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

(New page: ==Solution== To get the fraction, we need to find the number of people who got grades that are "acceptable" (C and D is "acceptable!?") over the total number of people.<br><br>Finding the ...)
 
Line 1: Line 1:
 +
==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?
 +
 +
<math>\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}</math>
 +
 
==Solution==
 
==Solution==
To get the fraction, we need to find the number of people who got grades that are "acceptable" (C and D is "acceptable!?") over the total number of people.<br><br>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 <math>5+4+3+3 = 15</math>.<br><br>So we know the top of the fraction is 15. Only 5 people got "unacceptable" scores, so there are <math>15 + 5 = 20</math> total score.<br><br><math>\frac{15}{20}</math> is our fraction. Of course, we're not done. You will usually have to reduce your fraction to lowest terms.<br><br><math>\frac{15}{20} = \frac{3}{4}</math>.<br><br>(C) is the answer
+
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 <math>5+4+3+3 = 15</math>.
 +
 
 +
So we know the top of the fraction is 15. Only 5 people got "unacceptable" scores, so there are <math>15 + 5 = 20</math> scores.
 +
 
 +
<math>\frac{15}{20}=\frac{3}{4}</math> is our fraction, so <math>\boxed{\text{C}}</math> is the answer.
 +
 
 +
==See Also==
 +
 
 +
[[1985 AJHSME Problems]]

Revision as of 21:45, 12 January 2009

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.

See Also

1985 AJHSME Problems