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 in 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:

Accepted Answer

E. Let $x, y, z$ be the number of students getting 6, 8, 20 marks. $6x + 8y + 20z = 504$. Mode is $8 \implies y = x + 2z \implies x = y - 2z$. Substitute $x$: $6(y - 2z) + 8y + 20z = 504 \implies 14y + 8z = 504 \implies 7y + 4z = 252$. Since $x, y, z$ are integers and $y > x, z$, by trial $y = 32, z = 7 \implies x = 32 - 14 = 18$. Total students $= 18 + 32 + 7 = 57$.

Distribution of Marks and Mode

Explanation

Practise this under exam conditions