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MBA CET 2026 Slot 4 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.

Q101EasyRatios
If A : B = 2 : 3 and B : C = 4 : 5, find the ratio A : C.
  1. 2:5
  2. 8:15
  3. 3:5
  4. 8:12
  5. 6:15
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Answer: B official key

Approach: Chain two ratios by making the common term (B) equal.

Solution:
A : B = 2 : 3 and B : C = 4 : 5.
Make B equal: LCM of 3 and 4 = 12.
A : B = 8 : 12 and B : C = 12 : 15.
So A : B : C = 8 : 12 : 15 ⇒ A : C = 8 : 15.

Why the other options are wrong:
A (2:5) – simply multiplies the outer terms, ignoring B.
C (3:5) – uses B's term from the second ratio wrongly.
D (8:12) – this is A : B, not A : C.
E (6:15) – multiplies 2 × 3 instead of 2 × 4.

Cetking shortcut: A : C = (2 × 4) : (3 × 5) = 8 : 15 — multiply the 'A-side' numerators and the 'C-side' denominators directly.

Answer: B

Q102EasyAges
The ratio of the present ages of a mother and her daughter is 7:1. Four years ago, the ratio of their ages was 19:1. What will be the mother's age 4 years from now?
  1. 42 years
  2. 38 years
  3. 46 years
  4. 36 years
  5. 40 years
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Answer: C official key

Approach: Ages in ratio — form one equation from the past ratio.

Solution:
Present ages: mother = 7x, daughter = x.
4 years ago: (7x − 4)/(x − 4) = 19/1.
7x − 4 = 19x − 76 ⇒ 12x = 72 ⇒ x = 6.
Mother now = 42 years (daughter 6; 4 years ago 38 : 2 = 19 : 1 ✓).
Mother 4 years from now = 42 + 4 = 46 years.

Why the other options are wrong:
A (42) – mother's present age; the question asks 4 years later.
B (38) – mother's age 4 years ago.
D, E (36, 40) – do not satisfy both ratios.

Cetking shortcut: Options test: the present age must be a multiple of 7 and (age − 4) : (age/7 − 4) = 19 : 1; 42 → 38 : 2 = 19 : 1 instantly.

Answer: C

Q103EasyNumbers
Find the difference between the place value and the face value of the digit 7 in the number 4782.
  1. 700
  2. 7
  3. 693
  4. 782
  5. 707
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Answer: C official key

Approach: Place value = digit × its positional value; face value = the digit itself.

Solution:
In 4782, 7 is in the hundreds place.
Place value = 7 × 100 = 700.
Face value = 7.
Difference = 700 − 7 = 693.

Why the other options are wrong:
A (700) – place value only.
B (7) – face value only.
D (782) – the number after the 4, not a place value.
E (707) – adds instead of subtracts.

Cetking shortcut: Difference = digit × (place − 1) = 7 × 99 = 693.

Answer: C

Q104EasyRemainders
What is the units digit of the product 15 * 22 * 37 * 44?
  1. 2
  2. 5
  3. 0
  4. 4
  5. 8
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Answer: C official key

Approach: Units digit of a product depends only on the units digits of the factors.

Solution:
Units digits: 5, 2, 7, 4.
5 × 2 = 10 → units digit 0.
Anything × (number ending in 0) ends in 0.
Units digit of 15 × 22 × 37 × 44 = 0.

Why the other options are wrong:
A, D, E (2, 4, 8) – come from ignoring the 5 × 2 pair.
B (5) – 5 × odd ends in 5, but here 5 meets an even number.

Cetking shortcut: A factor ending in 5 together with any even factor always gives units digit 0.

Answer: C

Q105EasyHCF LCM
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, and 12 seconds, respectively. In 30 minutes, how many times do they toll together?
  1. 15
  2. 16
  3. 30
  4. 31
  5. 12
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Answer: B official key

Approach: Bells toll together at multiples of the LCM of their intervals; count the start too.

Solution:
LCM(2, 4, 6, 8, 10, 12) = 2³ × 3 × 5 = 120 s = 2 min.
In 30 min they toll together every 2 min: 30 / 2 = 15 times after the start.
Add the toll at the start (time 0): 15 + 1 = 16.

Why the other options are wrong:
A (15) – forgets the toll at the beginning.
C, D (30, 31) – assume they meet every minute.
E (12) – wrong LCM (150 s).

Cetking shortcut: 'Commence together' = add 1 to (total time ÷ LCM).

Answer: B

Q106EasyHCF LCM
Three numbers are in the ratio of 3:4:5, and their least common multiple (LCM) is 2400. Their highest common factor (HCF) is
  1. 40
  2. 80
  3. 120
  4. 200
  5. 15
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Answer: A official key

Approach: For numbers in ratio a:b:c with a, b, c co-prime, numbers = ax, bx, cx; HCF = x and LCM = x × LCM(a, b, c).

Solution:
Numbers = 3x, 4x, 5x (3, 4, 5 pairwise co-prime).
LCM = x × LCM(3, 4, 5) = 60x.
60x = 2400 ⇒ x = 40.
HCF = x = 40 (numbers 120, 160, 200).

Why the other options are wrong:
B (80), C (120), D (200) – 120 and 200 are actual numbers, not the HCF; 80 is 2x.
E (15) – no basis.

Cetking shortcut: HCF = LCM ÷ (3 × 4 × 5) = 2400 ÷ 60 = 40.

Answer: A

Q107EasyPercentages
Solve: 15% of 200 + 20% of 150 =
  1. 50
  2. 55
  3. 60
  4. 65
  5. 70
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Answer: C official key

Approach: Direct percentage of a number.

Solution:
15% of 200 = 0.15 × 200 = 30.
20% of 150 = 0.20 × 150 = 30.
Sum = 30 + 30 = 60.

Cetking shortcut: x% of y = y% of x: 15% of 200 = 200% of 15 = 30.

Answer: C

Q108EasySimplification
Solve: 0.2 * 0.2 + 0.3 * 0.3 + 2 * 0.2 * 0.3
  1. 0.025
  2. 0.25
  3. 0.5
  4. 0.05
  5. 0.6
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Answer: B official key

Approach: Recognise the identity a² + b² + 2ab = (a + b)².

Solution:
a = 0.2, b = 0.3.
0.2 × 0.2 + 0.3 × 0.3 + 2 × 0.2 × 0.3 = (0.2 + 0.3)².
= (0.5)² = 0.25.
Check: 0.04 + 0.09 + 0.12 = 0.25.

Why the other options are wrong:
A (0.025) – decimal placed wrongly.
C (0.5) – forgets to square a + b.
D (0.05), E (0.6) – arithmetic slips.

Cetking shortcut: Spot a² + b² + 2ab at a glance → (a + b)²; no multiplication needed.

Answer: B

Q109EasyFactors
What is the smallest number by which 675 should be multiplied to obtain a perfect cube?
  1. 2
  2. 3
  3. 5
  4. 6
  5. 9
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Answer: C official key

Approach: Prime-factorise; every prime's power must be a multiple of 3 for a perfect cube.

Solution:
675 = 25 × 27 = 3³ × 5².
3³ is already a cube; 5² needs one more 5.
Multiply by 5: 675 × 5 = 3375 = 15³.

Why the other options are wrong:
A (2), D (6) – introduce a new prime 2.
B (3) – 3⁴ would spoil the cube.
E (9) – makes 3⁵.

Cetking shortcut: Group primes in triples; the missing members form the multiplier.

Answer: C

Q110EasyNumbers
Which of the following numbers is a perfect square?
  1. 4410
  2. 2362
  3. 7921
  4. 5688
  5. 1253
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Answer: C official key

Approach: Perfect squares end in 0, 1, 4, 5, 6, 9 and have a matching square root near √N.

Solution:
2362 (ends in 2), 5688 (ends in 8), 1253 (ends in 3) – squares never end in 2, 3, 8 → rejected.
4410 – ends in a single 0; a square ending in 0 must end in 00 → rejected.
7921: 89² = (90 − 1)² = 8100 − 180 + 1 = 7921 ✓.

Why the other options are wrong:
A (4410) – odd number of trailing zeros.
B, D, E – last digits 2, 8, 3 are impossible for squares.

Cetking shortcut: Last-digit test removes four options in seconds; verify the survivor with (90 − 1)².

Answer: C

Q111EasyAverages
Calculate the average of the first 50 natural numbers.
  1. 25
  2. 25.5
  3. 26
  4. 50.5
  5. 51
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Answer: B official key

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

Solution:
Sum of 1 to 50 = 50 × 51 / 2 = 1275.
Average = 1275 / 50 = 25.5.
Or directly (50 + 1)/2 = 25.5.

Why the other options are wrong:
A (25) – uses n/2.
C (26) – rounding up.
D (50.5), E (51) – use n + 1 or (n + 1) without halving.

Cetking shortcut: In any AP the average = (first + last)/2 = (1 + 50)/2.

Answer: B

Q112EasyAverages
A batsman has a certain average of runs for 11 innings. In the 12th inning, he makes a score of 90 runs, thereby increasing his average by 5. His new average is:
  1. 30
  2. 40
  3. 45
  4. 50
  5. 35
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Answer: E official key

Approach: Total runs after 12 innings = 12 × new average; set it equal to old total + 90.

Solution:
Let the old average (11 innings) = x.
New average = x + 5 over 12 innings.
12(x + 5) = 11x + 90 ⇒ 12x + 60 = 11x + 90 ⇒ x = 30.
New average = 30 + 5 = 35.

Why the other options are wrong:
A (30) – the old average.
B, C, D (40, 45, 50) – do not balance the totals.

Cetking shortcut: New average = score − (old innings × increase) = 90 − 11 × 5 = 35.

Answer: E

Q113EasyRemainders
A number when divided by 5 leaves a remainder of 3. What is the remainder when the square of that number is divided by 5?
  1. 1
  2. 2
  3. 3
  4. 4
  5. 0
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Answer: D official key

Approach: Remainders multiply: rem(N²) = rem(r²).

Solution:
N = 5k + 3, so N leaves remainder 3.
N² leaves the same remainder as 3² = 9 when divided by 5.
9 = 5 × 1 + 4 ⇒ remainder 4.
Check: N = 8, N² = 64 = 5 × 12 + 4 ✓.

Why the other options are wrong:
C (3) – assumes the remainder stays the same.
A, B, E – arithmetic slips.

Cetking shortcut: Pick the smallest such number (3): 3² = 9 → remainder 4.

Answer: D

Q114EasySimplification
Which of the following fractions is the smallest?
  1. 7/6
  2. 7/9
  3. 4/5
  4. 5/7
  5. 9/11
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Answer: D official key

Approach: Compare fractions by decimals or by 'gap from 1'.

Solution:
7/6 = 1.17 (greater than 1).
7/9 = 0.778, 4/5 = 0.800, 5/7 = 0.714, 9/11 = 0.818.
Smallest = 5/7 = 0.714.

Why the other options are wrong:
A (7/6) – more than 1, the largest.
B (7/9 = 0.778), C (4/5 = 0.8), E (9/11 = 0.818) – all above 0.714.

Cetking shortcut: For fractions of the form (n − k)/n, compare the gap k/n: 5/7 is short by 2/7 ≈ 0.29, the biggest gap → smallest fraction.

Answer: D

Q115EasyAges
The ratio of the ages of A and B is 3:4. After 10 years, the ratio of their ages will be 4:5. What is the sum of their present ages?
  1. 60 years
  2. 70 years
  3. 80 years
  4. 50 years
  5. 90 years
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Answer: B official key

Approach: Ages in ratio — the future ratio gives one equation.

Solution:
Present ages: A = 3x, B = 4x.
(3x + 10)/(4x + 10) = 4/5.
15x + 50 = 16x + 40 ⇒ x = 10.
Ages 30 and 40 (after 10 years 40 : 50 = 4 : 5 ✓). Sum = 70.

Why the other options are wrong:
A (60), C (80), D (50), E (90) – sum must be 7x and satisfy the future ratio; only 70 does.

Cetking shortcut: Sum must be a multiple of 7 (3x + 4x) → only 70 among options.

Answer: B

Q116EasyAverages
The average age of 15 students is 18 years. If the teacher's age is added, the average increases by 1 year. What is the teacher's age?
  1. 32 years
  2. 34 years
  3. 35 years
  4. 33 years
  5. 30 years
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Answer: B official key

Approach: Sum = average × count; new person's value = new sum − old sum.

Solution:
Sum of 15 students = 15 × 18 = 270.
With teacher: 16 people, average 19 ⇒ sum = 16 × 19 = 304.
Teacher's age = 304 − 270 = 34 years.

Why the other options are wrong:
E (30), A (32), D (33) – D adds 15 to the old average 18; others are arithmetic slips.
C (35) – uses 16 × 1 added to the new average (16 + 19).

Cetking shortcut: Teacher = new average + (old count × increase) = 19 + 15 × 1 = 34.

Answer: B

Q117EasyPercentages
The price of sugar is increased by 20%. By how much percent must a household reduce its consumption so as not to increase the expenditure?
  1. 20.33%
  2. 16.66%
  3. 25.50%
  4. 15.33%
  5. 10.66%
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Answer: B official key

Approach: Expenditure = price × consumption; keep the product constant.

Solution:
New price = 1.2 × old.
Consumption must become 1/1.2 = 5/6 of old.
Reduction = 1 − 5/6 = 1/6 = 16.66%.
Formula: r/(100 + r) × 100 = 20/120 × 100 = 16.66%.

Why the other options are wrong:
A (20.33%) – assumes the cut equals the rise.
C, D, E – arithmetic slips.

Cetking shortcut: 20% = 1/5 rise ⇒ cut = 1/(5 + 1) = 1/6 = 16.66%.

Answer: B

Q118EasyPercentages
A man spends 35% of his income on food, 25% on children's education, and 80% of the remaining on house rent. What percent of his income is he left with?
  1. 8%
  2. 10%
  3. 12%
  4. 15%
  5. 6%
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Answer: A official key

Approach: Successive spending — apply the 80% to the remaining income only.

Solution:
Food + education = 35% + 25% = 60% ⇒ remaining = 40%.
Rent = 80% of 40% = 32%.
Left = 40% − 32% = 8% of income.

Why the other options are wrong:
E (6%), B (10%), C (12%), D (15%) – come from taking 80% of the whole income or mixing bases.

Cetking shortcut: Left = 40% × 20% = 8%.

Answer: A

Q119EasyProfit Loss
By selling 12 oranges for a rupee, a man loses 20%. How many for a rupee should he sell to gain 20%?
  1. 8
  2. 10
  3. 15
  4. 16
  5. 9
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Answer: A official key

Approach: Fix the cost of one orange from the loss case, then price it for the gain.

Solution:
SP of 1 orange = Re 1/12 at 20% loss ⇒ CP = (1/12)/0.8 = 5/48.
For 20% gain: SP = 1.2 × 5/48 = 1/8.
So he must sell 8 oranges for a rupee.

Why the other options are wrong:
C (15), D (16) – sell more per rupee, i.e. a bigger loss.
B (10), E (9) – do not reach 20% gain.

Cetking shortcut: Count × (1 + profit%) is constant: 12 × 0.8 = n × 1.2 ⇒ n = 8.

Answer: A

Q120MediumProfit Loss
A man bought an article and sold it at a gain of 5%. If he had bought it at 5% less and sold it for Rs. 1 less, he would have made a profit of 10%. The cost price of the article is:
  1. Rs. 100
  2. Rs. 150
  3. Rs. 200
  4. Rs. 250
  5. Rs. 300
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Answer: C official key

Approach: Set CP = 100x and compare the two scenarios.

Solution:
Original: CP = 100x, SP = 105x.
New: CP = 95x, SP = 105x − 1, profit 10% ⇒ SP = 1.1 × 95x = 104.5x.
105x − 1 = 104.5x ⇒ 0.5x = 1 ⇒ x = 2.
CP = Rs. 200. Check: 190 → 209 = 10% profit ✓.

Why the other options are wrong:
A, B, D, E – on testing, the two SPs do not differ by Rs. 1.

Cetking shortcut: Difference in the two SPs = 105% − (110% of 95%) = 105% − 104.5% = 0.5% of CP = Rs. 1 ⇒ CP = 200.

Answer: C

Q121EasyPartnership
A, B, and C enter into a partnership with a capital in which A's contribution is Rs. 10,000. If out of a total profit of Rs. 1000, A gets Rs. 500 and B gets Rs. 300, then C's capital is:
  1. Rs. 4000
  2. Rs. 5000
  3. Rs. 6000
  4. Rs. 9000
  5. Rs. 4500
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Answer: A official key

Approach: Same time period ⇒ profit ratio = capital ratio.

Solution:
C's profit = 1000 − 500 − 300 = Rs. 200.
Profit ratio A : B : C = 500 : 300 : 200 = 5 : 3 : 2.
A's 5 parts = Rs. 10,000 ⇒ 1 part = Rs. 2,000.
C's capital = 2 × 2,000 = Rs. 4,000.

Why the other options are wrong:
C (6000) – this is B's capital.
B (5000) – A's half, not C's.
D, E – no basis in the ratio.

Cetking shortcut: C's capital = 10,000 × 200/500 = Rs. 4,000.

Answer: A

Q122EasyPartnership
A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined them after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
  1. 3 : 5 : 2
  2. 3 : 5 : 5
  3. 6 : 10 : 5
  4. 3 : 7 : 8
  5. 6 : 10 : 3
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Answer: C official key

Approach: Profit ratio = capital × time.

Solution:
Capitals: A = 3k, B = 5k, C = 5k (equal to B).
Time: A and B 12 months, C 6 months.
A : B : C = 3 × 12 : 5 × 12 : 5 × 6 = 36 : 60 : 30 = 6 : 10 : 5.

Why the other options are wrong:
A (3:5:2), B (3:5:5) – ignore that C invested only for 6 months or scale wrongly.
D (3:7:8), E (6:10:3) – wrong time/amount for C.

Cetking shortcut: Halve C's capital for half the time: 3 : 5 : 2.5 = 6 : 10 : 5.

Answer: C

Q123EasyMandays
If 12 carpenters working 6 hours a day can make 460 chairs in 24 days, how many chairs will 18 carpenters make in 36 days working 8 hours a day?
  1. 1260
  2. 1320
  3. 1400
  4. 1380
  5. 1200
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Answer: D official key

Approach: Work ∝ men × days × hours: M₁D₁H₁/W₁ = M₂D₂H₂/W₂.

Solution:
Case 1: 12 × 24 × 6 = 1728 man-hours for 460 chairs.
Case 2: 18 × 36 × 8 = 5184 man-hours.
5184/1728 = 3 ⇒ chairs = 460 × 3 = 1380.

Why the other options are wrong:
A (1260), B (1320), C (1400), E (1200) – miss one of the factors (e.g. 18/12 × 36/24 = 2.25 alone).

Cetking shortcut: Multiply the factor changes: (18/12) × (36/24) × (8/6) = 1.5 × 1.5 × 4/3 = 3.

Answer: D

Q124EasyMandays
A contractor undertook to do a certain piece of work in 9 days. He employed a certain number of men, but 6 of them were absent from the very first day, and the rest could finish the work in 15 days. The number of men originally employed were:
  1. 12
  2. 15
  3. 18
  4. 20
  5. 24
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Answer: B official key

Approach: Total man-days of work stay constant.

Solution:
Let x men be employed: work = 9x man-days.
Actually (x − 6) men took 15 days: work = 15(x − 6).
9x = 15x − 90 ⇒ 6x = 90 ⇒ x = 15.

Why the other options are wrong:
A (12) – with 6 present, 6 × 15 = 90 ≠ 108.
C (18), D (20), E (24) – 9x ≠ 15(x − 6).

Cetking shortcut: Absent men × 15 days = extra days (15 − 9) × x ⇒ 6 × 15 = 6x ⇒ x = 15.

Answer: B

Q125MediumPipes and Cisterns
A cistern has a leak which would empty it in 8 hours. A tap is turned on which admits 6 liters a minute into the cistern and it is now emptied in 12 hours. What is the capacity of the cistern?
  1. 8000 liters
  2. 8240 liters
  3. 8640 liters
  4. 9000 liters
  5. 9600 liters
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Answer: C official key

Approach: Net rate = leak rate − inlet rate; find the tap's rate as a fraction of the tank.

Solution:
Leak alone empties in 8 h ⇒ −1/8 per hour.
With tap on, it empties in 12 h ⇒ net −1/12 per hour.
Tap rate = 1/8 − 1/12 = 1/24 tank per hour ⇒ tap fills the tank in 24 h.
Tap gives 6 L/min = 360 L/h ⇒ capacity = 24 × 360 = 8640 L.

Why the other options are wrong:
A, B, D, E – result from wrong time (e.g. 20 h or 25 h) or not converting minutes to hours.

Cetking shortcut: Tap time = (8 × 12)/(12 − 8) = 24 h; capacity = 24 × 60 × 6 = 8640.

Answer: C

Q126EasyPipes and Cisterns
A tank is filled in 5 hours by three pipes A, B, and C. The pipe C is twice as fast as B, and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
  1. 20 hours
  2. 25 hours
  3. 30 hours
  4. 35 hours
  5. 40 hours
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Answer: D official key

Approach: Use efficiencies: total work = combined efficiency × time.

Solution:
Let A = 1 unit/h ⇒ B = 2, C = 4 (C twice B, B twice A).
Combined = 7 units/h; tank filled in 5 h ⇒ capacity = 35 units.
A alone = 35/1 = 35 hours.

Why the other options are wrong:
A (20), B (25), C (30), E (40) – wrong efficiency ratios (e.g. 1 : 2 : 3).

Cetking shortcut: A alone time = 5 × (1 + 2 + 4)/1 = 35 h.

Answer: D

Q127MediumTime & Work
A, B, and C can do a piece of work in 20, 30, and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
  1. 10 days
  2. 12 days
  3. 15 days
  4. 16 days
  5. 18 days
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Answer: C official key

Approach: Work in 3-day cycles using LCM units.

Solution:
Total work = LCM(20, 30, 60) = 60 units. Efficiencies: A = 3, B = 2, C = 1 per day.
Days 1 and 2: A alone = 3 + 3 = 6 units.
Day 3: A + B + C = 3 + 2 + 1 = 6 units.
One 3-day cycle = 12 units ⇒ 60/12 = 5 cycles = 15 days (finishes exactly at the end of a cycle).

Why the other options are wrong:
A (10), B (12) – assume help every day or every second day.
D (16), E (18) – add extra days after the work is already done.

Cetking shortcut: Cycle work = 1/5 of the job ⇒ 5 cycles × 3 days = 15.

Answer: C

Q128MediumTime & Work
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
  1. 4
  2. 5
  3. 6
  4. 7
  5. 8
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Answer: A official key

Approach: Equate total work from both groups to find the man : boy efficiency.

Solution:
10(6M + 8B) = 2(26M + 48B) ⇒ 60M + 80B = 52M + 96B ⇒ 8M = 16B ⇒ 1M = 2B.
Work = 10 × (6 × 2 + 8)B = 10 × 20B = 200 boy-days.
15 men + 20 boys = 30B + 20B = 50B.
Time = 200/50 = 4 days.

Why the other options are wrong:
B (5) – uses 1M = 1.5B or similar slip.
C, D, E – wrong efficiency relation.

Cetking shortcut: Convert everyone into boys (1M = 2B) and divide total boy-days by the new strength.

Answer: A

Q129EasyTrains Boats
A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from the opposite direction in 6 seconds. The speed of the second train is:
  1. 48 km/hr
  2. 54 km/hr
  3. 66 km/hr
  4. 82 km/hr
  5. 90 km/hr
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Answer: D official key

Approach: Opposite directions ⇒ relative speed = sum; distance = sum of lengths.

Solution:
Distance = 108 + 112 = 220 m; time = 6 s.
Relative speed = 220/6 = 110/3 m/s = 110/3 × 18/5 = 132 km/h.
Speed of second train = 132 − 50 = 82 km/h.

Why the other options are wrong:
A (48), B (54), C (66) – use relative speed as a difference or forget the conversion.
E (90) – arithmetic slip.

Cetking shortcut: 220 m in 6 s ⇒ 220 × 3.6/6 = 132 km/h; subtract 50.

Answer: D

Q130MediumTrains Boats
A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 km/hr and 4 km/hr and passes them completely in 9 and 10 seconds respectively. The length of the train is:
  1. 45 m
  2. 50 m
  3. 54 m
  4. 72 m
  5. 60 m
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Answer: B official key

Approach: Same direction ⇒ relative speed = difference; train length is the same in both cases.

Solution:
Let train speed = v km/h. Relative speeds: (v − 2) and (v − 4) km/h.
L = (v − 2) × 5/18 × 9 = (v − 4) × 5/18 × 10.
9v − 18 = 10v − 40 ⇒ v = 22 km/h.
L = 20 × 5/18 × 9 = 50 m (check: 18 × 5/18 × 10 = 50 m ✓).

Why the other options are wrong:
A (45), C (54), D (72), E (60) – do not give the same length from both persons.

Cetking shortcut: Test options: 50 m in 9 s = 20 km/h relative and in 10 s = 18 km/h relative — the 2 km/h gap matches the walkers' 4 − 2.

Answer: B

Q131EasyMixtures
A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:
  1. 400 kg
  2. 500 kg
  3. 600 kg
  4. 640 kg
  5. 700 kg
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Answer: C official key

Approach: Alligation on profit percentages.

Solution:
Profits: 8% and 18%; mean 14%.
Quantity at 8% : quantity at 18% = (18 − 14) : (14 − 8) = 4 : 6 = 2 : 3.
Quantity at 18% = 3/5 × 1000 = 600 kg.

Why the other options are wrong:
A (400 kg) – this is the part sold at 8%.
B (500 kg) – would give an average of 13%.
D, E – average would exceed 14%.

Cetking shortcut: 14% is closer to 18% than to 8% ⇒ more than half (600) is at 18%.

Answer: C

Q132MediumMixtures
A vessel is filled with liquid, 3 parts of which are water and 5 parts 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. 2/7
  4. 1/7
  5. 1/5
Show answerHide answer

Answer: E official key

Approach: Replacement: only syrup is removed-and-not-replaced; track the syrup fraction.

Solution:
Syrup = 5/8 of the vessel. Draw off fraction f and replace with water.
Syrup left = 5/8 × (1 − f). Need it = 1/2.
1 − f = (1/2)/(5/8) = 4/5 ⇒ f = 1/5.

Why the other options are wrong:
A (1/3), B (1/4), C (2/7), D (1/7) – leave syrup at 5/12, 15/32, 25/56, 15/28 of the vessel, not 1/2.

Cetking shortcut: Fraction to replace = 1 − (target syrup / present syrup) = 1 − (4/8 ÷ 5/8) = 1/5.

Answer: E

Q133EasyInterest
A sum of money at simple interest doubles itself in 8 years. What is the rate of interest per annum?
  1. 10.50%
  2. 12%
  3. 12.50%
  4. 15%
  5. 20%
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Answer: C official key

Approach: Simple interest: I = PRT/100.

Solution:
Doubles ⇒ interest = principal ⇒ I = P.
P = P × R × 8/100 ⇒ R = 100/8 = 12.5%.

Why the other options are wrong:
A, B, D, E – give interest ≠ P in 8 years (e.g. 12% → 96%).

Cetking shortcut: SI doubling: R × T = 100 ⇒ R = 100/8.

Answer: C

Q134EasyInterest
What will be the compound interest on Rs. 1000 at 10% per annum for 1 year, when the interest is compounded half-yearly?
  1. Rs. 100
  2. Rs. 110
  3. Rs. 105
  4. Rs. 102.50
  5. Rs. 112.50
Show answerHide answer

Answer: D official key

Approach: Half-yearly compounding: halve the rate, double the periods.

Solution:
Rate per half-year = 5%; periods = 2.
Amount = 1000 × 1.05² = 1000 × 1.1025 = 1102.50.
CI = 1102.50 − 1000 = Rs. 102.50.

Why the other options are wrong:
A (100) – simple interest / annual compounding.
B, C, E – arithmetic slips.

Cetking shortcut: CI for 2 periods at r% = 2r + r²/100 = 10 + 0.25 = 10.25% of 1000.

Answer: D

Q135EasyQuadrilaterals
The diagonals of a rhombus are 24 cm and 10 cm. The perimeter of the rhombus is:
  1. 52 cm
  2. 56 cm
  3. 68 cm
  4. 72 cm
  5. 60 cm
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Answer: A official key

Approach: Diagonals of a rhombus bisect each other at right angles.

Solution:
Half-diagonals = 12 cm and 5 cm.
Side = √(12² + 5²) = √169 = 13 cm.
Perimeter = 4 × 13 = 52 cm.

Why the other options are wrong:
B, C, D, E – use full diagonals or add them instead of using Pythagoras.

Cetking shortcut: 5–12–13 triplet: half-diagonals 5 and 12 ⇒ side 13.

Answer: A

Q136EasyMensuration
Three solid cubes of sides 3 cm, 4 cm, and 5 cm are melted to form a new solid cube. The side of the new cube is:
  1. 6 cm
  2. 8 cm
  3. 9 cm
  4. 12 cm
  5. 7 cm
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Answer: A official key

Approach: Melting conserves volume.

Solution:
Total volume = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³.
New side = ∛216 = 6 cm.

Why the other options are wrong:
B, C, D, E – add sides (12) or cube-roots incorrectly.

Cetking shortcut: Remember 3³ + 4³ + 5³ = 6³.

Answer: A

Q137EasyTime Speed Distance
In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:
  1. 10 m
  2. 20 m
  3. 30 m
  4. 40 m
  5. 60 m
Show answerHide answer

Answer: B official key

Approach: Races with a head start — compare the times to finish.

Solution:
A runs 500 − 140 = 360 m at speed 3 ⇒ time = 120 units.
In 120 units B runs 4 × 120 = 480 m.
When A finishes, B is 500 − 480 = 20 m short ⇒ A wins by 20 m.

Why the other options are wrong:
A (10), C (30), D (40), E (60) – wrong time base (e.g. using 500 m for A).

Cetking shortcut: Distance ratio = speed ratio: B covers 360 × 4/3 = 480 m while A covers 360 m.

Answer: B

Q138MediumTime Speed Distance
Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?
  1. 36 m
  2. 24 m
  3. 60 m
  4. 72 m
  5. 48 m
Show answerHide answer

Answer: E official key

Approach: Constant speeds ⇒ speed ratios chain: (A/B) × (B/C) = (A/C).

Solution:
Let the race length = L. When A finishes, B has run L − 12 and C has run L − 18.
When B finishes, C has run L − 8 ⇒ C/B = (L − 8)/L.
Also C/B = (L − 18)/(L − 12).
(L − 8)/L = (L − 18)/(L − 12) ⇒ L² − 20L + 96 = L² − 18L ⇒ 2L = 96 ⇒ L = 48 m.
Check: A : B : C = 48 : 36 : 30; B/C = 36/30 = 48/40 ✓.

Why the other options are wrong:
A (36), B (24), C (60), D (72) – fail the chained ratio check.

Cetking shortcut: B beats C by 6 m over the last L − 12 m and by 8 m over L m: 6/(L − 12) = 8/L ⇒ L = 48.

Answer: E

Q139EasyProbability
In a single throw of two dice, find the probability of getting a sum of 7 or 11.
  1. 1/6
  2. 1/9
  3. 7/36
  4. 5/36
  5. 2/9
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Answer: E official key

Approach: Classical probability with two dice: 36 outcomes.

Solution:
Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways.
Sum 11: (5,6), (6,5) = 2 ways.
Mutually exclusive ⇒ favourable = 8; P = 8/36 = 2/9.

Why the other options are wrong:
A (1/6) – sum 7 only.
B (1/9) – 4/36, undercounts.
C (7/36), D (5/36) – miscounted pairs.

Cetking shortcut: Remember: P(sum 7) = 6/36 and P(sum 11) = 2/36.

Answer: E

Q140EasyPnC
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
  1. 5
  2. 10
  3. 15
  4. 20
  5. 25
Show answerHide answer

Answer: D official key

Approach: Divisible by 5 ⇒ units digit is 0 or 5; only 5 is available.

Solution:
Units digit fixed = 5 (1 way).
Hundreds digit: any of the remaining 5 digits.
Tens digit: any of the remaining 4 digits.
Total = 5 × 4 × 1 = 20.

Why the other options are wrong:
A (5) – fills only one free place.
C (15), E (25) – allow repetition or wrong count.
B (10) – uses combinations (5C2).

Cetking shortcut: Fix the restricted place first, then fill the rest: 5P2 = 20.

Answer: D

Q141EasyBar Graph

Questions 141–145 share this

Bar graph. With reference to this graph answer the next 5 questions.

Quarterly Sales Performance by Product (Sales ('000s))

Data read from chart:

QuarterProduct AProduct BProduct C
Q1453055
Q2605040
Q3556570
Q4807560
Figure for MBA CET 2026 Slot 4 question 141
What is the average sales of Product B across all four quarters?
  1. 50
  2. 55
  3. 60
  4. 65
  5. 52.5
Show answerHide answer

Answer: B official key

Approach: Average = sum of the four quarterly values ÷ 4.

Solution:
Product B: Q1 = 30, Q2 = 50, Q3 = 65, Q4 = 75 (in '000s).
Sum = 30 + 50 + 65 + 75 = 220.
Average = 220/4 = 55.

Why the other options are wrong:
A (50) – Q2 value only.
C (60), D (65) – read another product or a single quarter.
E (52.5) – arithmetic slip.

Cetking shortcut: Pair the ends: (30 + 75) + (50 + 65) = 105 + 115 = 220.

Answer: B

Q142EasyBar Graph

Refer to the passage above ↑

Which quarter witnessed the highest total sales for all three products combined?
Figure for MBA CET 2026 Slot 4 question 142
  1. Q1
  2. Q2
  3. Q3
  4. Q4
  5. Q3 and Q4 have equal sales
Show answerHide answer

Answer: D official key

Approach: Add the three bars of each quarter and compare.

Solution:
Q1 = 45 + 30 + 55 = 130.
Q2 = 60 + 50 + 40 = 150.
Q3 = 55 + 65 + 70 = 190.
Q4 = 80 + 75 + 60 = 215 → highest.

Why the other options are wrong:
A, B – clearly lower totals.
C (Q3) – 190 < 215.
E – Q3 and Q4 are not equal (190 vs 215).

Cetking shortcut: Q4 has the two tallest bars (80, 75) and its third bar (60) is only 10 below Q3's 70 — Q4 wins by inspection.

Answer: D

Q143EasyBar Graph

Refer to the passage above ↑

What is the percentage increase in sales of Product A from Q1 to Q4?
Figure for MBA CET 2026 Slot 4 question 143
  1. 75%
  2. 80%
  3. 77.78%
  4. 43.75%
  5. 56.25%
Show answerHide answer

Answer: C official key

Approach: Percentage increase = (new − old)/old × 100.

Solution:
Product A: Q1 = 45, Q4 = 80.
Increase = 80 − 45 = 35.
% increase = 35/45 × 100 = 7/9 × 100 = 77.78%.

Why the other options are wrong:
D (43.75%) – divides by the new value 80 (35/80).
A (75%), B (80%) – rounding or wrong base.
E (56.25%) – wrong values.

Cetking shortcut: 35/45 = 7/9 = 77.78% (memorise ninths: 1/9 = 11.11%).

Answer: C

Q144EasyBar Graph

Refer to the passage above ↑

What is the ratio of the total sales of Product C to the total sales of Product A?
Figure for MBA CET 2026 Slot 4 question 144
  1. 4:11
  2. 15:16
  3. 16:15
  4. 7:11
  5. 8:15
Show answerHide answer

Answer: B official key

Approach: Find each product's annual total, then simplify the ratio.

Solution:
Product C total = 55 + 40 + 70 + 60 = 225.
Product A total = 45 + 60 + 55 + 80 = 240.
C : A = 225 : 240 = 15 : 16 (divide by 15).

Why the other options are wrong:
C (16:15) – ratio reversed (A : C).
A, D, E – not supported by the totals.

Cetking shortcut: Both totals are multiples of 15: 225 = 15 × 15, 240 = 15 × 16.

Answer: B

Q145EasyBar Graph

Refer to the passage above ↑

In which quarter did Product B have the lowest share of the total sales?
Figure for MBA CET 2026 Slot 4 question 145
  1. Q1
  2. Q2
  3. Q3
  4. Q4
  5. Q1 and Q2
Show answerHide answer

Answer: A official key

Approach: Share = Product B ÷ quarter total; compare fractions.

Solution:
Q1: 30/130 = 23.1%.
Q2: 50/150 = 33.3%.
Q3: 65/190 = 34.2%.
Q4: 75/215 = 34.9%.
Lowest share = Q1.

Why the other options are wrong:
B, C, D – each above 33%.
E – Q1 and Q2 shares differ (23.1% vs 33.3%).

Cetking shortcut: In Q1, B (30) is less than a quarter of 130; in every other quarter B is about one-third of the total.

Answer: A

Q146EasyPie Chart

Questions 146–150 share this

Answer the next 5 questions based on the pie chart shown here.

Directions: Answer the questions based on the pie chart provided, which depicts the budget allocation of a startup company for the financial year 2024.

Data read from chart:

CategoryShare of budget
Salaries35%
Marketing25%
R&D15%
Infrastructure15%
Miscellaneous10%
Figure for MBA CET 2026 Slot 4 question 146
If the total budget of the company is Rs. 20 Crores, what is the difference between the amount spent on 'Salaries' and 'Infrastructure'?
  1. Rs. 3 Crores
  2. Rs. 4 Crores
  3. Rs. 5 Crores
  4. Rs. 2.5 Crores
  5. Rs. 3.5 Crores
Show answerHide answer

Answer: B official key

Approach: Convert the percentage difference into an amount.

Solution:
Salaries = 35%, Infrastructure = 15%.
Difference = 20% of the budget.
20% of Rs. 20 crore = Rs. 4 crore.

Why the other options are wrong:
A (3 Cr) – uses 15% difference.
C (5 Cr) – uses Marketing 25%.
D, E – arithmetic slips.

Cetking shortcut: Subtract percentages first, then convert once: 20% × 20 = 4.

Answer: B

Q147EasyPie Chart

Refer to the passage above ↑

The expenditure on 'R&amp;D' is what percent less than the expenditure on 'Marketing'?
Figure for MBA CET 2026 Slot 4 question 147
  1. 10%
  2. 40%
  3. 60%
  4. 25%
  5. 33.33%
Show answerHide answer

Answer: B official key

Approach: 'Percent less than' uses the compared-to value (Marketing) as base.

Solution:
Marketing = 25%, R&D = 15%.
Difference = 10.
% less = 10/25 × 100 = 40%.

Why the other options are wrong:
A (10%) – the percentage-point difference, not percent.
E (33.33%) – uses a wrong base (10/30).
C (60%) – R&D as % of Marketing.
D (25%) – Marketing's share.

Cetking shortcut: R&D/Marketing = 15/25 = 60% ⇒ 40% less.

Answer: B

Q148EasyPie Chart

Refer to the passage above ↑

What is the central angle corresponding to the sector for 'Salaries'?
Figure for MBA CET 2026 Slot 4 question 148
  1. 108 degrees
  2. 120 degrees
  3. 135 degrees
  4. 126 degrees
  5. 144 degrees
Show answerHide answer

Answer: D official key

Approach: Central angle = share % × 360°.

Solution:
Salaries = 35%.
Angle = 0.35 × 360° = 126°.

Why the other options are wrong:
A (108°) – 30%.
B (120°) – one-third.
C (135°) – 37.5%.
E (144°) – 40%.

Cetking shortcut: 1% = 3.6° ⇒ 35 × 3.6 = 126°.

Answer: D

Q149EasyPie Chart

Refer to the passage above ↑

If the company decides to cut the 'Miscellaneous' budget by half and add that amount to 'R&amp;D', what will be the new percentage share of 'R&amp;D'?
Figure for MBA CET 2026 Slot 4 question 149
  1. 18%
  2. 25%
  3. 20%
  4. 22.5%
  5. 17.5%
Show answerHide answer

Answer: C official key

Approach: Budget total is unchanged — just move share from one sector to another.

Solution:
Miscellaneous = 10%; half of it = 5%.
This 5% is moved to R&D: 15% + 5% = 20%.
Total stays 100%, so the new R&D share = 20%.

Why the other options are wrong:
E (17.5%) – adds half of half.
A (18%), D (22.5%) – wrong amount moved.
B (25%) – adds the whole 10%.

Cetking shortcut: Transfers inside the same total don't change the base, so add percentages directly.

Answer: C

Q150EasyPie Chart

Refer to the passage above ↑

What is the ratio of expenditure on 'Marketing' and 'Miscellaneous' together to the expenditure on 'Salaries'?
Figure for MBA CET 2026 Slot 4 question 150
  1. 3:5
  2. 1:1
  3. 5:7
  4. 4:5
  5. 7:5
Show answerHide answer

Answer: B official key

Approach: Add the two shares and compare with Salaries.

Solution:
Marketing + Miscellaneous = 25% + 10% = 35%.
Salaries = 35%.
Ratio = 35 : 35 = 1 : 1.

Why the other options are wrong:
C (5:7) – Marketing alone to Salaries.
E (7:5) – Salaries to Marketing.
A, D – not supported by the data.

Cetking shortcut: Ratios of amounts = ratios of percentages; no need for the total budget.

Answer: B

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