1995 AHSME Problems/Problem 1

Problem

Kim earned scores of 87,83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will


$\mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} }$

Solution

The average of the first three test scores is $\frac{88+83+87}{3}=86$. The average of all four exams is $\frac{87+83+88+90}{4}=87$. It increased by one point. $\mathrm{(B)}$

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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