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MBA CET 2025 Slot 5 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.

Q101EasyNumbers

Which of the following is a prime number?

  1. 517
  2. 559
  3. 571
  4. 533
  5. 539
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Answer: C

Approach: A number is prime if no prime up to its square root divides it.

Solution:
Check each option against small primes:
517 = 11 × 47 (5 − 1 + 7 = 11, divisible by 11) → composite
559 = 13 × 43 → composite
533 = 13 × 41 → composite
539 = 7 × 77 = 7² × 11 → composite
571: √571 ≈ 23.9; not divisible by 2, 3 (digit sum 13), 5, 7, 11, 13, 17, 19, 23 → prime

Cetking shortcut: Kill options fast with the 11-test (alternating digit sum) and 7/13 tests; the survivor is the prime.

Answer: C

Q102EasySequences and Series

Find the sum of first 90 even natural numbers.

  1. 8180
  2. 8190
  3. 9180
  4. 8100
  5. 8150
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Answer: B

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

Solution:
Even numbers: 2 + 4 + … + 180 = 2(1 + 2 + … + 90)
= 2 × 90 × 91/2 = 90 × 91
= 8190

Cetking shortcut: n(n+1) directly: 90 × 91 = 8190. Option D (8100 = 90²) is the sum of the first 90 odd numbers — a trap.

Answer: B

Q103EasyFactors

How many distinct prime factors are there in 9900?

  1. 7
  2. 4
  3. 11
  4. 5
  5. 15
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Answer: B

Approach: Prime factorise and count the different primes.

Solution:
9900 = 99 × 100
99 = 3² × 11; 100 = 2² × 5²
9900 = 2² × 3² × 5² × 11
Distinct primes: 2, 3, 5, 11 → 4

Why the other options are wrong:
A (7), C (11), D (5), E (15): confuse distinct primes with total prime factors (2+2+2+1 = 7) or with number of factors.

Answer: B

Q104EasyHCF LCM

Find the HCF of 405, 585, 765 and 900.

  1. 35
  2. 45
  3. 55
  4. 65
  5. 25
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Answer: B

Approach: HCF = product of common primes with their lowest powers.

Solution:
405 = 3⁴ × 5
585 = 3² × 5 × 13
765 = 3² × 5 × 17
900 = 2² × 3² × 5²
Common: 3² and 5 → HCF = 9 × 5 = 45

Cetking shortcut: All four numbers end in 5 or 0 and have digit sum divisible by 9 → divisible by 45. Only 45 among options works (55, 65, 35 fail on 405).

Answer: B

Q105EasyHCF LCM

Two numbers are in the ratio 4:5. Their LCM is 240. Find the sum of numbers.

  1. 108
  2. 180
  3. 120
  4. 105
  5. 125
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Answer: A

Approach: Numbers in ratio a:b (co-prime) are ax and bx; their LCM is abx.

Solution:
Let numbers be 4x and 5x (4, 5 co-prime)
LCM = 4 × 5 × x = 20x = 240 → x = 12
Numbers: 48 and 60
Sum = 9x = 108

Cetking shortcut: Sum = (4 + 5) × (240/20) = 9 × 12 = 108.

Answer: A

Q106EasyAges

Raju got married 8years ago. His present age is 6/5 times his age at the time of his marriage. Raju’s sister was 10 years younger than him at the time of his marriage. The present age of Raju's sister is?

  1. 30 years
  2. 35 years
  3. 38 years
  4. 36 years
  5. 40 years
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Answer: C

Approach: Set the age at marriage as x and translate the fractional relation.

Solution:
Age at marriage = x; present age = x + 8
x + 8 = (6/5)x → 5x + 40 = 6x → x = 40
Raju now = 48
Sister at marriage = 40 − 10 = 30
Sister now = 30 + 8 = 38 years

Why the other options are wrong:
A (30): sister's age at the marriage, not now.
D/E: arithmetic slips; B (35): mixes present and past.

Cetking shortcut: (6/5)x − x = 8 → x/5 = 8 → x = 40 instantly.

Answer: C

Q107EasyEquations Basics

An employer engaged a servant with free boarding and lodging for one year with the condition that the servant will be given Rs.15000 and a cycle at the end of the year. The servant agreed but served the employer only for 9 months and thus received Rs. 11000 and a cycle. The price of the cycle is:

  1. Rs.1100
  2. Rs.1000
  3. Rs.1200
  4. Rs.1500
  5. Rs.1300
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Answer: B

Approach: Total yearly pay = cash + cycle; pay for 9 months = 3/4 of it.

Solution:
Let cycle = ₹x. Full-year pay = 15000 + x
9 months = 3/4 year → (3/4)(15000 + x) = 11000 + x
45000 + 3x = 44000 + 4x
x = ₹1000

Cetking shortcut: Missing 3 months cost him 15000 − 11000 = 4000 cash, which is 1/4 of the yearly pay → yearly pay = 16000 → cycle = 16000 − 15000 = ₹1000.

Answer: B

Q108EasyMensuration

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

  1. 45 cm
  2. 50 cm
  3. 55 cm
  4. 53 cm
  5. 48 cm
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Answer: B

Approach: Square side = L − 4 = W + 3; equate square area to rectangle area.

Solution:
L − 4 = W + 3 → L = W + 7
(W + 3)² = W(W + 7)
W² + 6W + 9 = W² + 7W → W = 9
L = 16
Perimeter = 2(16 + 9) = 50 cm

Cetking shortcut: Check options via L + W = perimeter/2: 25 → W = 9, L = 16; square side 12, 12² = 144 = 16 × 9. ✓

Answer: B

Q109EasyAlgebra

Which of the following is a factor of x³ - 19x + 30 ?

  1. x + 2
  2. x + 1
  3. x - 2
  4. x - 1
  5. x – 7
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Answer: C

Approach: Factor theorem: (x − a) is a factor if f(a) = 0.

Solution:
f(x) = x³ − 19x + 30
x + 2 → f(−2) = −8 + 38 + 30 = 60 ≠ 0
x + 1 → f(−1) = −1 + 19 + 30 = 48 ≠ 0
x − 2 → f(2) = 8 − 38 + 30 = 0 ✓
(Full factorisation: (x − 2)(x − 3)(x + 5))

Why the other options are wrong:
A: f(−2) = 60
B: f(−1) = 48
D: f(1) = 12
E: f(7) = 343 − 133 + 30 = 240

Answer: C

Q110EasyAlgebra

If x+1 is a factor of 2x³ + ax² + 2bx + 1: then find the values of a and b given that 2a - 3b = 4

  1. 5, 3
  2. 5, 2
  3. 3, 4
  4. 2, 7
  5. 3, 6
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Answer: B

Approach: Factor theorem gives one equation; combine with the given linear equation.

Solution:
x + 1 is a factor → f(−1) = 0
2(−1)³ + a(1) + 2b(−1) + 1 = 0 → a − 2b − 1 = 0 → a = 2b + 1
Substitute in 2a − 3b = 4: 4b + 2 − 3b = 4 → b = 2
a = 5 → (a, b) = (5, 2)

Cetking shortcut: Test options in 2a − 3b = 4: only (5, 2) gives 10 − 6 = 4; confirm a − 2b = 1 ✓.

Answer: B

Q111EasyPercentages

In measuring the side of a square, an error of 5% in excess is made. The error % in the calculated area is,

  1. 10.75%
  2. 10.25%
  3. 10.50%
  4. 10.00%
  5. 11.25%
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Answer: B

Approach: Area varies as side², so a +5% side error gives successive +5% and +5%.

Solution:
New side = 1.05s
New area = (1.05)²s² = 1.1025s²
Error in area = 10.25%

Why the other options are wrong:
D (10%): simply doubling 5% ignores the 0.25% cross term.

Cetking shortcut: Successive % formula: a + b + ab/100 = 5 + 5 + 25/100 = 10.25%.

Answer: B

Q112EasyPercentages

In a factory producing parts of an automobile, the parts manufactured on the shop floor are required to go through three quality checks, each conducted after a specific part of the processing on the raw material is completed Only parts that are not rejected at one stage are put through the subsequent stages of production and testing. If average rejection rates at these three testing machines during a month are 10%, 5% and 2% respectively, then what is the effective rejection rate for the whole plant?

  1. 17%
  2. 15.25%
  3. 16.21%
  4. 16.76%
  5. 17.2%
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Answer: C

Approach: Rejections at successive stages apply to the survivors — multiply pass rates.

Solution:
Take 100 parts
After stage 1 (10% rejected): 90
After stage 2 (5% rejected): 0.95 × 90 = 85.5
After stage 3 (2% rejected): 0.98 × 85.5 = 83.79
Effective rejection = 100 − 83.79 = 16.21%

Why the other options are wrong:
A (17%): simply adds 10 + 5 + 2, ignoring that later stages see fewer parts.
B, D, E: computation slips.

Cetking shortcut: Pass rate = 0.9 × 0.95 × 0.98 = 0.8379; rejection = 16.21%.

Answer: C

Q113EasyProfit Loss

A shopkeeper bought 150 calculators at the rate of Rs 250 per calculator. He spent Rs 2500 on transportation and packing If the marked price of a calculator is Rs. 320 per calculator and the shopkeeper gives a discount of 5% on the marked price, then what will be the percentage profit gained by the shopkeeper?

  1. 12%
  2. 13%
  3. 14%
  4. 15%
  5. 16%
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Answer: C

Approach: Profit % = (Total SP − Total CP)/Total CP × 100, with overheads added to CP.

Solution:
CP = 150 × 250 + 2500 = 37,500 + 2,500 = ₹40,000
SP per unit = 320 × 0.95 = ₹304
Total SP = 150 × 304 = ₹45,600
Profit = ₹5,600 → 5600/40000 × 100 = 14%

Cetking shortcut: Per unit: CP = 250 + 2500/150 = 266.67; SP = 304; 304/266.67 = 1.14 → 14%.

Answer: C

Q114EasyProfit Loss

An uneducated retailer marks all his goods at 50% above the cost price and thinking that he will still make 25% profit, offers a discount of 25% on the marked price. What is his actual profit on the sales?

  1. 12%
  2. 12.5%
  3. 15%
  4. 10%
  5. 11%
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Answer: B

Approach: Markup followed by discount: net factor = 1.5 × 0.75.

Solution:
CP = 100 → MP = 150
Discount 25% of 150 = 37.5 → SP = 112.5
Profit = 12.5 on CP 100 → 12.5%

Why the other options are wrong:
A/D/E: arithmetic slips; C (15%): wrong net factor.

Cetking shortcut: Successive: +50% then −25% → 50 − 25 − (50 × 25)/100 = 12.5%.

Answer: B

Q115EasyProfit Loss

On selling a pen at 5% loss and a book at 15% gain Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain then he gains Rs. 13. The actual price of the book (in Rs.) is :

  1. 100
  2. 60
  3. 80
  4. 75
  5. 95
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Answer: C

Approach: Write gain/loss on each item as a % of its CP and solve the two equations.

Solution:
Let CP of pen = P, book = B
Case 1: −0.05P + 0.15B = 7
Case 2: 0.05P + 0.10B = 13
Add: 0.25B = 20 → B = 80
(Check: P = 100 → −5 + 12 = 7 ✓; 5 + 8 = 13 ✓)

Cetking shortcut: The pen terms are equal and opposite, so simply add the two equations: 25% of B = 20 → B = 80.

Answer: C

Q116EasyRatios

Aman divides his property so that his son’s share to his wife's and the wife's share to his daughter are both in the ratio 3:1. If the daughter gets 10,000 less than the son, find the total worth of the property.

  1. 20,000
  2. 16,000
  3. 15,750
  4. 16,250
  5. 20,500
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Answer: D

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

Solution:
Son : Wife = 3 : 1 = 9 : 3
Wife : Daughter = 3 : 1
Son : Wife : Daughter = 9 : 3 : 1
Son − Daughter = 8 units = ₹10,000 → 1 unit = ₹1,250
Total = 13 units = 13 × 1,250 = ₹16,250

Cetking shortcut: Total must be a multiple of 13 × 1250; only 16,250 = 13 × 1250 fits.

Answer: D

Q117EasyMandays

A hostel has provisions for 250 students for 35 days. After 5 days, a fresh batch of 25 students was admitted to the hostel. Again after 10 days, a batch of 25 students left the hostel. How long will the remaining provisions survive?

  1. 17 days
  2. 19 days
  3. 21 days
  4. 15 days
  5. 25 days
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Answer: B

Approach: Mandays (student-days) are conserved: total provisions = students × days.

Solution:
Total = 250 × 35 = 8750 student-days
First 5 days: 250 × 5 = 1250
Next 10 days: 275 × 10 = 2750
Remaining = 8750 − 4000 = 4750
Now 250 students → 4750/250 = 19 days

Cetking shortcut: After 5 days, 30 days of food for 250 remain. 25 extra students for 10 days eat 250 student-days = 1 day for 250. 30 − 10 − 1 = 19 days.

Answer: B

Q118EasyRatios

In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with the remaining meal?

  1. 20
  2. 30
  3. 40
  4. 50
  5. 60
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Answer: B

Approach: Equivalence: 120 men's meal = 200 children's meal.

Solution:
1 man ≡ 200/120 = 5/3 children
150 children eat 150/200 = 3/4 of the meal
Remaining 1/4 of the meal feeds 1/4 × 120 = 30 men

Cetking shortcut: Children left unfed = 50 → men = 50 × 120/200 = 30.

Answer: B

Q119MediumTime & Work

A man and a boy working together can complete a work in 24 days. If for the last 6 days, the man alone does the work, then it is completed in 26 days. How long will the boy take to complete the work alone?

  1. 72 days
  2. 64 days
  3. 84 days
  4. 24 days
  5. 32 days
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Answer: A

Approach: Compare the two schedules to get the man:boy efficiency ratio.

Solution:
Schedule 1: 24 days together → work = 24M + 24B
Schedule 2: 20 days together + 6 days man alone → 26M + 20B
24M + 24B = 26M + 20B → 4B = 2M → M = 2B
Work = 24(2B + B) = 72B → boy alone = 72 days

Cetking shortcut: Last 6 days by man = last 4 days by both → the boy's 4 days of work = man's 2 extra days → M = 2B; boy alone = 24 × 3 = 72.

Answer: A

Q120MediumPartnership

Two friends P and Q started a business investing in the ratio of 5 :6. R joined them after six months investing an amount equal to that of Q's. At the end of the year, 20% profit was earned which was equal to Rs. 98,000. What was the amount invested by R ?

  1. Rs 1,05,000
  2. Rs 1,55,000
  3. Rs. 2,05,000
  4. Rs. 2,10,000
  5. Rs 1,95,000
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Answer: D

Approach: Profit share ∝ capital × time; a 20% annual return on R's money for half a year gives R's share.

Solution:
P : Q : R capitals = 5x : 6x : 6x; months = 12 : 12 : 6
Capital-time = 60x : 72x : 36x = 10 : 12 : 6
R's share of ₹98,000 = 6/28 × 98,000 = ₹21,000
R's money earned 20% per annum for only 6 months → 10% of capital
R's investment = 21,000/0.10 = ₹2,10,000
Check: P = 1,75,000 → 35,000; Q = 2,10,000 → 42,000; R → 21,000; total 98,000 ✓

Why the other options are wrong:
A (1,05,000): treats R's ₹21,000 as a full-year 20% return — that is only R's half-year-equivalent capital.
B, C, E: do not fit the 5:6:6 structure.

Cetking shortcut: Annualised capital = 98,000/0.2 = 4,90,000 = 5x + 6x + 3x = 14x → x = 35,000 → R = 6x = 2,10,000.

Answer: D

Q121EasyPartnership

A and B rent a pasture for 10 months. A puts in 80 cows for 7 months and leaves. How many cows can B put in for the remaining 3 months, if he pays half as much rent again as A?

  1. 250
  2. 280
  3. 275
  4. 200
  5. 120
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Answer: B

Approach: Rent is shared in the ratio of cow-months; 'half as much again' means 1.5 times.

Solution:
A's cow-months = 80 × 7 = 560
B pays 1.5 × A's rent → B's cow-months = 1.5 × 560 = 840
B's cows for 3 months = 840/3 = 280

Why the other options are wrong:
A/C/D/E: misread 'half as much again' as 0.5× (gives 93.3) or use 10 months instead of 3.

Cetking shortcut: Cows = 80 × 7 × 1.5/3 = 280 in one line.

Answer: B

Q122EasyMixtures

An empty fuel tank of a car was filled with A type petrol. When the tank was half-empty, it was filled with B type petrol. Again when the tank was half empty, it was filled with A type petrol. When the tank was half-empty again, it was filled with B type petrol. What is the percentage of A type petrol present in the tank?

  1. 35%
  2. 37%
  3. 40%
  4. 37.5%
  5. 32.5%
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Answer: D

Approach: Each 'half empty then refill' keeps half of what was there and adds the new type.

Solution:
Tank = 1. Fill A: A = 1
Half out, fill B: A = 0.5, B = 0.5
Half out, fill A: A = 0.25 + 0.5 = 0.75, B = 0.25
Half out, fill B: A = 0.375, B = 0.125 + 0.5 = 0.625
A = 37.5%

Cetking shortcut: Work backwards: last fill is B (50%), so A = ½ × (A before) = ½ × 75% = 37.5%.

Answer: D

Q123EasyMixtures

Alcohol costs Rs.3.50 per litre and kerosene oil costs Rs. 2.50 per litre. In what proportion should these be mixed so that the resulting mixture may be Rs. 2.75 per litre?

  1. 1:2
  2. 1:4
  3. 1:5
  4. 1:3
  5. 1:1
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Answer: D

Approach: Alligation: quantity ratio = (dearer − mean) : (mean − cheaper), inverted to the cheaper/dearer items.

Solution:
Alcohol ₹3.50, kerosene ₹2.50, mean ₹2.75
Alcohol : Kerosene = (2.75 − 2.50) : (3.50 − 2.75)
= 0.25 : 0.75 = 1 : 3

Why the other options are wrong:
E (1:1) would give mean ₹3.00; A (1:2) gives ₹2.83; B (1:4) gives ₹2.70; C (1:5) gives ₹2.67.

Cetking shortcut: Mean 2.75 is one-quarter of the way from 2.50 to 3.50 → only 1 part of the dearer liquid in 4 → 1 : 3.

Answer: D

Q124MediumMixtures

Three containers have their volumes in the ratio 3:4:5 They are full of mixtures of milk and water. The mixtures contain milk and water in the ratio of (4:1),(3:1) and (5:2) respectively The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is

  1. 89:71
  2. 191:72
  3. 157:71
  4. 157:53
  5. 151:48
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Answer: D

Approach: Find milk and water in each container on a common base, then add.

Solution:
Volumes 3 : 4 : 5; take them as 3k, 4k, 5k and scale by 35 to clear fractions: 105, 140, 175
C1 (4:1): milk 84, water 21
C2 (3:1): milk 105, water 35
C3 (5:2): milk 125, water 50
Total milk = 314, water = 106 → 157 : 53

Cetking shortcut: Milk = 3×(4/5) + 4×(3/4) + 5×(5/7) = 2.4 + 3 + 3.57 = 8.971 out of 12 = 0.7476 milk fraction; 157/(157 + 53) = 157/210 = 0.7476 ✓ (151:48 gives 0.759, so compute, don't eyeball).

Answer: D

Q125EasyInterest

Aman pays 40 times the annual rent to purchase a building. The rate % per annum he derives from his investment is

  1. 25%
  2. 2.5%
  3. 50%
  4. 5%
  5. 7.5%
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Answer: B

Approach: Rate of return = annual income ÷ investment × 100.

Solution:
Annual rent = R → price = 40R
Return = R/40R × 100 = 2.5%

Why the other options are wrong:
D (5%) would need 20 years' rent; A (25%) misplaces the decimal.

Cetking shortcut: Rate = 100/40 = 2.5%.

Answer: B

Q126MediumInterest

Mr. Dua invested money in two schemes, A and B offering compound interest @ 8 p.c.p.a. and 12 p.c.p.a. respectively. If the total amount of interest accrued through two schemes together in two years was Rs 4538.00 and the total amount invested was Rs 20,000, what was the amount invested in Scheme A?

  1. Rs 6000
  2. Rs. 6250
  3. Rs. 6500
  4. Rs. 16000
  5. Rs. 16500
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Answer: B

Approach: CI for 2 years at r% = P[(1 + r/100)² − 1]; split the principal as x and 20000 − x.

Solution:
8% for 2 years → 16.64% of x
12% for 2 years → 25.44% of (20000 − x)
0.1664x + 0.2544(20000 − x) = 4538
5088 − 0.088x = 4538 → 0.088x = 550 → x = 6250
Check: 1040 + 0.2544 × 13750 = 1040 + 3498 = 4538 ✓

Cetking shortcut: If all ₹20,000 were at 12%, interest = 5088. Each rupee moved to 8% loses 0.088. Loss = 550 → 550/0.088 = 6250 (alligation on effective 2-year rates).

Answer: B

Q127EasyInterest

A sum of money is borrowed and paid back in two annual installments of Rs 882 each, allowing 5% compound interest. The sum borrowed was

  1. Rs. 1600
  2. Rs. 1620
  3. Rs. 1640
  4. Rs. 1680
  5. Rs. 1650
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Answer: C

Approach: Sum borrowed = present value of the instalments at 5% CI.

Solution:
PV of 1st instalment = 882/1.05 = 840
PV of 2nd instalment = 882/1.05² = 882/1.1025 = 800
Sum = 840 + 800 = ₹1640

Cetking shortcut: 882 = 2 × 441 = 2 × 21²; with 1.05 = 21/20, PVs are 882 × 20/21 = 840 and 840 × 20/21 = 800.

Answer: C

Q128EasyTime & Work

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs 3000 as wages for the whole work, what are the daily wages A, B and C respectively (in rs ? ) :

  1. 300,200,250
  2. 200,300,250
  3. 300,200,400
  4. 300,250,500
  5. 300,400,250
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Answer: A

Approach: Wages are split in the ratio of work done; daily wage = share ÷ days worked.

Solution:
A's 5 days = 5/10 = 1/2 of the work
B's 5 days = 5/15 = 1/3
C's 2 days = 1 − 1/2 − 1/3 = 1/6
Shares: A 1500, B 1000, C 500
Daily: A 1500/5 = 300, B 1000/5 = 200, C 500/2 = 250

Cetking shortcut: A's daily wage must be 300 and B's 200 (only A, C, D, E start with 300); C's work 1/6 in 2 days → 500/2 = 250 → A.

Answer: A

Q129EasyTime & Work

A is thrice as efficient as B. Working together, they complete the work in 3 days. If working alone, B takes 8 days more than A to finish the whole work. What is the number of days taken by A to finish the whole work alone?

  1. 6
  2. 4
  3. 8
  4. 12
  5. 10
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Answer: B

Approach: Time is inversely proportional to efficiency.

Solution:
Efficiency A : B = 3 : 1 → time A : B = 1 : 3
Difference = 2 units = 8 days → 1 unit = 4
A = 4 days, B = 12 days
Check together: 1/4 + 1/12 = 1/3 → 3 days ✓

Why the other options are wrong:
D (12) is B's time, not A's.
A/C/E fail the 'together in 3 days' check.

Answer: B

Q130EasyPipes and Cisterns

A, B and C are three taps connected to a tank. A and B together can fill the tank in 6 h, B and C together can fill it in 10 h and A and C together can fill it in 7(1/2) h. In how much time would all three take to fill the tank?

  1. 6 h
  2. 5 h
  3. 10 h
  4. 12 h
  5. 8 h
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Answer: B

Approach: Add the three pairwise rates: it counts each tap twice.

Solution:
A + B = 1/6, B + C = 1/10, A + C = 1/7.5 = 2/15
Sum = 5/30 + 3/30 + 4/30 = 12/30 = 2/5
2(A + B + C) = 2/5 → A + B + C = 1/5
Time = 5 hours

Cetking shortcut: LCM 30 tank: pairwise rates 5, 3, 4 units/h; sum 12 = twice the total → 6 units/h → 30/6 = 5 h.

Answer: B

Q131EasyPipes and Cisterns

A pump can be operated both for filling a tank and for emptying it. The capacity of the tank is 2400 m³ The emptying capacity of the pump is 10 m³ per minute higher than its filling capacity. Consequently, the pump needs 8 minutes less to empty the tank than to fill it Find the filling capacity of the pump

  1. 50 m³/min
  2. 56 m³/min
  3. 55 m³/min
  4. 52 m³/min
  5. 54 m³/min
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Answer: A

Approach: Time = volume ÷ rate; set up the 8-minute difference.

Solution:
Filling rate = y m³/min, emptying = y + 10
2400/y − 2400/(y + 10) = 8
2400 × 10 = 8y(y + 10) → y(y + 10) = 3000
y² + 10y − 3000 = 0 → (y − 50)(y + 60) = 0 → y = 50
Check: 2400/50 = 48 min, 2400/60 = 40 min, difference 8 ✓

Cetking shortcut: Option test: need y(y + 10) = 3000 → 50 × 60 = 3000 ✓ immediately.

Answer: A

Q132MediumTime Speed Distance

A road from A to B is 11.5 km long: first goes uphill, then crosses a plain, and then goes downhill. A person walking from A to B covered this road in 2 h 54 min, and the return journey took him 3 h 6 min. His speed uphill is 3 km/h, on the plain 4 km/h and downhill is 5 km/h. What is the length of the plain part of the journey?

  1. 4.5 km
  2. 5 km
  3. 5.5 km
  4. 6 km
  5. 4 km
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Answer: E

Approach: Uphill one way is downhill the other way; add the two journeys so hill parts combine.

Solution:
Let uphill (A→B) = x, plain = y, downhill (A→B) = z; x + y + z = 11.5
A→B: x/3 + y/4 + z/5 = 2.9 h (2 h 54 min)
B→A: x/5 + y/4 + z/3 = 3.1 h (3 h 6 min)
Add: (1/3 + 1/5)(x + z) + y/2 = 6 → (8/15)(11.5 − y) + y/2 = 6
× 30: 16(11.5 − y) + 15y = 180 → 184 − y = 180 → y = 4 km
(Also x = 3, z = 4.5: 1 + 1 + 0.9 = 2.9 ✓)

Cetking shortcut: Adding the two trips removes the direction issue — every hill km costs 8/15 h round trip, every plain km 1/2 h. If all 11.5 km were hill: 11.5 × 8/15 = 6.133 h; actual 6 h; each plain km saves 1/30 h → 0.133 × 30 = 4 km.

Answer: E

Q133EasyTime Speed Distance

In a 100 m race, A beats B by 10m and C by 13m. In a race of 180 m , B will beat C by

  1. 5.4 m
  2. 5 m
  3. 6 m
  4. 4 m
  5. 4.4 m
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Answer: C

Approach: Distances covered in the same time are in the ratio of speeds.

Solution:
When A runs 100, B runs 90 and C runs 87
So B : C = 90 : 87 = 30 : 29
When B runs 180, C runs 180 × 29/30 = 174
B beats C by 180 − 174 = 6 m

Why the other options are wrong:
A (5.4 m): uses 3 m per 100 m instead of 3 m per 90 m.
B, D, E: arithmetic slips; only 6 m satisfies the 30 : 29 speed ratio.

Cetking shortcut: B beats C by 3 m in 90 m → 1 m per 30 m → 6 m in 180 m.

Answer: C

Q134EasyTrains Boats

Two trains 130 m and 110m long are going in the same direction. The faster train takes one minute to pass the other completely. If they are moving in opposite directions, they pass each other completely in 3 seconds. Find the speed of each train?

  1. 42 m/s, 38 m/s
  2. 36 m/s, 42 m/s
  3. 38 m/s, 36 m/s
  4. 40 m/s, 36 m/s
  5. 42 m/s, 36 m/s
Show answerHide answer

Answer: A

Approach: Crossing distance = sum of lengths; relative speed is difference (same direction) or sum (opposite).

Solution:
Total length = 130 + 110 = 240 m
Same direction: F − S = 240/60 = 4 m/s
Opposite: F + S = 240/3 = 80 m/s
F = 42 m/s, S = 38 m/s

Why the other options are wrong:
E (42, 36) and D (40, 36): sum 78/76, not 80.
B, C: difference not 4 or sum wrong.

Cetking shortcut: Options must sum to 80 and differ by 4: only (42, 38).

Answer: A

Q135EasyTrains Boats

Aman rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:

  1. 1 km/hr
  2. 1.5 km/hr
  3. 2 km/hr
  4. 3 km/hr
  5. 1.75 km/hr
Show answerHide answer

Answer: A

Approach: Equal time → distance ∝ speed; then use the total-time equation.

Solution:
4/(x + y) = 3/(x − y) → x = 7y
Downstream = 8y, upstream = 6y
48/8y + 48/6y = 14 → 6/y + 8/y = 14 → y = 1 km/h

Cetking shortcut: Downstream : upstream speed = 4 : 3 = 8 : 6 → stream = (8 − 6)/2 = 1 unit; total time 48/8 + 48/6 = 14 ✓ with unit = 1.

Answer: A

Q136MediumTime Speed Distance

A ship 156 km from the shore springs a leak which admits 2 1/3 metric tons of water in 6 1/2 minutes. A quantity of 68 metric tons would suffice to sink it, but the pumps can throw out 15 metric tons in an hour. The average rate of sailing so that it just reaches the shore as it begins to sink should be

  1. 16 km/hr
  2. 18 km/hr
  3. 15 km/hr
  4. 14 km/hr
  5. 17 km/hr
Show answerHide answer

Answer: C

Approach: Net inflow = leak rate − pump rate; ship must reach shore before 68 t accumulates.

Solution:
Leak: (7/3) t in (13/2) min → 14/39 t/min → 840/39 = 280/13 t/h
Net inflow = 280/13 − 15 = 85/13 t/h
Time to sink = 68 ÷ 85/13 = 68 × 13/85 = 10.4 h
Speed = 156/10.4 = 15 km/h

Why the other options are wrong:
A/B/D/E: rounding the leak rate early or forgetting the pump gives wrong times (e.g. ignoring pump → 3.16 h).

Cetking shortcut: Keep fractions: 68 × 13/85 = 4 × 13/5 = 10.4 h exactly → 156/10.4 = 15.

Answer: C

Q137EasyQuadrilaterals

Instead of walking along two adjacent sides of a rectangular fields a boy took a short cut along the diagonal and saved the distance equal to half of the longer side. Then the ratio of the shorter side to the longer side is

  1. 2/3
  2. 3/4
  3. 1/5
  4. 1/4
  5. 2/5
Show answerHide answer

Answer: B

Approach: Saving = (a + b) − diagonal = b/2, where b is the longer side.

Solution:
a + b − √(a² + b²) = b/2 → √(a² + b²) = a + b/2
Square: a² + b² = a² + ab + b²/4
3b²/4 = ab → a/b = 3/4

Why the other options are wrong:
A (2/3): 2 + 3 − √13 ≈ 1.39 ≠ 1.5.
C/D/E: savings don't equal half the longer side.

Cetking shortcut: Try 3-4-5: walking 3 + 4 = 7, diagonal 5, saved 2 = half of 4 ✓.

Answer: B

Q138EasyPolygons

The interior angle of a regular polygon exceeds the exterior angle by 132°. Then the number of sides in the polygon is

  1. 15
  2. 18
  3. 20
  4. 12
  5. 25
Show answerHide answer

Answer: A

Approach: Interior + exterior = 180°, and exterior angle = 360°/n for a regular polygon.

Solution:
Let exterior = x → interior = 180 − x
(180 − x) − x = 132 → x = 24°
n = 360/24 = 15

Why the other options are wrong:
B (18) gives exterior 20°, difference 140°.
C (20) → 18°, difference 144°.
D (12) → 30°, difference 120°.
E (25) → 14.4°, difference 151.2°.

Cetking shortcut: Exterior = (180 − 132)/2 = 24° → n = 15.

Answer: A

Q139MediumMensuration

Two sides of a plot measure 32m and 24m and the angle between them a perfect right angle. The other two sides measure 25m each and the other three angles are not right angles. What is the area of the plot? (in m²)

Figure for MBA CET 2025 Slot 5 question 139
  1. 534
  2. 634
  3. 684
  4. 456
  5. 658
Show answerHide answer

Answer: C

Approach: Split the plot along the diagonal into a right triangle and an isosceles triangle.

Solution:
Right triangle legs 24 and 32 → diagonal = 40 (8 × 3-4-5)
Area 1 = ½ × 24 × 32 = 384
Isosceles triangle 25, 25, base 40 → height = √(25² − 20²) = 15
Area 2 = ½ × 40 × 15 = 300
Total = 384 + 300 = 684 m²

Cetking shortcut: Both triangles are triplets: 24-32-40 and 15-20-25 → no square roots needed.

Answer: C

Q140EasyMensuration

The length of a room is double its breadth. The cost of coloring the ceiling at Rs 25 per sq. meter is Rs 5000 and the cost of painting the four walls at Rs. 240 per sq metre is Rs. 64800. Find the height of the room:

  1. 4 m
  2. 4.5 m
  3. 5 m
  4. 3.5 m
  5. 3 m
Show answerHide answer

Answer: B

Approach: Ceiling = l × b; four walls = 2(l + b)h.

Solution:
Ceiling area = 5000/25 = 200 m² → 2x² = 200 → x = 10 → b = 10, l = 20
Wall area = 64800/240 = 270 m²
2(20 + 10)h = 270 → 60h = 270 → h = 4.5 m

Why the other options are wrong:
A (4 m) → walls 240 m² → ₹57,600.
C (5 m) → 300 m² → ₹72,000.

Answer: B

Q141EasyMensuration

The trunk of a tree is a right cylinder 1.5m in radius and 10m high. What is the volume of the timber which remains when the trunk is trimmed just enough to reduce it to a rectangular parallelopiped on a square base?

  1. 40 m³
  2. 45 m³
  3. 35 m³
  4. 25 m³
  5. 50 m³
Show answerHide answer

Answer: B

Approach: The largest square cross-section inside a circle has its diagonal equal to the diameter.

Solution:
Diameter = 2 × 1.5 = 3 m = diagonal of the square base
Side² = (diagonal²)/2 = 9/2 = 4.5 m²
Volume = 4.5 × 10 = 45 m³

Cetking shortcut: Square area = d²/2 directly — no need to find the side.

Answer: B

Q142EasyMensuration

A garden is 24 m long and 14 m wide. There is a path 1 m wide outside the garden along its sides If the path is to be constructed with square marble tiles 20 cm*20 cm, find the number of tiles required to cover the path?

  1. 1500
  2. 2000
  3. 1800
  4. 2200
  5. 2400
Show answerHide answer

Answer: B

Approach: Path area = outer rectangle − garden; tiles = path area ÷ tile area.

Solution:
Outer = (24 + 2) × (14 + 2) = 26 × 16 = 416 m²
Garden = 24 × 14 = 336 m²
Path = 80 m²
Tile = 0.2 × 0.2 = 0.04 m²
Tiles = 80/0.04 = 2000

Why the other options are wrong:
Forgetting the four 1 m x 1 m corner squares gives 76 m² → 1900 tiles; double-counting them gives other wrong totals. Tile area is 0.04 m², i.e. 25 tiles per m².

Cetking shortcut: Outer path area = 2w(l + b) + 4w² = 2(38) + 4 = 80 m²; tiles = 80 × 25 = 2000.

Answer: B

Q143MediumClocks & Calendars

Find the day of the week on 26 January 1950

  1. Monday
  2. Tuesday
  3. Wednesday
  4. Thursday
  5. Friday
Show answerHide answer

Answer: D

Approach: Odd-days method: count odd days up to the date, then map 0 = Sunday, 1 = Monday, …

Solution:
26 Jan 1950 = 1949 complete years + 26 days
1600 years → 0 odd days; 300 years → 1 odd day
49 years = 12 leap + 37 ordinary → 12 × 2 + 37 × 1 = 61 → 61 mod 7 = 5
26 days of January → 26 mod 7 = 5
Total = 0 + 1 + 5 + 5 = 11 → 11 mod 7 = 4 → Thursday

Cetking shortcut: Known anchor: India's first Republic Day (26 Jan 1950) was a Thursday.

Answer: D

Q144EasyClocks & Calendars

An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

  1. 144 degrees
  2. 160 degrees
  3. 180 degrees
  4. 90 degrees
  5. 125 degrees
Show answerHide answer

Answer: C

Approach: The hour hand moves 30° per hour.

Solution:
8 a.m. to 2 p.m. = 6 hours
Rotation = 6 × 30° = 180°

Why the other options are wrong:
A/B/D/E: not multiples of 30° for whole hours (except D = 3 hours).

Answer: C

Q145EasyClocks & Calendars

On January 12, 1980, it was a Saturday. The day of the week on January 12, 1979 was

  1. Monday
  2. Tuesday
  3. Wednesday
  4. Thursday
  5. Friday
Show answerHide answer

Answer: E

Approach: Going back one year shifts the day back by the number of odd days in between.

Solution:
Jan 12, 1979 → Jan 12, 1980 contains Feb 1979 only (Feb 29, 1980 comes after Jan 12)
365 days = 52 weeks + 1 odd day
So Jan 12, 1979 = Saturday − 1 = Friday

Why the other options are wrong:
D (Thursday): wrongly counts 1980 as leap for this interval (2 odd days).

Answer: E

Q146HardPnC

Three Englishmen and three Frenchmen work for the same company Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

  1. 10
  2. 9
  3. 8
  4. 7
  5. 6
Show answerHide answer

Answer: C

Approach: Treat it as a gossip problem with a restriction: only the bilingual Englishman (E3) can talk to both groups. Use the 4-call core trick so that only 8 calls are needed.

Solution:
Englishmen E1, E2, E3 (E3 knows French); Frenchmen F1, F2, F3.
1. E1–E3 (English) and 2. E2–E3 (English): E3 now knows E1, E2, E3.
3. F1–F2 (French): they know F1, F2.
4. E3–F3 (French): both know E1, E2, E3, F3.
5. F1–E3 (French): both now know all 6 secrets.
6. F2–F3 (French): both now know all 6 secrets.
7. E3–E1 and 8. E3–E2 (English): E1 and E2 now know all 6.
Every call uses a language both speakers know, and everyone knows all six secrets after 8 calls.
8 is also the minimum: for 6 people at least 2n − 4 = 8 calls are needed even with no language restriction.

Why the other options are wrong:
A (10) and B (9): possible but not minimal — e.g. 'gather at E3 then spread' takes 9.
D (7), E (6): fewer than the 2n − 4 = 8 lower bound, so impossible.

Answer: C

Q147EasyPnC

In a chess competition involving some boys and girls of a school: every student had to play exactly one game with every other student It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

  1. 215
  2. 230
  3. 200
  4. 180
  5. 250
Show answerHide answer

Answer: C

Approach: Games among n players = nC2; mixed games = boys × girls.

Solution:
Girls: G(G − 1)/2 = 45 → G(G − 1) = 90 = 10 × 9 → G = 10
Boys: B(B − 1)/2 = 190 → B(B − 1) = 380 = 20 × 19 → B = 20
Boy-girl games = 20 × 10 = 200

Cetking shortcut: Recognise 45 = 10C2 and 190 = 20C2 instantly.

Answer: C

Q148MediumProbability

It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they will win alternately is:

  1. 5/72
  2. 5/36
  3. 7/72
  4. 7/36
  5. 1/36
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Answer: B

Approach: 'Win alternately' over 3 games means the sequences A-B-A or B-A-B (independent games).

Solution:
P(A wins) = 6/12 = 1/2, P(B wins) = 4/12 = 1/3 (draw = 1/6)
P(A, B, A) = 1/2 × 1/3 × 1/2 = 1/12
P(B, A, B) = 1/3 × 1/2 × 1/3 = 1/18
Total = 1/12 + 1/18 = 3/36 + 2/36 = 5/36

Why the other options are wrong:
A (5/72): halves the result wrongly.
E (1/36)/C/D: count only one sequence or mis-add.

Answer: B

Q149EasyCircles

Aman riding a bicycle completes one lap of a circular field along its circumference at the speed of 14.4 km/hr in 1 minute 28 seconds. What is the area of the field?

  1. 7958 sq mt
  2. 9856 sq mt
  3. 8842 sq mt
  4. 8958 sq mt
  5. 7856 sq mt
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Answer: B

Approach: Distance in one lap = circumference; then area = πr².

Solution:
Speed = 14.4 × 5/18 = 4 m/s
Time = 1 min 28 s = 88 s → circumference = 352 m
2 × 22/7 × r = 352 → r = 56 m
Area = 22/7 × 56 × 56 = 9856 m²

Cetking shortcut: 352 = 2πr with π = 22/7 → r = 352 × 7/44 = 56; area = 22 × 8 × 56 = 9856.

Answer: B

Q150EasyEquations Basics

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at the rate 4 kmph and partly on bicycle at the rate 9 kmph. The distance travelled on foot is

  1. 16 km
  2. 14 km
  3. 15 km
  4. 13 km
  5. 18 km
Show answerHide answer

Answer: A

Approach: Total time = time on foot + time on bicycle.

Solution:
Foot = x km, bicycle = 61 − x km
x/4 + (61 − x)/9 = 9
× 36: 9x + 4(61 − x) = 324 → 5x + 244 = 324 → x = 16 km

Why the other options are wrong:
B (14 km): 3.5 h on foot + 47 km/9 ≠ 9 h.
Others fail the time check.

Cetking shortcut: Alligation on time: if all 9 h by cycle → 81 km; 20 km extra; each foot-hour loses 5 km → 4 h on foot → 16 km.

Answer: A

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