Difference between revisions of "1993 AJHSME Problems/Problem 15"
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==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> |
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+ | ==See Also== | ||
+ | {{AJHSME box|year=1993|num-b=14|num-a=16}} |
Revision as of 21:48, 22 December 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 & . Then . What we are trying to find is . Solving,
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 |