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

Latest revision as of 06:26, 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.

However, $\textbf{D}$ isn't right, as then $5$ would be $\frac{1}{5}$ of the total, which clearly is wrong as from the

diagram, we could tell it's less than $5 \times 5$ because $5$ is the maximum value shown, and some values don't reach $5$ ,

and there are exactly $5$ grades. Thus, the answer is $\boxed{\textbf{(C)} \frac{3}{4}}$

~lovelearning999

Video Solution

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

~savannahsolver

@@ Line 40: / Line 40: @@
 <math>\frac{15}{20}=\frac{3}{4}</math> is our fraction, so <math>\boxed{\text{C}}</math> is the answer.
+==Solution 2 (Educated Guess/Answer Choices)==
+We could tell it's clearly not <math>\frac{1}{2}</math>, so that eliminates <math>\textbf{A}</math>.
+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>.
+<math>\textbf{E}</math> is way too close, so we can guess that <math>\textbf{E}</math> isn't the answer.
+However, <math>\textbf{D}</math> isn't right, as then <math>5</math> would be <math>\frac{1}{5}</math> of the total, which clearly is wrong as from the
+diagram, we could tell it's less than <math>5 \times 5</math> because <math>5</math> is the maximum value shown, and some values don't reach <math>5</math>,
+and there are exactly <math>5</math> grades. Thus, the answer is <math>\boxed{\textbf{(C)}  \frac{3}{4}}</math>
+~lovelearning999
+==Video Solution==
+https://youtu.be/5KwEd-yul00
+~savannahsolver
 ==See Also==

1985 AJHSME (Problems • Answer Key • Resources)
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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