Difference between revisions of "1993 AJHSME Problems/Problem 15"
(→Problem) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | Say that the four numbers are <math>a, b, c,</math> & <math> | + | Say that the four numbers are <math>a, b, c,</math> & <math>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} = \boxed{\mathrm{(A)}\ 81}</cmath> |
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1993|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:11, 4 July 2013
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 
& . 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)
![\[\frac{a+b+c}{3} = \boxed{\mathrm{(A)}\ 81}\]](http://latex.artofproblemsolving.com/7/5/f/75f77f2548165dabed62ecd4085330a634f3c43e.png)
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
