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1986 AJHSME Problems/Problem 12

Revision as of 18:08, 24 January 2009 by Waffle (talk | contribs) (→‎Solution)

Problem

The table below displays the grade distribution of the $30$ 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]

$\text{(A)}\ 12\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 33\frac{1}{3}\% \qquad \text{(D)}\ 40\% \qquad \text{(E)}\ 50\%$

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..

$\frac{2 + 4 + 5 + 1}{2 + 2 + 1 + 1 + 4 + 3 + 1 + 3 + 5 + 2 + 1 + 1 + 1 + 2 + 1}$

Which simplifies to...

$\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40 \%$

Note that I did not simplify $\frac{4}{10}$ to $\frac{2}{5}$, because that would just be a time wasting step, because I can easily go from $\frac{4}{10}$ to a percent anyway.

See Also

1986 AJHSME Problems

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