◆ QA · Arithmetic

Percentages, formulas + CAT PYQs

An A-to-Z formula sheet, a graded confidence-building set, and real CAT previous-year questions, every answer independently solved and code-verified.

12formula blocks
18CAT PYQs
★★★priority
Formula Sheet CAT PYQs

Formula & Concept Sheet (A → Z)

Everything from first principles to the exam shortcuts and traps.

1The basics
  • x% of N = (x/100)·N → 35% of 80 = 28
  • What % is A of B? = (A/B)·100 → 60 of 240 = 25%
  • Find the whole: x% of N = A ⇒ N = A·100/x → 40% of N = 60 ⇒ 150
  • Symmetry: x% of y = y% of x → 16% of 25 = 25% of 16 = 4
2Conversions to memorize
Frac%Frac%
1/250%1/812.5%
1/333.33%3/837.5%
2/366.67%5/862.5%
1/425%7/887.5%
1/616.67%1/911.11%
1/714.28%1/119.09%
3Percentage change
  • % change = (New − Old)/Old × 100.
  • Multiplying factor: +r% → ×(1+r/100); −r% → ×(1−r/100).
  • Find original: Old = New / (1 ± r/100). e.g. 12100 at 10%/yr, 2 yrs ago = 10000.
  • % point ≠ % change: 30%→40% is +10 points but a 33.3% increase.
4Successive changes
  • a% then b%: net = a + b + ab/100 (with signs).
  • +10% then −10% ⇒ −1%;  +20% then +20% ⇒ +44%.
  • Many changes ⇒ multiply the MFs (1.15 × 1.15 × 0.90).
  • Equal up-then-down by x% ⇒ loss = x²/100 %.
5“x% more / less” & reverse
  • A is x% more than B ⇒ A = B(1+x/100).
  • A is x% more ⇒ B is x/(100+x) less. (25% more ⇒ 20% less)
  • A is x% less ⇒ B is x/(100−x) more. (20% less ⇒ 25% more)
6Product-constancy
  • Price × Consumption fixed. Price up r% ⇒ cut consumption r/(100+r).
  • Sugar +25% ⇒ consume 20% less.
  • Price down r% ⇒ consume r/(100−r) more.
7Compound growth / decline
  • Final = Initial × (1 ± r/100)ⁿ; different rates ⇒ multiply MFs.
  • Per-capita = Total/Population ⇒ pop MF = (1+growth)/(1+per-capita growth).
8Mixtures / concentration
  • Track the part that stays constant (pulp, pure metal, solute).
  • Grapes: 20 kg fresh (10% pulp) ⇒ 2 kg pulp ⇒ dry (80% pulp) = 2.5 kg.
9Profit / loss / discount
  • Profit% = (SP−CP)/CP×100; SP = CP(1+profit%/100).
  • Discount% = (MP−SP)/MP×100; SP = MP(1−discount%/100).
  • Markup 25% then discount 20% ⇒ 1.25×0.80 = 1.0 (no profit).
10Standard CAT templates
  • Pass/fail: (pass% − scored%) of max = shortfall marks.
  • Two groups: part + part = total; weighted-% = total-%.
  • Income−expenditure−savings: apply separate MFs, recompute savings.
  • Votes: work in % of valid votes; gap% of total = margin.
11Speed shortcuts
  • 10% = shift decimal; 5% = half of 10%; 1% = shift twice. 35% = 25%+10%.
  • Use fractions for ugly %s: 12.5% = 1/8, 16.67% = 1/6, 37.5% = 3/8.
12Common traps
  • “x% more” ≠ reverse “x% less”, use x/(100±x).
  • Percentage points vs percentage change.
  • “of the rest” ⇒ apply next % to the remaining base.
  • Successive changes are not additive, multiply MFs.

Revision Set, 20 questions

Easy → moderate, built to warm up and gain confidence. Aim 100% on Level 1, 80%+ on Level 2, then build speed on Level 3.

Level 1 · Warm-up basics

EasyB1
What is 35% of 80?
Show solution
28  (0.35 × 80)
EasyB2
60 is what percent of 240?
Show solution
25%  (60/240 × 100)
EasyB3
A number increased by 20% becomes 96. Find the original number.
Show solution
80  (n × 1.2 = 96)
EasyB4
A price rises from Rs. 250 to Rs. 300. What is the percentage increase?
Show solution
20%  ((300−250)/250 × 100)
EasyB5
A shirt priced at Rs. 800 is given a 15% discount. Find the new price.
Show solution
Rs. 680  (800 × 0.85)
EasyB6
A quantity is increased by 10% and then decreased by 10%. What is the net percentage change?
Show solution
1% decrease  (1.1 × 0.9 = 0.99; equal up-then-down always loses x²/100 %)
EasyB7
Express 5/8 as a percentage.
Show solution
62.5%  (5/8 = 0.625; recall 1/8 = 12.5%)
EasyB8
40% of a number is 60. Find the number.
Show solution
150  (60 / 0.40)

Level 2 · Easy CAT applications

EasyE1
Fresh grapes contain 90% water while dry grapes contain 20% water. What is the weight of dry grapes obtained from 20 kg of fresh grapes?
  • (A) 2 kg
  • (B) 2.5 kg
  • (C) 2.4 kg
  • (D) 10 kg
Show solution
(B) 2.5 kg. Pulp is constant. Fresh: 10% of 20 = 2 kg pulp. In dry grapes pulp is 80% ⇒ dry weight = 2/0.8 = 2.5 kg.
EasyCAT 2023 · Slot 2E2
In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employees is
  • (A) 4 : 5
  • (B) 6 : 5
  • (C) 5 : 6
  • (D) 5 : 4
Show solution
(A) 4 : 5. Take 100 employees, total salary S. Manufacturing: 20 share S/6 ⇒ avg S/120. Others: 80 share 5S/6 ⇒ avg S/96. Ratio = 96 : 120 = 4 : 5.
EasyCAT 2021 · Slot 3E3
In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be
  • (A) 86
  • (B) 84
  • (C) 80
  • (D) 78
Show solution
(B) 84. Won so far = 12. (12 + 0.6x)/(40 + x) = 0.5 ⇒ x = 80. Winning 90%: 12 + 0.9 × 80 = 84.
EasyE4
The price of sugar increases by 25%. By what percentage must a family reduce its consumption of sugar so that the expenditure on sugar remains the same?
  • (A) 25%
  • (B) 20%
  • (C) 16.67%
  • (D) 22.5%
Show solution
(B) 20%. Required cut = r/(100+r) = 25/125 = 20%.
EasyE5
A student scored 30% of the maximum marks in an examination and failed by 50 marks. If the pass mark is 40% of the maximum marks, find the maximum marks.
  • (A) 400
  • (B) 450
  • (C) 500
  • (D) 600
Show solution
(C) 500. Gap = (40 − 30)% = 10% of max = 50 ⇒ max = 500.
EasyE6
The population of a town increases by 10% every year. If the present population is 12100, what was the population two years ago?
  • (A) 10000
  • (B) 9800
  • (C) 11000
  • (D) 9900
Show solution
(A) 10000. 12100 / (1.1)² = 10000.

Level 3 · Moderate CAT

ModerateCAT 2022M1
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
  • (A) 30
  • (B) 27.5
  • (C) 25
  • (D) 20
Show solution
(C) 25%. (1+a)(1+2a) = 1.875 ⇒ 2a² + 3a − 0.875 = 0 ⇒ a = 0.25.
ModerateCAT 2025 · Slot 1M2
In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was
  • (A) 55
  • (B) 47
  • (C) 48
  • (D) 49
Show solution
(A) 55. 0.6g = 0.4b + 8 ⇒ 3g − 2b = 40; b, g multiples of 5, b > 10. Smallest: b = 25, g = 30 ⇒ total 55.
ModerateCAT 2023 · Slot 1M3
The salaries of three friends Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is
  • (A) 26%
  • (B) 30%
  • (C) 28%
  • (D) 25%
Show solution
(A) 26%. After yr1: 6, 7.5, 8.4. Yr2: Sita 8.4, Mita 10.5. Gita = mean ⇒ Gita = 9.45. Hike = (9.45 − 7.5)/7.5 = 26%.
ModerateCAT 2025 · Slot 1M4
Kamala divided her investment of Rs 100000 between stocks, bonds, and gold. Her investment in bonds was 25% of her investment in gold. With annual returns of 10%, 6%, 8% on stocks, bonds, and gold, respectively, she gained a total amount of Rs 8200 in one year. The amount, in rupees, that she gained from the bonds, was
Show solution
Rs. 900. B = 0.25G ⇒ G = 4B, S = 100000 − 5B. 0.10(100000−5B) + 0.06B + 0.08(4B) = 8200 ⇒ B = 15000 ⇒ gain = 6% × 15000 = 900.
ModerateCAT 2022 · Slot 2M5
In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was:
  • (A) 62800
  • (B) 50240
  • (C) 40192
  • (D) 60288
Show solution
(A) 62800. Other three split 70% as 1:2:3 ⇒ 11.67%, 23.33%, 35%. Winner 35%, runner-up 30% ⇒ gap 5% of cast = 2512 ⇒ cast = 50240 ⇒ registered = 50240/0.8 = 62800.
ModerateM6
A man spends 75% of his income. If his income increases by 20% and his expenditure increases by 10%, by what percentage do his savings increase?
  • (A) 40%
  • (B) 45%
  • (C) 50%
  • (D) 55%
Show solution
(C) 50%. Income 100, spend 75, save 25. New: income 120, spend 82.5, save 37.5 ⇒ increase = (37.5 − 25)/25 = 50%.

Practice questions generated · up to 100

Original easy-hard warm-up drills (not CAT PYQs). Pick the levels, generate a set, reveal answers.

More CAT Questions, 18 actual PYQs

Straight from CAT papers 2017-2025. Full question text; every answer solved & code-verified. marks the hard ones.

ModerateCAT 2017 · Slot 11
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
  • (A) 35
  • (B) 50
  • (C) 60
  • (D) 65
Show solution
(D) 65. Take boys 100, girls 200. Admitted = 0.30·200 + 0.45·100 = 105. Not admitted = 195/300 = 65%.
ModerateCAT 2017 · Slot 22
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
  • (A) 13600
  • (B) 13000
  • (C) 13400
  • (D) 14000
Show solution
(B) 13000. Export = 0.85 · 0.80 · T = 0.68T = 8840 ⇒ T = 13000.
ModerateCAT 2017 · Slot 23
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
  • (A) 16
  • (B) 13
  • (C) 10
  • (D) 7
Show solution
(C) 10. Per-capita = production/population ⇒ 1.40/(1+p) = 1.27 ⇒ p ≈ 10.2% ≈ 10%.
ModerateCAT 2018 · Slot 14
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?
  • (A) N ≤ 200.
  • (B) 243 ≤ N ≤ 252.
  • (C) 201 ≤ N ≤ 242.
  • (D) N ≥ 253
Show solution
(B) 243 ≤ N ≤ 252. 36 is 68% short of pass ⇒ 36 = 0.32·pass ⇒ pass = 112.5 = 0.45N ⇒ N = 250.
ModerateCAT 2019 · Slot 15
Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is
  • (A) 60
  • (B) 80
  • (C) 70
  • (D) 75
Show solution
(C) 70. Post-review = 60% of M. 0.72M = (0.6M + 35) + 7 ⇒ 0.12M = 42 ⇒ M = 350. Pass = 0.6·350 + 35 = 245 ⇒ 70%.
ModerateCAT 2019 · Slot 1 · TITA6
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
Show solution
20. 20% gap = 30 ⇒ total 150 (girls 90, boys 60). Passers = 102, of which 30 boys ⇒ 72 girls pass ⇒ 18 don't ⇒ 18/90 = 20%.
ModerateCAT 2019 · Slot 17
The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala's income goes down by 4% and Bimala's goes up by 10%, then the percentage by which Kamala's income would exceed Bimala's is nearest to
  • (A) 31
  • (B) 29
  • (C) 28
  • (D) 32
Show solution
(A) 31. Amala 120 ⇒ Bimala 100, Kamala 150. New: Kamala 144, Bimala 110 ⇒ (144 − 110)/110 ≈ 30.9% ≈ 31%.
ModerateCAT 2019 · Slot 28
In 2010, a library contained a total of 11500 books in two categories - fiction and nonfiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the nonfiction category. How many fiction books were in the library in 2015?
  • (A) 6160
  • (B) 6600
  • (C) 6000
  • (D) 5500
Show solution
(B) 6600. F + NF = 11500; 1.1F + 1.12NF = 12760 ⇒ F = 6000 ⇒ 2015 fiction = 1.1·6000 = 6600.
ModerateCAT 2019 · Slot 2 · TITA9
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
Show solution
80. B = 72/0.9 = 80; C = 80/1.25 = 64; D = 64/0.8 = 80.
ModerateCAT 2020 · Slot 110
In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to :
  • (A) 59
  • (B) 55
  • (C) 62
  • (D) 66
Show solution
(D) 66. Literate 65, young-literate 16.25 ⇒ illiterate 35, young-illiterate = 28 − 16.25 = 11.75 ⇒ old-illiterate 23.25 ⇒ 23.25/35 ≈ 66%.
ModerateCAT 2020 · Slot 211
In May, John bought the same amount of rice and the same amount of wheat as he had bought in April, but spent ₹150 more due to price increase of rice and wheat by 20% and 12%, respectively. If John had spent ₹450 on rice in April, then how much did he spend on wheat in May?
  • (A) ₹560
  • (B) ₹570
  • (C) ₹590
  • (D) ₹580
Show solution
(A) ₹560. Rice extra = 90 ⇒ wheat extra = 60 = 12% ⇒ wheat April = 500 ⇒ May = 500·1.12 = 560.
ModerateCAT 2020 · Slot 312
In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is
  • (A) 439
  • (B) 357
  • (C) 399
  • (D) 417
Show solution
(C) 399. Asha − Bishnu = 57 marks = 12% of max ⇒ max = 475 ⇒ Geeta = 0.84·475 = 399.
ModerateCAT 2020 · TITA13
A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is
Show solution
250. 0.4w = 0.5b, w + b = 450 ⇒ b = 200, w = 250. Metallic = 100 + 100 = 200 ⇒ non-metallic = 250.
ModerateCAT 2024 · Slot 114
In September, the incomes of Kamal, Amal and Vimal are in the ratio 8 : 6 : 5. They rent a house together, and Kamal pays 15%, Amal pays 12% and Vimal pays 18% of their respective incomes to cover the total house rent in that month. In October, the house rent remains unchanged while their incomes increase by 10%, 12% and 15%, respectively. In October, the percentage of their total income that will be paid as house rent, is nearest to
  • (A) 14.84
  • (B) 12.75
  • (C) 13.26
  • (D) 15.18
Show solution
(C) 13.26. Rent = 1.2k + 0.72k + 0.9k = 2.82k. Oct income = 8.8k + 6.72k + 5.75k = 21.27k ⇒ 2.82/21.27 ≈ 13.26%.
ModerateCAT 2025 · Slot 315
The monthly sales of a product from January to April were 120, 135, 150 and 165 units, respectively. The cost price of the product was Rs. 240 per unit, and a fixed marked price was used for the product in all the four months. Discounts of 20%, 10% and 5% were given on the marked price per unit in January, February and March, respectively, while no discounts were given in April. If the total profit from January to April was Rs. 138825, then the marked price per unit, in rupees, was
  • (A) 520
  • (B) 525
  • (C) 510
  • (D) 515
Show solution
(B) 525. Revenue = (120·0.8 + 135·0.9 + 150·0.95 + 165)·MP = 525·MP. Cost = 240·570 = 136800. 525·MP − 136800 = 138825 ⇒ MP = 525.
Hard CAT 2021 · Slot 316
The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is
  • (A) 68.75
  • (B) 68.25
  • (C) 69.25
  • (D) 68.50
Show solution
(A) 68.75. From 1980 total +25%: 1.4M + 1.2F = 1.25(M+F) ⇒ F = 3M. 1990 female = 1.25·1.2F = 1.5F; male = female/2 = 0.75F. Total 1990 = 2.25F = 6.75M; total 1970 = 4M ⇒ (6.75 − 4)/4 = 68.75%.
Hard CAT 2023 · Slot 317
The population of a town in 2020 was 1,00,000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was
  • (A) 73,000
  • (B) 75,000
  • (C) 74,000
  • (D) 72,000
Show solution
(A) 73,000. Need (1 − y/100)(1 + (y+10)/100) > 1 ⇒ y² + 10y < 1000 ⇒ max y = 27 ⇒ 2021 pop = 100000·0.73 = 73000.
Hard CAT 2023 · Slot 3 · TITA18
A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
Show solution
340. Let total T. Sold = 0.2T + 96 + 0.4·apples = 0.5T, with apples = 0.75T − 240 and bananas = 240 − 0.15T. Integer/positivity constraints ⇒ T multiple of 20, 320 < T < 1600 ⇒ smallest T = 340 (mangoes 136, bananas 189, apples 15).

CAT 2024 & 2025, recent

Fresh questions distributed from the real CAT 2024 & CAT 2025 papers into this chapter.

ModerateCAT 2024 · Slot 3R1
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
  • (A) 27.5
  • (B) 30
  • (C) 25
  • (D) 20
Show solution
(C) 25. (1 + r)(1 + 2r) = 1.875 ⇒ 2r² + 3r − 0.875 = 0 ⇒ r = 0.25. First increment = 25%.
Hard CAT 2025 · Slot 1R2
A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to
  • (A) 29
  • (B) 23
  • (C) 27
  • (D) 25
Show solution
(C) 27. Start 60 L acid. Remove 20% → 48. Replace 10% with acid: 48 × 0.9 + 200 × 0.10 = 63.2. Remove 15% → 53.72. Final % = 53.72/200 ≈ 26.9 ≈ 27.
ModerateCAT 2025 · Slot 1 · TITAR3
Kamala divided her investment of Rs 100000 between stocks, bonds, and gold. Her investment in bonds was 25% of her investment in gold. With annual returns of 10%, 6%, 8% on stocks, bonds, and gold, respectively, she gained a total amount of Rs 8200 in one year. The amount, in rupees, that she gained from the bonds, was
Show solution
900. gold = g, bonds = g/4, stocks = 100000 − 1.25g. 0.10(100000−1.25g) + 0.06(g/4) + 0.08g = 8200 → g = 60000, bonds = 15000, gain = 900.
ModerateCAT 2025 · Slot 3R4
The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
  • (A) 45000
  • (B) 50000
  • (C) 54000
  • (D) 40000
Show solution
(C) 54000. 5x + 25y = 30 × 60000 = 1800000. After raise: 6x + 25y = 1800000 × 1.05 = 1890000 ⇒ x = 90000 ⇒ 25y = 1350000 ⇒ y = 54000.