1986 AJHSME Problems/Problem 12

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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?

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

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 $\frac{2 + 4 + 5 + 1}{2 + 2 + 1 + 0 + 0 + 1 + 4 + 3 + 0 + 0 + 1 + 3 + 5 + 2 + 0 + 0 + 0 + 1 + 1 + 1 + 0 + 0 + 2 + 1 + 0}$

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

$\boxed{\text{D}}$

Note: As the problem tells us there are 30 students, the denominator calculation was unnecessary

1986 AJHSME (Problems • Resources)
Preceded by Problem 11	Followed by Problem 13
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

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1986 AJHSME Problems/Problem 12

From AoPSWiki

Problem

Solution

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