Question

Five students, Alec, Bryn, Igor, Molly and Suzy, too! en exam The exern was marked out of 30.

The highest mark awarded was 26.
.
The lowest mark was 7.
O The average (mean) mark was 16.
Molly achieved 8 marks more than Bryn.
• Igor got exactly three times the mark that Alec achieved.
What mark did each student achieve?

Answer 1

There are 3 possibilities:

`\begin{tabular}{lccc}Alec&7&7&8\\Bryn&18&9&7\\Igor&21&21&24\\Molly&26&17&15\\Suzy&8&26&26\end{tabular}`

Explanation:

This logic puzzle makes use of the relations between numbers and their mean.

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If two scores in the range of 7 to 26 have a ratio of exactly 3, they must be either 7 and 21, or 8 and 24. In order for the mean to be exactly 16, the total of the 5 scores must be 5×16 = 80.

one solution

If the 3:1 range involves scores of 21 and 7, then we know ...

Alec got 7

Igor got 21

The total of these and the score of 26 is 54. The remaining two scores must total 80-54 = 26. We know that there must be two scores that have a difference of 8, so there are two possibilities:

Molly got 26 and Bryn got 18, leaving Suzy with a score of 8.

Molly got 17 and Bryn got 9, leaving Suzy with a score of 26.

The possibilities here are ...

Alec 7, Bryn 18, Igor 21, Molly 26, Suzy 8

Alec 7, Bryn 9, Igor 21, Molly 17, Suzy 26

another solution

If the 3:1 range involves scores of 24 and 8, then we know ...

Alec got 8

Igor got 24

The total of these and the score of 26 is 58. The remaining two scores must total 80-58 = 22. That two scores have a difference of 8 leaves open one possibility: the remaining two scores are 7 and 15. (If one were 18=26-8, the other would be 22-18=4, too low for the given range.)

Alec 8, Bryn 7, Igor 24, Molly 15, Suzy 26

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Possible marks are ...

`\begin{tabular}{lccc}Alec&7&7&8\\Bryn&18&9&7\\Igor&21&21&24\\Molly&26&17&15\\Suzy&8&26&26\end{tabular}`

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Additional comment

When two numbers are required to have a specific sum (s) and a specific difference (d), the two numbers are (s+d)/2 and (s-d)/2.