Free Quantitative Aptitude questions with answers for MBA CET

Arithmetic, algebra, geometry, number systems, permutation and combination, probability and data interpretation - the quant questions that decide your MBA CET percentile. These 117 questions are also useful for CAT, MAT and ATMA. Tap “Show answer” under any question for the correct option and the explanation.

Question 1
Find the missing term: 100, 152, 382, 1339, ?, 33153.25
- (A)6025.5
- (B)6027.5
- (C)6028.5
- (D)6029.5
- (E)6024.5
Show answer and explanation
Correct answer: (B) 6027.5
Multiply successive terms by 1.5, 2.5, 3.5, 4.5 and 5.5, respectively, and add 2. Thus 1339 × 4.5 + 2 = 6027.5.
Question 2
An article is marked 50% above its cost price and sold after successive discounts of 20% and 15%. If the profit is ₹400, what is the cost price?
- (A)₹15,000
- (B)₹20,000
- (C)₹25,000
- (D)₹30,000
- (E)₹40,000
Show answer and explanation
Correct answer: (B) ₹20,000
Let CP = 100. MP = 150. SP = 150 × 0.80 × 0.85 = 102. Profit = 2% of CP. Therefore 2% of CP = ₹400, giving CP = ₹20,000.
Question 3
The diameter of a circle is 10.5 cm. A sphere has a radius equal to twice this diameter. Find the volume of the sphere. Use π = 22/7.
- (A)30,804 cm³
- (B)35,664 cm³
- (C)38,808 cm³
- (D)40,132 cm³
- (E)42,504 cm³
Show answer and explanation
Correct answer: (C) 38,808 cm³
Sphere radius = 2 × 10.5 = 21 cm. Volume = 4/3 × π × 21³ = 4/3 × 22/7 × 9261 = 38,808 cm³.
Question 4
Train A travels at 36 km/h and crosses a pole in 20 seconds. Train B is 100 m longer than A and travels at 18 km/h. How long will B take to cross a 500 m platform?
- (A)140 s
- (B)150 s
- (C)160 s
- (D)180 s
- (E)200 s
Show answer and explanation
Correct answer: (C) 160 s
36 km/h = 10 m/s, so A's length = 10 × 20 = 200 m. B's length = 300 m. Distance to cross platform = 300 + 500 = 800 m. B's speed = 5 m/s. Time = 800/5 = 160 s.
Question 5
Find the value of ?: 80% of 140 = 25 × 8 − ? + 12 × 4
- (A)136
- (B)137
- (C)146
- (D)147
- (E)156
Show answer and explanation
Correct answer: (A) 136
80% of 140 = 112. Hence 112 = 200 − ? + 48 = 248 − ?. Therefore ? = 248 − 112 = 136.
Question 6
P and Q invest in the ratio 2:3. P withdraws after 9 months. If their profit ratio is 3:8, for how many months was Q's money invested?
- (A)12
- (B)14
- (C)16
- (D)18
- (E)20
Show answer and explanation
Correct answer: (C) 16
Profit is proportional to capital × time. Thus (2 × 9)/(3 × T) = 3/8. Solving gives 16 = T.
Question 7
Water and milk are mixed in the ratio 3:7. From 200 litres of mixture, 25% is transferred to another vessel. If 10 litres of water is then added, how much water is in the second vessel?
- (A)18 L
- (B)20 L
- (C)22 L
- (D)25 L
- (E)28 L
Show answer and explanation
Correct answer: (D) 25 L
Transferred mixture = 25% of 200 = 50 L. Water in it = 50 × 3/10 = 15 L. Adding 10 L water gives 25 L.
Question 8
A man rows 52 km upstream and 42 km downstream in 10 hours. He also rows 60 km upstream and 66 km downstream in 13 hours. What is his speed in still water?
- (A)6 km/h
- (B)8 km/h
- (C)10 km/h
- (D)12 km/h
- (E)14 km/h
Show answer and explanation
Correct answer: (C) 10 km/h
Let still-water speed = b and stream speed = s. Then 52/(b−s)+42/(b+s)=10 and 60/(b−s)+66/(b+s)=13. Solving the simultaneous equations gives b = 10 km/h.
Question 9
A sum amounts to ₹2,480 in 2 years and ₹3,200 in 5 years at simple interest. Find the annual rate of interest.
- (A)8%
- (B)10%
- (C)12%
- (D)14%
- (E)16%
Show answer and explanation
Correct answer: (C) 12%
Interest for 3 years = 3200 − 2480 = ₹720, so annual interest = ₹240. Interest for 2 years = ₹480, hence principal = 2480 − 480 = ₹2,000. Rate = 240/2000 × 100 = 12%.
Question 10
A trader marks an item 40% above cost price and offers a 25% discount. What is the resulting profit percentage?
- (A)2.5%
- (B)5%
- (C)7.5%
- (D)10%
- (E)12.5%
Show answer and explanation
Correct answer: (B) 5%
Let CP = 100. MP = 140. SP after 25% discount = 140 × 0.75 = 105. Profit = 5, so profit percentage = 5%.
Question 11
A can complete a job in 12 days and B can complete it in 18 days. If they work together, how many days will they take?
- (A)6 days
- (B)7.2 days
- (C)8 days
- (D)9 days
- (E)10 days
Show answer and explanation
Correct answer: (B) 7.2 days
Combined daily work = 1/12 + 1/18 = 5/36. Therefore time = 36/5 = 7.2 days.
Question 12
The average of 8 numbers is 24. If one number, 38, is replaced by 22, what is the new average?
- (A)21
- (B)22
- (C)23
- (D)24
- (E)25
Show answer and explanation
Correct answer: (B) 22
Original total = 8 × 24 = 192. New total = 192 − 38 + 22 = 176. New average = 176/8 = 22.
Question 13
The incomes of A and B are in the ratio 5:7, while their expenditures are in the ratio 3:5. If each saves ₹2,000, what is A's income?
- (A)₹4,000
- (B)₹5,000
- (C)₹6,000
- (D)₹7,000
- (E)₹8,000
Show answer and explanation
Correct answer: (B) ₹5,000
Let incomes be 5x and 7x, and expenditures be 3y and 5y. Since 5x−3y = 2000 and 7x−5y = 2000, subtracting gives 2x = 2y, so x = y. Then 5x−3x = 2000, giving x = 1000. A's income = ₹5,000.
Question 14
If x + 1/x = 5, find x² + 1/x².
- (A)21
- (B)23
- (C)25
- (D)27
- (E)29
Show answer and explanation
Correct answer: (B) 23
Squaring, (x + 1/x)² = x² + 2 + 1/x². Thus 25 = x² + 1/x² + 2, so the required value is 23.
Question 15
A vehicle must cover 180 km in 30 hours. It covers half the distance in 2/5 of the total time. What speed is required for the remaining distance in the remaining time?
- (A)5 km/h
- (B)7.5 km/h
- (C)10 km/h
- (D)12 km/h
- (E)15 km/h
Show answer and explanation
Correct answer: (A) 5 km/h
Half the distance = 90 km. Time used = 2/5 × 30 = 12 hours. Remaining time = 18 hours. Required speed = 90/18 = 5 km/h.
Question 16
The average age of a class of 20 students is 15 years. If the age of the teacher is included, which is 30 years, what is the new average age?
- (A)18.6 years
- (B)15.7 years
- (C)21 years
- (D)30 years
Show answer and explanation
Correct answer: (B) 15.7 years
Total age of 20 students = 20 × 15 = 300 years. New total with teacher = 300 + 30 = 330 years. New average = 330 / 21 ≈ 15.7 years.
Question 17
A company sells 100 widgets in a month. If 20% of the widgets are defective, how many defective widgets did the company sell?
- (A)2
- (B)4
- (C)10
- (D)20
Show answer and explanation
Correct answer: (D) 20
Defective widgets = 20% of 100 = 20.
Question 18
Rohan sold the stock of Company X for ₹7,200 and incurred a loss of 40%. At what price should he have sold the stock to have gained a profit of 25%?
- (A)₹12,800
- (B)₹16,400
- (C)₹13,800
- (D)₹15,000
Show answer and explanation
Correct answer: (D) ₹15,000
SP = ₹7,200 is 60% of CP (since loss is 40%), so CP = 7,200 / 0.6 = ₹12,000. Desired SP for 25% profit = 12,000 × 1.25 = ₹15,000.
Question 19
A person purchased 15 pumpkins each for ₹20. Two pumpkins were rotten hence he disposed them, and the remaining was sold at ₹25. Find the rate of loss or profit.
- (A)25/3% profit
- (B)25/3% loss
- (C)25/4% loss
- (D)25/4% profit
Show answer and explanation
Correct answer: (A) 25/3% profit
CP = 15 × ₹20 = ₹300. Pumpkins sold = 13, SP = 13 × ₹25 = ₹325. Profit % = (25/300) × 100 = 25/3% ≈ 8.33%.
Question 20
If a company sells a bike with a marked price of ₹78,000 and gives a discount of 5% on ₹60,000 and 3% on the remaining amount, then the actual price charged by the company for a bike (in ₹) is:
- (A)74,460
- (B)75,520
- (C)75,850
- (D)75,952
Show answer and explanation
Correct answer: (A) 74,460
Discount on first ₹60,000 = 5% of 60,000 = ₹3,000. Remaining ₹18,000 at 3% = ₹540. Total discount = ₹3,540. Final price = 78,000 − 3,540 = ₹74,460.

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