Difference between revisions of "1986 AJHSME Problems/Problem 12"
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This is just a lot of adding. We just need to find the number of those who DID get the same on both tests, over (fraction-wise) the number of students who took both tests, TOTAL, regardless of whether they got the same or not. | This is just a lot of adding. We just need to find the number of those who DID get the same on both tests, over (fraction-wise) the number of students who took both tests, TOTAL, regardless of whether they got the same or not. | ||
| − | So, we have | + | So, we have <cmath>\frac{2 + 4 + 5 + 1}{2 + 2 + 1 + 0 + 0 + 1 + 4 + 3 + 0 + 0 + 1 + 3 + 5 + 2 + 0 + 0 + 0 + 1 + 1 + 1 + 0 + 0 + 2 + 1 + 0}</cmath> |
| − | < | + | Which simplifies to <cmath>\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40 \%</cmath> |
| − | + | <math>\boxed{\text{D}}</math> | |
| − | |||
| − | <math>\ | ||
| − | |||
| − | |||
==See Also== | ==See Also== | ||
[[1986 AJHSME Problems]] | [[1986 AJHSME Problems]] | ||
Revision as of 18:20, 24 January 2009
Problem
The table below displays the grade distribution of the 
students in a mathematics class on the last two tests. For example, exactly one student received a 'D' on Test 1 and a 'C' on Test 2 (see circled entry). What percent of the students received the same grade on both tests?
![[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label("0",(2.5,.2),N); label("0",(3.5,.2),N); label("2",(4.5,.2),N); label("1",(5.5,.2),N); label("0",(6.5,.2),N); label("0",(2.5,1.2),N); label("0",(3.5,1.2),N); label("1",(4.5,1.2),N); label("1",(5.5,1.2),N); label("1",(6.5,1.2),N); label("1",(2.5,2.2),N); label("3",(3.5,2.2),N); label("5",(4.5,2.2),N); label("2",(5.5,2.2),N); label("0",(6.5,2.2),N); label("1",(2.5,3.2),N); label("4",(3.5,3.2),N); label("3",(4.5,3.2),N); label("0",(5.5,3.2),N); label("0",(6.5,3.2),N); label("2",(2.5,4.2),N); label("2",(3.5,4.2),N); label("1",(4.5,4.2),N); label("0",(5.5,4.2),N); label("0",(6.5,4.2),N); label("F",(1.5,.2),N); label("D",(1.5,1.2),N); label("C",(1.5,2.2),N); label("B",(1.5,3.2),N); label("A",(1.5,4.2),N); label("A",(2.5,5.2),N); label("B",(3.5,5.2),N); label("C",(4.5,5.2),N); label("D",(5.5,5.2),N); label("F",(6.5,5.2),N); label("Test 1",(-.5,5.2),N); label("Test 2",(2.6,6),N); [/asy]](http://latex.artofproblemsolving.com/4/5/6/456c4e5a5c3b5f6db20d14b617ecd66f31f0ad66.png)
Solution
This is just a lot of adding. We just need to find the number of those who DID get the same on both tests, over (fraction-wise) the number of students who took both tests, TOTAL, regardless of whether they got the same or not.
So, we have ![\[\frac{2 + 4 + 5 + 1}{2 + 2 + 1 + 0 + 0 + 1 + 4 + 3 + 0 + 0 + 1 + 3 + 5 + 2 + 0 + 0 + 0 + 1 + 1 + 1 + 0 + 0 + 2 + 1 + 0}\]](http://latex.artofproblemsolving.com/9/9/7/997daf0d64490419ca4bca5189873e55e97a4b95.png)
Which simplifies to ![\[\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40 \%\]](http://latex.artofproblemsolving.com/6/7/0/670d59e76b1a54370833ae32bddcea88c597e249.png)
