Difference between revisions of "1995 AHSME Problems/Problem 1"

(See also)
Line 10: Line 10:
 
==See also==
 
==See also==
 
{{AHSME box|year=1995|before=First question|num-a=2}}
 
{{AHSME box|year=1995|before=First question|num-a=2}}
 +
{{MAA Notice}}

Revision as of 12:58, 5 July 2013

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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