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MBA CET 2026 Attempt 2 Slot 1 question paper with solutions

200 questions · Quantitative Aptitude 50 · Verbal Ability 50 · Logical Reasoning 75 · Abstract Reasoning 25. Tap Show answer on any question for the answer and Cetking's solution. Where Cetking’s answer differs from the official CET key, both are shown with the reasoning.

Q101EasyHCF LCM

Find the HCF and LCM of 0.36, 0.72, and 1.62.

  1. 1.8 and 6.48
  2. 0.18 and 6.48
  3. 0.18 and 64.8
  4. 1.8 and 64.8
  5. 0.18 and 0.648
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Answer: B official key

Approach: HCF/LCM of decimals: equalise decimal places, work with integers, put the decimal back.

Solution:
Write as 36, 72, 162 hundredths (two decimal places).
36 = 2²·3², 72 = 2³·3², 162 = 2·3⁴.
HCF = 2·3² = 18 → 0.18.
LCM = 2³·3⁴ = 648 → 6.48.

Cetking shortcut: HCF must divide 0.36, so it cannot be 1.8 — kills A and D instantly; LCM must be a multiple of 1.62 and 0.72 → 6.48 (= 4 × 1.62).

Answer: B

Q102EasyPercentages

The population of a town is 18,000. It increases by 15% in the first year, 10% in the second year, and 25% in the third year. What will be the population after 3 years?

  1. 27,600
  2. 28,125
  3. 28,463
  4. 29,000
  5. 29,463
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Answer: C official key

Approach: Successive percentage growth: multiply the growth factors.

Solution:
After year 1: 18,000 × 1.15 = 20,700.
After year 2: 20,700 × 1.10 = 22,770.
After year 3: 22,770 × 1.25 = 28,462.5 ≈ 28,463.

Cetking shortcut: 1.15 × 1.10 = 1.265; × 1.25 = 1.58125; 18,000 × 1.58125 = 28,462.5.

Answer: C

Q103EasyRemainders

A number when divided by 256 gives a remainder 73.
When the same number is divided by 16, what will be the remainder?

  1. 1
  2. 7
  3. 9
  4. 11
  5. 13
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Answer: C official key

Approach: Remainder with a divisor that divides the original divisor.

Solution:
N = 256k + 73.
256 = 16 × 16, so 256k leaves remainder 0 when divided by 16.
So remainder = 73 mod 16; 73 = 16 × 4 + 9 → remainder 9.

Cetking shortcut: If the new divisor is a factor of the old one, just divide the old remainder by the new divisor.

Answer: C

Q104EasySurds & Indices

If x = (√10 + √6)/(√10 − √6) and y = (√10 − √6)/(√10 + √6), find the value of x² + y².

  1. 54
  2. 58
  3. 62
  4. 64
  5. 66
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Answer: C official key

Approach: Reciprocal pair: x + y and xy, then x² + y² = (x + y)² − 2xy.

Solution:
y = 1/x, so xy = 1.
x + y = [(√10+√6)² + (√10−√6)²] / (10 − 6) = 2(10+6)/4 = 32/4 = 8.
x² + y² = 8² − 2 × 1 = 62.

Cetking shortcut: For (√a+√b)/(√a−√b) + reciprocal: x + y = 2(a+b)/(a−b) = 32/4 = 8 directly.

Answer: C

Q105EasyNumbers

If the dates of birth of four persons are prime numbers, find the maximum possible average of the sum of their dates of birth.

  1. 25.5
  2. 27.2
  3. 26.4
  4. 31
  5. 25
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Answer: D official key

Approach: Maximise each value: the largest prime date of a month is 31.

Solution:
Dates of birth lie between 1 and 31; the largest prime date is 31.
The question does not require the four dates to be different, so each person can be born on the 31st.
Sum = 31 × 4 = 124; average = 124 / 4 = 31.

Why the other options are wrong:
A (25.5) is the average if the four dates had to be distinct (31, 29, 23, 19) — no such condition is given. B, C, E are not achievable maxima.

Cetking shortcut: Maximum average can never exceed the maximum single value — and it equals it when all values can be the same.

Answer: D

Q106EasyMixtures

Chemical A costs Rs.1000 per kg and chemical B costs Rs.150 per kg.
A chemist mixes the two chemicals and sells the mixture at Rs.840 per kg, thereby making a 20% profit.

In what ratio are chemical A and chemical B mixed?

  1. 11:6
  2. 6:11
  3. 10:15
  4. 10:1.5
  5. 100:84
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Answer: A official key

Approach: Find cost price of mixture from SP and profit, then alligation.

Solution:
CP of mixture = 840 / 1.20 = Rs 700 per kg.
Alligation: A (1000) and B (150) around mean 700.
A : B = (700 − 150) : (1000 − 700) = 550 : 300 = 11 : 6.

Why the other options are wrong:
B reverses the ratio; C, D, E do not reduce to 11 : 6.

Cetking shortcut: Always convert SP to CP before alligation; 840 itself is a trap mean.

Answer: A

Q107EasyRatios

In a manufacturing company, three distinct types of gadgets—A, B, and C—were produced in January in the ratio 3:4:5. Due to varying market demands in February, the company decided to retain the production level of A unchanged, double the production of B, and increase the production of C by fifty percent, resulting in a total combined output of 5402 units across all three types.

Considering these adjustments, determine the number of units produced in February for each type of gadget.

  1. A = 886, B = 2346, C = 2190
  2. A = 876, B = 2336, C = 2187
  3. A = 876, B = 2336, C = 2190
  4. A = 875, B = 2350, C = 2182
  5. A = 890, B = 2360, C = 2282
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Answer: C official key

Approach: Ratio with a multiplier k; apply the changes and solve for k.

Solution:
January: A = 3k, B = 4k, C = 5k.
February: A = 3k, B = 8k, C = 7.5k → total 18.5k.
18.5k = 5402 → k = 292.
A = 876, B = 2336, C = 7.5 × 292 = 2190.

Why the other options are wrong:
B (2187) and others fail the total 5402 or the 3 : 8 : 7.5 ratio.

Cetking shortcut: Only C sums to 5402 with A divisible by 3 and B = 8/3 of A — check sums: 876 + 2336 + 2190 = 5402.

Answer: C

Q108MediumPartnership

Amar, Akbar, and Anthony start a business, each investing Rs 30,000. After 4 months, Amar withdraws Rs 8,000, Akbar withdraws Rs 5,000, and Anthony invests Rs 10,000 more. At the end of the year, the total profit is Rs 1,05,000. What is Anthony's share of the profit?

  1. 29,432
  2. 31,818
  3. 43,750
  4. 31,500
  5. 43,500
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Answer: C official key

Approach: Partnership: profit ∝ capital × time.

Solution:
Amar: 30,000 × 4 + 22,000 × 8 = 2,96,000.
Akbar: 30,000 × 4 + 25,000 × 8 = 3,20,000.
Anthony: 30,000 × 4 + 40,000 × 8 = 4,40,000.
Ratio 296 : 320 : 440 = 37 : 40 : 55; total 132.
Anthony = 55/132 × 1,05,000 = 5/12 × 1,05,000 = 43,750.

Cetking shortcut: Work in thousands: first 4 months are equal (120 each), so compare only the extra: 176, 200, 320 → add 120 back → 296 : 320 : 440; 440/1056 = 5/12.

Answer: C

Q109EasyNumber Series (Quant)

Find the missing number in the series:

7, 22, 24, 73, 75, ?

  1. 150
  2. 226
  3. 225
  4. 95
  5. 81
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Answer: B official key

Approach: Alternating operations series: ×3 + 1, then +2.

Solution:
7 × 3 + 1 = 22
22 + 2 = 24
24 × 3 + 1 = 73
73 + 2 = 75
75 × 3 + 1 = 226

Why the other options are wrong:
C (225) is ×3 without +1; A (150) is ×2; D, E break the pattern.

Cetking shortcut: Jumps of +2 every alternate term signal a two-operation cycle; the other step is ×3 + 1.

Answer: B

Q110EasyNumber Series (Quant)

Find the missing number in the following series:

16, 40, 140, 630, ?, 22522.5

  1. 3485
  2. 3380
  3. 3930
  4. 3465
  5. 3845
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Answer: D official key

Approach: Multiplier series with the multiplier rising by 1.

Solution:
16 × 2.5 = 40
40 × 3.5 = 140
140 × 4.5 = 630
630 × 5.5 = 3465
Check: 3465 × 6.5 = 22,522.5 ✓

Cetking shortcut: A fractional last term (22522.5) hints at ×(n.5) multipliers; verify with the last term.

Answer: D

Q111EasyInterest

At what annual interest rate, compounded yearly, will $80,000 amount to $88,200 in 2 years?

  1. 4%
  2. 4.5%
  3. 5%
  4. 5.5%
  5. 6 %
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Answer: C official key

Approach: Compound interest: A = P(1 + r)ⁿ.

Solution:
88,200 / 80,000 = 1.1025.
(1 + r)² = 1.1025 → 1 + r = 1.05.
r = 5%.

Cetking shortcut: Know squares: 1.05² = 1.1025, 1.04² = 1.0816, 1.06² = 1.1236.

Answer: C

Q112EasyTrains Boats

A jogger running at 12 km/hr along a railway track is 300 m ahead of the engine of a 150 m long train running at 48 km/hr in the same direction.
In how much time will the train pass the jogger?

  1. 20 Seconds
  2. 25 Seconds
  3. 30 Seconds
  4. 45 Seconds
  5. 50 Seconds
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Answer: D official key

Approach: Relative speed in the same direction; distance = gap + train length.

Solution:
Relative speed = 48 − 12 = 36 km/hr = 36 × 5/18 = 10 m/s.
To pass the jogger the train covers the 300 m gap + its own 150 m = 450 m.
Time = 450 / 10 = 45 s.

Why the other options are wrong:
C (30 s) ignores the train length; A, B, E come from wrong distance or unconverted speed.

Cetking shortcut: 36 km/hr = 10 m/s is a must-know conversion.

Answer: D

Q113EasyTrains Boats

The speed of a boat in still water is 8 km/hr, and the speed of the stream is 2 km/hr.
A man rows to a place 30 km and then returns to the starting point.
Find the total time taken.

  1. 6 hours
  2. 7.5 hours
  3. 8 hours
  4. 8.5 hours
  5. 9 hours
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Answer: C official key

Approach: Boats and streams: downstream = b + s, upstream = b − s.

Solution:
Downstream speed = 8 + 2 = 10 km/hr → 30/10 = 3 h.
Upstream speed = 8 − 2 = 6 km/hr → 30/6 = 5 h.
Total = 3 + 5 = 8 hours.

Cetking shortcut: Total time = 2bd/(b² − s²) = 2 × 8 × 30 / 60 = 8 h.

Answer: C

Q114MediumMensuration

The length of a rectangular field is 25 meters 7 decimeters 5 centimeters, and its breadth is 18 meters 4 decimeters 6 centimeters.
If a square path of uniform width 35 cm is constructed inside the rectangle along its boundary, find the area of the path.

  1. 28.64 m²
  2. 29.82 m²
  3. 30.46 m²
  4. 31.75 m²
  5. 32.50 m²
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Answer: C official key

Approach: Inner path area = outer rectangle − inner rectangle (each side reduced by 2w).

Solution:
L = 25 m 7 dm 5 cm = 25.75 m; B = 18 m 4 dm 6 cm = 18.46 m; w = 0.35 m.
Inner rectangle = (25.75 − 0.70) × (18.46 − 0.70) = 25.05 × 17.76 = 444.888 m².
Outer = 25.75 × 18.46 = 475.345 m².
Path area = 475.345 − 444.888 = 30.457 ≈ 30.46 m².

Cetking shortcut: Inner path area = 2w(L + B) − 4w² = 0.7 × 44.21 − 0.49 = 30.947 − 0.49 = 30.46 m².

Answer: C

Q115EasyCircles

Four identical coins, each of radius (x+1) cm, are placed on a table such that each coin touches two others and their centers lie at the vertices of a square.
Find the area of the unoccupied region between the coins.
(π=3.14)

Figure for MBA CET 2026 Attempt 2 Slot 1 question 115
  1. (x² + 2x + 1) × (4 − 3.14)
  2. (x+1)² × (2 + π/2)
  3. (x+1)² × (1 - π/4) × 4
  4. (2x+2)² - (π(x+1)²)
  5. 4 (x² + 2x + 1)
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Answer: A official key

Approach: Unoccupied region = square joining centres − four quarter circles.

Solution:
Let r = x + 1. Adjacent coins touch, so the side of the square = 2r.
Square area = 4r². Each corner holds a quarter coin; 4 quarters = one full circle = πr².
Unoccupied area = 4r² − πr² = r²(4 − π).
With π = 3.14 and r² = x² + 2x + 1: (x² + 2x + 1)(4 − 3.14).

Why the other options are wrong:
B adds instead of subtracting; E ignores the coins. C and D are algebraically the same as r²(4 − π), but A is the form that uses the given π = 3.14.

Cetking shortcut: Region between 4 touching equal circles = (4 − π)r² ≈ 0.86r² — memorise it.

Answer: A

Q116EasyTables

Questions 116–120 share this

Study the table carefully and answer the questions that follow.

The table below shows the export value of selected major commodities from India during the period April–October 2024–25. The export values are expressed in US$ million.

CommodityExport Value
Engineering Goods67,426
Petroleum Products39,068
Electronic Goods19,068
Gems & Jewellery17,160
Drugs & Pharmaceuticals17,032

If the total export value of the given commodities is taken as 100%, which of the following commodities accounts for approximately 12% of the total exports?

  1. Engineering Goods
  2. Petroleum Products
  3. Electronic Goods
  4. Gems & Jewellery
  5. Drugs & Pharmaceuticals
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Answer: C official key

Approach: Find the total, then 12% of it, and match.

Solution:
Total = 67,426 + 39,068 + 19,068 + 17,160 + 17,032 = 1,59,754.
12% of 1,59,754 ≈ 19,170.
Electronic Goods = 19,068 → 19,068/1,59,754 ≈ 11.9% ≈ 12%.

Why the other options are wrong:
A ≈ 42%, B ≈ 24%, D ≈ 10.7%, E ≈ 10.7%.

Cetking shortcut: 10% of total ≈ 16,000, so 12% ≈ 19,200 — only Electronic Goods is near.

Answer: C

Q117EasyTables

Refer to the passage above ↑

What is the percentage share of Engineering Goods in the total exports of the selected commodities (approx.)?

  1. 38.5%
  2. 40.9%
  3. 42.2%
  4. 45.6%
  5. 52.3 %
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Answer: C official key

Approach: Percentage share = part / total × 100.

Solution:
Total = 1,59,754.
Engineering share = 67,426 / 1,59,754 × 100 ≈ 42.2%.

Cetking shortcut: 40% of 1,59,754 = 63,902; remaining 3,524 ≈ 2.2% → 42.2%.

Answer: C

Q118EasyTables

Refer to the passage above ↑

By how much (in US$ million) do Petroleum Products exports exceed the combined exports of Gems & Jewellery and Drugs & Pharmaceuticals?

  1. 4,876
  2. 4,904
  3. 5,012
  4. 5,146
  5. 5,346
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Answer: A official key

Approach: Simple difference from the table.

Solution:
Gems & Jewellery + Drugs & Pharma = 17,160 + 17,032 = 34,192.
Petroleum − combined = 39,068 − 34,192 = 4,876.

Cetking shortcut: Units digit: 0 + 2 = 2, and 8 − 2 = 6 → only A ends in 6.

Answer: A

Q119MediumTables

Refer to the passage above ↑

If exports of Petroleum Products decline by 8% and exports of Gems & Jewellery increase by 12%, what will be the net change (increase/decrease) in total exports of the five commodities (in US$ million)?

  1. Decrease of 1,058
  2. Decrease of 1,066
  3. Increase of 1,196
  4. Decrease of 1,248
  5. Increase of 1,058
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Answer: B official key

Approach: Net change = sum of individual changes; other commodities unchanged.

Solution:
Petroleum change = −8% of 39,068 = −3,125.44.
Gems change = +12% of 17,160 = +2,059.20.
Net = −3,125.44 + 2,059.20 = −1,066.24 → decrease of about 1,066.

Cetking shortcut: Sign check first: 8% of ~39k (≈3.1k) > 12% of ~17k (≈2.1k), so it is a decrease — eliminates C and E.

Answer: B

Q120MediumTables

Refer to the passage above ↑

If the total export value of all five commodities increases by 12%, and this entire increase is distributed only between Engineering Goods and Petroleum Products in the ratio of their original export values, what will be the increase in the export value of Petroleum Products (approximately, in US$ million)?

  1. 4,500
  2. 7,020
  3. 8,500
  4. 9,702
  5. 11720
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Answer: B official key

Approach: Distribute the increase in proportion to the two original values.

Solution:
Total = 1,59,754; 12% increase = 19,170.48.
Engineering + Petroleum = 67,426 + 39,068 = 1,06,494.
Petroleum's part = 19,170.48 × 39,068 / 1,06,494 ≈ 19,170 × 0.367 ≈ 7,033.
Nearest option ≈ 7,020.

Cetking shortcut: Petroleum : Engineering ≈ 39 : 67 ≈ 37% of the pair; 37% of 19,170 ≈ 7,000 → B.

Answer: B

Q121EasyTime Speed Distance

A thief starts running at a constant speed of 5 m/s. A policeman starts chasing the thief 2 seconds later at a speed of 10 m/s.
After how much time (from the moment the policeman starts running) will the policeman catch the thief?

  1. 1 second
  2. 2 seconds
  3. 3 seconds
  4. 4 seconds
  5. 5 second
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Answer: B official key

Approach: Chase problem: head start ÷ relative speed.

Solution:
In the first 2 s the thief covers 5 × 2 = 10 m.
Relative speed = 10 − 5 = 5 m/s.
Time to close 10 m = 10 / 5 = 2 s (from when the policeman starts).

Cetking shortcut: Policeman is twice as fast, so he catches up in the same time as the head-start time (2 s).

Answer: B

Q122EasyClocks & Calendars

January 1, 2010 was a Friday.
What was the day of the week on January 1, 2017?

  1. Monday
  2. Tuesday
  3. Saturday
  4. Sunday
  5. Friday
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Answer: D official key

Approach: Odd days: each ordinary year adds 1 day, each leap year 2.

Solution:
From 1 Jan 2010 to 1 Jan 2017 = years 2010–2016 = 7 years.
Leap years in between: 2012 and 2016 → 2 leap years, 5 ordinary.
Odd days = 5 × 1 + 2 × 2 = 9 → 9 mod 7 = 2.
Friday + 2 = Sunday.

Cetking shortcut: Odd days = number of years + number of leap years = 7 + 2 = 9 → 2.

Answer: D

Q123EasyAges

Two years ago, the ratio of Neha’s age to Pooja’s age was 5 : 6.
Four years from now, this ratio will become 7 : 8.
What is Pooja’s present age?

  1. 17 years
  2. 20 years
  3. 26 years
  4. 28 years
  5. 30 years
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Answer: B official key

Approach: Ages with two ratios: take the earlier ages as 5k and 6k.

Solution:
Two years ago: Neha = 5k, Pooja = 6k.
Four years from now (6 years later): (5k + 6)/(6k + 6) = 7/8.
40k + 48 = 42k + 42 → k = 3.
Pooja now = 6k + 2 = 18 + 2 = 20 years.

Cetking shortcut: Option check: Pooja 20 → Neha must be 17 (15 : 18 = 5 : 6 two years ago); in 4 years 21 : 24 = 7 : 8 ✓.

Answer: B

Q124EasyPercentages

The difference between two numbers is 1660.
If 7.5% of one number is equal to 12.5% of the other number, find the larger number.

  1. 1992
  2. 2075
  3. 2490
  4. 4150
  5. 3150
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Answer: D official key

Approach: Convert the percentage equality into a ratio.

Solution:
7.5% of a = 12.5% of b → a/b = 12.5/7.5 = 5/3.
Difference = 5 − 3 = 2 parts = 1660 → 1 part = 830.
Larger number = 5 × 830 = 4150.

Cetking shortcut: The smaller percentage goes with the larger number.

Answer: D

Q125EasyNumber Series (Quant)

Find the odd number in the following series:

216, 214, 211, 206, 200, 188

  1. 206
  2. 188
  3. 211
  4. 200
  5. 214
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Answer: D official key

Approach: Differences follow consecutive primes.

Solution:
Differences should be −2, −3, −5, −7, −11.
216 − 2 = 214, 214 − 3 = 211, 211 − 5 = 206, 206 − 7 = 199, 199 − 11 = 188.
The series shows 200 instead of 199, so 200 is the odd one.

Why the other options are wrong:
A, B, C, E all fit the prime-difference pattern.

Cetking shortcut: 188 fits only from 199 (199 − 11), which confirms 200 is wrong.

Answer: D

Q126EasyNumber Series (Quant)

Find the odd one out from the following numbers:

510, 254, 126, 64, 30, 14, 6

  1. 254
  2. 126
  3. 64
  4. 30
  5. 6
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Answer: C official key

Approach: Each term = (previous − 2) ÷ 2, i.e. terms are 2ⁿ − 2.

Solution:
510 = 2⁹ − 2, 254 = 2⁸ − 2, 126 = 2⁷ − 2, next should be 2⁶ − 2 = 62.
Then 30 = 2⁵ − 2, 14 = 2⁴ − 2, 6 = 2³ − 2.
64 breaks the pattern (should be 62).

Cetking shortcut: All numbers except 64 are 2 less than a power of 2.

Answer: C

Q127EasySimplification

Find the value of the following equation:

30% of 40% of (2/5) of 2500

  1. 500
  2. 400
  3. 360
  4. 120
  5. 109
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Answer: D official key

Approach: Fraction-of chain: multiply step by step.

Solution:
2/5 of 2500 = 1000.
40% of 1000 = 400.
30% of 400 = 120.

Cetking shortcut: 0.3 × 0.4 × 0.4 × 2500 = 0.048 × 2500 = 120.

Answer: D

Q128EasyPipes and Cisterns

Eight taps are fitted to a water tank. Some are inlet taps that fill the tank, while the remaining are outlet taps that empty it.
Each inlet tap can fill the tank in 12 hours, and each outlet tap can empty the tank in 36 hours.
When all the taps are opened together, the tank is filled in 3 hours.

How many of the taps are inlet (water) taps?

  1. 3
  2. 4
  3. 5
  4. 6
  5. 7
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Answer: C official key

Approach: Net work rate: inlets positive, outlets negative.

Solution:
Let n inlets, (8 − n) outlets.
n/12 − (8 − n)/36 = 1/3.
Multiply by 36: 3n − (8 − n) = 12 → 4n = 20 → n = 5.

Cetking shortcut: Option check: 5 inlets = 5/12, 3 outlets = 3/36 = 1/12 → net 4/12 = 1/3 ✓.

Answer: C

Q129EasySequences and Series

Find the sum of all numbers divisible by 6 between 100 and 400.

  1. 11,850
  2. 12,300
  3. 12,450
  4. 12,600
  5. 12,900
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Answer: C official key

Approach: AP sum of multiples in a range.

Solution:
First multiple of 6 above 100 = 102; last below 400 = 396.
Number of terms = (396 − 102)/6 + 1 = 50.
Sum = 50/2 × (102 + 396) = 25 × 498 = 12,450.

Cetking shortcut: Sum = 6 × (17 + 18 + … + 66) = 6 × 50 × 83/2 = 12,450.

Answer: C

Q130MediumSimplification

Arrange the following fractions in descending order:

(97/101), (48/50), (89/93), (76/79)

  1. (48/50) > (97/101) > (76/79) > (89/93)
  2. (97/101) > (48/50) > (89/93) > (76/79)
  3. (48/50) > (76/79) > (97/101) > (89/93)
  4. (76/79) > (97/101) > (48/50) > (89/93)
  5. (97/101) > (76/79) > (48/50) > (89/93)
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Answer: D official key

Approach: Compare fractions close to 1 by their gap from 1: smaller gap = larger fraction.

Solution:
1 − 97/101 = 4/101 ≈ 0.0396
1 − 48/50 = 2/50 = 0.04
1 − 89/93 = 4/93 ≈ 0.0430
1 − 76/79 = 3/79 ≈ 0.0380
Smallest gap → largest: 76/79 > 97/101 > 48/50 > 89/93.

Why the other options are wrong:
A, B, C, E put 48/50 or 97/101 ahead of 76/79, which has the smallest gap from 1.

Cetking shortcut: For a/b just below 1, compare (b − a)/b — the 'complement' method avoids long division.

Answer: D

Q131MediumTime Speed Distance

In a 1-kilometer race, A, B, and C are three participants.
A can give B, a start of 50 meters and C a start of 69 meters.

How many meters start can B, give C in the same race?

  1. 15 meters
  2. 18 meters
  3. 20 meters
  4. 21 meters
  5. 23 meters
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Answer: C official key

Approach: Race starts: convert to distances run in the same time.

Solution:
When A runs 1000 m, B runs 950 m and C runs 931 m.
So B : C speed = 950 : 931.
When B runs 1000 m, C runs 1000 × 931/950 = 980 m.
B can give C a start of 1000 − 980 = 20 m.

Why the other options are wrong:
B (19 m) is the naive 69 − 50 subtraction; A, D, E do not fit the ratio.

Cetking shortcut: 931/950 = 0.98 exactly (950 × 0.98 = 931) → 20 m.

Answer: C

Q132EasyEquations Basics

Lata has the same number of sisters as brothers.
Her brother Shyam has twice as many sisters as brothers.

How many children are there in the family?

  1. 3
  2. 5
  3. 6
  4. 7
  5. 8
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Answer: D official key

Approach: Count siblings excluding the person themselves.

Solution:
Let g girls and b boys.
Lata (a girl): sisters = g − 1, brothers = b → g − 1 = b.
Shyam (a boy): sisters = g, brothers = b − 1 → g = 2(b − 1).
g = 2(g − 2) → g = 4, b = 3 → total = 7.

Cetking shortcut: Option check: 4 girls + 3 boys — Lata sees 3 sisters, 3 brothers; Shyam sees 4 sisters, 2 brothers ✓.

Answer: D

Q133EasyProbability

A card is drawn from a set of playing cards. What is the probability that card drawn is a king or a queen?

  1. 1/2
  2. 1/4
  3. 2/13
  4. 5/13
  5. 1/13
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Answer: C official key

Approach: Simple probability with a standard 52-card deck.

Solution:
Kings = 4, Queens = 4 → favourable = 8.
P = 8/52 = 2/13.

Cetking shortcut: Each rank is 1/13; two ranks → 2/13.

Answer: C

Q134EasyProbability

A card is drawn from a set of playing cards. What is the probability that a card of red suit?

  1. 1/2
  2. 1/4
  3. 1/13
  4. 2/13
  5. 13/25
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Answer: A official key

Approach: Simple probability with a standard 52-card deck.

Solution:
Red suits: hearts and diamonds = 26 cards.
P = 26/52 = 1/2.

Cetking shortcut: Two of four suits are red → 1/2.

Answer: A

Q135EasyProbability

A card is drawn from a set of playing cards. What is the probability that a card having a number less than 9?

  1. 1/2
  2. 1/4
  3. 7/13
  4. 8/13
  5. 2/13
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Answer: C official key

Approach: Count numbered cards below 9 in a 52-card deck.

Solution:
Number cards less than 9: 2, 3, 4, 5, 6, 7, 8 → 7 per suit (Ace is not treated as a number card).
Favourable = 7 × 4 = 28.
P = 28/52 = 7/13.

Why the other options are wrong:
D (8/13) counts the Ace as 1; A, B, E do not match the count.

Cetking shortcut: 7 of the 13 ranks qualify → 7/13 directly.

Answer: C

Q136EasyInterest

A man invested ₹18,000 in ₹50 shares of a company at a 20% premium. If the company declares a 12% dividend at the end of the year, how much dividend will he receive?

  1. ₹1,440
  2. ₹1,500
  3. ₹1,600
  4. ₹1,800
  5. ₹2,160
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Answer: D official key

Approach: Shares: buy at market value, dividend on face value.

Solution:
Market price = 50 × 1.20 = Rs 60.
Number of shares = 18,000 / 60 = 300.
Dividend per share = 12% of 50 = Rs 6.
Total dividend = 300 × 6 = Rs 1,800.

Why the other options are wrong:
E (2,160) applies 12% to the amount invested; A, B, C come from wrong share counts or prices.

Cetking shortcut: Dividend yield on investment = 12% × 50/60 = 10% → 10% of 18,000 = 1,800.

Answer: D

Q137MediumPartnership

Two partners invest ₹1,80,000 and ₹1,20,000 respectively in a business. They agree that 50% of the total profit will be divided equally between them, and the remaining profit will be treated as interest on capital.
If one partner receives ₹600 more than the other, what is the total profit earned?

  1. ₹6,000
  2. ₹7,000
  3. ₹7,500
  4. ₹8,000
  5. ₹9,000
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Answer: A official key

Approach: Split profit into an equal part and a capital-ratio part.

Solution:
Let total profit = P. Half (P/2) is shared equally — no difference.
Other half is shared in 1,80,000 : 1,20,000 = 3 : 2.
Difference = (3 − 2)/5 × P/2 = P/10.
P/10 = 600 → P = Rs 6,000.

Cetking shortcut: Only the capital-ratio half creates the difference: 1/5 of half = 1/10 of total.

Answer: A

Q138EasyNumbers

How many times does the digit 2 appear in all the natural numbers from 1 to 200 (both inclusive)?

  1. 39
  2. 40
  3. 41
  4. 42
  5. 44
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Answer: C official key

Approach: Count digit occurrences place by place.

Solution:
1–99: units place 2 appears 10 times (2, 12, …, 92); tens place 10 times (20–29) → 20.
100–199: same pattern → 20.
200: one more 2 → 1.
Total = 20 + 20 + 1 = 41.

Cetking shortcut: Every 100 numbers contain 20 twos (10 units + 10 tens); add the hundreds digit only for 200.

Answer: C

Q139EasySimplification

Find the value of the following equation:

1.43 + 6 (1/4) − 3 (1/5) + 0.88 − (2.5 × 0.32)

  1. 45.6
  2. 4.56
  3. 0.456
  4. 5.46
  5. 5.13
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Answer: B official key

Approach: Convert mixed fractions to decimals and simplify (BODMAS).

Solution:
6 1/4 = 6.25; 3 1/5 = 3.2; 2.5 × 0.32 = 0.8.
1.43 + 6.25 − 3.2 + 0.88 − 0.8
= 7.68 − 3.2 + 0.88 − 0.8 = 4.48 + 0.08 = 4.56.

Cetking shortcut: Group: (1.43 + 0.88) + (6.25 − 3.2 − 0.8) = 2.31 + 2.25 = 4.56.

Answer: B

Q140EasySimplification

Find the value of

999 (1/7) + 999 (2/7) + 999 (3/7) + 999 (4/7) + 999 (5/7) + 999 (6/7)

  1. 5994
  2. 5995
  3. 5997
  4. 6000
  5. 6003
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Answer: C official key

Approach: Split each mixed number into whole and fraction parts.

Solution:
Whole parts: 999 × 6 = 5994.
Fractions: (1 + 2 + 3 + 4 + 5 + 6)/7 = 21/7 = 3.
Total = 5994 + 3 = 5997.

Cetking shortcut: Fraction pairs 1/7 + 6/7, 2/7 + 5/7, 3/7 + 4/7 each make 1 → 3.

Answer: C

Q141EasySurds & Indices

Find the value of x if

√(1.44/x) = 0.1

  1. 1.44
  2. 12
  3. 144
  4. 14.4
  5. 1.04
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Answer: C official key

Approach: Square both sides to remove the root.

Solution:
√(1.44/x) = 0.1 → 1.44/x = 0.01.
x = 1.44 / 0.01 = 144.

Cetking shortcut: Dividing by 0.01 = multiplying by 100.

Answer: C

Q142MediumMensuration

A cylinder is surmounted by a cone at one end and a hemisphere at the other end. The common radius is 3.5 cm, the height of the cylinder is 6.5 cm, and the total height of the structure is 12.8 cm.
The volume V of the structure lies between:

  1. 370 cm³ to 380 cm³
  2. 380 cm³ to 390 cm³
  3. 390 cm³ to 400 cm³
  4. 420 cm³ to 440 cm³
  5. 480 cm³ to 520 cm³
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Answer: A official key

Approach: Composite solid: cone + cylinder + hemisphere with common radius.

Solution:
r = 3.5 cm; hemisphere height = r = 3.5 cm.
Cone height = 12.8 − 6.5 − 3.5 = 2.8 cm.
πr² = 22/7 × 12.25 = 38.5.
Cylinder = 38.5 × 6.5 = 250.25; Cone = 1/3 × 38.5 × 2.8 = 35.93.
Hemisphere = 2/3 × 38.5 × 3.5 = 89.83.
V = 250.25 + 35.93 + 89.83 ≈ 376.0 cm³ → between 370 and 380.

Cetking shortcut: V = πr²(h_cyl + h_cone/3 + 2r/3) = 38.5 × (6.5 + 0.933 + 2.333) = 38.5 × 9.77 ≈ 376.

Answer: A

Q143EasyPolygons

An angle that is less than 360° and more than 180° is called

  1. Acute Angle
  2. Obtuse Angle
  3. Reflex Angle
  4. Complete Angle
  5. Complementary angle
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Answer: C official key

Approach: Angle classification by size.

Solution:
Acute < 90°, right = 90°, obtuse 90°–180°, straight = 180°.
Between 180° and 360° → reflex angle; 360° → complete angle.

Why the other options are wrong:
A (< 90°) and B (90°–180°) are too small; D is exactly 360°; E is a pair summing to 90°.

Answer: C

Q144MediumMensuration

A solid lead sphere of radius 8 cm is melted and recast into two types of smaller spherical balls:

  • 40% of the molten lead is used to make balls of radius 1 cm.
  • The remaining 60% is used to make balls of radius 2 cm.

During recasting, 10% of the total lead is wasted.
How many balls of radius 1 cm can be made?

  1. 184
  2. 192
  3. 204
  4. 216
  5. 256
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Answer: A official key

Approach: Volume conservation: number of balls = usable volume ÷ volume of one ball.

Solution:
Sphere volume ∝ r³: big sphere = 8³ = 512 units (unit = 4/3·π).
Lead for 1 cm balls = 40% of 512 = 204.8 units.
10% wastage applies → usable = 90% of 204.8 = 184.32 units.
Each 1 cm ball = 1³ = 1 unit → 184 balls (whole balls only).

Why the other options are wrong:
C (204) ignores the 10% wastage; B, D, E use wrong shares of the lead.

Cetking shortcut: Work in r³ units — 4/3 π cancels everywhere.

Answer: A

Q145EasyEquations Basics

When a certain amount is distributed equally among 14 girls, each girl receives ₹160 more than what she would receive if the same amount were distributed equally among 18 girls.
If ₹1,120 is kept aside before distribution and the remaining amount is then distributed among 14 girls, each girl now receives ₹80 less than before.
Find the original total amount.

  1. ₹8,640
  2. ₹9,600
  3. ₹10,080
  4. ₹11,520
  5. ₹12,520
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Answer: C official key

Approach: Equal distribution: difference in shares gives the total.

Solution:
T/14 − T/18 = 160 → T(18 − 14)/252 = 160 → T = 160 × 63 = 10,080.
Check: 10,080/14 = 720; (10,080 − 1,120)/14 = 8,960/14 = 640 = 720 − 80 ✓.

Cetking shortcut: 1,120/14 = 80 — the second statement is automatically true; use only the first. Also total must be divisible by 14 and 18 (LCM 126): 10,080 = 126 × 80.

Answer: C

Q146EasyPie Chart

Questions 146–150 share this

Study the chart and answer the questions:

The pie chart given here represents the domestic expenditure of a family in percent. Study the chart and answer the following questions if the total monthly income of the family is Rs. 33,650

Data read from chart (Expenditure of Family):

ItemPercent
Food25%
House Rent18%
Entertainment9%
Child Education23%
Medical13%
Provident Fund12%
Figure for MBA CET 2026 Attempt 2 Slot 1 question 146

House rent per month is

  1. Rs. 6000
  2. Rs. 6152
  3. Rs. 6057
  4. Rs. 6048
  5. Rs. 7740
Show answerHide answer

Answer: C official key

Approach: Pie percentage to absolute value.

Solution:
House Rent = 18% of 33,650.
10% = 3,365; 8% = 2,692; total = 6,057.

Cetking shortcut: 1% = 336.5 → 18 × 336.5 = 6,057.

Answer: C

Q147EasyPie Chart

Refer to the passage above ↑

Annual Saving in the form of provident funds

Figure for MBA CET 2026 Attempt 2 Slot 1 question 147
  1. Rs. 48456
  2. Rs. 48540
  3. Rs. 44856
  4. Rs. 45480
  5. Rs. 46480
Show answerHide answer

Answer: A official key

Approach: Monthly PF from the pie, then annualise.

Solution:
PF = 12% of 33,650 = 4,038 per month.
Annual = 4,038 × 12 = 48,456.

Why the other options are wrong:
B, D shuffle digits (48,540 / 45,480); C, E are miscalculations.

Cetking shortcut: 12% × 12 months = 144% of monthly income: 1.44 × 33,650 = 48,456.

Answer: A

Q148EasyPie Chart

Refer to the passage above ↑

After provident fund deduction and payment of house rent,total monthly income of family remains:

Figure for MBA CET 2026 Attempt 2 Slot 1 question 148
  1. Rs. 23545
  2. Rs. 24435
  3. Rs. 23555
  4. Rs. 25355
  5. Rs. 26355
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Answer: C official key

Approach: Remaining share = 100% − (Rent + PF).

Solution:
Rent + PF = 18% + 12% = 30%.
Remaining = 70% of 33,650 = 23,555.

Why the other options are wrong:
A (23,545) and B (24,435) are near-miss digit traps; D, E are wrong percentages.

Cetking shortcut: 33,650 − 30% (10,095) = 23,555.

Answer: C

Q149EasyPie Chart

Refer to the passage above ↑

Total amount per month,the family spends on food and entertainment combined together is

Figure for MBA CET 2026 Attempt 2 Slot 1 question 149
  1. Rs. 11432
  2. Rs. 11441
  3. Rs. 12315
  4. Rs. 12443
  5. Rs. 13443
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Answer: B official key

Approach: Add the two percentages, then convert.

Solution:
Food + Entertainment = 25% + 9% = 34%.
34% of 33,650 = 11,441.

Why the other options are wrong:
A (11,432) is a near-miss; C, D, E use wrong percentages.

Cetking shortcut: 1% = 336.5; × 34 = 11,441.

Answer: B

Q150EasyPie Chart

Refer to the passage above ↑

What is the difference between medical and entertainment expenses?

Figure for MBA CET 2026 Attempt 2 Slot 1 question 150
  1. 1346
  2. 1386
  3. 2019
  4. 2355
  5. 3365
Show answerHide answer

Answer: A official key

Approach: Difference of two pie percentages, then convert.

Solution:
Medical − Entertainment = 13% − 9% = 4%.
4% of 33,650 = 1,346.

Why the other options are wrong:
C (2,019) = 6%; D (2,355) = 7%; E (3,365) = 10%; B is a near-miss.

Cetking shortcut: 1% = 336.5 → × 4 = 1,346.

Answer: A

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