Difference between revisions of "1993 AJHSME Problems/Problem 15"
Mrdavid445 (talk | contribs) (Created page with "==Problem== The arithmetic mean (average) of four numbers is <math>85</math>. If the largest of these numbers is <math>97</math>, then the mean of the remaining three numbers i...") |
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<math>\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3</math> | <math>\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3</math> | ||
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| + | ==Solution== | ||
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| + | Say that the four numbers are <math>a, b, c,</math> & <math>97</math>, with <math>a \le b \le c \le 97</math>. Then <math>\frac{a+b+c+97}{4} = 85</math>. What we are trying to find is <math>\frac{a+b+c}{3}</math>. Solving, <cmath>\frac{a+b+c+97}{4} = 85</cmath> <cmath>a+b+c+97 = 340</cmath> <cmath>a+b+c = 243</cmath> <cmath>\frac{a+b+c}{3} = 81 \Rightarrow \mathrm{(A)}</cmath> | ||
Revision as of 14:26, 1 May 2012
Problem
The arithmetic mean (average) of four numbers is 
. If the largest of these numbers is 
, then the mean of the remaining three numbers is
Solution
Say that the four numbers are 
& , with 
. Then 
. What we are trying to find is 
. Solving, ![\[\frac{a+b+c+97}{4} = 85\]](http://latex.artofproblemsolving.com/b/4/6/b462f498d6513ef72710b2e38700c73c2ba0653c.png)
![\[a+b+c+97 = 340\]](http://latex.artofproblemsolving.com/9/7/6/976916ecc429c246a91ad3f990758ab41496a0b3.png)
![\[a+b+c = 243\]](http://latex.artofproblemsolving.com/9/6/e/96e4481a2489bef2e59dd2346998dbf08b3f1bb9.png)
![\[\frac{a+b+c}{3} = 81 \Rightarrow \mathrm{(A)}\]](http://latex.artofproblemsolving.com/c/5/1/c51c4a7d728720fed7fe89b59a7b8c260e91581b.png)
