1985 AJHSME Problems/Problem 17

Problem

If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$, then your score on the seventh test was

$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$

Solution

Solution 1

If the average score of the first six is $84$, then the sum of those six scores is $6\times 84=504$.

The average score of the first seven is $85$, so the sum of the seven is $7\times 85=595$

Taking the difference leaves us with just the seventh score, which is $595-504=91$, so the answer is $\boxed{\text{D}}$

Solution 2

Let's remove the condition that the average of the first seven tests is $85$, and say the 7th test score was a $0$. Then, the average of the first seven tests would be \[\frac{6\times 84}{7}=72\]

If we increase the seventh test score by $7$, the average will increase by $\frac{7}{7}=1$. We need the average to increase by $85-72=13$, so the seventh test score is $7\times 13=91$ more than $0$, which is clearly $91$. This is choice $\boxed{\text{D}}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions


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