Sample Previous Year Questions: Averages | Quantitative for GMAT PDF Download

Q1. A book club’s average monthly reading is 3 books per member for 10 members. If two members increase their reading to 6 books each, what must the remaining members’ average reading be to keep the overall club average at 3 books per member?
A. 2.75
B. 2.25
C. 2.90
D. 3.0
E. 3.15
Ans: Option B

Explanation:

Original total = 10 × 3 = 30 books.

New target total = 10 × 3 = 30 books.

Two members now read 6 + 6 = 12; remaining eight must sum to 30 − 12 = 18.

Required average = 18 / 8 = 2.25.

Q2. The average salary of 8 managers is $85,000. Two managers receive raises, changing their salaries from $80,000 to $90,000 and $95,000 to $105,000. What is the new average salary?

A. $86,000
B. $87,500
C. $88,000
D. $87,500
E. $90,550
Ans: Option D

Explanation:

Original sum = 8 × 85,000 = 680,000.

Increase = (90,000 − 80,000) + (105,000 − 95,000) = 10,000 + 10,000 = 20,000.

New sum = 700,000.

New average = 700,000 / 8 = 87,500.

Q3. A sequence of numbers in arithmetic progression has an average of 50 across the first 9 terms. If the first term is 32, find the common difference.

A. 2.7
B. 3.5
C. 4.5
D. 5.8
E. 6.3
Ans: Option C

Explanation:

Sum = 9 × 50 = 450.

Sum formula = (first + ninth) × 9 / 2 = (32 + [32 + 8d]) × 9/2 = (64 + 8d) × 9/2 = (64 + 8d) ×4.5.

(64 + 8d) × 4.5 = 450 
→ 64 + 8d = 450/4.5 = 100 
→ 8d = 36 
→ d = 4.5

Q4. A student’s grade is based on 4 tests: 3 midterms (each weighted 1) and 1 final (weighted 2). The midterm scores are 70, 80, and 90. What score must the student achieve on the final to have an overall average of 85?

A) 80.8
B) 85.3
C) 97.5
D)92.5
E) 100.2

Ans: D) 92.5

Explanation:
Total weight units:
– Each midterm counts as 1, so 3 midterms = 3 units
– Final counts as 2 units
– Total = 3 + 2 = 5 units

Sum of midterm scores = 70 + 80 + 90 = 240

To get an overall average of 85 over 5 units, the weighted total must be 85 × 5 = 425

Let F be the final exam score. Then:
weighted total = midterm sum + (final score × final weight)
240 + (F × 2) = 425

Solve for F:
2F = 425 – 240 = 185
F = 185 ÷ 2 = 92.5

Q5. A course grade is based on 3 assignments: 2 quizzes (each weighted 20%) and 1 final exam (weighted 60%). If a student scores 70 and 80 on the quizzes and 90 on the final exam, what is their course grade?
​A) 82
B) 84
C) 86
D) 88
E) 90
Ans: B) 84

Explanation:

1. Understand the weights:
   • Each quiz has a weight of 20% (0.2), and the final exam has a weight of 60% (0.6).
   • Total weight = 0.2 + 0.2 + 0.6 = 1 (or 100%).
2. Calculate the weighted contributions:
   • First quiz: 70 × 0.2 = 14
   • Second quiz: 80 × 0.2 = 16
   • Final exam: 90 × 0.6 = 54
3. Sum the weighted scores:
   • Total = 14 + 16 + 54 = 84
4. Alternative method (weighted average formula):
   • Weighted average = (Quiz 1 × Weight 1 + Quiz 2 × Weight 2 + Final × Weight 3) ÷ Total weight
   • = (70 × 0.2 + 80 × 0.2 + 90 × 0.6) ÷ 1 = 14 + 16 + 54 = 84

Q6. A student scored an average of 80 points on her first 4 math tests. What score must she achieve on her 5th test to have an overall average of 82?

Possible Answers:A) 84
B) 86
C) 88
D) 90
E) 92

Ans: D) 90

Explanation:

Sum of first 4 test scores = 4 × 80 = 320
For an overall average of 82 over 5 tests, total points needed = 5 × 82 = 410
Let S be the 5th test score. Then:
320 + S = 410

Solve for S:
S = 410 − 320 = 90

Q7. A retailer’s average daily sales over 30 days is $5,000. On day 31, sales drop to $3,000. What is the new average sales over the 31 days?
A. $4,935
B. $4,900
C. $4,870
D. $4,850
E. $4,820
Ans: A, $4,935

Explanation:

Original sum = 30 × 5,000 = 150,000.

New sum = 150,000 + 3,000 = 153,000.

New count = 31.

New average = 153,000 / 31 ≈ 4,935.48 ≈ $4935

Q8. In a dataset, the average is 50. When three values are removed, the average of the remaining values rises to 52. If the sum of removed values is 114, how many values were originally in the dataset?

A. 15
B. 18
C. 20
D. 22
E. 21
Ans: E, 21

Explanation:

Let original count = n; original sum = 50n.

Remaining count = n − 3; remaining sum = 50n − 114.

New average = (50n − 114) / (n − 3) = 52.

50n − 114 = 52n − 156.

52n − 50n = 156 − 114 
→ 2n = 42
 → n = 21.

 Q9. A sequence of numbers has mean 100. If each number is increased by 20%, what is the new mean?

A. 100
B. 110
C. 120
D. 140
E. 160

Ans: C, 120

Explanation:

Original mean = 100

Increasing each value by 20% means multiplying each by 1.20

New mean = original mean × 1.20 = 100 × 1.20 = 120

Q10.The average of $n$n items is 40. When 5 items with average 52 are removed, the new average of the remaining items becomes 35. Find $n$n.

A. 12
B. 15
C. 17
D. 20
E. 22
Ans: C, 17

Explanation:

Original total sum = 40 × n = 40n

Sum of the 5 removed items = 5 × 52 = 260

Total of the remaining items = 40n – 260; remaining count = n – 5

New average condition: (40n – 260) ÷ (n – 5) = 35

Multiply both sides by (n – 5):
40n – 260 = 35(n – 5) = 35n – 175

Rearrange:
40n – 35n = 260 – 175
5n = 85

Divide by 5:
n = 85 ÷ 5 = 17