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MBA CET 2025 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.

Q101MediumRemainders

After the division of a number successively by 3, 4 and 7, the remainders obtained are 2,1 and 4 respectively. What will be the remainder if 84 divides the same number?

  1. 50
  2. 48
  3. 53
  4. 80
  5. 72
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Answer: C

Approach: Successive division — reconstruct the number from the last quotient.

Solution:
Let the last quotient be k. Work from the innermost division outwards.
Third division (by 7) leaves 4: the quotient before it = 7k + 4.
Second division (by 4) leaves 1: the quotient before it = 4(7k + 4) + 1 = 28k + 17.
First division (by 3) leaves 2: N = 3(28k + 17) + 2 = 84k + 53.
Note 3 × 4 × 7 = 84, so N = 84k + 53 ⇒ remainder on dividing by 84 = 53.
Check with k = 0: N = 53 → 53 ÷ 3 = 17 r 2; 17 ÷ 4 = 4 r 1; 4 ÷ 7 = 0 r 4. ✓

Cetking shortcut: Remainder by the product of divisors = r1 + d1·r2 + d1·d2·r3 = 2 + 3×1 + 12×4 = 53.

Answer: C

Solution figure for question 101Solution figure for question 101Solution figure for question 101
Q102EasyRatios

At a college football game, 4/5 of the seats on the lower deck of the stadium were sold. If 1/4 of all the seating in the stadium is located on the lower deck, and if 2/3 of all the seats in the stadium were sold, what is the fraction of unsold seats on the lower deck to the total number of unsold seats?

  1. 3/20
  2. 1/20
  3. 1/2
  4. 1/5
  5. 2/5
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Answer: A

Approach: Assume a convenient total and convert each fraction to counts.

Solution:
Let total seats = 60 (LCM of 4, 5, 3).
Lower deck = 1/4 × 60 = 15 seats; sold = 4/5 × 15 = 12 → unsold on lower deck = 3.
Total sold = 2/3 × 60 = 40 → total unsold = 20.
Required fraction = 3/20.

Why the other options are wrong:
B (1/20): this is unsold lower-deck seats as a fraction of all seats, not of unsold seats.
C, D, E: do not follow from the counts 3 and 20.

Cetking shortcut: Take total = 60 (LCM of denominators) — all counts become integers instantly.

Answer: A

Solution figure for question 102Solution figure for question 102
Q103MediumNumbers

Find the smallest number, which must be subtracted from 306452 to make it a perfect square.

  1. 1980
  2. 1748
  3. 1396
  4. 1558
  5. 1290
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Answer: B

Approach: Perfect squares — estimate the root, then test which option leaves a square.

Solution:
553² = 305809 and 554² = 306916, so 306452 lies between them. Strictly, the smallest subtraction is 306452 − 305809 = 643, but 643 is not an option, so we test the options.
306452 − 1748 = 304704 = 552² (552² = 302500 + 2200 + 4 = 304704). ✓
Other options: 306452 − 1980 = 304472, − 1396 = 305056, − 1558 = 304894, − 1290 = 305162 — all lie strictly between 552² = 304704 and 553² = 305809, so none is a perfect square.
Hence the only choice that gives a perfect square is 1748.

Why the other options are wrong:
A, C, D, E: the results (304472, 305056, 304894, 305162) are not squares.

Cetking shortcut: Only the option giving 552² = 304704 works; the result must end in …04, which spots it fast.

Answer: B

Q104EasyHCF LCM

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

  1. 94
  2. 74
  3. 240
  4. 364
  5. 184
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Answer: D

Approach: LCM + common remainder, then the divisibility condition.

Solution:
LCM(6, 9, 15, 18) = 90, so the number = 90k + 4.
Test k = 1, 2, 3, 4: 94, 184, 274, 364.
Divisible by 7? 94 = 7×13 + 3 ✗; 184 = 7×26 + 2 ✗; 274 = 7×39 + 1 ✗; 364 = 7×52 ✓.
Least such multiple of 7 = 364.

Why the other options are wrong:
A (94), E (184): of the form 90k + 4 but not multiples of 7.
B (74): leaves remainder 2 on division by 6, so fails.
C (240): leaves remainder 0 on division by 6.

Cetking shortcut: 90k + 4 ≡ 6k + 4 (mod 7) = 0 ⇒ 6k ≡ 3 ⇒ k ≡ 4 → number = 364.

Answer: D

Q105EasyHCF LCM

A shopkeeper has three kinds of sugar: 184 kg, 230 kg and 276 kg. He wants to store them in a minimum number of bags of equal size without mixing. Find the required number of bags

  1. 18
  2. 15
  3. 17
  4. 20
  5. 16
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Answer: B

Approach: Biggest equal bag = HCF ⇒ fewest bags.

Solution:
Largest bag size = HCF(184, 230, 276).
184 = 2³ × 23, 230 = 2 × 5 × 23, 276 = 2² × 3 × 23 → HCF = 2 × 23 = 46 kg.
Bags = 184/46 + 230/46 + 276/46 = 4 + 5 + 6 = 15.

Cetking shortcut: Differences 46 and 46 hint the HCF directly: 230 − 184 = 46 and 46 divides all three.

Answer: B

Solution figure for question 105Solution figure for question 105
Q106MediumAverages

The average mark of a class of n students is 64. If eight new students with an average mark of 73 join the class, the new average of the entire class will be a whole number. Find the number of students in the class (given that n lies between 25 and 60)

  1. 18
  2. 36
  3. 72
  4. 27
  5. 45
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Answer: B

Approach: Average shift — extra marks of the newcomers spread over the whole class.

Solution:
Total marks = 64n + 8 × 73 = 64n + 584.
New average = (64n + 584)/(n + 8) = 64 + 72/(n + 8).
For a whole number, (n + 8) must divide 72: possible n + 8 = 36, 72, 24, 18, …
With 25 < n < 60, n + 8 lies between 33 and 68 ⇒ n + 8 = 36, so n = 28.
The class (old + new students) therefore has 36 students; new average = 64 + 72/36 = 66.
Option B (36) is the strength of the entire class after the new students join.

Why the other options are wrong:
A (18), C (72): divisors of 72 but n + 8 = 18 or 72 gives n outside 25–60.
D (27), E (45): 72/(27) and 72/(45) are not whole numbers.

Cetking shortcut: Newcomers bring 8 × 9 = 72 extra marks; whole-class size must be a divisor of 72 between 33 and 68 → 36.

Answer: B

Solution figure for question 106Solution figure for question 106
Q107EasyAverages

The age of the captain of a cricket team of 11 players is 25 years, the wicket keeper is 3 years older than the captain. If the ages of these two are excluded, the average age of the remaining players of the team becomes 1 year less than the average age of the whole team. What is the average age of the whole team?

  1. 25
  2. 22
  3. 26
  4. 28
  5. 24
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Answer: B

Approach: Split the team total into excluded + remaining.

Solution:
Let the team average = x ⇒ team total = 11x.
Captain = 25, wicket-keeper = 28 ⇒ their total = 53.
Remaining 9 players: average x − 1 ⇒ total 9(x − 1).
11x = 53 + 9x − 9 ⇒ 2x = 44 ⇒ x = 22.

Cetking shortcut: Removing two members lowers the average by 1 for 9 players: 53 − 2x = −9 ⇒ the two exceed the average by 9 in total ⇒ x = (53 − 9)/2 = 22.

Answer: B

Solution figure for question 107Solution figure for question 107
Q108EasyAverages

The average number of printing errors in a book of 512 pages is 4. If the total number of printing errors on the first 302 pages is 1208, then what is the average number of printing errors on the remaining pages?

  1. 2
  2. 3
  3. 5
  4. 4
  5. 6
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Answer: D

Approach: Total = average × number.

Solution:
Total errors = 512 × 4 = 2048.
Errors in remaining pages = 2048 − 1208 = 840.
Remaining pages = 512 − 302 = 210.
Average = 840/210 = 4.

Cetking shortcut: First 302 pages average 1208/302 = 4, equal to the overall average ⇒ the rest must also average 4.

Answer: D

Solution figure for question 108Solution figure for question 108
Q109MediumEquations Basics

A woman sells to the first customer half of her stock of apples and half an apple, to the second customer she sells half her remaining stock and half an apple, and so on to the third and to the fourth customer. She finds that she now has 15 apples left. How many apples did she have before she started selling them?

  1. 250
  2. 255
  3. 254
  4. 270
  5. 350
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Answer: B

Approach: Reverse the operations from the end.

Solution:
After a sale: left = x/2 − ½ ⇒ before the sale x = 2(left + ½) = 2 × left + 1.
After 4th customer: 15 ⇒ before 4th = 31.
Before 3rd = 2 × 31 + 1 = 63.
Before 2nd = 2 × 63 + 1 = 127.
Before 1st = 2 × 127 + 1 = 255.

Cetking shortcut: Each step maps L → 2L + 1, i.e. (L + 1) doubles: 16 × 2⁴ = 256 ⇒ start = 255.

Answer: B

Solution figure for question 109Solution figure for question 109Solution figure for question 109
Q110MediumEquations Basics

A boy has enough money to buy 20 exercise books. If each book cost 25 paise less, he could buy two more books and still have 70 paise left. How much money did he have originally?

  1. 35
  2. 42
  3. 48
  4. 52
  5. 50
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Answer: C

Approach: Form one linear equation in the price.

Solution:
Let price per book = p paise ⇒ money = 20p paise.
New price = (p − 25) paise; 22 books + 70 paise left: 22(p − 25) + 70 = 20p.
22p − 550 + 70 = 20p ⇒ 2p = 480 ⇒ p = 240 paise.
Money = 20 × 240 = 4800 paise = Rs. 48.
Check: 22 × 215 = 4730; 4800 − 4730 = 70 paise. ✓

Cetking shortcut: Options ÷ 20 give the price; only Rs. 48 → Rs. 2.40 per book works: Rs. 2.15 × 22 = 47.30, leaving 70 paise.

Answer: C

Solution figure for question 110Solution figure for question 110
Q111MediumNumbers

The digit at unit place of a two-digit number is increased by 100% and the digit at ten places of the same number is increased by 50%. The new number thus formed is 19 more than the original number. What is the original number?

  1. 22
  2. 25
  3. 27
  4. 29
  5. 21
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Answer: D

Approach: Place-value algebra.

Solution:
Original number = 10a + b.
Units digit +100% → 2b; tens digit +50% → 1.5a. New number = 10(1.5a) + 2b = 15a + 2b.
15a + 2b = 10a + b + 19 ⇒ 5a + b = 19.
a = 2 ⇒ b = 9 → 29; a = 3 ⇒ b = 4 → 34 (not an option).
Check 29: tens 2 → 3, units 9 → 18 ⇒ new value 30 + 18 = 48 = 29 + 19. ✓

Why the other options are wrong:
A (22): 5×2 + 2 = 12 ≠ 19.
B (25): 15 ≠ 19.
C (27): 17 ≠ 19.
E (21): 11 ≠ 19.

Cetking shortcut: Just test 5 × (tens) + (units) = 19 on the options — only 29 gives 10 + 9 = 19.

Answer: D

Solution figure for question 111Solution figure for question 111Solution figure for question 111
Q112EasyPercentages

The price of sugar was increased by 32%. A family reduces its consumption so that the expenditure on sugar is up by 10% only. If the total consumption of sugar before the price rises was 10 kg per month, then the consumption of sugar per month at present (in kg) is:

  1. 8.5
  2. 9
  3. 8.33
  4. 8.6
  5. 8.2
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Answer: C

Approach: Expenditure = Price × Consumption.

Solution:
Let original price = P; original expenditure = 10P.
New price = 1.32P; new expenditure = 1.10 × 10P = 11P.
New consumption = 11P/1.32P = 11/1.32 = 8.33 kg.

Why the other options are wrong:
A, B, D, E: none equals 11/1.32 = 8.33.

Cetking shortcut: Consumption factor = 1.10/1.32 = 5/6 ⇒ 10 × 5/6 = 8.33 kg.

Answer: C

Solution figure for question 112Solution figure for question 112
Q113EasyPercentages

In a housing society, 30% of residents are men over the age of 18 and 40% are women over the age of 18. If there are 24 children living in the housing society, then how many total residents live in the society?

  1. 75
  2. 80
  3. 72
  4. 100
  5. 32
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Answer: B

Approach: Complement percentage gives the count.

Solution:
Adults = 30% + 40% = 70% ⇒ children = 30%.
30% of total = 24 ⇒ total = 24 × 100/30 = 80.

Cetking shortcut: 10% = 8 ⇒ 100% = 80.

Answer: B

Solution figure for question 113
Q114EasyPercentages

The price of an item was increased by 10%. This reduced the monthly total sales by 20%. What is the overall effect on the value of monthly sales?

  1. 12% decrease
  2. 10% increase
  3. 8% decrease
  4. 10% decrease
  5. 13% decrease
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Answer: A

Approach: Successive percentage change.

Solution:
Value factor = 1.10 × 0.80 = 0.88.
Change = 0.88 − 1 = −12% ⇒ 12% decrease.
Formula check: 10 − 20 + (10 × −20)/100 = −10 − 2 = −12%.

Why the other options are wrong:
D (10% decrease): simple addition, ignores the cross term.
C, E, B: do not match 1.1 × 0.8.

Cetking shortcut: a + b + ab/100 = 10 − 20 − 2 = −12.

Answer: A

Solution figure for question 114
Q115EasyProfit Loss

Two musical instruments were purchased for Rs. 8,000. The first was sold at a profit of 40% and the second at a loss of 40%. If the sale price was the same in both cases, then the cost price of the cheaper instrument (in Rs.) was:

  1. 2500
  2. 2400
  3. 2200
  4. 2000
  5. 2300
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Answer: B

Approach: Same SP ⇒ CP ratio = inverse of multiplying factors.

Solution:
Let the costs be x (profit 40%) and y (loss 40%).
Same SP: 1.4x = 0.6y ⇒ x : y = 0.6 : 1.4 = 3 : 7.
x + y = 8000 ⇒ x = 3/10 × 8000 = 2400, y = 5600.
Cheaper instrument = Rs. 2400.

Cetking shortcut: CP ratio = 60 : 140 = 3 : 7 ⇒ 8000 × 3/10 = 2400.

Answer: B

Solution figure for question 115
Q116EasyProfit Loss

Five kg of butter was bought by a shopkeeper for ₹300. One kg becomes unsaleable. He sells the remaining in such a way that, on the whole, he incurs a loss of 10%. At what price (in Rs. per kg) was the butter sold?

  1. 67.50
  2. 52.5
  3. 72.5
  4. 60
  5. 55
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Answer: A

Approach: Total SP from overall loss, then per-unit price.

Solution:
Cost = Rs. 300; overall loss 10% ⇒ total SP = 0.9 × 300 = Rs. 270.
Saleable butter = 5 − 1 = 4 kg.
SP per kg = 270/4 = Rs. 67.50.

Why the other options are wrong:
E (55) / B (52.5): dividing over 5 kg or wrong loss base.
D (60): that is the cost price per kg.

Answer: A

Solution figure for question 116
Q117MediumProfit Loss

A man sold two steel chairs for Rs. 500 each. On one, he gains 20% and on the other, he loses 12%. How much does he gain or lose in the whole transaction?

  1. 1.5% gain
  2. 1.5% loss
  3. 2% gain
  4. 2% loss
  5. 1% loss
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Answer: A

Approach: Compute both cost prices from the same SP.

Solution:
CP₁ = 500/1.20 = 416.67.
CP₂ = 500/0.88 = 568.18.
Total CP = 984.85; total SP = 1000.
Gain = 15.15 ⇒ gain% = 15.15/984.85 × 100 ≈ 1.54% ≈ 1.5% gain.

Why the other options are wrong:
B, D, E: give a loss, but total CP is below total SP.
C (2% gain): over-estimates; exact value ≈ 1.54%.

Cetking shortcut: Take SP = 1320 (LCM-friendly): CP₁ = 1100, CP₂ = 1500 ⇒ CP total 2600, SP 2640 ⇒ gain 40/2600 ≈ 1.54%.

Answer: A

Solution figure for question 117Solution figure for question 117Solution figure for question 117
Q118EasyProfit Loss

A manufacturer sells a pair of glasses to a wholesale dealer at a profit of 18%. The wholesaler sells the same to a retailer at a profit of 20%. The retailer in turn sells them to a customer for Rs. 30.09, thereby earning a profit of 25%. What is the cost price of the pair of glasses to the manufacturer?

  1. 15
  2. 20
  3. 17
  4. 18
  5. 16
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Answer: C

Approach: Successive multiplying factors, reversed.

Solution:
Customer price = CP × 1.18 × 1.20 × 1.25.
1.18 × 1.20 = 1.416; × 1.25 = 1.77.
CP = 30.09/1.77 = Rs. 17.

Cetking shortcut: Test the options: 17 × 1.77 = 30.09 ✓.

Answer: C

Solution figure for question 118Solution figure for question 118
Q119MediumMensuration

A rectangular paper, when folded into two congruent parts has a perimeter of 34 cm for each part folded along one set of sides and the same is 38 cm when folded along the other set of sides. What is the area of the paper (in sq. cms)?

  1. 130
  2. 140
  3. 150
  4. 160
  5. 145
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Answer: B

Approach: Two linear equations from the folded halves.

Solution:
Let the sheet be l × b.
Fold across the length: half is (l/2) × b ⇒ perimeter = l + 2b = 34.
Fold across the breadth: half is l × (b/2) ⇒ perimeter = 2l + b = 38.
Adding: 3(l + b) = 72 ⇒ l + b = 24. Subtracting: l − b = 4 ⇒ l = 14, b = 10.
Area = 14 × 10 = 140 sq. cm.

Cetking shortcut: Add the two perimeters: 3(l + b) = 72 ⇒ l + b = 24, then subtract for l − b = 4.

Answer: B

Solution figure for question 119Solution figure for question 119Solution figure for question 119Solution figure for question 119
Q120MediumMixtures

Three containers have their volumes in the ratio 3 : 4 : 5. The containers are filled with a mixture of milk and water. The containers have milk and water in the ratio 4 : 1, 3 : 1 and 5 : 2 respectively. The contents of these three mixtures are poured into a fourth container. The ratio of milk and water in the fourth container is?

  1. 4:1
  2. 157:53
  3. 160:53
  4. 151:48
  5. 5:2
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Answer: B

Approach: Absolute quantities of each component, then add.

Solution:
Volumes 3 : 4 : 5.
Milk = 3 × 4/5 + 4 × 3/4 + 5 × 5/7 = 12/5 + 3 + 25/7.
Water = 3 × 1/5 + 4 × 1/4 + 5 × 2/7 = 3/5 + 1 + 10/7.
Multiply by 35: milk = 84 + 105 + 125 = 314; water = 21 + 35 + 50 = 106.
Ratio = 314 : 106 = 157 : 53.

Why the other options are wrong:
A (4:1), E (5:2): just one container's ratio.
C, D: arithmetic slips in the LCM step.

Cetking shortcut: Total = 12 parts; water total ≈ 106/35 ≈ 3.03 ⇒ milk : water ≈ 2.96 : 1 — only 157:53 (≈ 2.96) fits.

Answer: B

Solution figure for question 120Solution figure for question 120Solution figure for question 120Solution figure for question 120
Q121MediumPartnership

Two friends, P and Q started a business investing in the ratio 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. 98000. What was the amount invested by R?

  1. 190000
  2. 200000
  3. 210000
  4. 250000
  5. 180000
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Answer: C

Approach: Partnership — time-weighted capital.

Solution:
Let P, Q invest 5x and 6x for 12 months; R invests 6x for 6 months.
Capital employed over the year (in year-equivalents) = 5x + 6x + 6x × 6/12 = 14x.
Profit = 20% of capital employed = 0.2 × 14x = 2.8x = 98000.
x = 98000/2.8 = 35000.
R's investment = 6x = 6 × 35000 = Rs. 2,10,000.
Check: R's capital-months 210000 × 6 = 12.6 lakh out of 58.8 lakh total ⇒ R's profit share = 98000 × 3/14 = 21000 (10% of his amount for half a year at 20% p.a.). ✓

Why the other options are wrong:
A, B, D, E: do not give 6x with 2.8x = 98000.

Cetking shortcut: Year-equivalent capital = 14 units; 20% of 14 = 2.8 units = 98000 ⇒ 1 unit = 35000 ⇒ R = 6 units.

Answer: C

Solution figure for question 121Solution figure for question 121Solution figure for question 121Solution figure for question 121
Q122MediumMixtures

A cane contains a mixture of two liquids, 'A' and 'B' in the ratio 7 : 5. When 9 liters of mixture are drawn off and the cane is filled with 'B', the ratio of 'A' and 'B' becomes 7 : 9 liters. How many liters of Liquid 'A' was initially contained by the cane?

  1. 15.25
  2. 18
  3. 21
  4. 17.5
  5. 16.75
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Answer: C

Approach: Replacement — component removed in proportion.

Solution:
Let A = 7k, B = 5k (total 12k).
9 L drawn: A removed = 9 × 7/12 = 21/4, B removed = 9 × 5/12 = 15/4.
After refilling with 9 L of B: A = 7k − 21/4, B = 5k − 15/4 + 9 = 5k + 21/4.
(7k − 21/4) : (5k + 21/4) = 7 : 9 ⇒ 63k − 189/4 = 35k + 147/4 ⇒ 28k = 84 ⇒ k = 3.
Initial A = 7 × 3 = 21 litres.

Why the other options are wrong:
A, B, D, E: the equation fixes k = 3 uniquely, so A must be 7 × 3 = 21; e.g. A = 18 gives final ratio 12.75 : 18.11, not 7 : 9.

Cetking shortcut: A is unchanged except for the proportional removal: total after = 12k and A share falls from 7/12 to 7/16 ⇒ A after = 7k − 21/4 = 7/16 × 12k ⇒ k = 3.

Answer: C

Q123EasyMensuration

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?

  1. 152600
  2. 154600
  3. 153600
  4. 152500
  5. 153500
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Answer: C

Approach: Distance = speed × time, then rectangle with ratio sides.

Solution:
12 km/hr = 12000 m/60 min = 200 m/min.
Perimeter = 200 × 8 = 1600 m.
Sides 3x and 2x ⇒ 2(3x + 2x) = 10x = 1600 ⇒ x = 160.
Length = 480 m, breadth = 320 m.
Area = 480 × 320 = 153600 sq. m.

Why the other options are wrong:
A, B, D, E: near-miss values; 480 × 320 is exactly 153600.

Cetking shortcut: Area = 6x² with x = 160 ⇒ 6 × 25600 = 153600.

Answer: C

Solution figure for question 123Solution figure for question 123Solution figure for question 123Solution figure for question 123Solution figure for question 123
Q124MediumPartnership

A, B and C start a business by investing Rs.2000, Rs.3000 and Rs.4000 respectively. But B increases his investment to Rs.4000 after 4 months and C withdraws Rs.1000 at the end of 9 months. What is A's share out of a total profit of Rs. 8475 earned in a year?

  1. 1600
  2. 1800
  3. 1500
  4. 1400
  5. 1200
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Answer: B

Approach: Capital × months for each partner.

Solution:
A: 2000 × 12 = 24000.
B: 3000 × 4 + 4000 × 8 = 12000 + 32000 = 44000.
C: 4000 × 9 + 3000 × 3 = 36000 + 9000 = 45000.
Ratio = 24 : 44 : 45, total 113.
A's share = 24/113 × 8475 = 24 × 75 = Rs. 1800.

Why the other options are wrong:
A, C, D, E: none equals 24/113 × 8475 = 1800 (ignoring the changes in B's and C's capital gives a different ratio).

Cetking shortcut: 8475 = 113 × 75 — once you see that, A's share = 24 × 75 = 1800.

Answer: B

Solution figure for question 124Solution figure for question 124Solution figure for question 124
Q125EasyTriangles

In a triangle ABC, points P, Q and R are the mid-points of the sides AB, BC and CA respectively. If the area of the triangle ABC is 20 sq. units, what is the area of the triangle PQR?

  1. 10 Sq units
  2. 12 Sq units
  3. 5 Sq units
  4. 4 Sq units
  5. 15 Sq units
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Answer: C

Approach: Mid-point theorem — medial triangle has 1/4 area.

Solution:
By the mid-point theorem, PQ, QR, RP are each half of the opposite side and parallel to it.
So triangle PQR is similar to ABC with ratio 1 : 2 ⇒ area ratio 1 : 4.
The four small triangles APR, PBQ, RQC and PQR are congruent.
Area of PQR = 20/4 = 5 sq. units.

Why the other options are wrong:
A (10): half, not a quarter.
B, D, E: not 20/4.

Answer: C

Solution figure for question 125
Q126EasyTrains Boats

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is

  1. 45km/hrs.
  2. 50km/hrs.
  3. 54km/hrs.
  4. 53km/hrs.
  5. 55km/hrs.
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Answer: B

Approach: Relative speed, same direction.

Solution:
Relative speed = 125/10 = 12.5 m/s = 12.5 × 18/5 = 45 km/hr.
Same direction ⇒ train speed − 5 = 45 ⇒ train speed = 50 km/hr.

Why the other options are wrong:
A (45): relative speed, forgot to add the man's 5 km/hr.
C, D, E: arithmetic slips.

Answer: B

Solution figure for question 126Solution figure for question 126
Q127EasyTrains Boats

The length of the bridge, which a train 130 metres long and travelling at, 45 km/hr can cross in 30 seconds, is

  1. 200m
  2. 230m
  3. 245m
  4. 250m
  5. 220m
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Answer: C

Approach: Train length + bridge length = speed × time.

Solution:
45 km/hr = 45 × 5/18 = 12.5 m/s.
Distance in 30 s = 12.5 × 30 = 375 m.
Bridge = 375 − 130 = 245 m.

Why the other options are wrong:
D (250), B (230), A (200), E (220): miscalculated distance or wrong subtraction.

Answer: C

Solution figure for question 127Solution figure for question 127
Q128MediumTrains Boats

Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is

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

Approach: Relative speed — opposite direction trains.

Solution:
Let speeds be x and y; lengths = 27x and 17y.
Opposite directions: (27x + 17y)/(x + y) = 23.
27x + 17y = 23x + 23y ⇒ 4x = 6y ⇒ x : y = 3 : 2.

Why the other options are wrong:
A, C, D, E: do not satisfy 4x = 6y.

Cetking shortcut: Ratio = (t₂ − t)/(t − t₁) = (23 − 17)/(27 − 23) = 6/4 = 3 : 2.

Answer: B

Solution figure for question 128Solution figure for question 128Solution figure for question 128
Q129MediumTrains Boats

A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

  1. 66km/hrs.
  2. 76km/hrs.
  3. 71km/hrs.
  4. 81km/hrs.
  5. 80km/hrs.
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Answer: D

Approach: Equal train length in both cases.

Solution:
4.5 km/hr = 1.25 m/s; 5.4 km/hr = 1.5 m/s. Let the train speed = v m/s.
Length = 8.4(v − 1.25) = 8.5(v − 1.5).
8.4v − 10.5 = 8.5v − 12.75 ⇒ 0.1v = 2.25 ⇒ v = 22.5 m/s.
v = 22.5 × 18/5 = 81 km/hr.
Check: length = 8.4 × 21.25 = 178.5 m = 8.5 × 21 ✓.

Why the other options are wrong:
A, B, C, E: substituting gives unequal train lengths.

Cetking shortcut: Test options in m/s: 81 km/hr = 22.5 m/s ⇒ 8.4 × 21.25 = 8.5 × 21 = 178.5 ✓.

Answer: D

Solution figure for question 129Solution figure for question 129Solution figure for question 129Solution figure for question 129Solution figure for question 129
Q130MediumTime & 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. 12
  2. 15
  3. 16
  4. 18
  5. 13
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Answer: B

Approach: LCM / units method with cycles.

Solution:
Total work = LCM(20, 30, 60) = 60 units.
Efficiencies: A = 3, B = 2, C = 1 units/day.
3-day cycle: day 1 = 3, day 2 = 3, day 3 = 3 + 2 + 1 = 6 ⇒ 12 units per 3 days.
60/12 = 5 cycles ⇒ 5 × 3 = 15 days (work ends exactly on a cycle boundary).

Why the other options are wrong:
A (12), C (16), D (18), E (13): wrong cycle count or wrong cycle output.

Answer: B

Solution figure for question 130Solution figure for question 130Solution figure for question 130Solution figure for question 130Solution figure for question 130
Q131MediumMandays

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. 3
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Answer: A

Approach: Mandays — convert everyone to boy-units.

Solution:
Let a man do M and a boy B units/day.
10(6M + 8B) = 2(26M + 48B) ⇒ 60M + 80B = 52M + 96B ⇒ 8M = 16B ⇒ M = 2B.
Work = 10(6 × 2B + 8B) = 10 × 20B = 200B.
15 men + 20 boys = 30B + 20B = 50B per day.
Days = 200B/50B = 4.

Why the other options are wrong:
B, C, D, E: do not equal 200/50.

Cetking shortcut: 1 man = 2 boys ⇒ first group = 20 boy-units × 10 days = 200; new group = 50 boy-units ⇒ 4 days.

Answer: A

Solution figure for question 131Solution figure for question 131Solution figure for question 131Solution figure for question 131
Q132MediumTime & Work

A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?

  1. 11:30 Am
  2. 12:30 Pm
  3. 11:00 Am
  4. 1:00 Pm
  5. 1:30 Pm
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Answer: D

Approach: Rate method with a machine switched off.

Solution:
Take work = 120 units (LCM of 8, 10, 12): P = 15, Q = 12, R = 10 units/hr.
9 A.M.–11 A.M. (2 hr), all three: 2 × 37 = 74 units.
Remaining = 120 − 74 = 46 units; Q + R = 22 units/hr.
Time = 46/22 ≈ 2.09 hr ≈ 2 hr 5 min.
Finish ≈ 11 A.M. + 2 hr 5 min ≈ 1:05 P.M. ≈ 1:00 P.M.

Why the other options are wrong:
A, C: before 11 A.M. the work is only 74/120 done.
B (12:30 P.M.): only 33 more units by then.
E (1:30 P.M.): overshoots by ~25 min.

Answer: D

Solution figure for question 132Solution figure for question 132Solution figure for question 132
Q133EasyTime & Work

A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in

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

Approach: Units method with pair rates.

Solution:
Work = 24 units. A+B = 3, B+C = 2, A+B+C = 4 units/day.
C = 4 − 3 = 1; A = 4 − 2 = 2.
A + C = 3 units/day ⇒ 24/3 = 8 days.

Why the other options are wrong:
B (6): that is the trio's time.
D (12): that is B + C's time.
A, E: not 24/3.

Cetking shortcut: C = 1/6 − 1/8 = 1/24 and A = 1/6 − 1/12 = 2/24 ⇒ A + C = 3/24 = 1/8 ⇒ 8 days.

Answer: C

Solution figure for question 133Solution figure for question 133Solution figure for question 133
Q134EasyAges

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

  1. 4
  2. 8
  3. 10
  4. 6
  5. 5
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Answer: A

Approach: AP sum of ages.

Solution:
Ages: x, x + 3, x + 6, x + 9, x + 12.
Sum = 5x + 30 = 50 ⇒ x = 4.
Youngest = 4 years.

Why the other options are wrong:
C (10): the middle child's age.
B, D, E: do not give sum 50.

Cetking shortcut: Middle child = 50/5 = 10 ⇒ youngest = 10 − 6 = 4.

Answer: A

Solution figure for question 134Solution figure for question 134
Q135EasyAges

Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?

  1. 24
  2. 27
  3. 26
  4. 29
  5. 30
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Answer: A

Approach: Ratio of ages with time shift.

Solution:
Sameer = 5x, Anand = 4x.
(5x + 3)/(4x + 3) = 11/9 ⇒ 45x + 27 = 44x + 33 ⇒ x = 6.
Anand = 4 × 6 = 24 years.

Why the other options are wrong:
E (30): Sameer's age.
B, C, D: not multiples of 4 that fit.

Cetking shortcut: Anand's age must be a multiple of 4: test 24 ⇒ Sameer 30 ⇒ 33 : 27 = 11 : 9 ✓.

Answer: A

Solution figure for question 135Solution figure for question 135Solution figure for question 135Solution figure for question 135
Q136EasyAges

The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years)

  1. 8,20,28 respectively
  2. 16,28,36 respectively
  3. 20,35,45 respectively
  4. 18,32,49 respectively
  5. 8,30,45 respectively
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Answer: B

Approach: Ratio ages with a past-sum condition.

Solution:
Present ages 4x, 7x, 9x ⇒ sum 20x.
Eight years ago: 20x − 24 = 56 ⇒ 20x = 80 ⇒ x = 4.
Ages = 16, 28, 36 years.

Why the other options are wrong:
A, D, E: not in ratio 4 : 7 : 9.
C (20, 35, 45): right ratio, but sum 8 years ago = 76, not 56.

Cetking shortcut: Only options in ratio 4:7:9 are B and C; B's sum 80 − 24 = 56 ✓.

Answer: B

Solution figure for question 136Solution figure for question 136
Q137MediumPnC

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

  1. 700
  2. 695
  3. 756
  4. 823
  5. 633
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Answer: C

Approach: Combinations — case-wise addition.

Solution:
3 men + 2 women: 7C3 × 6C2 = 35 × 15 = 525.
4 men + 1 woman: 7C4 × 6C1 = 35 × 6 = 210.
5 men: 7C5 = 21.
Total = 525 + 210 + 21 = 756.

Why the other options are wrong:
A, B, D, E: miss a case or miscount 6C2 = 15.

Answer: C

Solution figure for question 137Solution figure for question 137
Q138EasyPnC

In how many ways can the letters of the word 'LEADER' be arranged?

  1. 360
  2. 240
  3. 180
  4. 210
  5. 270
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Answer: A

Approach: Permutations with identical letters.

Solution:
LEADER: 6 letters with E repeated twice (L, A, D, R once).
Arrangements = 6!/2! = 720/2 = 360.

Why the other options are wrong:
B, C, D, E: not 720/2.

Answer: A

Solution figure for question 138
Q139EasyPnC

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

  1. 6000
  2. 5075
  3. 5040
  4. 3500
  5. 4090
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Answer: C

Approach: nPr for ordered selections.

Solution:
LOGARITHMS has 10 distinct letters (L, O, G, A, R, I, T, H, M, S).
4-letter words without repetition = 10P4 = 10 × 9 × 8 × 7 = 5040.

Why the other options are wrong:
A, B, D, E: not 10 × 9 × 8 × 7.

Answer: C

Solution figure for question 139Solution figure for question 139
Q140EasySurds & Indices

The cube root of 0.000216 is:

  1. 0.6
  2. 0.06
  3. 0.006
  4. 0.16
  5. 0.016
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Answer: B

Approach: Cube root via powers of 10.

Solution:
0.000216 = 216 × 10⁻⁶.
∛216 = 6 and ∛10⁻⁶ = 10⁻².
∛0.000216 = 6 × 10⁻² = 0.06.
Check: 0.06³ = 0.000216 ✓.

Why the other options are wrong:
A (0.6): cube = 0.216.
C (0.006): cube = 0.000000216.
D, E: 16 is not a cube root of 216.

Answer: B

Q141MediumTime Speed Distance

In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is

  1. 1 hrs.
  2. 2 hrs.
  3. 3 hrs.
  4. 4 hrs.
  5. 2.5 hrs.
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Answer: A

Approach: Fixed distance — speed down, time up.

Solution:
Let the usual speed = s km/hr; scheduled time = 600/s.
600/(s − 200) − 600/s = 1/2.
600 × 200 = ½ × s(s − 200) ⇒ s² − 200s − 240000 = 0 ⇒ (s − 600)(s + 400) = 0 ⇒ s = 600 km/hr.
Scheduled duration of the flight = 600/600 = 1 hr (the slowed flight took 1 hr + 30 min = 1.5 hr).
The duration asked (scheduled flight time) = 1 hr.

Why the other options are wrong:
B (2 hrs): s = 300 ⇒ 600/100 − 2 = 4 hr extra, not 0.5.
C, D, E: likewise give extra time ≠ 30 min.

Cetking shortcut: Test options as the scheduled time t: 600/t − 600/(t + 0.5) must be 200 ⇒ t = 1: 600 − 400 = 200 ✓.

Answer: A

Solution figure for question 141Solution figure for question 141Solution figure for question 141Solution figure for question 141Solution figure for question 141
Q142EasyTrigonometry

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

  1. 270m
  2. 273m
  3. 173m
  4. 275m
  5. 175m
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Answer: B

Approach: tan θ = height / base.

Solution:
Ship at 30°: distance = 100 × cot 30° = 100√3 ≈ 173.2 m.
Ship at 45°: distance = 100 × cot 45° = 100 m.
Opposite sides ⇒ distance between ships = 173.2 + 100 = 273.2 ≈ 273 m.

Why the other options are wrong:
C (173 m): only one ship's distance.
A, D, E: rounding or arithmetic slips; 100(√3 + 1) = 273.2.

Answer: B

Solution figure for question 142Solution figure for question 142
Q143EasyMensuration

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is:

  1. 3.6kg
  2. 3.69kg
  3. 3.696kg
  4. 3.96kg
  5. 3.66kg
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Answer: C

Approach: Hollow cylinder volume × density.

Solution:
External radius R = 8/2 = 4 cm; internal radius r = 4 − 1 = 3 cm.
Volume = 22/7 × (16 − 9) × 21 = 22/7 × 7 × 21 = 462 cm³.
Weight = 462 × 8 = 3696 g = 3.696 kg.

Why the other options are wrong:
A, B, D, E: rounding variants; exact value is 3.696 kg.

Answer: C

Solution figure for question 143Solution figure for question 143
Q144MediumTrains Boats

A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

  1. 2:1
  2. 2:3
  3. 8:3
  4. 3:5
  5. 5:8
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Answer: C

Approach: Speed ∝ 1/time for a fixed distance.

Solution:
Upstream time = 8 h 48 min = 8.8 hr; downstream time = 4 hr.
Downstream : upstream speed = 8.8 : 4 = 11 : 5.
b + c = 11k, b − c = 5k ⇒ b = 8k, c = 3k.
Boat : current = 8 : 3.

Why the other options are wrong:
A, B, D, E: do not give (b + c)/(b − c) = 11/5.

Cetking shortcut: b : c = (11 + 5)/2 : (11 − 5)/2 = 8 : 3.

Answer: C

Solution figure for question 144Solution figure for question 144Solution figure for question 144
Q145MediumTrains Boats

A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is

  1. 2mph
  2. 2.5mph
  3. 3mph
  4. 4mph
  5. 3.2mph
Show answerHide answer

Answer: A

Approach: Boats and streams — time difference equation.

Solution:
36/(10 − s) − 36/(10 + s) = 1.5.
36 × 2s = 1.5(100 − s²) ⇒ 72s = 150 − 1.5s² ⇒ s² + 48s − 100 = 0.
(s + 50)(s − 2) = 0 ⇒ s = 2 mph.
Check: 36/8 − 36/12 = 4.5 − 3 = 1.5 hr = 90 min ✓.

Why the other options are wrong:
B (2.5): 36/7.5 − 36/12.5 = 1.92 hr.
C (3): 36/7 − 36/13 ≈ 2.37 hr.
D, E: even larger gaps.

Cetking shortcut: Option test is fastest: s = 2 gives 4.5 − 3 = 1.5 hr ✓.

Answer: A

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Q146EasyTrains Boats

A man can row three-quarters of a kilometer against the stream in 11 minutes 15 seconds and down the stream in 7 minutes 30 seconds. What is the speed of the man (in km/hr) in still water?

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

Answer: E

Approach: Average of downstream and upstream speeds.

Solution:
Upstream: 0.75 km in 11 min 15 s = 11.25 min ⇒ speed = 0.75 × 60/11.25 = 4 km/hr.
Downstream: 0.75 km in 7.5 min ⇒ speed = 0.75 × 60/7.5 = 6 km/hr.
Still water speed = (6 + 4)/2 = 5 km/hr.

Why the other options are wrong:
A (1 km/hr): that is the stream speed.
B, C, D: not (6 + 4)/2.

Answer: E

Solution figure for question 146Solution figure for question 146
Q147EasyPipes and Cisterns

Pipe A can fill a cistern in 37 minutes 30 seconds and Pipe B fills it in 45 minutes individually. If both pipes are opened together and pipe B is turned off after sometime, it takes half an hour for the cistern to be filled. After how much time pipe B is turned off?

  1. 8 mins
  2. 9 mins
  3. 10 mins
  4. 12 mins
  5. 6 mins
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Answer: B

Approach: Fill fraction by the full-time pipe first.

Solution:
A fills 1/37.5 = 2/75 per min; in 30 min A fills 30 × 2/75 = 4/5.
B fills the remaining 1/5 of the cistern.
B's time = (1/5) × 45 = 9 min.
So B was turned off after 9 minutes.

Why the other options are wrong:
A, C, D, E: give B's share ≠ 1/5.

Cetking shortcut: A alone in 30 min = 30/37.5 = 80% ⇒ B did 20% of 45 min = 9 min.

Answer: B

Solution figure for question 147Solution figure for question 147
Q148EasyPipes and Cisterns

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

  1. 10 min 20sec
  2. 12 min 20sec
  3. 14 min 40sec
  4. 14 min 20sec
  5. 10 min 40sec
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Answer: C

Approach: LCM units, two phases.

Solution:
Tank = 60 units; A = 4, B = 3 units/min.
First 4 min together: 4 × 7 = 28 units.
Remaining 32 units by B alone: 32/3 = 10 2/3 min = 10 min 40 s.
Total = 4 min + 10 min 40 s = 14 min 40 s.

Why the other options are wrong:
E (10 min 40 s): time for B alone after A stops.
A, B, D: arithmetic slips in 32/3.

Answer: C

Solution figure for question 148Solution figure for question 148Solution figure for question 148Solution figure for question 148
Q149EasyClocks & Calendars

The angle between the minute hand and the hour hand of a clock when the time is 4.20, is

  1. 3 degree
  2. 5 degree
  3. 8 degree
  4. 10 degree
  5. 12 degree
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Answer: D

Approach: Clock angle formula.

Solution:
At 4:20, H = 4, M = 20.
Angle = |30 × 4 − 5.5 × 20| = |120 − 110| = 10°.

Why the other options are wrong:
A, B, C, E: not |120 − 110|.

Cetking shortcut: Minute hand at 4 exactly (120°); hour hand has moved 20 × 0.5 = 10° past 4 ⇒ 10°.

Answer: D

Solution figure for question 149Solution figure for question 149Solution figure for question 149
Q150MediumTime Speed Distance

A and B take part in 100m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

  1. 4.14kmph
  2. 4.5kmph
  3. 4.32kmph
  4. 4.9kmph
  5. 4.11kmph
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Answer: A

Approach: Races — head start and time margin.

Solution:
A's speed = 5 km/hr = 25/18 m/s ⇒ A's time for 100 m = 100 × 18/25 = 72 s.
B starts 8 m ahead ⇒ runs 92 m, and finishes 8 s after A ⇒ B's time = 80 s.
B's speed = 92/80 = 1.15 m/s = 1.15 × 18/5 = 4.14 km/hr.

Why the other options are wrong:
B, C, D, E: use 100 m or 72/80 s wrongly.

Cetking shortcut: B's speed = 5 × (92/100) × (72/80) = 5 × 0.92 × 0.9 = 4.14 km/hr.

Answer: A

Solution figure for question 150Solution figure for question 150

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