Difference between revisions of "1986 AJHSME Problems/Problem 12"

(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==
  
{{Solution}}
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We need to find the number of those who did get the same on both tests over 30 (the number of students in the class).
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So, we have <cmath>\frac{2 + 4 + 5 + 1}{30}</cmath>
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Which simplifies to <cmath>\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40 \%</cmath>
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<math>\boxed{\text{D}}</math>
  
 
==See Also==
 
==See Also==
  
[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:16, 3 July 2013

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

We need to find the number of those who did get the same on both tests over 30 (the number of students in the class).

So, we have \[\frac{2 + 4 + 5 + 1}{30}\]

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

$\boxed{\text{D}}$

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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