Question 16

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:

Solution

Let $$x, y, z$$ be the number of students getting 6, 8 and 20 marks respectively.

=> $$6x + 8y + 20z = 504$$ -------------Eqn(I)

Since, the mode is 8, => Most populated category = $$y$$

=> $$y = x + 2z$$

=> $$x = y - 2z$$ -------------Eqn(II)

Substituting it in eqn(I), we get : 

=> $$(6y - 12z) + 8y + 20z = 504$$

=> $$14y + 8z = 504$$

By hit and trial, since $$x,y,z$$ are integers, we get : $$y = 32$$ and $$z = 7$$

Putting it in Eqn(II), => $$x = 32 - 2 \times 7 = 18$$

$$\therefore$$ Total students = $$x + y + z = 18 + 32 + 7 = 57$$

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