Difference between revisions of "1995 AHSME Problems/Problem 1"
(New page: ==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 <math> \mathrm{(A) \ ...) |
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The average of the first three test scores is <math>\frac{88+83+87}{3}=86</math>. The average of all four exams is <math>\frac{87+83+88+90}{4}=87</math>. It increased by one point. <math>\mathrm{(B)}</math> | The average of the first three test scores is <math>\frac{88+83+87}{3}=86</math>. The average of all four exams is <math>\frac{87+83+88+90}{4}=87</math>. It increased by one point. <math>\mathrm{(B)}</math> | ||
| − | ==See | + | ==See also== |
| + | {{AHSME box|year=1995|before=First question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest 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
Solution
The average of the first three test scores is 
. The average of all four exams is 
. It increased by one point. 
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| 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. 
