Difference between revisions of "1985 AJHSME Problems/Problem 5"

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==Solution 2 (Educated Guess/Answer Choices)==
 
==Solution 2 (Educated Guess/Answer Choices)==
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We could tell it's clearly not <math>\frac{1}{2}</math>, so that eliminates <math>\textbf{A}</math>.
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There is <math>5+4+3+3+5=20</math> people, so we cannot simplify to <math>\frac{2}{3}</math>, so it isn't <math>\textbf{B}</math>.
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<math>\textbf{E}</math> is way too close, so we can guess that <math>\textbf{E}</math> isn't the answer.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 06:19, 3 October 2024

Problem

[asy] unitsize(13); draw((0,0)--(20,0)); draw((0,0)--(0,15)); draw((0,3)--(-1,3)); draw((0,6)--(-1,6)); draw((0,9)--(-1,9)); draw((0,12)--(-1,12)); draw((0,15)--(-1,15)); fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black); fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black); fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black); fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black); fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black); label("A",(2.5,-.5),S); label("B",(4.5,-.5),S); label("C",(6.5,-.5),S); label("D",(8.5,-.5),S); label("F",(10.5,-.5),S); label("Grade",(15,-.5),S); label("$1$",(-1,3),W); label("$2$",(-1,6),W); label("$3$",(-1,9),W); label("$4$",(-1,12),W); label("$5$",(-1,15),W); [/asy]

The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory?

$\text{(A)}\ \frac{1}{2} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{4}{5} \qquad \text{(E)}\ \frac{9}{10}$

Solution

To get the fraction, we need to find the number of people who got grades that are "satisfactory" over the total number of people.

Finding the number of people who got acceptable grades is pretty easy. 5 people got A's, 4 people got B's, 3 people got C's and 3 people got D's. Adding this up, we just have $5+4+3+3 = 15$.

So we know the top of the fraction is 15. Only 5 people got "unacceptable" scores, so there are $15 + 5 = 20$ scores.

$\frac{15}{20}=\frac{3}{4}$ is our fraction, so $\boxed{\text{C}}$ is the answer.

Solution 2 (Educated Guess/Answer Choices)

We could tell it's clearly not $\frac{1}{2}$, so that eliminates $\textbf{A}$.

There is $5+4+3+3+5=20$ people, so we cannot simplify to $\frac{2}{3}$, so it isn't $\textbf{B}$.

$\textbf{E}$ is way too close, so we can guess that $\textbf{E}$ isn't the answer.

Video Solution

https://youtu.be/5KwEd-yul00

~savannahsolver

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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