 + x100 > 1 / (100−(x+10)/100)
(100+x)/100 > 100/(110−x)
(100 + x) (110 – x) > 100 * 100
11,000 + 110x – 100x – x2 > 10000
1000 + 10x – x2 > 0
X2 – 10x – 1000 < 0
X2 – 10x + 25 < 1000 + 25
(x – 5)2 < 1025
x – 5 < 32
x < 37
Hence, the answer is 10% to 37%.
Ques 10: Two people are putting trees in a field. After a while, a third worker is added, and only half as many trees are planted. How many trees can the second worker plant as a percentage of how many trees the first worker planted if the second worker is 1/3 as efficient as the first and third workers together?
- 65%
- 60%
- 70%
- 75%
Click Here for the Answer
Ans: B. Half as large means an increase of 50%.
=> 3rd worker=( 1stWorker+2ndworker)/2-----------------------Equation (1)
Also given, number of trees planted by 1st + 3rd = 3 * 2nd --------Equation (2)
This problem can be solved easily using options:
a) If trees planted by 2nd worker = 65 1st worker = 100 3rd worker = 82.5 (from Equation(1))
Putting in Equation (2) => 182.5 = 3 * 65 = 185 (Not satisfied)
b) If trees planted by 2nd worker = 60
1st worker = 100
3rd worker = 80 (from Equation(1))
Putting in Equation(2) => 180 = 3 * 60 = 180 (Satisfied)
The second worker can plant 60% of the number of trees planted by first worker
Hence, the answer is 60%
Ques 11: A solid steel cube is melted and recast so that its length, width, and height all change by 10%, 10%, and -20%, respectively. How much does this cube's surface area change?
- No Change
- +1%
- +2%
- -1%
Click Here for the Answer
Ans: D. Let the surface area of each side of initial cube be 100.
The surface area of the resulting cuboid would be given by 2(lb + bh + lh)
So, ∆lb = 100 + 10% ---> 110 + 10% ---> 121
∆bh = 100 + 10% ---> 110 - 20% ---> 88
∆lh = 100 + 10% ---> 110 - 20% ---> 88
Therefore, new surface area = 2 (121 + 88 + 88) = 594
% change in surface area = a * (−6/600) ∗ 100 = −1 (Initial surface area = 6 sides x 100 = 600)
The percentage change in surface area of this cube is -1%.
Hence, the answer is -1%.
Ques 12: A person who lends money to the impoverished takes advantage of their plight by charging them an interest rate of fifty percent. But he is never discovered because he pays 20% of his total capital in bribes to the authorities (his initial investment plus his profit). If he has Rs 25,000 to invest at the beginning of the fourth year, how much bribe did he pay at the conclusion of the second year?
- Rs 4,900
- Rs 5,200
- Rs 5,000
- Rs 6,000
Click Here for the Answer
Ans: B. Let the moneylender has Rs 100 initially.
1st year ---> 100 + 50 (interest) ---> 150 – 20% bribe ---> 120
2nd year ---> 120 + 60 (interest) ---> 180 – 20% bribe ---> 144
3rd year ---> 144 + 72 (interest) ---> 216 – 20% bribe ---> 172.8 (This is the capital he will have at the beginning of 4th year)
So, if 172.8 --> 25,000
=> 36 (bribe given in 2nd year) = (25000/172.8) ∗ 36 which on simplifying becomes 25000/4.8. Thus greater than 5000 but less than 6000.
He gave Rs.5200 as bribe at the end of the second year.
Hence, the answer is Rs.5200.
Ques 13: The tensile strength of a substance A is a multiple of the quantities of a, b, c, and d used. What will be the overall change in A's tensile strength if the amounts of material a, b, c, and d are altered by 30%, -30%, -25%, and 25%, respectively?
- No Change
- +14.68%
- –14.68%
- Depends on the initial amount of a, b, c, d
Click Here for the Answer
Ans: C. Since, tensile strength of A = a * b * c * d, If we assume initial tensile strength to be 100, we can apply successive % changes to arrive at the final figure
100 -30% ---> 70 (Any of the change can be carried out first, the result would be same)
70 + 30% ---> 91
91 + 25% ---> 113.75
113.75 – 25% --> ~85.31
Therefore, percent change in tensile strength = -14.68%
The overall change in tensile strength of A will be -14.68%.
Hence, the answer is -14.68%.
Ques 14: Following is a question followed by three affirmative statements. Examine the statements to determine which one(s) are necessary to answer the question. What was the discount percentage offered?
- The table was sold for Rs 12650, resulting in a 26.5% profit.
- Without the discount, a 30% profit would have been realised.
- The cost of the table was Rs 10,000.
- Only ii and iii
- Only i and iii
- Any two of the above
- None of these
Click Here for the Answer
Ans: A. From statement – I
S.P. = Rs 12650, Profit = 26.5%
∴ C.P = 100126.5
* 12650 = 10000
From statement II
Mark Price = 130% of C.P. = Rs 13000
∴ From statement I and II
Discount = Rs (13000 - 12650) = Rs 350
Discount % = 35013000
* 100 and Therefore, can be calculated
∴ Statement I and II can give the answer. However, statement II and III together cannot.
Statement I and II can give the answer.
Hence, the answer is Only I and II.
Choice A is the correct answer.
Ques 15: A container contains 175 millilitres of water and 700 millilitres of alcohol. Gopal removes 10% of the mélange and replaces it with the same volume of water. The procedure is performed again. What is the current proportion of water in the mixture?
- 30.3
- 35.2
- 25.4
- 20.5...
Click Here for the Answer
Ans: B. Final quantity of Alcohol in the mixture= (700/(700+175) x (90/100)2x(700+175) = 567ml
So, the final volume of water in the mixture = 875-567=308ml
Hence, the water percentage in the mixture = (308/875) x 100= 35.2%
The given sample CAT Questions on Percentage can help the candidates to ace the quantitative aptitude section of the CAT Exam.
How to Prepare Percentage Questions for CAT?
Preparing for percentage questions is a crucial part of CAT exam preparation. Some tips for preparation of percentage for CAT exam are as follows:
- Understand the basic concepts of percentages, such as fractions, ratios, and percentages conversions.
- Solve a variety of percentage-based questions, including those based on profit and loss, interest, discounts, and taxes.
- Learn some shortcuts to save time during the exam. For example, to find a percentage of a number, you can multiply the number by the percentage and divide the result by 100.
- Carefully analyze each question to determine what percentage concept it requires.
- Set a time limit for each question, and move on if you are taking too long. Come back to difficult questions later, if time allows.
- Take as many mock tests as possible to improve your speed and accuracy.
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