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MBA CET 2026 Slot 1 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.

Q101EasySequences and Series

What is the value of
(37+38+39+···+86) ?

  1. 3075
  2. 3050
  3. 3070
  4. 3010
  5. 3087
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Answer: A official key

Approach: Sum of an AP = number of terms × average of first and last term.

Solution:
Number of terms = 86 − 37 + 1 = 50
Average = (37 + 86)/2 = 123/2 = 61.5
Sum = 50 × 61.5 = 3075

Cetking shortcut: 50 × 61.5 = 25 × 123 = 3075 — pairing terms (37+86 = 123, 25 pairs) gives the same result.

Answer: A

Q102EasyNumbers

The number 63_42_0 is a 7-digit number divisible by both 3 and 5. If the thousands digit is greater than the tens digit, then which of the following options gives the correct digits in the thousands place and the tens place respectively?

  1. 3,1
  2. 6,4
  3. 9,7
  4. 9,6
  5. 3,9
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Answer: D official key

Approach: Divisibility by 5 (last digit 0/5) and by 3 (digit sum multiple of 3), then the given order condition.

Solution:
Number: 6 3 _ 4 2 _ 0 → ends in 0, so divisible by 5 already.
Known digit sum = 6 + 3 + 4 + 2 + 0 = 15, so (first blank + second blank) must be a multiple of 3.
Check options: 3+1 = 4 ✗, 6+4 = 10 ✗, 9+7 = 16 ✗, 9+6 = 15 ✓, 3+9 = 12 ✓
Condition: first blank (called the thousands digit) > second blank (called the tens digit). 9 > 6 ✓, 3 > 9 ✗.
So the digits are 9 and 6.

Why the other options are wrong:
A (3,1), B (6,4), C (9,7): digit sum 19, 25, 31 — not divisible by 3.
E (3,9): divisible by 3, but 3 < 9 breaks the order condition.

Cetking shortcut: Since 15 is already a multiple of 3, just pick the pair whose sum is a multiple of 3 with the first digit larger.

Answer: D

Q103EasySequences and Series

What is the sum of the first 60 natural numbers?

  1. 1850
  2. 1890
  3. 1780
  4. 1875
  5. 1830
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Answer: E official key

Approach: Sum of first n natural numbers = n(n + 1)/2.

Solution:
n = 60
Sum = 60 × 61 / 2 = 30 × 61 = 1830

Cetking shortcut: Sum must be a multiple of 61 (and of 30): only 1830 = 30 × 61.

Answer: E

Q104EasyHCF LCM

Three positive integers are such that the difference between the largest and the smallest number is 180.
The HCF of the three numbers is the greatest possible, and the smallest number is 90.
If the middle number differs from the smallest number by 60, what is the HCF of the three numbers?

  1. 14
  2. 21
  3. 28
  4. 30
  5. 35
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Answer: D official key

Approach: The HCF of numbers also divides every difference between them.

Solution:
Smallest = 90, largest = 90 + 180 = 270, middle = 90 + 60 = 150
HCF(90, 150, 270): 90 = 2·3²·5, 150 = 2·3·5², 270 = 2·3³·5
Common part = 2 × 3 × 5 = 30
Check via differences: HCF(60, 180, 90) = 30 ✓

Why the other options are wrong:
14, 21, 28, 35 contain a factor of 7, which does not divide 90.

Cetking shortcut: HCF must divide 90, 60 and 180 — among the options only 30 divides all three.

Answer: D

Q105EasyHCF LCM

Three positive integers are in the ratio 3 : 5 : 7. The difference between the largest and the middle number is 56. The LCM of the smallest and the largest number is 588. What is the sum of the three numbers?

  1. 430
  2. 580
  3. 420
  4. 349
  5. 575
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Answer: C official key

Approach: Write numbers as 3k, 5k, 7k; use the difference to find k, verify with LCM.

Solution:
7k − 5k = 2k = 56 → k = 28
Numbers: 84, 140, 196
Check LCM(84, 196): 84 = 2²·3·7, 196 = 2²·7² → LCM = 2²·3·7² = 588 ✓
Sum = 15k = 15 × 28 = 420

Cetking shortcut: Sum = 15k and 2k = 56 → sum = 15 × 28 directly; the LCM is only a check.

Answer: C

Q106EasyHCF LCM

Three bells ring at intervals of 30 seconds, 45 seconds, and 60 seconds.
After how much time will they ring together again?

  1. 90 seconds
  2. 180 seconds
  3. 120 seconds
  4. 240 seconds
  5. 300 seconds
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Answer: B official key

Approach: Bells ring together again after the LCM of their intervals.

Solution:
30 = 2·3·5, 45 = 3²·5, 60 = 2²·3·5
LCM = 2²·3²·5 = 180 seconds

Why the other options are wrong:
90 is not a multiple of 60; 120 not a multiple of 45; 240 not of 45; 300 not of 45.

Answer: B

Q107EasyAverages

There are two Sections, namely, C and D of a Class, consisting of 30 and 50 students respectively. If the average weight of students of Section C is 45 kg and that of Section D is 50 kg, then find the average weight of students of the Class overall ?

  1. 48.125 kg
  2. 49.250 kg
  3. 40.120 kg
  4. 41.105 kg
  5. 38.115 kg
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Answer: A official key

Approach: Weighted average = total weight / total students.

Solution:
Section C total = 30 × 45 = 1350 kg
Section D total = 50 × 50 = 2500 kg
Overall = (1350 + 2500)/80 = 3850/80 = 48.125 kg

Why the other options are wrong:
The average must lie between 45 and 50, and closer to 50 because Section D has more students — so C, D and E (all below 45) are impossible. B (49.25) would need D to be far larger than 50 students against 30.

Cetking shortcut: Deviation method: 45 + 5 × 50/80 = 45 + 3.125 = 48.125.

Answer: A

Q108EasyAverages

In an examination, the average marks of all students was 72. After correcting mistakes, the average marks of 40 students decreased from 80 to 60, while the marks of the remaining students remained unchanged. As a result, the overall average of all students became 70. Find the total number of students.

  1. 450
  2. 410
  3. 428
  4. 400
  5. 475
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Answer: D official key

Approach: Total change in marks = change in average × number of students.

Solution:
40 students' average falls by 80 − 60 = 20 → total falls by 40 × 20 = 800
Overall average falls by 72 − 70 = 2 → total falls by 2n
2n = 800 → n = 400

Cetking shortcut: n = (40 × 20)/2 = 400 in one line.

Answer: D

Q109EasyAverages

A company has 12 workers and 6 managers. The average salary of all 18 employees is Rs. 900.

  • One worker whose salary was Rs. 800 is replaced by a new worker.
  • One manager whose salary was Rs. 1200 is also replaced by a new manager.

After these changes, the average salary of all 18 employees becomes Rs. 910.

If the new worker earns Rs. 950, What is the salary of the new manager?

  1. Rs. 1324
  2. Rs. 1190
  3. Rs. 1280
  4. Rs. 1150
  5. Rs. 1230
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Answer: E official key

Approach: Change in total salary = change in average × number of employees.

Solution:
Old total = 18 × 900 = 16200; new total = 18 × 910 = 16380 → increase = 180
Worker change: 950 − 800 = +150
Manager change must be 180 − 150 = +30
New manager = 1200 + 30 = Rs. 1230

Cetking shortcut: Total rise 18 × 10 = 180; worker accounts for 150, manager +30.

Answer: E

Q110MediumAges

The present ages of A, B, and C are in the ratio 6 : 8 : 10. Five years ago, the sum of the ages of A and B was equal to the age of C at that time. Find the total present age of A, B, and C.

  1. 114
  2. 120
  3. 100
  4. 145
  5. 180
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Official CET key: B · Cetking's answer: A — see why below

Approach: Ages in ratio → write as 6k, 8k, 10k and apply the past-age condition.

Solution:
As printed: (6k − 5) + (8k − 5) = 10k − 5 → 14k − 10 = 10k − 5 → k = 5/4 → total 24k = 30, which is not among the options.
The official key (120) corresponds to k = 5: A = 30, B = 40, C = 50.
This k satisfies the condition when it reads 'ten years ago, the sum of the ages of A and B equalled C's present age': (30 − 10) + (40 − 10) = 50 = C ✓ → 14k − 20 = 10k → k = 5.
Total = 24k = 24 × 5 = 120

Why the other options are wrong:
A (114), C (100), D (145): not multiples of 24, so they cannot be 24k for a whole-number k.
E (180): k = 7.5 (ages 45, 60, 75) fits neither the printed nor the corrected condition.

Cetking shortcut: Total = 24k, so the answer must be a multiple of 24 — only 120 (and 180) qualify; test k = 5 first.

Answer: B

Q111EasySimplification

Simplify:
((3⁴ × 2³ − 5³)/7) + √256 − (2/5) × (25/4) + 0.75 × 8

  1. 90.53
  2. 91.25
  3. 98.75
  4. 94.21
  5. 98.50
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Answer: D official key

Approach: BODMAS — evaluate powers and roots, then division/multiplication, then addition/subtraction.

Solution:
3⁴ × 2³ = 81 × 8 = 648; 648 − 5³ = 648 − 125 = 523
523/7 = 74.714
√256 = 16
(2/5) × (25/4) = 50/20 = 2.5
0.75 × 8 = 6
Total = 74.714 + 16 − 2.5 + 6 = 94.214 ≈ 94.21

Cetking shortcut: Only 523/7 is non-terminating; the rest add to 16 − 2.5 + 6 = 19.5, so answer = 74.71 + 19.5 = 94.21.

Answer: D

Q112MediumAges

When Ravi was born:

  • His father was 30 years older than his brother.
  • His mother was 26 years older than his sister.
  • His brother is 4 years older than his sister.

After 6 years, the average age of the family (Ravi, his brother, his sister, his father, and his mother) shall be 22 years. What was the age of his mother when Ravi was born?

  1. 40 years
  2. 30 years
  3. 35 years
  4. 46 years
  5. 48 years
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Answer: B official key

Approach: Age differences never change; build every age from the sister's age at Ravi's birth.

Solution:
At Ravi's birth let sister = s. Then brother = s + 4, father = (s + 4) + 30 = s + 34, mother = s + 26, Ravi = 0.
Sum at Ravi's birth = 0 + s + (s + 4) + (s + 34) + (s + 26) = 4s + 64
6 years later each of the 5 members is 6 years older: sum = 4s + 64 + 30 = 4s + 94
Average 22 → sum = 5 × 22 = 110 → 4s + 94 = 110 → s = 4
Mother's age at Ravi's birth = s + 26 = 30 years

Cetking shortcut: Total after 6 years = 110 → total at birth = 110 − 30 = 80 = 4s + 64 → s = 4 → mother = 30.

Answer: B

Q113EasyPercentages

In a town of 2000 people, 60% are males. Out of the males, 40% are literate. If 55% of the total population is literate, what percentage of females is literate?

  1. 67.5%
  2. 72.5%
  3. 70.5%
  4. 77.5%
  5. 80.5%
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Answer: D official key

Approach: Split population into males/females and subtract literate males from total literates.

Solution:
Males = 60% of 2000 = 1200; females = 800
Literate males = 40% of 1200 = 480
Total literate = 55% of 2000 = 1100
Literate females = 1100 − 480 = 620
Female literacy = 620/800 × 100 = 77.5%

Cetking shortcut: Alligation: 0.6 × 40 + 0.4 × f = 55 → 24 + 0.4f = 55 → f = 77.5.

Answer: D

Q114EasyPercentages

Ramesh spends 25% of his income on rent, 30% of the remaining on food, and 20% of the remaining on transport. If he saves ₹21,000, find the amount spent on transport.

  1. ₹4,250
  2. ₹5,550
  3. ₹5,250
  4. ₹6,200
  5. ₹5,150
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Answer: C official key

Approach: Successive percentages apply to the remaining amount each time.

Solution:
After rent: 75% of I remains
Food = 30% of 0.75I = 0.225I → remaining 0.525I
Transport = 20% of 0.525I = 0.105I → savings = 0.42I
0.42I = 21000 → I = 50000
Transport = 0.105 × 50000 = ₹5,250

Cetking shortcut: Savings = 0.75 × 0.70 × 0.80 = 0.42 of income; transport = savings × 0.2/0.8 = 21000/4 = 5250.

Answer: C

Q115EasyPercentages

The salary of P is 60% more than Q, while the salary of R is 20% less than the total salary of P and Q together.
If Q's salary is ₹30,000, find the salary of R.

  1. ₹62,400
  2. ₹60,500
  3. ₹61,700
  4. ₹60,850
  5. ₹72,400
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Answer: A official key

Approach: Convert each 'more/less than' into a multiplier.

Solution:
Q = 30000; P = 1.6 × 30000 = 48000
P + Q = 78000
R = 80% of 78000 = 62400

Cetking shortcut: R = 0.8 × 2.6Q = 2.08 × 30000 = 62400.

Answer: A

Q116EasyProfit Loss

The marked price of an article is ₹2500. Two successive discounts of 10% and 5% are allowed. Even then, the shopkeeper earns a 20% profit.
Find the cost price of the article.

  1. ₹2137.50
  2. ₹2257.50
  3. ₹1668.50
  4. ₹1752.75
  5. ₹1781.25
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Answer: E official key

Approach: Find SP after successive discounts, then CP = SP / (1 + profit%).

Solution:
SP = 2500 × 0.90 × 0.95 = 2250 × 0.95 = 2137.50
SP = 1.2 × CP → CP = 2137.50/1.2 = ₹1781.25

Why the other options are wrong:
A (₹2137.50) is the selling price, not the cost price.

Cetking shortcut: CP = 2500 × 0.9 × 0.95 / 1.2 = 2500 × 0.7125 = 1781.25.

Answer: E

Q117EasyProfit Loss

A shopkeeper marks an article 40% above its cost price. He offers a two-stage discount — first a 10% discount, and then an additional 5% discount on the reduced price. Find his overall profit percentage if the cost price of the article is Rs. 200.

  1. 26.8%
  2. 30%
  3. 25.4%
  4. 19.7%
  5. 14.5%
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Answer: D official key

Approach: Apply mark-up then successive discounts as multipliers on CP.

Solution:
MP = 200 × 1.4 = 280
After 10%: 280 × 0.9 = 252
After 5%: 252 × 0.95 = 239.40
Profit = 239.40 − 200 = 39.40 → 39.40/200 = 19.7%

Cetking shortcut: 1.4 × 0.9 × 0.95 = 1.197 → 19.7% profit.

Answer: D

Q118MediumProfit Loss

A person bought a book and a novel for ₹2800. He sold the book at 10% profit and the novel at 15.25% profit. If his overall profit was 13%, then what is the difference between the cost prices of the book and the novel?

  1. Rs. 200
  2. Rs. 400
  3. Rs. 350
  4. Rs. 450
  5. Rs. 389
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Answer: B official key

Approach: Overall profit is a weighted average of the two profit rates (alligation).

Solution:
Let book = b, novel = 2800 − b
0.10b + 0.1525(2800 − b) = 0.13 × 2800 = 364
427 − 0.0525b = 364 → b = 1200; novel = 1600
Difference = 1600 − 1200 = ₹400

Cetking shortcut: Alligation: book : novel = (15.25 − 13) : (13 − 10) = 2.25 : 3 = 3 : 4 → 1200 : 1600 → difference = 2800/7 = 400.

Answer: B

Q119EasyPipes and Cisterns

Three pipes X, Y and Z can fill a tank completely in 24 minutes, 12 minutes and 8 minutes respectively.
All three pipes are opened together when the tank is empty.

  • Pipe X fills solution A
  • Pipe Y fills solution B
  • Pipe Z fills solution C

After 4 minutes, what is the proportion of solution C in the tank?

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

Approach: Each pipe's share in the tank is proportional to its filling rate.

Solution:
Rates per minute: X = 1/24, Y = 1/12 = 2/24, Z = 1/8 = 3/24
Ratio X : Y : Z = 1 : 2 : 3
Share of C (from Z) = 3/(1 + 2 + 3) = 3/6 = 1/2
(After 4 min: 4/24 + 8/24 + 12/24 = 1 → tank exactly full, no overflow issue.)

Cetking shortcut: Times 24 : 12 : 8 → rates 1 : 2 : 3; the 4 minutes do not matter.

Answer: A

Q120EasyPipes and Cisterns

Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern by itself?

  1. 8 hours
  2. 6 hours
  3. 10 hours
  4. 7 hours
  5. 4 hours
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Answer: B official key

Approach: Combined rate = sum of individual rates; form a quadratic in A's time.

Solution:
Let A take a hours, B take a + 6 hours
1/a + 1/(a + 6) = 1/4
4(2a + 6) = a(a + 6) → a² − 2a − 24 = 0
(a − 6)(a + 4) = 0 → a = 6 hours

Cetking shortcut: Option test: A = 6 → B = 12 → 1/6 + 1/12 = 3/12 = 1/4 ✓ — quick to verify.

Answer: B

Q121EasyPipes and Cisterns

Two outlet pipes P and Q are connected to a full tank. Pipe P alone can empty the tank in 12 minutes and pipe Q alone can empty it in 18 minutes.
If both pipes are opened together, how long will it take to empty the tank completely?

  1. 6.2 minutes
  2. 5.4 minutes
  3. 6.8 minutes
  4. 7.2 minutes
  5. 4.5 minutes
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Official CET key: D · Cetking's answer: B — see why below

Approach: Two outlets work together → add their emptying rates.

Solution:
P empties 1/12 per min, Q empties 1/18 per min
Together = 1/12 + 1/18 = 3/36 + 2/36 = 5/36 per min
Time = 36/5 = 7.2 minutes

Why the other options are wrong:
Time must lie between 6 (if both were as fast as P: 12/2) and 9 (if both were as slow as Q: 18/2) → B (5.4) and E (4.5) are impossible; 6.2 and 6.8 do not equal 36/5.

Cetking shortcut: Two jobs together: T = ab/(a + b) = 12 × 18 / 30 = 7.2.

Answer: D

Q122MediumPartnership

A and B start a business together.

  • A invests ₹1,20,000 at the beginning.
  • After 4 months, B joins the business with ₹90,000.
  • After another 4 months, A adds ₹30,000 to his investment, while B withdraws ₹30,000 from his capital.

At the end of 1 year, in what ratio should A and B share the profit?

  1. 10:7
  2. 9:8
  3. 13:5
  4. 14:9
  5. 14:5
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Official CET key: C · Cetking's answer: A — see why below

Approach: Profit share ∝ capital × time; split each partner's investment into periods.

Solution:
A: ₹1,20,000 for 8 months = 9,60,000; then ₹1,50,000 for 4 months = 6,00,000 → total 15,60,000
B: joins at month 4 with ₹90,000 for 4 months = 3,60,000; then ₹60,000 for 4 months = 2,40,000 → total 6,00,000
A : B = 15,60,000 : 6,00,000 = 156 : 60 = 13 : 5

Why the other options are wrong:
A (10:7), B (9:8), D (14:9), E (14:5) come from wrong periods — e.g. giving B 12 months or ignoring the mid-year top-up/withdrawal.

Cetking shortcut: Work in ₹10,000 units and months: A = 12×8 + 15×4 = 156, B = 9×4 + 6×4 = 60 → 13 : 5.

Answer: C

Q123EasyPartnership

A invests ₹6,000 in a business for 8 months, and B invests ₹9,000 for 12 months. If the total profit at the end of the year is ₹3,000, find A's share.

  1. ₹923.08
  2. ₹933.07
  3. ₹813.06
  4. ₹863.02
  5. ₹683.05
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Answer: A official key

Approach: Profit share ∝ capital × time.

Solution:
A = 6000 × 8 = 48,000; B = 9000 × 12 = 1,08,000
A : B = 48 : 108 = 4 : 9
A's share = 3000 × 4/13 = ₹923.08

Cetking shortcut: 13 parts of 3000 → one part ≈ 230.77; A = 4 parts = 923.08.

Answer: A

Q124EasyPartnership

A, B, and C start a business with capitals in the ratio 3 : 4 : 5.

  • A invests for 12 months,
  • B invests for 9 months,
  • C invests for 6 months.

If the total profit at the end of the year is ₹36,000, find C's share.

  1. ₹10,582.24
  2. ₹11,682.21
  3. ₹12,772.34
  4. ₹14,792.37
  5. ₹10,588.24
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Answer: E official key

Approach: Profit share ∝ capital ratio × months.

Solution:
A = 3 × 12 = 36, B = 4 × 9 = 36, C = 5 × 6 = 30
Total = 102 parts
C's share = 36000 × 30/102 = 36000 × 5/17 = ₹10,588.24

Why the other options are wrong:
A (₹10,582.24) is a near-miss decoy: 180000/17 = 10588.235…, so the digits must end in 588.

Cetking shortcut: 36 : 36 : 30 = 6 : 6 : 5 → C = 5/17 of 36000.

Answer: E

Q125EasyClocks & Calendars

A clock shows 7 O'clock in the morning. By how much angle will the hours hand rotate when the clock shows 11:30 O'clock in the morning?

  1. 40°
  2. 125°
  3. 135°
  4. 150°
  5. 90°
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Answer: C official key

Approach: The hour hand moves 30° per hour (0.5° per minute).

Solution:
From 7:00 to 11:30 = 4.5 hours
Rotation = 4.5 × 30° = 135°

Cetking shortcut: Hour hand: 0.5° per minute → 270 min × 0.5 = 135°.

Answer: C

Q126MediumTime & Work
A can complete a work in 12 days and B can complete the same work in 18 days. A and B start the work together, but after 4 days, A leaves the work. C then joins B and together they finish the remaining work in 5 days. If the total wages paid for the work is ₹7200 and wages are divided in proportion to the work done, What is the C's share of the wages?
  1. Rs. 1675
  2. Rs. 1079
  3. Rs. 1200
  4. Rs. 1145
  5. Rs. 1935
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Answer: C official key

Approach: Wages are split by the fraction of work each person does.

Solution:
A's rate = 1/12, B's rate = 1/18 per day
First 4 days (A + B): 4 × (1/12 + 1/18) = 4 × 5/36 = 5/9 done
Remaining work = 4/9, done by B + C in 5 days
B's part in those 5 days = 5/18
C's work = 4/9 − 5/18 = 8/18 − 5/18 = 3/18 = 1/6
C's wages = 7200 × 1/6 = ₹1200

Cetking shortcut: Only C's fraction matters: 4/9 − 5/18 = 1/6 → 7200/6.

Answer: C

Q127EasyTrains Boats
A train of length 100 metres is moving at a speed of 72 km/h. A man is walking in the same direction at a speed of 6 km/h. How much time will the train take to completely pass the man?
  1. 6.45 seconds
  2. 5.45 seconds
  3. 3.25 seconds
  4. 7.35 seconds
  5. 4.23 seconds
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Answer: B official key

Approach: Same direction → relative speed = difference; distance = train length.

Solution:
Relative speed = 72 − 6 = 66 km/h = 66 × 5/18 = 55/3 m/s
Time = 100 ÷ (55/3) = 300/55 = 5.45 s

Cetking shortcut: Time = length × 18/(5 × relative km/h) = 100 × 18/330 = 5.45 s.

Answer: B

Q128EasyTrains Boats
A train 150 m long is running at 54 km/h. It crosses a bridge in 20 seconds. Find the length of the bridge.
  1. 250 m
  2. 340 m
  3. 150 m
  4. 190 m
  5. 140 m
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Official CET key: C · Cetking's answer: A — see why below

Approach: Crossing a bridge: distance = train length + bridge length.

Solution:
54 km/h = 54 × 5/18 = 15 m/s
Distance in 20 s = 15 × 20 = 300 m
Bridge = 300 − 150 = 150 m

Why the other options are wrong:
A (250 m): not consistent — 150 + 250 = 400 m would need 26.7 s at 15 m/s.

Cetking shortcut: 15 m/s × 20 s = 300 m total; subtract the train.

Answer: C

Q129MediumTrains Boats
Two trains of length 350m and 250m completely cross each other when running in same and opposite direction in 10 sec. and 5 sec. respectively. Find the speed of faster train in kmph?
  1. 328 km/h
  2. 324 km/h
  3. 454 km/h
  4. 226 km/h
  5. 594 km/h
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Answer: B official key

Approach: Same direction → difference of speeds; opposite → sum; distance = sum of lengths.

Solution:
Total length = 350 + 250 = 600 m
Same direction: a − b = 600/10 = 60 m/s
Opposite direction: a + b = 600/5 = 120 m/s
Faster train a = (60 + 120)/2 = 90 m/s = 90 × 18/5 = 324 km/h

Cetking shortcut: Faster = half of (sum + difference) = 90 m/s → × 3.6 = 324 km/h.

Answer: B

Q130MediumMixtures
A 30-litre mixture of milk and water is in the ratio 2 : 1. 15 litres of the mixture is removed and replaced with pure milk. This process is repeated one more time. What is the final ratio of milk to water in the mixture?
  1. 10:1
  2. 12:5
  3. 8:3
  4. 11:1
  5. 15:7
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Answer: D official key

Approach: Track milk and water after each removal (half the mixture) and refill with milk.

Solution:
Start: milk 20 L, water 10 L
Remove 15 L (half): milk 10, water 5; add 15 milk → milk 25, water 5
Remove 15 L (half): milk 12.5, water 2.5; add 15 milk → milk 27.5, water 2.5
Milk : water = 27.5 : 2.5 = 11 : 1

Cetking shortcut: Water is never added, so it halves each time: 10 → 5 → 2.5. Milk = 30 − 2.5 = 27.5 → 11 : 1.

Answer: D

Q131EasyMixtures
Tea worth ₹120 per kg and ₹150 per kg are mixed with a third variety of tea in the ratio 1 : 2 : 3. If the cost price of the mixture is ₹165 per kg, what is the cost price per kg of the third variety of tea?
  1. ₹150 per kg
  2. ₹290 per kg
  3. ₹190 per kg
  4. ₹180 per kg
  5. ₹100 per kg
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Answer: C official key

Approach: Weighted average of prices with weights 1 : 2 : 3.

Solution:
(1 × 120 + 2 × 150 + 3 × x)/6 = 165
120 + 300 + 3x = 990 → 3x = 570 → x = 190

Cetking shortcut: Deviations from 165: −45 × 1 + (−15) × 2 = −75 must be balanced by 3(x − 165) = 75 → x = 190.

Answer: C

Q132EasyMixtures
A container contains 60 litres of milk. Some milk is replaced by water twice. Each time 12 litres of mixture is replaced. After the two replacements, What is the final ratio of milk to water?
  1. 16:9
  2. 18:7
  3. 17:3
  4. 11:4
  5. 15:4
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Answer: A official key

Approach: Repeated replacement: milk left = initial × (1 − x/V)ⁿ.

Solution:
Milk left = 60 × (1 − 12/60)² = 60 × (4/5)² = 60 × 16/25 = 38.4 L
Water = 60 − 38.4 = 21.6 L
Milk : water = 38.4 : 21.6 = 16 : 9

Cetking shortcut: Milk fraction = (4/5)² = 16/25 → milk : water = 16 : 9 directly.

Answer: A

Q133MediumMensuration
A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the cost of painting the toy at the rate of ₹5 per 1000 cm².
  1. ₹0.52
  2. ₹0.83
  3. ₹0.43
  4. ₹0.29
  5. ₹0.37
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Answer: A official key

Approach: Painted surface = curved surface of cone + curved surface of hemisphere (shared base not painted).

Solution:
r = 6/2 = 3 cm, cone height h = 4 cm → slant l = √(3² + 4²) = 5 cm
Cone CSA = πrl = π × 3 × 5 = 15π
Hemisphere CSA = 2πr² = 2π × 9 = 18π
Total = 33π = 33 × 3.1416 ≈ 103.67 cm²
Cost = 103.67 × 5/1000 = ₹0.518 ≈ ₹0.52

Cetking shortcut: 33π ≈ 103.7 cm²; ₹5 per 1000 cm² = ₹0.005 per cm² → ≈ ₹0.52.

Answer: A

Q134EasyMensuration
A triangular pennant has a base of 10 cm and a height of 8 cm. The same amount of cloth is used to make a rectangular banner whose length is 5 cm. Find the breadth of the rectangle.
  1. 10 cm
  2. 8 cm
  3. 12 cm
  4. 6 cm
  5. 14 cm
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Answer: B official key

Approach: Equal cloth → equal area: ½ × base × height = length × breadth.

Solution:
Triangle area = ½ × 10 × 8 = 40 cm²
Rectangle: 5 × b = 40 → b = 8 cm

Answer: B

Q135EasyMensuration
A cylindrical tank has a diameter of 10 m and a height of 6 m. Find the volume of water it can hold.
  1. 481.73 m³
  2. 351.49 m³
  3. 471.24 m³
  4. 341.74 m³
  5. 321.63 m³
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Answer: C official key

Approach: Volume of cylinder = πr²h.

Solution:
r = 10/2 = 5 m, h = 6 m
V = π × 25 × 6 = 150π = 150 × 3.1416 = 471.24 m³

Cetking shortcut: 150π: 150 × 3.14 = 471 → only 471.24 matches.

Answer: C

Q136MediumClocks & Calendars
What day of the week was 15th August, 2010?
  1. Sunday
  2. Monday
  3. Tuesday
  4. Thursday
  5. Saturday
Show answerHide answer

Answer: A official key

Approach: Odd-days method: count odd days from a known reference (multiple of 400 years = 0 odd days).

Solution:
Years 1–2000 complete → 0 odd days (2000 is a multiple of 400)
2001–2009: 9 years incl. 2 leap years (2004, 2008) → 9 + 2 = 11 → 4 odd days
2010 up to 15 Aug: Jan 3, Feb 0, Mar 3, Apr 2, May 3, Jun 2, Jul 3, Aug (15) 1 → 17 → 3 odd days
Total = 0 + 4 + 3 = 7 → 0 odd days → Sunday

Cetking shortcut: Anchor on Independence Day: 15 Aug 1947 was a Friday; 63 years later with 16 leap days (1948–2008) → 79 → 2 odd days → Friday + 2 = Sunday.

Answer: A

Q137MediumClocks & Calendars
If 15th August 2012 was a Wednesday, then what day of the week was 26th January 1998?
  1. Saturday
  2. Monday
  3. Tuesday
  4. Wednesday
  5. Friday
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Answer: B official key

Approach: Count odd days between the two dates and move backward from the known day.

Solution:
26 Jan 1998 → 26 Jan 2012 = 14 years with 3 leap days (Feb 2000, 2004, 2008) → 14 + 3 = 17 → 3 odd days
26 Jan 2012 → 15 Aug 2012: Jan 5, Feb 29 (leap) 1, Mar 3, Apr 2, May 3, Jun 2, Jul 3, Aug 15 → 1 → total 20 → 6 odd days
Total = 3 + 6 = 9 → 2 odd days
15 Aug 2012 is 2 days ahead of 26 Jan 1998 → Wednesday − 2 = Monday

Cetking shortcut: Move back 2 weekdays from Wednesday → Monday.

Answer: B

Q138EasyClocks & Calendars
An accurate clock shows 3 o'clock in the afternoon. Through how many degrees will the hour hand rotate when the clock shows 9 o'clock of the same evening?
  1. 140°
  2. 170°
  3. 180°
  4. 110°
  5. 100°
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Answer: C official key

Approach: Hour hand moves 30° per hour.

Solution:
3 pm to 9 pm = 6 hours
Rotation = 6 × 30° = 180°

Answer: C

Q139EasyClocks & Calendars
What is the angle between the hour and minute hands of a clock at 8:00?
  1. 110°
  2. 150°
  3. 180°
  4. 190°
  5. 120°
Show answerHide answer

Answer: E official key

Approach: Angle = |30H − 5.5M|.

Solution:
At 8:00: H = 8, M = 0 → 30 × 8 = 240°
Smaller angle = 360° − 240° = 120°

Cetking shortcut: Minute hand at 12, hour hand at 8 → 4 hour-gaps × 30° = 120°.

Answer: E

Q140MediumProbability
A book has 15 pages: 1 cover page, 9 theory pages (fixed order), and 5 drawing pages. The cover page is always first. The drawing pages must remain together but can be in any order. What is the probability that the pages are arranged according to the requirement?
  1. 10·8! / 14!
  2. 10·5! / 14!
  3. 10·8! / 10!
  4. 8·5! / 13!
  5. 10·8! / 7!
Show answerHide answer

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

Approach: Probability = favourable arrangements / total arrangements, with the cover fixed first.

Solution:
Total: cover is first, other 14 distinct pages in any order → 14!
Favourable: tie the 5 drawing pages into one block → 9 theory pages + 1 block = 10 units
Theory pages have a fixed order, so only the block's position varies: 10 positions
Drawing pages inside the block: 5!
Favourable = 10 × 5!
P = 10 · 5! / 14!

Why the other options are wrong:
A (10·8!/14!): 8! has no meaning here — only the 5 drawing pages permute.
C and E: denominator must be 14! (all arrangements after the cover).
D (8·5!/13!): wrong count of units and total.

Cetking shortcut: 'Fixed order' items act as identical slots: choose the block's slot (10 ways) × 5!.

Answer: B

Q141EasyCaselet

Questions 141–145 share this

Directions: Study the following information carefully and answer the question that follows.

Natasha wants to pursue her B. Tech from ADC University, India, but to be able to afford it, she has to take an education loan. The loan agreement guaranteed to pay 80% of all her expenses. This way she only had to bear the remaining costs. She had to pay the rent for her new apartment was Rs. 550 per month. She then paid her tuition fee for the current semester worth Rs. 25000. On an average she spent Rs. 340 on utilities and groceries per month. Given that, Natasha's course lasted a total of two years (comprising of 2 semesters per year) and the bank gave 80% of the total expenses of two years at the beginning of her course

How much did the bank have to pay in total for two years on behalf of Natasha?
  1. Rs. 90308
  2. Rs. 85428
  3. Rs. 97088
  4. Rs. 90288
  5. Rs. 92286
Show answerHide answer

Answer: C official key

Approach: Total 2-year expenses (24 months, 4 semesters) × 80%.

Solution:
Rent = 550 × 24 = 13,200
Tuition = 25,000 × 4 semesters = 1,00,000
Utilities & groceries = 340 × 24 = 8,160
Total = 1,21,360
Bank pays 80% = 0.8 × 1,21,360 = Rs. 97,088

Cetking shortcut: Monthly items together = 890 × 24 = 21,360; + 1,00,000 tuition; × 0.8.

Answer: C

Q142EasyCaselet

Refer to the passage above ↑

If the bank charges simple interest at the rate of 9% per annum, then find the total interest amount that Natasha paid after 2 years. (Assume she pays off the entire loan after 2 years of completion of course)
  1. Rs. 17075.84
  2. Rs. 17005.48
  3. Rs. 17975.84
  4. Rs. 16845.48
  5. Rs. 17475.84
Show answerHide answer

Answer: E official key

Approach: Simple interest = P × R × T / 100 on the loan amount.

Solution:
Principal = loan from Q1 = Rs. 97,088
SI for 2 years at 9% = 97,088 × 9 × 2/100 = 97,088 × 0.18 = Rs. 17,475.84

Why the other options are wrong:
A and C are the correct value shifted by −400 and +500; B and D are unrelated decoys — only 0.18 × 97,088 = 17,475.84 is exact.

Cetking shortcut: 18% of 97,088: 10% = 9,708.8, 8% = 7,767.04 → 17,475.84.

Answer: E

Q143EasyCaselet

Refer to the passage above ↑

The annual expenditure on utilities is what percent less than the annual expenditure on rent?
  1. 28.15%
  2. 38.18%
  3. 37%
  4. 42.10%
  5. 40.28%
Show answerHide answer

Answer: B official key

Approach: Percent less = (rent − utilities)/rent × 100.

Solution:
Annual rent = 550 × 12 = 6,600
Annual utilities (& groceries) = 340 × 12 = 4,080
Percent less = (6,600 − 4,080)/6,600 × 100 = 2,520/6,600 × 100 = 38.18%

Why the other options are wrong:
Dividing by the wrong base (utilities, 2,520/4,080 = 61.8%) or mis-subtracting gives the other values — 'less than rent' means divide by rent.

Cetking shortcut: Monthly figures give the same %: 210/550 = 38.18%.

Answer: B

Q144EasyCaselet

Refer to the passage above ↑

Natasha gets an internship for a period of 3 months. The company where she'll be doing internship pays Rs. 12000 per month. The utilities and rent for these 3 months is approximately what percentage of the total amount she will earn from the internship?
  1. 7.41%
  2. 5.41%
  3. 17.41%
  4. 15.41%
  5. 12.41%
Show answerHide answer

Answer: A official key

Approach: Expense as a percentage of income for the same 3 months.

Solution:
Rent + utilities for 3 months = (550 + 340) × 3 = 2,670
Internship earnings = 12,000 × 3 = 36,000
% = 2,670/36,000 × 100 = 7.41%

Cetking shortcut: Monthly: 890/12,000 = 7.41% — the 3 months cancel.

Answer: A

Q145EasyCaselet

Refer to the passage above ↑

Natasha decides to live with her relatives for 6 months so she will not have to pay for rent and utilities. How much shall she save on rent and utilities if she decides to stay with her relatives for 6 months?
  1. Rs. 8340
  2. Rs. 3640
  3. Rs. 5340
  4. Rs. 8940
  5. Rs. 7970
Show answerHide answer

Answer: C official key

Approach: Saving = 6 months of rent + utilities.

Solution:
Monthly = 550 + 340 = 890
6 months = 890 × 6 = Rs. 5,340

Answer: C

Q146MediumPie Chart

Questions 146–150 share this

Study the following Pie Chart carefully and answer the question that follows :

The pie chart given below represents percentage of number of students in five different schools viz. A, B, C, D and E. The total number of students in all the schools is 1,68,000.

Data read from chart:

SchoolPercentage of students
A28%
B15%
C24%
D18%
E15%

The below table shows the ratio of girls and boys in the given five schools.

SchoolGirls : Boys
A3 : 4
B2 : 3
C5 : 3
D1 : 2
E4 : 1
Figure for MBA CET 2026 Slot 1 question 146
Find the ratio of number of girls from each school.
  1. 1:2:2:4:5
  2. 3:5:7:9:8
  3. 4:2:5:2:4
  4. 1:2:3:2:1
  5. 5:2:6:7:4
Show answerHide answer

Answer: C official key

Approach: Girls in each school = school total × girls' share; reduce to simplest ratio.

Solution:
Totals (of 1,68,000): A 28% = 47,040; B 15% = 25,200; C 24% = 40,320; D 18% = 30,240; E 15% = 25,200
Girls: A 3/7 × 47,040 = 20,160; B 2/5 × 25,200 = 10,080; C 5/8 × 40,320 = 25,200; D 1/3 × 30,240 = 10,080; E 4/5 × 25,200 = 20,160
Divide by 5,040: 4 : 2 : 5 : 2 : 4

Cetking shortcut: Work in % only: A 28 × 3/7 = 12, B 15 × 2/5 = 6, C 24 × 5/8 = 15, D 18 × 1/3 = 6, E 15 × 4/5 = 12 → 12:6:15:6:12 = 4:2:5:2:4.

Answer: C

Q147EasyPie Chart

Refer to the passage above ↑

If 14.28% students of school A change their school and go to school D, what will be the percentage change in the number of students of school D.
Figure for MBA CET 2026 Slot 1 question 147
  1. 11.11%
  2. 22.22%
  3. 33.33%
  4. 12.18%
  5. 14.15%
Show answerHide answer

Answer: B official key

Approach: 14.28% = 1/7; move that many students from A to D and find D's % rise.

Solution:
A = 47,040 → 1/7 × 47,040 = 6,720 students move
D = 30,240 → new D = 36,960
% change = 6,720/30,240 × 100 = 22.22%

Cetking shortcut: In % terms: 28 × 1/7 = 4 points added to D's 18 → 4/18 = 22.22%.

Answer: B

Q148EasyPie Chart

Refer to the passage above ↑

What is the ratio of average number of students from School A and D to that of the average number of students from School B, C and E?
Figure for MBA CET 2026 Slot 1 question 148
  1. 18 : 23
  2. 14 : 25
  3. 23 : 18
  4. 8 : 9
  5. 27 : 19
Show answerHide answer

Answer: C official key

Approach: Compare averages using the pie percentages (same total cancels).

Solution:
Average of A and D = (28 + 18)/2 = 23%
Average of B, C, E = (15 + 24 + 15)/3 = 54/3 = 18%
Ratio = 23 : 18

Why the other options are wrong:
A (18 : 23) is the reversed ratio.

Cetking shortcut: No need for absolute numbers — percentages of the same total give the ratio directly.

Answer: C

Q149EasyPie Chart

Refer to the passage above ↑

The number of students in School E is what percentage of the difference between the number of students in School C and School B?
Figure for MBA CET 2026 Slot 1 question 149
  1. 156.76%
  2. 186.67%
  3. 60%
  4. 166.67%
  5. 74%
Show answerHide answer

Answer: D official key

Approach: Express E as a percentage of (C − B), working on pie percentages.

Solution:
E = 15%; C − B = 24% − 15% = 9%
15/9 × 100 = 166.67%

Why the other options are wrong:
B (186.67%) and A (156.76%) are near-miss decoys; C (60%) is the inverse (9/15).

Cetking shortcut: 15/9 = 5/3 = 166.67%.

Answer: D

Q150MediumPie Chart

Refer to the passage above ↑

Find the average number of boys in all the schools combined?
Figure for MBA CET 2026 Slot 1 question 150
  1. 15784
  2. 16464
  3. 17840
  4. 16465
  5. 15400
Show answerHide answer

Answer: B official key

Approach: Boys in each school = school total × boys' share; average over 5 schools.

Solution:
A 4/7 × 47,040 = 26,880
B 3/5 × 25,200 = 15,120
C 3/8 × 40,320 = 15,120
D 2/3 × 30,240 = 20,160
E 1/5 × 25,200 = 5,040
Total boys = 82,320 → average = 82,320/5 = 16,464

Why the other options are wrong:
D (16,465) is a one-off decoy; the exact average is 16,464.

Cetking shortcut: Total girls (from Q1) = 85,680 → boys = 1,68,000 − 85,680 = 82,320 → ÷ 5 = 16,464.

Answer: B

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