Veronica has 6 marks on her report card.

The mean of the 6 marks is 74 .

The mode of the 6 marks is 76 .

The median of the 6 marks is 76 .

The lowest mark is 50 .

The highest mark is 94 .

Only one mark appears twice and no mark appears more than twice.

Assuming all of her marks are integers, the number of possibilities for her second lowest mark is
(A) 17
(B) 16
(C) 25
(D) 18
(E) 24
Veronica has 6 marks on her report card.

The mean of the 6 marks is 74 .

The mode of the 6 marks is 76 .

The median of the 6 marks is 76 .

The lowest mark is 50 .

The highest mark is 94 .

Only one mark appears twice and no mark appears more than twice.

Assuming all of her marks are integers, the number of possibilities for her second lowest mark is
(A) 17
(B) 16
(C) 25
(D) 18
(E) 24

## Solution

Since the mode of Veronica's 6 marks is 76, and only one mark appears more than once (and no marks appear more than twice), then two of the marks must be 76. This tells us that four of her marks were $50,76,76,94$.

Since the median of her marks is 76 and she has six marks in total (that is, an even number of marks), then the two marks of 76 must be 3rd and 4th when the marks are arranged in increasing order. Let the second lowest mark be $M$, and the second highest be $N$. So the second lowest mark $M$ is between (but not equal to) 50 and 76, and the second highest mark $N$ is between (but not equal to) 76 and 94. We still need to use the fact that the mean of Veronica's marks is 74 , so

$$
\begin{aligned}
\frac{50+M+76+76+N+94}{6} & =74 \\
M+N+296 & =444 \\
M+N & =148 \\
M & =148-N
\end{aligned}
$$

We know already that $M$ is one of 51 through 75, but the possibilities for $N$ and the equation (*) restrict these possibilities further.

Since $N$ can be any of 77 through 93 , there are exactly 17 possibilities for $N$. The largest value of $M$ corresponds to $N=77$ (ie. $M=71$ ) and the smallest value for $M$ is when $N=93$ (ie. $M=55$ ). Thus the possibilities for $M$ are 55 through 71, ie. there are 17 possibilities in total for $M$, the second smallest mark.

ANSwER: (A)