Difference between revisions of "1986 AJHSME Problems/Problem 12"
5849206328x (talk | contribs) (New page: ==Problem== The table below displays the grade distribution of the <math>30</math> students in a mathematics class on the last two tests. For example, exactly one student received a 'D' ...) |
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==Solution== | ==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.. | ||
| + | |||
| + | <math>\frac{2 + 4 + 5 + 1}{2 + 2 + 1 + 1 + 4 + 3 + 1 + 3 + 5 + 2 + 1 + 1 + 1 + 2 + 1}</math> | ||
| + | |||
| + | Which simplifies to... | ||
| + | |||
| + | <math>\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40%</math> | ||
| + | |||
| + | Note that I did not simplify <math>\frac{4}{10}</math> to <math>\frac{2}{5}</math>, because that would just be a time wasting step, because I can easily go from <math>\frac{4}{10}</math> to a percent anyway. | ||
| + | |||
==See Also== | ==See Also== | ||
[[1986 AJHSME Problems]] | [[1986 AJHSME Problems]] | ||
Revision as of 18:07, 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..
Which simplifies to...
$\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40%$ (Error compiling LaTeX. Unknown error_msg)
Note that I did not simplify 
to 
, because that would just be a time wasting step, because I can easily go from to a percent anyway.
