Difference between revisions of "1985 AJHSME Problems/Problem 5"
(→Solution) |
(→Solution 2 (Educated Guess/Answer Choices)) |
||
Line 42: | Line 42: | ||
==Solution 2 (Educated Guess/Answer Choices)== | ==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. | ||
==Video Solution== | ==Video Solution== |
Revision as of 06:19, 3 October 2024
Contents
Problem
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?
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 .
So we know the top of the fraction is 15. Only 5 people got "unacceptable" scores, so there are scores.
is our fraction, so is the answer.
Solution 2 (Educated Guess/Answer Choices)
We could tell it's clearly not , so that eliminates .
There is people, so we cannot simplify to , so it isn't .
is way too close, so we can guess that isn't the answer.
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.