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MBA CET 2026 Slot 3 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.

Q101MediumHCF LCM

The HCF of 2472, 1284, and a third number N is 12.

If their LCM is 2³ × 3² × 5 × 103 × 107, then which of the following could be the value of N?

  1. 2² × 3² × 7
  2. 2² × 3³ × 103
  3. 2² × 3² × 5
  4. 3³ × 103
  5. 2² × 3²
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Answer: C official key

Approach: HCF takes the lowest power of each common prime; LCM takes the highest power of every prime.

Solution:
2472 = 2³ × 3 × 103 and 1284 = 2² × 3 × 107.
HCF of all three = 12 = 2² × 3 → N must be divisible by 2² × 3, and N's power of 2 is at most 3.
LCM = 2³ × 3² × 5 × 103 × 107. The two given numbers supply 2³, 103, 107 but only 3¹ and no 5.
So N must supply 3² (exactly, not 3³) and 5¹, and must not bring any new prime.
Minimum such N = 2² × 3² × 5 = 180 → matches option C.

Why the other options are wrong:

  • A: contains 7 and no 5 → LCM would have 7 and lack 5.
  • B: 3³ makes LCM contain 3³, and 5 is missing.
  • D: no factor 2² → HCF would become 3, not 12; also 3³ and no 5.
  • E: no 5 → LCM would lack 5.

Cetking shortcut: Look only for the prime missing from the two given numbers (here 5) plus the exact power 3² — only C has both.

Answer: C

Q102EasyHCF LCM
The smallest perfect square number that is exactly divisible by 2, 3, 4, 6, 9, 18, 30, and 60 is k², where k is a natural number.What is the sum of all prime factors of k?
  1. 8
  2. 9
  3. 10
  4. 11
  5. 12
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Answer: C official key

Approach: Least perfect square divisible by a set = LCM of the set made to have all even exponents.

Solution:
LCM(2, 3, 4, 6, 9, 18, 30, 60) = 2² × 3² × 5 = 180.
Exponent of 5 is odd → multiply by 5: 180 × 5 = 900 = 30².
k = 30 = 2 × 3 × 5.
Sum of prime factors of k = 2 + 3 + 5 = 10.

Cetking shortcut: Primes appearing anywhere in the list are 2, 3, 5 — k must contain each, so the sum is 10 directly.

Answer: C

Q103EasyRemainders
Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.
  1. 2
  2. 4
  3. 5
  4. 7
  5. 9
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Answer: B official key

Approach: Remainder of a sum = remainder of the sum of individual remainders.

Solution:
Sum = 73 + 75 + 78 + 57 + 197 = 480.
34 × 14 = 476.
480 − 476 = 4.

Cetking shortcut: Remainders by 34: 5 + 7 + 10 + 23 + 27 = 72 → 72 − 68 = 4.

Answer: B

Q104EasyEquations Basics
In a sanctuary, there are horses and ostriches. Each animal has one head. The total number of heads is 86 and the total number of legs is 250. How many ostriches are there?
  1. 45
  2. 46
  3. 47
  4. 48
  5. 49
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Answer: C official key

Approach: Heads–legs problem: two linear equations in two unknowns.

Solution:
Let horses = h, ostriches = o.
Heads: h + o = 86. Legs: 4h + 2o = 250.
Halve legs: 2h + o = 125 → subtract heads equation: h = 39.
o = 86 − 39 = 47.

Cetking shortcut: If all 86 were ostriches → 172 legs; extra 250 − 172 = 78 legs ÷ 2 per horse = 39 horses → ostriches = 47.

Answer: C

Q105EasyEquations Basics

There are two examination halls, X and Y.

Statement I:
If 15 students are shifted from hall X to hall Y, then the number of students in both halls, becomes equal.

Statement II:
If 25 students are shifted from hall Y to hall X, then the number of students in hall X becomes twice the number of students in hall Y.

Based on the above statements, answer the following question:

The number of students in halls, X and Y, respectively, is:

  1. 90, 60
  2. 105, 75
  3. 120, 90
  4. 135, 105
  5. 140, 60
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Answer: D official key

Approach: Translate each transfer statement into a linear equation and solve.

Solution:
Let hall X = x, hall Y = y.
I: x − 15 = y + 15 → x − y = 30.
II: x + 25 = 2(y − 25) → x = 2y − 75.
Substitute: 2y − 75 − y = 30 → y = 105, x = 135.
Check: 135+25 = 160 = 2 × 80 = 2 × (105−25). ✓

Cetking shortcut: Only options with x − y = 30 survive (A, B, C, D); test II on them — only 135, 105 gives 160 = 2 × 80.

Answer: D

Q106EasySequences and Series
The sum of three numbers in a GP is 26 and their product is 216. What are the three numbers?
  1. 16, 6, 4
  2. 4, 6, 16
  3. 9, 6, 11
  4. 2, 6, 18
  5. 5, 7, 14
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Answer: D official key

Approach: Three numbers in GP: take them as a/r, a, ar so the product is a³.

Solution:
Product a³ = 216 → a = 6.
Sum: 6/r + 6 + 6r = 26 → 6/r + 6r = 20 → 3r² − 10r + 3 = 0.
r = 3 or 1/3 → numbers 2, 6, 18.
Check: 2 + 6 + 18 = 26, 2 × 6 × 18 = 216. ✓

Why the other options are wrong:

  • A, B: 4, 6, 16 sums to 26 but 6/4 ≠ 16/6 (not a GP); product 384.
  • C: 9, 6, 11 is not a GP; product 594.
  • E: 5, 7, 14 is not a GP; product 490.

Cetking shortcut: Middle term of a 3-term GP = cube root of the product = 6; then check which option has 6² = first × third (2 × 18 = 36).

Answer: D

Q107EasyNumber Series (Quant)

Find the next number in the given series:

650, 601, 565, 540, 524, ?

  1. 520
  2. 518
  3. 515
  4. 514
  5. 512
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Answer: C official key

Approach: Difference series — differences are consecutive perfect squares.

Solution:
Differences: 650→601 = −49, 601→565 = −36, 565→540 = −25, 540→524 = −16.
These are −7², −6², −5², −4² → next is −3² = −9.
524 − 9 = 515.

Cetking shortcut: Whenever differences are 49, 36, 25… think squares immediately.

Answer: C

Q108EasyNumber Series (Quant)

Find the missing number in the following series:

94, 118, 158, ?, 286, 374

  1. 198
  2. 206
  3. 214
  4. 226
  5. 232
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Answer: C official key

Approach: Second-difference series — the differences grow by a constant 16.

Solution:
Known differences: 118 − 94 = 24, 158 − 118 = 40 (+16).
Next differences: 56, 72, 88.
Missing term = 158 + 56 = 214.
Check: 214 + 72 = 286, 286 + 88 = 374. ✓

Cetking shortcut: Use the end: 374 − 286 = 88, so the difference before is 72 → 286 − 72 = 214.

Answer: C

Q109EasySimplification
Find the value of (1 − 1/3)(1 − 1/4)(1 − 1/5)...(1 − 1/100)
  1. 99/100
  2. 2/100
  3. 1
  4. 101/100
  5. 101/100
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Answer: B official key

Approach: Telescoping product: (1 − 1/n) = (n − 1)/n.

Solution:
Product = (2/3)(3/4)(4/5)…(99/100).
Every numerator cancels the previous denominator.
Only the first numerator (2) and last denominator (100) remain → 2/100 (= 1/50).

Why the other options are wrong:

  • A: 99/100 is just the last factor.
  • C: the product of fractions less than 1 cannot be 1.
  • D, E: 101/100 is greater than 1 — impossible for a product of proper fractions (D and E are identical in the paper).

Cetking shortcut: First numerator over last denominator.

Answer: B

Q110EasyAlgebra

If

x + 1/x = 2,

then the value of

(2x² + 2)/(3x² + 5x + 3)

is:

  1. 4/11
  2. 1/2
  3. 1 (3/4)
  4. 2 (1/3)
  5. 1
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Answer: A official key

Approach: x + 1/x = 2 forces x = 1 (for real x).

Solution:
x + 1/x = 2 → x² − 2x + 1 = 0 → (x − 1)² = 0 → x = 1.
(2x² + 2)/(3x² + 5x + 3) = (2 + 2)/(3 + 5 + 3) = 4/11.

Cetking shortcut: Divide numerator and denominator by x: 2(x + 1/x) / (3(x + 1/x) + 5) = 4/(6 + 5) = 4/11.

Answer: A

Q111EasySurds & Indices

Given that

√13 = 3.605 and √130 = 11.40

find the value of

√1.3 + √1300 + √0.013

  1. 36.164
  2. 36.304
  3. 37.164
  4. 37.304
  5. 38.304
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Answer: D official key

Approach: Shift the decimal in pairs: √(13 × 100ⁿ) and √(130 × 100ⁿ) only move the decimal point.

Solution:
√1.3 = √(130/100) = √130 / 10 = 11.40 / 10 = 1.140.
√1300 = √(13 × 100) = 10√13 = 36.05.
√0.013 = √(130/10000) = √130 / 100 = 0.114.
Sum = 1.140 + 36.05 + 0.114 = 37.304.

Cetking shortcut: Count digits in pairs from the decimal point: 1.3 → 130 family, 1300 → 13 family, 0.013 → 130 family. No √10 needed.

Answer: D

Q112EasyAverages
The average of the first ten odd prime numbers excluding the smallest and the largest among them is
  1. 15.5
  2. 16.2
  3. 17.5
  4. 18.4
  5. 19.2
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Answer: A official key

Approach: List the primes, drop the end values, then average.

Solution:
First ten odd primes: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.
Exclude smallest 3 and largest 31.
Sum of remaining 8 = 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 124.
Average = 124 / 8 = 15.5.

Cetking shortcut: Total of all ten = 158; 158 − 34 = 124; 124/8 = 15.5.

Answer: A

Q113EasyAverages
The average salary per head of all the workers in a company is ₹9000 The average salary of 15 officers is ₹5000 and the average salary per head of the remaining workers is ₹10,000 Find the total number of workers in the company.
  1. 75
  2. 80
  3. 50
  4. 40
  5. 30
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Answer: A official key

Approach: Weighted average / alligation of two groups.

Solution:
Let total workers = n.
9000n = 15 × 5000 + (n − 15) × 10000.
9000n = 75000 + 10000n − 150000 → 1000n = 75000 → n = 75.

Cetking shortcut: Alligation: officers are 4000 below average, others 1000 above → officers : others = 1 : 4 → 15 : 60 → total 75.

Answer: A

Q114EasyRatios

If A : B = 3 : 4
B : C = 8 : 10
C : D = 15 : 17

and if A+B+C+D=212

then the value of D is ?

  1. 68
  2. 65
  3. 76
  4. 85
  5. 90
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Answer: A official key

Approach: Chain the ratios by making the common term equal.

Solution:
A : B = 3 : 4 = 6 : 8 and B : C = 8 : 10 → A : B : C = 6 : 8 : 10.
C : D = 15 : 17; make C = 30 (LCM of 10 and 15): A : B : C = 18 : 24 : 30 and C : D = 30 : 34.
A : B : C : D = 18 : 24 : 30 : 34, total 106 parts.
1 part = 212 / 106 = 2 → D = 34 × 2 = 68.

Cetking shortcut: 212 = 2 × 106, so the answer must be 2 × (D's parts); only 68 = 2 × 34 fits once C is scaled to 30.

Answer: A

Q115EasyRatios
Mr Sharma owns 2505 rice bags and distributes them among his three sons Aman Rohit and Suresh in a certain ratio Out of the total bags received by each of them Aman sells 30 bags Rohit donates 30 bags and Suresh loses 25 bags Now the ratio of bags with them is 46 : 41 : 34 respectively How many bags did Rohit receive from his father.
  1. 840
  2. 850
  3. 870
  4. 890
  5. 900
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Answer: B official key

Approach: Write the final quantities as ratio multiples k, then add back what was removed.

Solution:
After changes: Aman = 46k, Rohit = 41k, Suresh = 34k.
Original: Aman = 46k + 30, Rohit = 41k + 30, Suresh = 34k + 25.
Total: 121k + 85 = 2505 → 121k = 2420 → k = 20.
Rohit received 41 × 20 + 30 = 850.

Cetking shortcut: 2505 − 85 = 2420 is divisible by 121 (= 46 + 41 + 34) → k = 20 at once.

Answer: B

Q116EasyAges
Five years ago, the average age of P and Q was 15 years. After 5 more years, the average age of P, Q, and R will be 25 years. If R is currently 5 years younger than Q, what will be the age of R after 10 years?
  1. 20
  2. 25
  3. 30
  4. 35
  5. 40
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Official CET key: C · Cetking's answer: E — see why below

Approach: Convert every average into a total at a fixed time point, then shift ages carefully.

Solution:
5 years ago: P + Q = 2 × 15 = 30 → now P + Q = 40.
5 years from now: (P + Q + 10) + (R + 5) = 3 × 25 = 75.
50 + R + 5 = 75 → R = 20 (present age).
Check with R = Q − 5: Q = 25, P = 15 → consistent.
R after 10 years = 20 + 10 = 30.

Why the other options are wrong:

  • E (40): does not satisfy the totals; R is 20 now, so 30 after 10 years.
  • A (20): R's present age, not after 10 years.
  • B, D: do not satisfy the totals.

Cetking shortcut: Total of P, Q at +5 years = 30 + 2 × 10 = 50; three-person total 75 → R at +5 = 25 → R at +10 = 30.

Answer: C

Q117EasyPercentages
An engineering college has only four batches that contain 20, 40, 60 and 80 students respectively. The pass percentage of these classes, are 10%, 20%, 30% and 40% respectively. Find the pass percentage of the entire college.
  1. 25%
  2. 30%
  3. 28%
  4. 32%
  5. 35%
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Answer: B official key

Approach: Overall pass % = total passed / total students (weighted, not simple average).

Solution:
Total students = 20 + 40 + 60 + 80 = 200.
Passed = 10% of 20 + 20% of 40 + 30% of 60 + 40% of 80 = 2 + 8 + 18 + 32 = 60.
Pass % = 60/200 × 100 = 30%.

Cetking shortcut: Weights 1 : 2 : 3 : 4 → (10 + 40 + 90 + 160)/10 = 30%.

Answer: B

Q118EasyMixtures
Anita mixes two mixtures of milk and water. She mixes 40 litres of the first mixture containing 20% water and 60 litres of the second mixture containing 30% water. Find the percentage of water in the final mixture.
  1. 24%
  2. 26%
  3. 28%
  4. 30%
  5. 32%
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Answer: B official key

Approach: Add the actual water quantities and divide by total volume.

Solution:
Water in first = 20% of 40 = 8 L.
Water in second = 30% of 60 = 18 L.
Total water = 26 L in 100 L → 26%.

Cetking shortcut: Weighted average: (40 × 20 + 60 × 30)/100 = 26%.

Answer: B

Q119EasyPercentages
A machine costs ₹3,20,000 at present. If its value depreciates at the rate of 5% per annum, what will be the value of the machine after 6 years?
  1. ₹2,45,760
  2. ₹2,52,000
  3. ₹2,35,230
  4. ₹2,30,400
  5. ₹2,60,000
Show answerHide answer

Answer: C official key

Approach: Depreciation: Value = P(1 − r/100)ⁿ.

Solution:
Value = 3,20,000 × (0.95)⁶.
(0.95)² = 0.9025; (0.95)³ = 0.857375; (0.95)⁶ = 0.857375² ≈ 0.73509.
Value ≈ 3,20,000 × 0.73509 ≈ ₹2,35,229 ≈ ₹2,35,230.

Cetking shortcut: Binomial: (1 − 0.05)⁶ ≈ 1 − 0.30 + 0.0375 − 0.0025 ≈ 0.735 → 3,20,000 × 0.735 ≈ 2,35,200 → C.

Answer: C

Q120MediumPercentages
The price of raw materials has increased by 15%. Labour cost has also increased from 25% of the cost of raw material to 30% of the cost of raw material. By how much percentage should the usage of raw materials be reduced so as to keep the total cost the same?
  1. 15.2 %
  2. 16.4 %
  3. 18.6 %
  4. 19.8 %
  5. 20 %
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Answer: B official key

Approach: Fix the old total cost; let usage of raw material become a fraction u and keep total cost equal.

Solution:
Old: raw material cost = 100, labour = 25% of 100 = 25, total = 125.
New price of raw material = 115 per old unit of usage; labour = 30% of raw material cost.
New total = 115u + 0.30 × 115u = 149.5u.
149.5u = 125 → u = 0.8361.
Reduction in usage = (1 − 0.8361) × 100 ≈ 16.4%.

Cetking shortcut: Reduction = 1 − 125/149.5 = 24.5/149.5 ≈ 16.4%.

Answer: B

Q121EasyProfit Loss
A dishonest shopkeeper uses a 900 gram weight instead of a 1 kilogram weight. He sells goods per kilogram at the same price at which he buys one kilogram. What is his profit percentage?
  1. 10%
  2. 11 (1/9)%
  3. 12 (1/2)%
  4. 15%
  5. 9 %
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Answer: B official key

Approach: False weight: Profit % = error / (true value given) × 100.

Solution:
Let CP of 1 kg = ₹100.
He gives 900 g (cost ₹90) but charges ₹100.
Profit = 10 on cost 90 → 10/90 × 100 = 11 1/9 %.

Cetking shortcut: Profit % = shortfall / weight actually given = 100/900 = 1/9 = 11.11%.

Answer: B

Q122EasyProfit Loss
A person sold an article for ₹3600 and got a profit of 20%. Had he sold the article for ₹3150, how much profit would he have got?
  1. 4 %
  2. 5 %
  3. 6 %
  4. 7 %
  5. 8 %
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Answer: B official key

Approach: Find CP from the first sale, then profit % for the new SP.

Solution:
CP = 3600 / 1.2 = ₹3000.
New SP = 3150 → profit = 150.
Profit % = 150/3000 × 100 = 5%.

Cetking shortcut: 3150/3600 = 7/8 of the old SP → 1.2 × 7/8 = 1.05 → 5% profit.

Answer: B

Q123EasyInterest
A certain sum amounts to ₹575 at 5% simple interest in the same time in which ₹750 amounts to ₹840 at 4% simple interest. What is the original sum?
  1. ₹480
  2. ₹500
  3. ₹520
  4. ₹540
  5. ₹560
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Answer: B official key

Approach: Use the second loan to find time, then reverse simple interest for the first.

Solution:
Second case: SI = 840 − 750 = 90 = 750 × 4 × t/100 → t = 3 years.
First case: Amount = P(1 + 5 × 3/100) = 1.15P = 575.
P = 575 / 1.15 = ₹500.

Cetking shortcut: Test options: 500 × 1.15 = 575 ✓ directly.

Answer: B

Q124EasyTrains Boats
A train crosses a pole in 8 seconds. If the length of the train is 200 metres, find the speed of the train.
  1. 18 km/h
  2. 72 km/h
  3. 90 km/h
  4. 25 m/s
  5. 30 m/s
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Answer: C official key

Approach: Crossing a pole: distance = length of the train.

Solution:
Speed = 200 m / 8 s = 25 m/s.
In km/h: 25 × 18/5 = 90 km/h.

Why the other options are wrong:

  • A, B: wrong conversions (18 and 72 km/h = 5 and 20 m/s).
  • D: 25 m/s is numerically the same speed as 90 km/h; the official key takes the km/h form.
  • E: 30 m/s = 108 km/h.

Cetking shortcut: m/s → km/h: multiply by 3.6.

Answer: C

Q125EasyTime Speed Distance
A person covers half of his journey at 30 km/h, one-third at 40 km/h, and the remaining part at 20 km/h.What is his average speed for the whole journey?
  1. 26 km/h
  2. 27 km/h
  3. 28 km/h
  4. 30 km/h
  5. 32 km/h
Show answerHide answer

Answer: D official key

Approach: Average speed = total distance / total time (not the mean of speeds).

Solution:
Take total distance = 6 units (LCM of 2 and 3): parts 3, 2, 1.
Time = 3/30 + 2/40 + 1/20 = 0.1 + 0.05 + 0.05 = 0.2.
Average speed = 6 / 0.2 = 30 km/h.

Cetking shortcut: Better: take distance = 120 km → 60/30 + 40/40 + 20/20 = 2 + 1 + 1 = 4 h → 120/4 = 30.

Answer: D

Q126EasyPipes and Cisterns
Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes?
  1. 8 minutes
  2. 9 minutes
  3. 7 minutes
  4. 15 minutes
  5. 24 minutes
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Answer: A official key

Approach: Work done by each pipe = rate × time it stays open.

Solution:
A works all 18 min: 18/24 = 3/4 of the tank.
B must fill the remaining 1/4: x/32 = 1/4 → x = 8 min.
B should be closed after 8 minutes.

Cetking shortcut: Remaining fraction × B's time = 1/4 × 32 = 8.

Answer: A

Q127EasyPipes and Cisterns
A cistern is normally filled in 6 hours but takes 4 hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in how much time?
  1. 15 hours
  2. 16 hours
  3. 12 hours
  4. 20 hours
  5. 24 hours
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Answer: A official key

Approach: Leak rate = filling rate without leak − net rate with leak.

Solution:
Pipe alone: 1/6 per hour; with leak the tank takes 10 h → 1/10 per hour.
Leak rate = 1/6 − 1/10 = (5 − 3)/30 = 1/15.
The leak empties a full tank in 15 hours.

Cetking shortcut: Leak time = (6 × 10)/(10 − 6) = 60/4 = 15 h.

Answer: A

Q128MediumTime & Work
Three diggers dug a ditch of 324 m deep in six days working simultaneously. During one shift, the third digger digs as many metres more than the second as the second digs more than the first. The third digger’s work in 10 days is equal to the first digger’s work in 14 days. How many metres does the first digger dig per shift?
  1. 15 m
  2. 18 m
  3. 21 m
  4. 12 m
  5. 30 m
Show answerHide answer

Answer: A official key

Approach: Rates in AP: first = x, second = x + d, third = x + 2d; use the 10-day/14-day equality.

Solution:
10(x + 2d) = 14x → 20d = 4x → x = 5d.
Rates: 5d, 6d, 7d → together 18d per day (one shift per day).
6 days: 6 × 18d = 108d = 324 → d = 3.
First digger = 5d = 15 m per shift.

Why the other options are wrong:

  • B (18 m): second digger's rate (6d).
  • C (21 m): third digger's rate (7d).
  • D, E: do not satisfy 10 × third = 14 × first.

Cetking shortcut: Third : first = 14 : 10 = 7 : 5 → rates 5 : 6 : 7; daily total 324/6 = 54 = 18 parts → 1 part = 3 → first = 15.

Answer: A

Q129EasyTrains Boats
If the speed of a boat in still water is 8 km/h and the rate of stream is 4 km/h, what is the upstream speed of the boat?
  1. 12 km/h
  2. 8 km/h
  3. 6 km/h
  4. 4 km/h
  5. 3 km/h
Show answerHide answer

Answer: D official key

Approach: Upstream speed = boat speed − stream speed.

Solution:
Upstream = 8 − 4 = 4 km/h.

Why the other options are wrong:

  • A: 12 km/h is the downstream speed (8 + 4).

Answer: D

Q130MediumClocks & Calendars
In a 24-hour period, the hour and minute hands of a clock are at right angles to each other excluding all positions where the hands overlap or are directly opposite. How many such right-angle positions occur?
  1. 40
  2. 42
  3. 44
  4. 46
  5. 48
Show answerHide answer

Answer: C official key

Approach: In 12 hours the hands are at right angles 22 times (not 24); double it for a day.

Solution:
In 12 hours the minute hand gains 11 full rounds on the hour hand.
Each round produces 2 right angles → 22 right angles in 12 hours.
In 24 hours: 22 × 2 = 44.
Overlaps and straight-line positions are separate events and are not counted anyway.

Why the other options are wrong:

  • E (48): assumes 2 per hour × 24 — ignores that the hands gain only 22 rounds in a day.
  • A, B, D: miscounts.

Cetking shortcut: Right angles = 44, straight (opposite) = 22, coincide = 22 in a day — memorise this triplet.

Answer: C

Q131EasyClocks & Calendars
In the year 2026, Republic Day of India falls on a Monday. On which day of the week will Children’s Day in India fall in the same year?
  1. Thursday
  2. Friday
  3. Saturday
  4. Sunday
  5. Monday
Show answerHide answer

Answer: C official key

Approach: Count odd days from 26 January to 14 November (2026 is not a leap year).

Solution:
Odd days after 26 Jan: Jan 5, Feb 0, Mar 3, Apr 2, May 3, Jun 2, Jul 3, Aug 3, Sep 2, Oct 3, Nov 14 → 0.
Total = 5 + 0 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 0 = 26 → 26 mod 7 = 5.
(Check: 292 days = 41 weeks + 5 days.)
Monday + 5 = Saturday.

Cetking shortcut: Month codes: 26 Jan Monday → 14 Nov is 292 days later; 292 = 287 + 5 → Monday + 5 = Saturday.

Answer: C

Q132EasyProbability
A bag contains 5 red and 7 green balls. Two balls are drawn one after the other, with replacement of the first ball before drawing the second. What is the probability that both balls drawn are red?
  1. 5/12
  2. 25/144
  3. 5/24
  4. 35/144
  5. 1/2
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Answer: B official key

Approach: With replacement the draws are independent: multiply probabilities.

Solution:
P(red) = 5/12 on each draw.
P(both red) = 5/12 × 5/12 = 25/144.

Why the other options are wrong:

  • A: probability of a single red.
  • C, D: mix of without-replacement / red-green products (35/144 = one red and one green in a fixed order).

Answer: B

Q133EasyProbability
In a class, 30% of the students passed Mathematics, 20% passed Science, and 10% passed both subjects. If a student is selected at random, what is the probability that the student has passed Mathematics or Science?
  1. 2/5
  2. 3/4
  3. 5/3
  4. 3/10
  5. 1/2
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Answer: A official key

Approach: Addition rule: P(M ∪ S) = P(M) + P(S) − P(M ∩ S).

Solution:
P(M ∪ S) = 0.30 + 0.20 − 0.10 = 0.40 = 2/5.

Why the other options are wrong:

  • B, E: too large; would require fewer students in both.
  • C: 5/3 > 1 — impossible as a probability.
  • D: 3/10 is P(Maths) alone.

Answer: A

Q134MediumMensuration
The volume V of a pyramid varies jointly as its height h and the area of its base .A pyramid has a base area of 48 dm² and a height of 150 cm, giving a volume of 240 dm³.Find the area of the base (in cm²) of another pyramid whose volume is 420 dm³ and whose height is 180 cm.
  1. 4200 cm²
  2. 7000 cm²
  3. 7200 cm²
  4. 8400 cm²
  5. 5600 cm²
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Answer: B official key

Approach: Joint variation: V = kAh. Keep all units in dm.

Solution:
150 cm = 15 dm, 180 cm = 18 dm.
240 = k × 48 × 15 → k = 240/720 = 1/3.
420 = (1/3) × A × 18 → A = 420/6 = 70 dm².
70 dm² = 70 × 100 cm² = 7000 cm².

Cetking shortcut: A₂ = A₁ × (V₂/V₁) × (h₁/h₂) = 48 × 7/4 × 5/6 = 70 dm²; 1 dm² = 100 cm².

Answer: B

Q135EasyMensuration
A tank measures 99 cm × 120 cm × 80 cm. It is to be filled using a cylindrical bucket of radius 12 cm and height 20 cm. If the bucket is filled up to 75% of its height each time, what is the minimum number of times the bucket must be used to completely fill the tank?
  1. 120
  2. 130
  3. 140
  4. 135
  5. 145
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Answer: C official key

Approach: Number of fills = tank volume / volume carried per fill (π = 22/7).

Solution:
Tank = 99 × 120 × 80 = 9,50,400 cm³.
Water per fill = π × 12² × (75% of 20 = 15) = (22/7) × 144 × 15 = 47520/7 cm³.
Fills = 950400 × 7 / 47520 = 140.

Cetking shortcut: 99 = 9 × 11 and 7 pairs with 22/7 — the numbers are built to cancel; 950400 ÷ 6788.57 = 140 exactly with π = 22/7.

Answer: C

Q136EasyMensuration
The diameters of two cones are equal. If their slant heights be in the ratio 5:7, find the ratio of their curved surface areas.
  1. 25 : 7
  2. 25 : 49
  3. 5 : 49
  4. 5 : 7
  5. 7 : 12
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Answer: D official key

Approach: Curved surface area of a cone = πrl.

Solution:
Equal diameters → equal r.
CSA₁ : CSA₂ = πrl₁ : πrl₂ = l₁ : l₂ = 5 : 7.

Why the other options are wrong:

  • B: squares the ratio (that would be for similar figures' areas).
  • A, C, E: no basis.

Answer: D

Q137EasyNumber Series (Quant)
Four of the following five number pairs are alike in a certain way. Which one does not belong to that group?

15 – 46, 12 – 37, 9 – 28, 8 – 33, 20 – 61

  1. 20–61
  2. 8–33
  3. 9–28
  4. 12–37
  5. 5 – 46
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Answer: B official key

Approach: Find the rule linking the two numbers of each pair: second = 3 × first + 1.

Solution:
15 → 46 (45 + 1), 12 → 37 (36 + 1), 9 → 28 (27 + 1), 20 → 61 (60 + 1).
8 → 3 × 8 + 1 = 25, but the pair shows 33 → odd one out.

Why the other options are wrong:

  • A, C, D, E: all follow ×3 + 1 (E is the stem's 15–46 pair, printed as '5 – 46' in the option).

Answer: B

Q138EasyNumber Series (Quant)
Four of the following five are alike in a certain way and so form a group. Which one of the following does not belong to that group?

123, 25, 341, 218, 62

  1. 62
  2. 218
  3. 341
  4. 25
  5. 123
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Answer: B official key

Approach: Compare each number with the nearby perfect cube.

Solution:
25 = 3³ − 2, 62 = 4³ − 2, 123 = 5³ − 2, 341 = 7³ − 2.
218 = 6³ + 2 → the only one with +2 → odd one out.

Why the other options are wrong:

  • A, C, D, E: all are n³ − 2.

Cetking shortcut: Numbers just below/above 27, 64, 125, 216, 343 → test cubes first.

Answer: B

Q139EasyPnC
Five men and four women are seated in a row such that all the four women always sit together.In how many different ways can they be seated?
  1. 720
  2. 4,320
  3. 8,640
  4. 17,280
  5. 34,560
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Answer: D official key

Approach: Women together → treat them as one block.

Solution:
Units: 5 men + 1 women-block = 6 → 6! = 720 ways.
Women inside the block: 4! = 24 ways.
Total = 720 × 24 = 17,280.

Why the other options are wrong:

  • A: forgets the internal 4! arrangement.
  • B, C: partial counts (e.g. 6! × 3!, 6! × 4!/2).
  • E: double-counts (×2).

Answer: D

Q140EasyPnC
How many words can be formed out of the letters of the word 'EDUCATION' such that vowels occupy the odd positions?
  1. 1440
  2. 2880
  3. 2840
  4. 2480
  5. 720
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Answer: B official key

Approach: Place vowels in their allowed positions, then consonants in the rest.

Solution:
EDUCATION: 9 distinct letters — vowels E, U, A, I, O (5), consonants D, C, T, N (4).
Odd positions 1, 3, 5, 7, 9 = 5 places → vowels: 5! = 120.
Remaining 4 even places → consonants: 4! = 24.
Total = 120 × 24 = 2880.

Why the other options are wrong:

  • A: halves the answer (counts only one group fully).
  • E: only 5! × 3! style partial count.
  • C, D: arithmetic misreads of 2880.

Cetking shortcut: 5 vowels exactly fill 5 odd slots → answer is simply 5! × 4!.

Answer: B

Q141EasyPie Chart

Questions 141–145 share this

Study the Pie Chart and answer the question that follow.

The pie chart represents the distribution of the annual agricultural yield of a certain place among different crops in terms of their central angles.

Data read from chart:

CropCentral angle
Rice108°
Wheat90°
Maize54°
Pulses36°
Sugarcane72°
Figure for MBA CET 2026 Slot 3 question 141
If the total annual agricultural production is 4800 tonnes, what is the difference between the production of Rice and Wheat?
  1. 120 tonnes
  2. 180 tonnes
  3. 240 tonnes
  4. 300 tonnes
  5. 360 tonnes
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Answer: C official key

Approach: Value = (central angle / 360°) × total.

Solution:
Rice − Wheat = 108° − 90° = 18°.
18/360 × 4800 = 4800/20 = 240 tonnes.

Cetking shortcut: Work with the angle difference directly: 1° = 4800/360 = 13.33 t → 18° = 240 t.

Answer: C

Q142MediumPie Chart

Refer to the passage above ↑

If the production of Sugarcane is increased by 20%, what will be its new percentage share (approx.) in total production, assuming others remain constant?
Figure for MBA CET 2026 Slot 3 question 142
  1. 20%
  2. 23%
  3. 25%
  4. 28 %
  5. 40 %
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Answer: B official key

Approach: Increase one sector and recompute its share over the NEW total.

Solution:
Sugarcane = 72/360 = 20% of total (take total = 100).
After 20% increase: 20 × 1.2 = 24; new total = 100 + 4 = 104.
New share = 24/104 × 100 ≈ 23.08% ≈ 23%.

Why the other options are wrong:

  • A: ignores the increase.
  • C: uses a 25% increase and/or forgets the total grows.
  • D, E: overstated shares.

Cetking shortcut: If a part p% grows by r%, new share = p(1 + r)/(100 + pr) → 24/104.

Answer: B

Q143EasyPie Chart

Refer to the passage above ↑

If Rice production decreases by 10% and Wheat production increases by 10%, which crop will have the highest production?
Figure for MBA CET 2026 Slot 3 question 143
  1. Rice
  2. Wheat
  3. Sugarcane
  4. Maize
  5. Pulses
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Answer: B official key

Approach: Apply the changes to the sector angles and compare.

Solution:
Rice: 108 × 0.9 = 97.2°.
Wheat: 90 × 1.1 = 99°.
Others unchanged: Sugarcane 72°, Maize 54°, Pulses 36°.
Wheat (99°) is now the highest.

Why the other options are wrong:

  • A: Rice drops to 97.2°, below Wheat.
  • C, D, E: 72°, 54°, 36° are far below.

Answer: B

Q144EasyPie Chart

Refer to the passage above ↑

The production of Rice, Wheat, and Sugarcane together exceeds the production of Maize and Pulses by what ratio?
Figure for MBA CET 2026 Slot 3 question 144
  1. 2 : 1
  2. 3 : 1
  3. 4 : 1
  4. 5 : 2
  5. 7 : 3
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Answer: B official key

Approach: Compare the summed angles of the two groups.

Solution:
Rice + Wheat + Sugarcane = 108 + 90 + 72 = 270°.
Maize + Pulses = 54 + 36 = 90°.
Ratio = 270 : 90 = 3 : 1.

Why the other options are wrong:

  • A (2 : 1): ratio of the excess (270 − 90 = 180) to 90 — the question asks for the ratio of the two totals.
  • C, D, E: do not match 270 : 90.

Cetking shortcut: The two groups make the whole 360°: 90° vs 270° → 1 : 3 instantly.

Answer: B

Q145EasyPie Chart

Refer to the passage above ↑

If the production of Pulses is decreased by 20% and the production of Sugarcane is increased by 20%, while the total production remains unchanged, which of the following will be true?
Figure for MBA CET 2026 Slot 3 question 145
  1. Pulses will have the lowest production
  2. Sugarcane will have higher production than Wheat
  3. Sugarcane will have equal production as Rice
  4. Pulses will have higher production than Maize
  5. Rice will lose its top position
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Answer: A official key

Approach: Apply the changes to the angles and test each statement.

Solution:
Pulses: 36 × 0.8 = 28.8°. Sugarcane: 72 × 1.2 = 86.4°.
Others: Rice 108°, Wheat 90°, Maize 54°.
Pulses (28.8°) is still the lowest → A is true.

Why the other options are wrong:

  • B: Sugarcane 86.4° < Wheat 90°.
  • C: 86.4° ≠ Rice 108°.
  • D: Pulses 28.8° < Maize 54°.
  • E: Rice 108° remains on top.

Cetking shortcut: Pulses was already the smallest and it only falls — A is true without calculating.

Answer: A

Q146EasyBar Graph

Questions 146–150 share this

Study the chart carefully and answer the questions that follow.

Chart title: Expenses of Company (stacked bar chart; legend: Salaries, Training, Overhead, Promotion, Other Expenses; years 2021 to 2025).

The given stacked bar chart shows the annual expenditure (in ₹ lakhs) of a company under five heads—Salaries, Training, Overhead, Promotion, and Other Expenses—for the years 2021 to 2025.

Data read from chart (₹ lakhs, segment values estimated from stack boundaries):

YearSalariesTrainingOverheadPromotionOther ExpensesTotal
2021205510545
202230105151575
20233015551065
2024155551040
202530101010565
Figure for MBA CET 2026 Slot 3 question 146
The average annual expenditure on Training from 2021 to 2025 is approximately what percentage of the average annual Salaries expenditure for the same period?
  1. 36 %
  2. 32 %
  3. 46%
  4. 50%
  5. 54%
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Answer: A official key

Approach: Ratio of averages over the same period = ratio of totals.

Solution:
Training: 5 + 10 + 15 + 5 + 10 = 45 → average 9.
Salaries: 20 + 30 + 30 + 15 + 30 = 125 → average 25.
9/25 × 100 = 36%.

Cetking shortcut: Skip averages: 45/125 = 36%.

Answer: A

Q147EasyBar Graph

Refer to the passage above ↑

The difference between Salaries and Training expenses was maximum in which year?
Figure for MBA CET 2026 Slot 3 question 147
  1. 2022 and 2025
  2. 2021 and 2024
  3. 2022 and 2024
  4. 2021 and 2023
  5. 2023 and 2025
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Answer: A official key

Approach: Read the Salaries and Training segments for each year and subtract.

Solution:
2021: 20 − 5 = 15; 2022: 30 − 10 = 20; 2023: 30 − 15 = 15; 2024: 15 − 5 = 10; 2025: 30 − 10 = 20.
Maximum difference 20 occurs in 2022 and 2025.

Why the other options are wrong:

  • B: 2021 (15), 2024 (10).
  • C: 2024 is the minimum.
  • D: 15 each.
  • E: 2023 gives only 15.

Answer: A

Q148EasyBar Graph

Refer to the passage above ↑

Based on the given data, in which year was the total expenditure highest?
Figure for MBA CET 2026 Slot 3 question 148
  1. 2021
  2. 2022
  3. 2023
  4. 2024
  5. 2025
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Answer: B official key

Approach: Total expenditure = height of the full stacked bar.

Solution:
Totals: 2021 = 45, 2022 = 75, 2023 = 65, 2024 = 40, 2025 = 65.
Highest = 2022 (75).

Cetking shortcut: Just compare bar tops — no segment reading needed.

Answer: B

Q149EasyBar Graph

Refer to the passage above ↑

The total expenditure on Training is equal to the total expenditure of which of the following years?
Figure for MBA CET 2026 Slot 3 question 149
  1. 2023
  2. 2024
  3. 2022
  4. 2021
  5. 2025
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Answer: D official key

Approach: Sum the Training segments and match with a year's bar total.

Solution:
Training total = 5 + 10 + 15 + 5 + 10 = 45.
Year totals: 2021 = 45, 2022 = 75, 2023 = 65, 2024 = 40, 2025 = 65.
45 matches 2021.

Answer: D

Q150MediumBar Graph

Refer to the passage above ↑

Considering only the years in which Training expenditure exceeds Promotion expenditure by at least ₹10 lakhs, what is the total expenditure on Overhead in those years?
Figure for MBA CET 2026 Slot 3 question 150
  1. ₹20 lakhs
  2. ₹30 lakhs
  3. ₹ 5 lakhs
  4. ₹10 lakhs
  5. ₹ 25 lakhs
Show answerHide answer

Official CET key: C · Cetking's answer: B — see why below

Approach: Filter the years by the condition, then add Overhead for those years.

Solution:
Training − Promotion: 2021: 5 − 10 = −5; 2022: 10 − 15 = −5; 2023: 15 − 5 = 10; 2024: 5 − 5 = 0; 2025: 10 − 10 = 0.
Only 2023 satisfies 'at least ₹10 lakh'.
Overhead in 2023 = 50 − 45 = ₹5 lakh.

Why the other options are wrong:

  • A, B, E: add Overhead of years that fail the condition (e.g. all five years = 30).
  • D: misreads the 2023 Overhead segment as 10 (or adds 2025's 10).

Cetking shortcut: Only one bar (2023) has a tall Training and a thin Promotion segment — read just that bar.

Answer: C

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