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MBA CET 2026 Slot 2 question paper with solutions

200 questions · Quantitative Aptitude 50 · Verbal Ability 50 · Logical Reasoning 75 · Abstract Reasoning 25. Tap Show answer on any question for the answer and Cetking's solution. Where Cetking’s answer differs from the official CET key, both are shown with the reasoning.

Q101EasySimplification

DIRECTION for the question: Solve the following question and mark the best possible option.

587 × 999 = ?
  1. 586413
  2. 587523
  3. 614823
  4. 615173
  5. 615976
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Answer: A official key

Approach: Write 999 as 1000 − 1 and distribute.

Solution: 587 × 999 = 587 × (1000 − 1)
= 587000 − 587
= 586413

Cetking shortcut: Last three digits of N × 999 (N < 1000) are 1000 − N = 413; leading part is N − 1 = 586 → 586413. Only option A ends in 413.

Answer: A

Q102EasyNumbers
How many 3-digit numbers are there in between 100 and 300, having first and the last digit as 2?
  1. 9
  2. 10
  3. 11
  4. 12
  5. 13
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Answer: B official key

Approach: Fix the first and last digit, count choices for the middle digit.

Solution: Number must be of the form 2 _ 2 and lie between 100 and 300.
Middle digit can be 0, 1, 2, …, 9 → 10 choices.
Numbers: 202, 212, 222, …, 292 → 10 numbers.

Why the other options are wrong:
A (9): forgets 202 (middle digit 0).
C, D, E: overcount; only 10 middle digits exist.

Answer: B

Q103EasyHCF LCM

DIRECTION for the question: Solve the following question and mark the best possible option.

The H.C.F. of 2923 and 3239 is?
  1. 37
  2. 47
  3. 73
  4. 79
  5. 22
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Answer: D official key

Approach: Euclid's division method.

Solution: 3239 − 2923 = 316, so HCF(3239, 2923) = HCF(2923, 316).
2923 = 316 × 9 + 79 (316 × 9 = 2844; 2923 − 2844 = 79).
316 = 79 × 4 + 0 → remainder 0.
HCF = 79. Check: 2923 = 79 × 37, 3239 = 79 × 41.

Cetking shortcut: The HCF must divide the difference 316 = 4 × 79. Among options only 79 divides 316 (and 37 does not divide 3239).

Answer: D

Q104EasySimplification

DIRECTION for the question: Solve the following question and mark the best possible option.

400 ÷ 0.1 × 0.2 =?
  1. 0.8
  2. 8
  3. 80
  4. 8000
  5. 800
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Answer: E official key

Approach: BODMAS — division and multiplication left to right.

Solution: 400 ÷ 0.1 = 4000 (dividing by 0.1 = multiplying by 10).
4000 × 0.2 = 800.

Why the other options are wrong:
A/B/C: decimal-place slips.
D (8000): uses × 2 instead of × 0.2.

Answer: E

Q105EasySimplification

DIRECTION for the question: Solve the following question and mark the best possible option.

Given 168 × 32 = 5376, then 5.376 ÷ 16.8 is equal to?
  1. 0.032
  2. 0.32
  3. 3.2
  4. 320
  5. 32
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Answer: B official key

Approach: Use 5376 ÷ 168 = 32 and adjust decimals.

Solution: 168 × 32 = 5376 ⇒ 5376 ÷ 168 = 32.
5.376 = 5376 ÷ 1000 and 16.8 = 168 ÷ 10.
5.376 ÷ 16.8 = (5376 ÷ 168) × (10 ÷ 1000) = 32 ÷ 100 = 0.32.

Cetking shortcut: Estimate: 5.4 ÷ 17 ≈ 0.3 → only 0.32 fits.

Answer: B

Q106EasyRatios

DIRECTION for the question: Solve the following question and mark the best possible option.

When a sum of money was distributed to 12 persons instead of 16 persons, each person got Rs. 400 more. What was the sum?
  1. Rs. 14400
  2. Rs. 16000
  3. Rs. 19200
  4. Rs. 14151
  5. Rs. 16987
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Answer: C official key

Approach: Equate the difference in per-head shares.

Solution: Let the sum be S. S/12 − S/16 = 400.
LCM 48: (4S − 3S)/48 = 400 ⇒ S/48 = 400.
S = 19200.

Cetking shortcut: The 4 removed people's money (4 × S/16) is now shared as 400 each by 12 people = 4800 ⇒ S/16 = 1200 ⇒ S = 19200.

Answer: C

Q107EasyEquations Basics

DIRECTION for the question: Solve the following question and mark the best possible option.

On a 26 question test, five points were deducted for each wrong answer and eight points were credited for each correct answer. If all the questions were answered, how many were correct if the score was zero?
  1. 6
  2. 9
  3. 10
  4. 13
  5. 17
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Answer: C official key

Approach: Linear equation in number of correct answers.

Solution: Let correct = x, wrong = 26 − x.
8x − 5(26 − x) = 0 ⇒ 13x = 130 ⇒ x = 10.
Check: 10 × 8 = 80 and 16 × 5 = 80 → score 0.

Cetking shortcut: For zero score, correct : wrong = 5 : 8 ⇒ correct = 5/13 × 26 = 10.

Answer: C

Q108EasyAverages

DIRECTION for the question: Solve the following question and mark the best possible option.

A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is
  1. 250
  2. 276
  3. 280
  4. 285
  5. 240
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Answer: D official key

Approach: Weighted average over Sundays and other days.

Solution: Sundays on days 1, 8, 15, 22, 29 → 5 Sundays, 25 other days.
Total = 5 × 510 + 25 × 240 = 2550 + 6000 = 8550.
Average = 8550 ÷ 30 = 285.

Why the other options are wrong:
E (240): ignores the Sunday visitors.
B (276): uses 4 Sundays and 26 other days → 8280/30 = 276.

Cetking shortcut: Average = 240 + 5 × 270 / 30 = 240 + 45 = 285.

Answer: D

Q109EasyTime Speed Distance

DIRECTION for the question: Solve the following question and mark the best possible option.

A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is?
  1. 35
  2. 37
  3. 37.5
  4. 40.5
  5. 40
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Answer: C official key

Approach: Harmonic mean for equal distances.

Solution: Average speed = 2 × 50 × 30 / (50 + 30) = 3000/80 = 37.5 km/hr.
Check: time = 150/50 + 150/30 = 3 + 5 = 8 h; 300/8 = 37.5.

Why the other options are wrong:
E (40): simple average — the classic trap.
A, B, D: calculation slips.

Answer: C

Q110EasyPercentages

DIRECTION for the question: Solve the following question and mark the best possible option.

The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average % increase in the population per year is?
  1. 4.37%
  2. 5%
  3. 6%
  4. 8.75%
  5. 8%
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Answer: B official key

Approach: Total percentage increase spread evenly over 10 years.

Solution: Increase = 2,62,500 − 1,75,000 = 87,500.
Total % increase = 87,500/1,75,000 × 100 = 50%.
Average per year = 50% ÷ 10 = 5%.

Why the other options are wrong:
D (8.75%): divides 87,500 by 10,000 (wrong base).
C, E: no valid working gives these; 'average' here is the simple per-year rate, not compound (≈4.1%).

Answer: B

Q111EasyPercentages

DIRECTION for the question: Solve the following question and mark the best possible option.

If p% of p is 36, then p is equal to
  1. 15
  2. 60
  3. 600
  4. 3600
  5. 36
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Answer: B official key

Approach: Convert 'p% of p' into an equation.

Solution: (p/100) × p = 36 ⇒ p² = 3600 ⇒ p = 60.
Check: 60% of 60 = 36.

Why the other options are wrong:
D (3600): stops at p².
C (600): 600% of 600 = 3600.
A, E: 15% of 15 = 2.25; 36% of 36 = 12.96.

Answer: B

Q112EasyProfit Loss

DIRECTION for the question: Solve the following question and mark the best possible option.

The profit earned by selling an article for Rs. 900 is double the loss incurred when the same article is sold for Rs. 450. At what price should the article be sold to make a 25% profit?
  1. Rs. 600
  2. Rs. 750
  3. Rs. 800
  4. Rs. 900
  5. Rs. 450
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Answer: B official key

Approach: Find CP from the profit–loss relation, then apply 25% markup.

Solution: Let CP = x. Profit at 900 = 900 − x; loss at 450 = x − 450.
900 − x = 2(x − 450) ⇒ 900 − x = 2x − 900 ⇒ 3x = 1800 ⇒ x = 600.
SP for 25% profit = 1.25 × 600 = 750.

Why the other options are wrong:
A (600): that is the CP.
C, D, E: 33%, 50% profit and a loss price respectively.

Cetking shortcut: CP divides 450 to 900 in ratio loss : profit = 1 : 2 ⇒ CP = 450 + 450/3 = 600.

Answer: B

Q113EasyProfit Loss

DIRECTION for the question: Solve the following question and mark the best possible option.

A fan is listed at Rs. 1500 and a discount of 20% is offered on the list price. What additional discount must be offered to the customer to bring the net price to Rs. 1104?
  1. 8%
  2. 10%
  3. 12%
  4. 15%
  5. 20%
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Official CET key: A · Cetking's answer: C — see why below

Approach: Successive discounts — second discount on the reduced price.

Solution: After 20% discount: 1500 × 0.8 = 1200.
Additional discount needed = 1200 − 1104 = 96.
Rate = 96/1200 × 100 = 8%.

Why the other options are wrong:
C (12%): would give 1200 × 0.88 = 1056.
B (10%): gives 1080.
D, E: far too large.

Answer: A

Q114EasyRatios

DIRECTION for the question: Solve the following question and mark the best possible option.

The expenses of A and B are in the ratio 2 : 3. A sum of Rs. 2800 is equally divided between them. If A saves Rs. 600, then B saves?
  1. Rs.200
  2. Rs.300
  3. Rs.400
  4. Rs.500
  5. Rs.600
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Official CET key: A · Cetking's answer: B — see why below

Approach: Income − expense = saving.

Solution: Each gets 2800 ÷ 2 = 1400.
A's expense = 1400 − 600 = 800.
A : B expense = 2 : 3 ⇒ B's expense = 800 × 3/2 = 1200.
B's saving = 1400 − 1200 = Rs. 200.

Why the other options are wrong:
B (300): no consistent working gives this.
E (600): assumes equal savings.
C, D: misuse the ratio on savings.

Answer: A

Q115EasyRatios

DIRECTION for the question: Solve the following question and mark the best possible option.

There are 420 coins consisting of one-rupee coins, 50-paise coins and 25-paise coins. If the ratio of their values is 2 : 3 : 5, then the number of one-rupee coins is
  1. 20
  2. 30
  3. 90
  4. 300
  5. 50
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Answer: B official key

Approach: Convert value ratio to number ratio.

Solution: Values = 2k, 3k, 5k rupees.
Counts: ₹1 coins = 2k; 50p coins = 3k ÷ 0.5 = 6k; 25p coins = 5k ÷ 0.25 = 20k.
2k + 6k + 20k = 28k = 420 ⇒ k = 15.
One-rupee coins = 2 × 15 = 30.

Why the other options are wrong:
D (300): number of 25p coins.
C (90): number of 50p coins.
A, E: arithmetic slips.

Answer: B

Q116EasyPartnership

DIRECTION for the question: Solve the following question and mark the best possible option.

Manick received Rs. 6000 as his share out of the total profit of Rs. 9000 which he and Raunaq earned at the end of one year. If Manick invested Rs. 20000 for 6 months, whereas Raunaq invested his amount for the whole year, what was the amount invested by Raunaq?
  1. Rs. 4000
  2. Rs. 5000
  3. Rs. 6000
  4. Rs. 10,000
  5. Rs. 9000
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Answer: B official key

Approach: Profit ratio = capital × months.

Solution: Raunaq's profit = 9000 − 6000 = 3000 ⇒ ratio 2 : 1.
Manick: 20000 × 6 = 120000. Raunaq: x × 12.
120000 : 12x = 2 : 1 ⇒ 24x = 120000 ⇒ x = 5000.

Why the other options are wrong:
D (10,000): ignores the time factor.
A, C, E: arithmetic slips.

Answer: B

Q117EasyPartnership

DIRECTION for the question: Solve the following question and mark the best possible option.

Arun, Kamal and Vinay invested ₹8000, ₹4000 and ₹8000 respectively in a business. Arun left after six months. If after eight months, there was a gain of ₹ 4005, then what will be the share of Kamal ?
  1. ₹890
  2. ₹1335
  3. ₹1602
  4. ₹1780
  5. ₹666
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Official CET key: A · Cetking's answer: B — see why below

Approach: Capital × time ratio.

Solution: Arun: 8000 × 6 = 48000; Kamal: 4000 × 8 = 32000; Vinay: 8000 × 8 = 64000.
Ratio = 48 : 32 : 64 = 3 : 2 : 4 (total 9 parts).
Kamal's share = 2/9 × 4005 = ₹890.

Why the other options are wrong:
B (₹1335): 4005/3 — treats shares as equal.
D (₹1780): Vinay's share (4/9).
C (₹1602): 2/5 of 4005.
E (₹666): not from any valid split.

Answer: A

Q118EasyRatios

DIRECTION for the question: Solve the following question and mark the best possible option.

The cost of 16 packets of salt, each weighing 900 grams, is Rs. 28. What will be the cost of 27 packets if each packet weighs 1 kg?
  1. Rs. 52.50
  2. Rs. 56
  3. Rs. 58.50
  4. Rs. 64.75
  5. Rs. 28.50
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Answer: A official key

Approach: Unitary method on total weight.

Solution: Weight bought = 16 × 900 g = 14,400 g for Rs. 28.
Required weight = 27 × 1000 g = 27,000 g.
Cost = 28 × 27000/14400 = 28 × 15/8 = Rs. 52.50.

Why the other options are wrong:
B (56): doubles 28.
C, D, E: calculation slips; 47.25 (packets only) is not offered.

Answer: A

Q119EasyTime & Work

DIRECTION for the question: Solve the following question and mark the best possible option.

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days will one cow eat one bag of husk?
  1. 10
  2. 20
  3. 40
  4. 35
  5. 45
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Answer: C official key

Approach: Unitary method (work = cows × days).

Solution: 40 cows eat 40 bags in 40 days ⇒ 40 cows eat 1 bag in 1 day.
1 cow eats 1 bag in 40 days.

Why the other options are wrong:
A, B, D, E: wrong scaling; time does not change when cows and bags reduce in the same ratio.

Answer: C

Q120EasyPipes and Cisterns

DIRECTION for the question: Solve the following question and mark the best possible option.

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
  1. 10 min 20 sec
  2. 11 min 45 sec
  3. 12 min 30 sec
  4. 14 min 40 sec
  5. 15 min 20 sec
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Answer: D official key

Approach: LCM-unit method.

Solution: Tank = LCM(15, 20) = 60 units. A = 4 units/min, B = 3 units/min.
First 4 min together: 7 × 4 = 28 units. Remaining = 32 units.
B alone: 32 ÷ 3 = 10 2/3 min = 10 min 40 sec.
Total = 4 min + 10 min 40 sec = 14 min 40 sec.

Why the other options are wrong:
A (10 min 20 sec) and others: miss the first 4 minutes or miscalculate the remainder.

Answer: D

Q121MediumPipes and Cisterns

DIRECTION for the question: Solve the following question and mark the best possible option.

Four pipes can fill a reservoir in 15, 20, 30 and 60 hours respectively. The first pipe was opened at 8 am, the second at 9 am, the third at 10 am and the fourth at 11 am. When will the reservoir be full?
  1. 1:00 pm
  2. 2:00 pm
  3. 2:30 pm
  4. 3:00 pm
  5. 4:00 pm
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Official CET key: D · Cetking's answer: B — see why below

Approach: LCM-unit method with staggered start times.

Solution: Tank = LCM(15, 20, 30, 60) = 60 units. Rates: 4, 3, 2, 1 units/hr.
8–9 am: 4 units. 9–10 am: 4 + 3 = 7 units. 10–11 am: 4 + 3 + 2 = 9 units.
Filled by 11 am = 4 + 7 + 9 = 20 units. Remaining = 40 units.
From 11 am all four pipes: 4 + 3 + 2 + 1 = 10 units/hr ⇒ 40 ÷ 10 = 4 hours.
Full at 11 am + 4 h = 3:00 pm.
Check (x = hours from 8 am = 7): 7/15 + 6/20 + 5/30 + 4/60 = (28 + 18 + 10 + 4)/60 = 1.

Why the other options are wrong:
B (2:00 pm) / C (2:30 pm): wrong combined rate (1/5) or rounding.
A, E: arithmetic slips.

Cetking shortcut: Let pipe 1 run x hours: 4x + 3(x − 1) + 2(x − 2) + (x − 3) = 60 ⇒ 10x − 10 = 60 ⇒ x = 7 ⇒ 8 am + 7 h = 3 pm.

Answer: D

Q122MediumTime & Work

DIRECTION for the question: Solve the following question and mark the best possible option.

A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
  1. 30 days
  2. 40 days
  3. 60 days
  4. 70 days
  5. 80 days
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Answer: C official key

Approach: Convert A's work into (A + B) work.

Solution: A works 16 days and B works 44 days: 16a + 44b = 1, with a + b = 1/30.
Write 16a + 44b = 16(a + b) + 28b = 16/30 + 28b = 1.
28b = 14/30 ⇒ b = 1/60.
B alone takes 60 days. (A alone: 1/30 − 1/60 = 1/60 → also 60 days.)

Why the other options are wrong:
A (30): that is A and B together.
B, D, E: wrong pairing of the 16 and 44 days.

Cetking shortcut: 16 days of A + 44 days of B = 16 days of (A + B) + 28 days of B. 16 of the 30 joint days are done, so 28 days of B = 14 joint days ⇒ B = 2 × 30 = 60 days.

Answer: C

Q123EasyTime & Work

DIRECTION for the question: Solve the following question and mark the best possible option.

Kim can do the work in 3 days, while David can do the same work in 2 days. Both of them finish the work together and get Rs. 150. What is the share of Kim in the amount?
  1. Rs. 30
  2. Rs. 60
  3. Rs. 70
  4. Rs. 75
  5. Rs. 90
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Answer: B official key

Approach: Share ∝ efficiency = 1/days.

Solution: Kim : David efficiency = 1/3 : 1/2 = 2 : 3.
Kim's share = 2/5 × 150 = Rs. 60.

Why the other options are wrong:
E (Rs. 90): uses days ratio 3 : 2 — the classic trap; that is David's share.
D (Rs. 75): equal split.
A, C: no valid ratio.

Answer: B

Q124EasyTime Speed Distance

DIRECTION for the question: Solve the following question and mark the best possible option.

If a person walked at 14 km/hr. instead of 10 km/hr., he would have walked 20 km more. The actual distance traveled by him is?
  1. 50 km
  2. 56 km
  3. 70 km
  4. 80 km
  5. 30 km
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Answer: A official key

Approach: Time is the same in both cases.

Solution: Extra distance = (14 − 10) × t = 20 ⇒ t = 5 hours.
Actual distance = 10 × 5 = 50 km.

Why the other options are wrong:
C (70 km): distance at 14 km/hr.
B, D, E: arithmetic slips.

Answer: A

Q125EasyTime Speed Distance

DIRECTION for the question: Solve the following question and mark the best possible option.

A train X starts from Meerut at 4 pm and reaches Ghaziabad at 5 pm. While another train Y starts from Ghaziabad at 4pm and reaches Meerut at 5:30pm. The two trains will cross each other at?
  1. 4.36 pm
  2. 4.42 pm
  3. 4.48 pm
  4. 4.50 pm
  5. 4.55 pm
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Answer: A official key

Approach: Relative speed with distance as 1 unit.

Solution: X takes 60 min, Y takes 90 min (4:00 to 5:30).
Let distance = 180 units. Speed X = 3 units/min, Y = 2 units/min.
Meeting time = 180 ÷ (3 + 2) = 36 min after 4 pm.
They cross at 4:36 pm.

Why the other options are wrong:
B–E: come from wrong times (e.g. averaging) — the formula gives exactly 36 min.

Cetking shortcut: T = (60 × 90)/(60 + 90) = 5400/150 = 36 min.

Answer: A

Q126EasyTime Speed Distance

DIRECTION for the question: Solve the following question and mark the best possible option.

Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speed is?

  1. 2:3
  2. 4:3
  3. 6:7
  4. 9:16
  5. 5:7
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Answer: B official key

Approach: Speed ratio = √(time of the other after meeting).

Solution: Howrah train: 9 h after meeting; Patna train: 16 h.
S1 : S2 = √16 : √9 = 4 : 3.

Why the other options are wrong:
D (9 : 16): takes the time ratio directly.
A (2 : 3): no valid working.
C, E: no valid working.

Answer: B

Q127EasyMixtures

DIRECTION for the question: Solve the following question and mark the best possible option.

In what ratio must water be mixed with milk to gain 16 (2/3)% on selling the mixture at cost price?

  1. 1:6
  2. 6:1
  3. 2:3
  4. 4:3
  5. 5:6
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Answer: A official key

Approach: Profit comes only from the free water.

Solution: Gain = 16 2/3% = 1/6.
If 6 litres of milk are bought, selling 7 litres at CP gives revenue for 7 litres → profit 1 litre's cost = 1/6 of cost.
So water : milk = 1 : 6.

Why the other options are wrong:
B (6 : 1): reversed ratio.
C, D, E: do not give a 1/6 gain.

Answer: A

Q128MediumMixtures

DIRECTION for the question: Solve the following question and mark the best possible option.

A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

  1. (1/3)
  2. (1/4)
  3. (1/5)
  4. (1/7)
  5. (3/5)
Show answerHide answer

Answer: C official key

Approach: Track the syrup, which is only removed, never added.

Solution: Take 8 litres: water 3, syrup 5.
Draw off x litres and add x water. Syrup left = 5 − (5/8)x.
Need syrup = 4 (half of 8): 5 − 5x/8 = 4 ⇒ x = 8/5 litres.
Fraction of mixture = (8/5) ÷ 8 = 1/5.

Why the other options are wrong:
E (3/5): would remove too much, leaving only 2 litres of syrup out of 8.
A, B, D: give syrup 3.33, 3.75, 4.29 out of 8 — not half.

Cetking shortcut: Syrup fraction 5/8 must become 1/2: (5/8)(1 − x) = 1/2 ⇒ 1 − x = 4/5 ⇒ x = 1/5.

Answer: C

Q129EasyInterest

DIRECTION for the question: Solve the following question and mark the best possible option.

The interest on a certain deposit at 4.5% p.a. is Rs. 202.50 in one year. How much will the additional interest in one year be on the same deposit, at 5% p.a.?

  1. Rs. 20.25
  2. Rs. 22.50
  3. Rs. 25
  4. Rs. 42.75
  5. Rs. 42.25
Show answerHide answer

Answer: B official key

Approach: Find principal, then apply the 0.5% extra rate.

Solution: P = 202.50 ÷ 0.045 = Rs. 4500.
Interest at 5% = 4500 × 0.05 = Rs. 225.
Additional interest = 225 − 202.50 = Rs. 22.50.

Why the other options are wrong:
A (20.25): 10% of 202.50.
D, E: wrongly add values.
C (25): 0.5% of 5000.

Cetking shortcut: Extra = 202.50 × 0.5/4.5 = 202.50/9 = 22.50.

Answer: B

Q130EasyInterest

DIRECTION for the question: Solve the following question and mark the best possible option.

A sum of money at simple interest amounts to Rs. 815 in 3 years and Rs. 854 in 4 years. The sum is?

  1. Rs. 650
  2. Rs. 690
  3. Rs. 698
  4. Rs. 700
  5. Rs. 750
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Answer: C official key

Approach: SI per year from the difference of amounts.

Solution: SI for 1 year = 854 − 815 = Rs. 39.
SI for 3 years = 3 × 39 = Rs. 117.
Principal = 815 − 117 = Rs. 698.

Why the other options are wrong:
B (690), D (700): close-number traps.
A, E: wrong interest count.

Answer: C

Q131EasyInterest

DIRECTION for the question: Solve the following question and mark the best possible option.

There is a 60% increase in the amount in 6 years at simple interest. What will be the compound interest of Rs. 12000 after 3 years at the same rate?

  1. Rs. 2160
  2. Rs. 3120
  3. Rs. 3972
  4. Rs. 6240
  5. Rs. 1200
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Answer: C official key

Approach: SI gives the rate; apply CI for 3 years.

Solution: 60% in 6 years at SI ⇒ rate = 10% p.a.
CI = 12000 × (1.1³ − 1) = 12000 × 0.331 = Rs. 3972.

Why the other options are wrong:
E (1200): just one year's interest.
A, B, D: wrong rate or wrong number of years; SI for 3 years would be 3600, and CI must be a bit more.

Cetking shortcut: CI for 3 years at 10% = 33.1% of P → 0.331 × 12000 = 3972.

Answer: C

Q132EasyMensuration

DIRECTION for the question: Solve the following question and mark the best possible option.

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?

  1. 46
  2. 81
  3. 126
  4. 252
  5. 37
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Answer: C official key

Approach: Identify which sides are painted.

Solution: Unpainted side = 9 ft, so the opposite side (9 ft) is painted along with the two other sides.
9 + 2b = 37 ⇒ b = 14 ft.
Area = 9 × 14 = 126 sq ft.

Why the other options are wrong:
D (252): (37 − 9) × 9, treating 28 as one side.
B (81): 9 × 9 square.
A, E: perimeter-type numbers, not area.

Answer: C

Q133EasyMensuration

DIRECTION for the question: Solve the following question and mark the best possible option.

A rectangular field is to be fenced on three sides, leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

  1. 34
  2. 40
  3. 68
  4. 88
  5. 38
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Answer: D official key

Approach: Find the other side from the area.

Solution: Other side = 680 ÷ 20 = 34 ft.
Fence = 20 + 34 + 34 = 88 ft.

Why the other options are wrong:
C (68): only the two 34-ft sides.
A (34): one side only.
B, E: arithmetic slips.

Answer: D

Q134EasyMensuration

DIRECTION for the question: Solve the following question and mark the best possible option.

A rectangular carpet has an area of 120 sq. meters and a perimeter of 46 meters. The length of its diagonal is?

  1. 15 meters
  2. 16 meters
  3. 17 meters
  4. 19 meters
  5. 20 meters
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Answer: C official key

Approach: Use (l + b)² − 2lb.

Solution: Perimeter 46 ⇒ l + b = 23. Area lb = 120.
d² = 23² − 2 × 120 = 529 − 240 = 289 ⇒ d = 17 m.
(Sides are 15 and 8: 15 × 8 = 120, 2(15 + 8) = 46.)

Why the other options are wrong:
A (15): that is the length.
B, D, E: not √289.

Cetking shortcut: 8-15-17 is a Pythagorean triplet — spot 15 × 8 = 120 instantly.

Answer: C

Q135EasyTime Speed Distance

DIRECTION for the question: Solve the following question and mark the best possible option.

In a 100 m race, A covers the whole distance in 36 seconds and B in 45 seconds. In this race, A beats B by

  1. 20 m
  2. 25 m
  3. 22.5 m
  4. 9 m
  5. 10 m
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Answer: A official key

Approach: Distance B covers in A's time.

Solution: B's speed = 100/45 m/s.
In 36 s B runs 36 × 100/45 = 80 m.
A beats B by 100 − 80 = 20 m.

Why the other options are wrong:
D (9 m): takes 45 − 36 seconds as metres.
B (25 m): uses A's speed × 9 s.
C, E: arithmetic slips.

Answer: A

Q136EasyTime Speed Distance

DIRECTION for the question: Solve the following question and mark the best possible option.

In a 200 m race, A beats B by 35 m or 7 seconds. A's time over the course is

  1. 40 sec
  2. 47 sec
  3. 33 sec
  4. 35 sec
  5. 20 sec
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Answer: C official key

Approach: Use the beat-margin to find B's speed.

Solution: B runs 35 m in 7 s ⇒ speed = 5 m/s.
B's time for 200 m = 200 ÷ 5 = 40 s.
A's time = 40 − 7 = 33 s.

Why the other options are wrong:
A (40 s): B's time.
B (47 s): adds 7 instead of subtracting.
D, E: no valid working.

Answer: C

Q137EasyPnC

DIRECTION for the question: Solve the following question and mark the best possible option.

In how many different ways can the letters of the word OPTICAL can be arranged in such a way that the vowels always come together?

  1. 120
  2. 720
  3. 2160
  4. 4320
  5. 6
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Answer: B official key

Approach: Block method.

Solution: OPTICAL has 7 distinct letters: vowels O, I, A (3); consonants P, T, C, L (4).
Vowel block + 4 consonants = 5 units → 5! = 120 ways.
Vowels inside the block: 3! = 6 ways.
Total = 120 × 6 = 720.

Why the other options are wrong:
A (120): forgets internal arrangement.
E (6): only the internal arrangement.
C, D: wrong block counts.

Answer: B

Q138EasyProbability

DIRECTION for the question: Solve the following question and mark the best possible option.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girl is:

  1. (21/46)
  2. (25/117)
  3. (1/50)
  4. (3/25)
  5. (21/31)
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Answer: A official key

Approach: Combinations.

Solution: Total = C(25, 3) = 25 × 24 × 23/6 = 2300.
Favourable = C(15, 2) × C(10, 1) = 105 × 10 = 1050.
P = 1050/2300 = 21/46.

Why the other options are wrong:
B, C, D, E: do not equal 1050/2300; e.g. D (3/25) ignores combinations.

Answer: A

Q139EasyProbability

DIRECTION for the question: Solve the following question and mark the best possible option.

Two dice are tossed. The probability that the total score is a prime number is

  1. (1/6)
  2. (1/2)
  3. (5/12)
  4. (7/9)
  5. (2/3)
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Answer: C official key

Approach: Count outcomes for each prime sum.

Solution: Sum 2: 1; Sum 3: 2; Sum 5: 4; Sum 7: 6; Sum 11: 2.
Favourable = 1 + 2 + 4 + 6 + 2 = 15.
P = 15/36 = 5/12.

Why the other options are wrong:
B (1/2): 18/36 — includes a non-prime sum.
A (1/6): only sum 7.
D, E: overcount.

Answer: C

Q140EasyNumber Series (Quant)

DIRECTION for the question: Find the next number in the sequence

7, 26, 63, 124, 215, 342, (---)

  1. 481
  2. 400
  3. 511
  4. 391
  5. 421
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Answer: C official key

Approach: Compare each term with a perfect cube.

Solution: 7 = 2³ − 1, 26 = 3³ − 1, 63 = 4³ − 1, 124 = 5³ − 1, 215 = 6³ − 1, 342 = 7³ − 1.
Next = 8³ − 1 = 512 − 1 = 511.

Why the other options are wrong:
A (481), E (421), D (391), B (400): none is 8³ − 1.

Cetking shortcut: Differences 19, 37, 61, 91, 127 rise by 18, 24, 30, 36 → next difference 169 → 342 + 169 = 511.

Answer: C

Q141EasyBar Graph

Questions 141–145 share this

DIRECTIONS for Questions: The bar-graph given below shows the percentage distribution of total expenditures of a company under various expense heads during 2023. Study the graph and answer the questions that follow:

Percentage Distribution of Total Expenditures of a Company

Data read from chart:

Expense headPercent Spent
Infrastructure20
Transport12.5
Advertisement15
Taxes10
Research and Development5
Salary20
Interest on Loans17.5
Figure for MBA CET 2026 Slot 2 question 141

The expenditure on the interest on loans is by what percentage more than the expenditure on transport?

  1. 5%
  2. 10%
  3. 20%
  4. 30%
  5. 40%
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Answer: E official key

Approach: Percentage-more formula on the bar values.

Solution: Interest on loans = 17.5%, Transport = 12.5% of total.
Difference = 5.
% more = 5/12.5 × 100 = 40%.

Why the other options are wrong:
A (5%): the percentage-point gap, not % more.
B, C, D: wrong base or misread bars.

Answer: E

Q142EasyBar Graph

Refer to the passage above ↑

What is the ratio of the total expenditure on infrastructure and transport to the total expenditure on taxes and interest on loans?

Figure for MBA CET 2026 Slot 2 question 142
  1. 5:4
  2. 8:7
  3. 9:7
  4. 13:11
  5. 5:9
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Answer: D official key

Approach: Add the shares, then simplify.

Solution: Infrastructure + Transport = 20 + 12.5 = 32.5.
Taxes + Interest on loans = 10 + 17.5 = 27.5.
Ratio = 32.5 : 27.5 = 65 : 55 = 13 : 11.

Why the other options are wrong:
A, B, C, E: come from misread bars (e.g. 20+15 : 10+17.5 gives 14 : 11, not offered).

Answer: D

Q143EasyBar Graph

Refer to the passage above ↑

If the expenditure on advertisements is Rs. 2.10 crores, then the difference between the expenditures on transport and taxes is:

Figure for MBA CET 2026 Slot 2 question 143
  1. Rs. 1.25 crores
  2. Rs. 95 lakhs
  3. Rs. 65 lakhs
  4. Rs. 35 lakhs
  5. Rs. 25 lakhs
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Answer: D official key

Approach: Use Advertisement as the scale.

Solution: Advertisement 15% = Rs. 2.10 crore ⇒ 1% = Rs. 0.14 crore.
Transport − Taxes = 12.5 − 10 = 2.5%.
Difference = 2.5 × 0.14 = Rs. 0.35 crore = Rs. 35 lakh.

Why the other options are wrong:
A, B, C, E: misread bars (e.g. transport as 10%) or crore–lakh unit slips.

Cetking shortcut: 2.5% is 1/6 of 15% → 2.10 crore ÷ 6 = 0.35 crore.

Answer: D

Q144EasyBar Graph

Refer to the passage above ↑

The total amount of expenditure of the company is how many times the expenditure on research and development?

Figure for MBA CET 2026 Slot 2 question 144
  1. 27
  2. 20
  3. 18
  4. 8
  5. 5
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Answer: B official key

Approach: Share of total.

Solution: R&D = 5% of total.
Total ÷ R&D = 100 ÷ 5 = 20 times.

Why the other options are wrong:
E (5): the R&D % itself.
A, C, D: misread bars.

Answer: B

Q145EasyBar Graph

Refer to the passage above ↑

If the interest on loans amounted to Rs. 2.45 crores, then the total amount of expenditure on advertisements, taxes and research and development is:

Figure for MBA CET 2026 Slot 2 question 145
  1. Rs. 7 crores
  2. Rs. 5.4 crores
  3. Rs. 4.2 crores
  4. Rs. 3 crores
  5. Rs. 2.4 crores
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Answer: C official key

Approach: Use Interest on loans as the scale.

Solution: Interest on loans 17.5% = Rs. 2.45 crore ⇒ total = 2.45 ÷ 0.175 = Rs. 14 crore.
Advertisement + Taxes + R&D = 15 + 10 + 5 = 30%.
Required = 30% of 14 = Rs. 4.2 crore.

Why the other options are wrong:
A (7 crore): 50% of total.
B, D, E: wrong combined share.

Cetking shortcut: 30/17.5 × 2.45 = 2.45 × 12/7 = 0.35 × 12 = 4.2.

Answer: C

Q146EasyPie Chart

Questions 146–150 share this

DIRECTIONS for Questions: The pie-chart provided below gives the distribution of land (in a village) under various food crops. Study the pie-chart carefully and answer the questions that follow.

Distribution of Areas (In Acres) under various food crops

Data read from chart (central angles):

CropAngle
Wheat72°
Rice72°
Others99°
Maize45°
Bajra18°
Jowar18°
Barley36°
Figure for MBA CET 2026 Slot 2 question 146

Which combination of three crops contributes to 50% of the total area under the food crops?

  1. Wheat, Barley and Jowar
  2. Rice, Wheat and Jowar
  3. Rice, Wheat and Barley
  4. Bajra , Maize and Rice
  5. Barley, Jowar and Rice
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Answer: C official key

Approach: Check which options sum to 180°.

Solution: Rice 72° + Wheat 72° + Barley 36° = 180° = 50%.

Why the other options are wrong:
A: 72 + 36 + 18 = 126°.
B: 72 + 72 + 18 = 162°.
D: 18 + 45 + 72 = 135°.
E: 36 + 18 + 72 = 126°.

Answer: C

Q147EasyPie Chart

Refer to the passage above ↑

If the total area under jowar was 1.5 million acres, then what was the area (in million acres) under rice?

Figure for MBA CET 2026 Slot 2 question 147
  1. 6
  2. 7.5
  3. 9
  4. 4.5
  5. 2
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Answer: A official key

Approach: Ratio of angles.

Solution: Rice : Jowar = 72° : 18° = 4 : 1.
Rice area = 4 × 1.5 = 6 million acres.

Why the other options are wrong:
C (9): uses 108°.
B, D, E: wrong angle ratio.

Answer: A

Q148EasyPie Chart

Refer to the passage above ↑

If the production of wheat is 6 times that of barley, then what is the ratio between the yield per acre of wheat and barley?

Figure for MBA CET 2026 Slot 2 question 148
  1. 3:2
  2. 3:1
  3. 12:1
  4. 2:3
  5. 3:5
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Answer: B official key

Approach: Yield = production ÷ area.

Solution: Area ratio Wheat : Barley = 72° : 36° = 2 : 1.
Production ratio = 6 : 1.
Yield ratio = (6/2) : (1/1) = 3 : 1.

Why the other options are wrong:
C (12 : 1): multiplies instead of divides.
A, D, E: inverted or wrong ratios.

Answer: B

Q149EasyPie Chart

Refer to the passage above ↑

If the yield per acre of rice was 50% more than that of barley, then the production of barley is what percentage of that of rice?

Figure for MBA CET 2026 Slot 2 question 149
  1. 30%
  2. 33.33%
  3. 35.5%
  4. 36%
  5. 50.65%
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Answer: B official key

Approach: Production = area × yield.

Solution: Let Barley yield = 2 units/acre ⇒ Rice yield = 3.
Barley production ∝ 36 × 2 = 72; Rice production ∝ 72 × 3 = 216.
Barley as % of Rice = 72/216 × 100 = 33.33%.

Why the other options are wrong:
E (50.65%): near the 50% you get from areas alone — ignores yield.
A, C, D: arithmetic slips.

Answer: B

Q150MediumPie Chart

Refer to the passage above ↑

If the total area goes up by 5%, and the area under wheat production goes up by 12%, then what will be the angle for wheat in the new pie-chart?

Figure for MBA CET 2026 Slot 2 question 150
  1. 62.4°
  2. 76.8°
  3. 80.6°
  4. 84.2°
  5. 12.4°
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Answer: B official key

Approach: Scale the share by relative growth.

Solution: Wheat's old share = 72/360 = 20%.
New share = 20% × 1.12/1.05 = 22.4%/1.05 ≈ 21.33%.
New angle = 72 × 1.12/1.05 = 80.64/1.05 = 76.8°.

Why the other options are wrong:
C (80.6°): ignores the 5% growth in total.
D (84.2°): adds growth instead of dividing.
A, E: wrong direction or nonsense.

Cetking shortcut: 1.12/1.05 = 16/15 → 72 × 16/15 = 76.8°.

Answer: B

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