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Averages – MBA CET questions with solutions

23 questions · CET 2025–2026 · Easy 20 · Medium 3 · Hard 0. Newest paper first. Full papers

Q107EasyCET 2026 · Attempt 2 Slot 2 · Q107
The average of 9 numbers is 15. If the average of the first five numbers is 13 and the average of the last five numbers is 17, then what is the fifth number ?
  1. 13
  2. 18
  3. 15
  4. 16
  5. 12
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Answer: C official key

Approach: Overlap in averages: the 5th number is counted in both groups.

Solution:
Sum of 9 numbers = 9 × 15 = 135.
First five sum = 5 × 13 = 65; last five sum = 5 × 17 = 85.
65 + 85 = 150 counts the 5th number twice.
5th number = 150 − 135 = 15.

Cetking shortcut: Deviation method: first five −2 each (−10), last five +2 each (+10) → 5th = 15 + 0 = 15.

Answer: C

Q108EasyCET 2026 · Attempt 2 Slot 2 · Q108
Zahir's marks were wrongly entered as 83 instead of 63 in a class test. Because of this mistake, the average marks for the class increased by 0.5. What is the number of students in the class ?
  1. 45
  2. 50
  3. 40
  4. 30
  5. 35
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Answer: C official key

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

Solution:
Wrong entry adds 83 − 63 = 20 to the total.
This raises the average by 0.5.
n × 0.5 = 20 → n = 40.

Cetking shortcut: Number of students = excess ÷ rise in average = 20 / 0.5 = 40.

Answer: C

Q109EasyCET 2026 · Attempt 2 Slot 2 · Q109
The average age of a cricket team of 11 players is reduced by 2 when two senior players aged 32 and 30 are replaced by 2 youngsters of equal age. What is the age of each youngster ?
  1. 22
  2. 20
  3. 24
  4. 26
  5. 25
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Answer: B official key

Approach: Replacement in average: total drop = drop in average × team size.

Solution:
Average falls by 2 for 11 players → total falls by 22.
Removed ages = 32 + 30 = 62.
Two youngsters together = 62 − 22 = 40 → each = 20.

Cetking shortcut: Each youngster = (62 − 22) / 2 = 20 in one line.

Answer: B

Q111EasyCET 2026 · Slot 4 · Q111
Calculate the average of the first 50 natural numbers.
  1. 25
  2. 25.5
  3. 26
  4. 50.5
  5. 51
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Answer: B official key

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

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

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

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

Answer: B

Q112EasyCET 2026 · Slot 4 · Q112
A batsman has a certain average of runs for 11 innings. In the 12th inning, he makes a score of 90 runs, thereby increasing his average by 5. His new average is:
  1. 30
  2. 40
  3. 45
  4. 50
  5. 35
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Answer: E official key

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

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

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

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

Answer: E

Q116EasyCET 2026 · Slot 4 · Q116
The average age of 15 students is 18 years. If the teacher's age is added, the average increases by 1 year. What is the teacher's age?
  1. 32 years
  2. 34 years
  3. 35 years
  4. 33 years
  5. 30 years
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Answer: B official key

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

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

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

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

Answer: B

Q112EasyCET 2026 · Slot 3 · Q112
The average of the first ten odd prime numbers excluding the smallest and the largest among them is
  1. 15.5
  2. 16.2
  3. 17.5
  4. 18.4
  5. 19.2
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Answer: A official key

Approach: List the primes, drop the end values, then average.

Solution:
First ten odd primes: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.
Exclude smallest 3 and largest 31.
Sum of remaining 8 = 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 124.
Average = 124 / 8 = 15.5.

Cetking shortcut: Total of all ten = 158; 158 − 34 = 124; 124/8 = 15.5.

Answer: A

Q113EasyCET 2026 · Slot 3 · Q113
The average salary per head of all the workers in a company is ₹9000 The average salary of 15 officers is ₹5000 and the average salary per head of the remaining workers is ₹10,000 Find the total number of workers in the company.
  1. 75
  2. 80
  3. 50
  4. 40
  5. 30
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Answer: A official key

Approach: Weighted average / alligation of two groups.

Solution:
Let total workers = n.
9000n = 15 × 5000 + (n − 15) × 10000.
9000n = 75000 + 10000n − 150000 → 1000n = 75000 → n = 75.

Cetking shortcut: Alligation: officers are 4000 below average, others 1000 above → officers : others = 1 : 4 → 15 : 60 → total 75.

Answer: A

Q108EasyCET 2026 · Slot 2 · Q108

DIRECTION for the question: Solve the following question and mark the best possible option.

A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is
  1. 250
  2. 276
  3. 280
  4. 285
  5. 240
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Answer: D official key

Approach: Weighted average over Sundays and other days.

Solution: Sundays on days 1, 8, 15, 22, 29 → 5 Sundays, 25 other days.
Total = 5 × 510 + 25 × 240 = 2550 + 6000 = 8550.
Average = 8550 ÷ 30 = 285.

Why the other options are wrong:
E (240): ignores the Sunday visitors.
B (276): uses 4 Sundays and 26 other days → 8280/30 = 276.

Cetking shortcut: Average = 240 + 5 × 270 / 30 = 240 + 45 = 285.

Answer: D

Q107EasyCET 2026 · Slot 1 · Q107

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

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

Approach: Weighted average = total weight / total students.

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

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

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

Answer: A

Q108EasyCET 2026 · Slot 1 · Q108

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

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

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

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

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

Answer: D

Q109EasyCET 2026 · Slot 1 · Q109

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

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

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

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

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

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

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

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

Answer: E

Q123EasyCET 2025 · Slot 6 · Q123
The average runs scored by 3 players is 160 per match. Their runs are in a ratio 4:6:2. What will be the average it the player who scored the minimum runs hits a century in the next match and the other two play with the same runs?
  1. 160.5
  2. 166.66
  3. 162.5
  4. 168.5
  5. 162.5
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Answer: B

Approach: Average = total runs / number of players; recompute the total for the next match.

Solution:
Total runs per match = 3 × 160 = 480; ratio 4 : 6 : 2 (12 parts) ⇒ 160, 240, 80.
Next match: lowest scorer (80) scores 100; the other two score the same 160 and 240.
New total = 160 + 240 + 100 = 500.
Average for the match = 500/3 = 166.66.

Why the other options are wrong:
C/E 162.5, A 160.5, D 168.5: none equals 500/3; combining both matches (980/6 = 163.33) is not what is asked and is not an option either.

Cetking shortcut: Only the lowest scorer changes by +20 ⇒ average rises by 20/3 = 6.66 ⇒ 160 + 6.66 = 166.66.

Answer: B

Q106MediumCET 2025 · Slot 4 · Q106

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
Q107EasyCET 2025 · Slot 4 · Q107

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
Q108EasyCET 2025 · Slot 4 · Q108

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
Q111EasyCET 2025 · Slot 3 · Q111
The average weight of 16 boys in a class is 50.25 kgs and that of the remaining 8 boys is 45.15 kgs. Find the average weight of all boys in the class
  1. 47.55
  2. 48
  3. 48.55
  4. 49.25
  5. 50.25
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Answer: C

Approach: Weighted average of two groups (50.25 and 45.15 are the average weights of each group).

Solution: Total of 16 boys = 16 × 50.25 = 804 kg.
Total of 8 boys = 8 × 45.15 = 361.2 kg.
Total of 24 boys = 804 + 361.2 = 1165.2 kg.
Average = 1165.2 ÷ 24 = 48.55 kg.

Cetking shortcut: Weights 16:8 = 2:1 → average = 45.15 + (2/3)(50.25 − 45.15) = 45.15 + 3.40 = 48.55.

Answer: C

Q112EasyCET 2025 · Slot 3 · Q112
A team of 8 persons joined a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points, the team scored?
  1. 588
  2. 645
  3. 665
  4. 672
  5. 555
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Answer: C

Approach: Total = average × count, then undo the change.

Solution: If the best scorer had 92, team total = 8 × 84 = 672.
He actually scored 85, i.e. 92 − 85 = 7 points less.
Actual team total = 672 − 7 = 665.

Answer: C

Q29EasyCET 2025 · Slot 2 · Q29

The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weight of all the boys in the class.

  1. 48.55 kg
  2. 47.55 kg
  3. 49.25 kg
  4. 50.25 kg
  5. 48.55 kg
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Answer: A

Approach: Weighted average = total weight ÷ total boys.

Solution:
16 × 50.25 = 804
8 × 45.15 = 361.2
Total = 1165.2 kg for 24 boys
Average = 1165.2 ÷ 24 = 48.55 kg.

Cetking shortcut: Gap = 50.25 − 45.15 = 5.1; weights are 16 : 8 = 2 : 1, so the average lies 1/3 of the gap below 50.25: 50.25 − 1.7 = 48.55.

Answer: A

Q102MediumCET 2025 · Slot 2 · Q102
Company of 40 employees. In 1925: first 30 average Rs 40 000; last 30 average Rs 60 000; overall average Rs 50 000. In 1926: first 10 ↑ 100 %, last 10 ↑ 200 %, others unchanged. What is new overall average bonus?
  1. 90 000
  2. 88 000
  3. 85 000
  4. 80 000
  5. 95 000
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Answer: E

Approach: Overlapping groups — 'first 30' and 'last 30' together count the middle 20 twice.

Solution:
Total (all 40) = 40 × 50,000 = 20,00,000.
First 30 total = 30 × 40,000 = 12,00,000; last 30 total = 30 × 60,000 = 18,00,000.
First 30 + last 30 = all 40 + middle 20 → 30,00,000 = 20,00,000 + middle 20 → middle 20 = 10,00,000.
First 10 = 12,00,000 − 10,00,000 = 2,00,000; last 10 = 18,00,000 − 10,00,000 = 8,00,000.
1926: first 10 up 100% → 4,00,000; last 10 up 200% → 24,00,000; middle 20 unchanged → 10,00,000.
New total = 38,00,000 → average = 38,00,000 / 40 = Rs 95,000.

Cetking shortcut: Increase = 100% of 2L + 200% of 8L = 18L; 18L/40 = 45,000 added to 50,000 → 95,000.

Answer: E

Q108EasyCET 2025 · Slot 2 · Q108
Team of 8 shooters. Best scored 85. If he had 92, avg = 84. Find actual total.
  1. 588
  2. 555
  3. 672
  4. 645
  5. 665
Show answerHide answer

Answer: E

Approach: Hypothetical total minus the extra points.

Solution:
If the best scored 92, total = 8 × 84 = 672.
He actually scored 85, i.e. 7 less → actual total = 672 − 7 = 665.

Answer: E

Q144MediumCET 2025 · Slot 2 · Q144
Take 6 distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers can be how much?
  1. 22.5
  2. 24.5
  3. 21.5
  4. 20
  5. 23.5
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Answer: A

Approach: Push the middle two numbers as high as the distinctness constraint allows.

Solution:
Order a < b < c < d < e < f. a + b = 28, e + f = 56.
To raise c, d, make e as large as possible: e < f → e = 27, f = 29.
Then d ≤ 26, c ≤ 25 → c + d = 51 (a, b e.g. 13, 15 stay below c).
Max average = (28 + 51 + 56)/6 = 135/6 = 22.5.

Cetking shortcut: e is at most 27 (just under half of 56), so the middle pair is 26 + 25.

Answer: A

Q32EasyCET 2025 · Slot 1 · Q32

The average age of three men is 50 years and their ages are in the proportion 3:5:7 respectively. Determine the age of the youngest man in the group ?

  1. 40 years
  2. 25 years
  3. 20 years
  4. 30 years
  5. 35 years
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Answer: D

Approach: Average → total → ratio split.

Solution:
Total age = 3 × 50 = 150.
Ratio 3 : 5 : 7 = 15 parts → 1 part = 10.
Youngest = 3 × 10 = 30 years.

Why the other options are wrong:
C 20 — wrong part value.
B 25, E 35, A 40 — not 3 parts of 150/15.

Cetking shortcut: Middle term of 3:5:7 equals the average (5 parts = 50) → 1 part = 10.

Answer: D

Solution figure for question 32

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