Difference between revisions of "1990 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | Let the numbers of boys and girls at Adams be <math>(A,a)</math> and the numbers of boys and girls at Baker be <math>(B,b)</math>. |
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| + | Then, reading down, we have <math>71A+76a=74(A+a)</math> and <math>81B+90b=84(B+b)</math>. | ||
| + | Reading across, we have <math>71A+81B=79(A+B)</math>. | ||
| + | |||
| + | Simplifying each in turn, <math>3A-2a=0 \implies A=2a/3</math> and <math>3B-6b=0 \implies B=2b</math> and <math>8A-2B=0 \implies 4A=B</math>. | ||
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| + | Thus <math>3b = 4a</math>, which implies that Adams has <math>\tfrac37</math> of the girls and Baker has <math>\tfrac47</math> of the girls. | ||
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| + | So the average score for all girls must be <math>\frac{3\cdot 76+4\cdot 90}7 = \frac{228+360}7 = 84</math> which is <math>\fbox{D}</math> | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1990|num-b=23|num-a= | + | {{AHSME box|year=1990|num-b=23|num-a=25}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:08, 4 February 2016
Problem
All students at Adams High School and at Baker High School take a certain exam. The average scores for boys, for girls, and for boys and girls combined, at Adams HS and Baker HS are shown in the table, as is the average for boys at the two schools combined. What is the average score for the girls at the two schools combined?
![$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{Average Scores}\\\hline Category&Adams&Baker&Adams\&Baker\\\hline Boys&71&81&79\\ Girls&76&90&?\\ Boys\&Girls&74&84& \\\hline \end{tabular}$](http://latex.artofproblemsolving.com/1/f/8/1f83de1aa18ebd28fe5642031ef4c8d748b07c58.png)
Solution
Let the numbers of boys and girls at Adams be 
and the numbers of boys and girls at Baker be 
.
Then, reading down, we have 
and 
.
Reading across, we have 
.
Simplifying each in turn, 
and 
and 
.
Thus 
, which implies that Adams has 
of the girls and Baker has 
of the girls.
So the average score for all girls must be 
which is 
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
