Question

A teacher noticed a strange distribution of marks in the exam. There were only three distinct scores: $6, 8,$ and $20$. The mode of the distribution was $8$. The sum of the scores of all the students was $504$. The number of students in the most populated category was equal to the sum of the number of students with lowest score and twice the number of students with the highest score. The total number of students in the class was:

• A. 50
• B. 51
• C. 53
• D. 56
• E. 57

Hint:

Translate “mode” and verbal relations into equations on counts, then enforce integrality (often by using divisibility) to find the unique feasible solution.

Answer 1

Answer: (E)

Step 1: Define variables.
Let x, y, z be the numbers of students scoring 6, 8, 20 respectively. Mode = 8 ⇒ y > x and y > z (all three scores occur).
Given: total sum 6x + 8y + 20z = 504.
“Most populated equals lowest plus twice highest” ⇒ y = x + 2z.

Step 2: Solve the system.
Substitute y = x + 2z:
6x + 8(x + 2z) + 20z = 504 ⇒ 14x + 36z = 504 ⇒ 7x + 18z = 252.
Hence x = (252 − 18z)/7 = 36 − (18/7)z.
For integer x, z must be a multiple of 7. Let z = 7k.
Then x = 36 − 18k, and y = x + 2z = 36 − 18k + 14k = 36 − 4k.
With x > 0, z > 0 and y > x, y > z, the only feasible k is k = 1 (since k = 0 gives a tie, k = 2 gives x = 0).

Step 3: Compute totals.
For k = 1: (x, y, z) = (18, 32, 7).
Total students N = x + y + z = 18 + 32 + 7 = 57.

Final Answer:
57