Difference between revisions of "1986 AHSME Problems/Problem 15"
(Created page with "==Problem== A student attempted to compute the average <math>A</math> of <math>x, y</math> and <math>z</math> by computing the average of <math>x</math> and <math>y</math>, and...") |
(Added a solution with explanation) |
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==Solution== | ==Solution== | ||
| − | + | The true average is <math>A=\frac{x+y+z}{3}</math>, and the student computed <math>B=\frac{\frac{x+y}{2}+z}{2}=\frac{x+y+2z}{4}</math>, so <math>B-A = \frac{2z-x-y}{12} = \frac{(z-x)+(z-y)}{12}</math>, which is always positive as <math>z>x</math> and <math>z>y</math>. Thus <math>B</math> is always greater than <math>A</math>, i.e. <math>\boxed{C}</math>. | |
== See also == | == See also == | ||
Latest revision as of 17:43, 1 April 2018
Problem
A student attempted to compute the average 
of 
and 
by computing the average of 
and 
,
and then computing the average of the result and . Whenever 
, the student's final result is
Solution
The true average is 
, and the student computed 
, so 
, which is always positive as 
and 
. Thus 
is always greater than , i.e. 
.
See also
| 1986 AHSME (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 • 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. 
