Difference between revisions of "1992 AJHSME Problems/Problem 13"
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Because there was an odd number of scores, <math> 91 </math> must be the middle score. Since there are two scores above <math> 91 </math> and <math> 94 </math> appears the most frequent (so at least twice) and <math> 94>91 </math>, <math> 94 </math> appears twice. Also, the sum of the five numbers is <math> 90 \times 5 =450 </math>. Thus, the sum of the lowest two scores is <math> 450-91-94-94= \boxed{\text{(B)}\ 171} </math>. | Because there was an odd number of scores, <math> 91 </math> must be the middle score. Since there are two scores above <math> 91 </math> and <math> 94 </math> appears the most frequent (so at least twice) and <math> 94>91 </math>, <math> 94 </math> appears twice. Also, the sum of the five numbers is <math> 90 \times 5 =450 </math>. Thus, the sum of the lowest two scores is <math> 450-91-94-94= \boxed{\text{(B)}\ 171} </math>. | ||
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1992|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:09, 4 July 2013
Problem
Five test scores have a mean (average score) of 
, a median (middle score) of 
and a mode (most frequent score) of 
. The sum of the two lowest test scores is
Solution
Because there was an odd number of scores, must be the middle score. Since there are two scores above and appears the most frequent (so at least twice) and 
, appears twice. Also, the sum of the five numbers is 
. Thus, the sum of the lowest two scores is 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
