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. $\mathrm{(B)}$

1995 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1995 AHSME Problems/Problem 1

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