Difference between revisions of "1955 AHSME Problems/Problem 7"

(Solution)
(Solution)
 
(One intermediate revision by the same user not shown)
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
Since the worker receives a <math>20</math>% cut in wages, his present wage would be <math>\frac{4}{5}</math> of the original wage before the cut. To regain his original pay he would have to obtain a raise of <math>1</math> <math>\div</math> <math>\frac{4}{5}</math> = <math>1</math><math>\frac{1}{4}</math>.
 
Since the worker receives a <math>20</math>% cut in wages, his present wage would be <math>\frac{4}{5}</math> of the original wage before the cut. To regain his original pay he would have to obtain a raise of <math>1</math> <math>\div</math> <math>\frac{4}{5}</math> = <math>1</math><math>\frac{1}{4}</math>.
Therefore, the worker would have to get a <math>\fbox{{\bf(B)} 25\%}</math> raise.
+
 
 +
Therefore, the worker would have to get a <math>\fbox{{\bf(B)} 25\%}</math> raise to acquire his original pay.
 +
 
 +
Solution by awesomechoco
 +
 
 +
== See Also ==
 +
{{AHSME box|year=1955|num-b=6|num-a=8}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 22:17, 9 July 2018

Problem

If a worker receives a $20$% cut in wages, he may regain his original pay exactly by obtaining a raise of:

$\textbf{(A)}\ \text{20\%}\qquad\textbf{(B)}\ \text{25\%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{\%}\qquad\textbf{(D)}\ \textdollar{20}\qquad\textbf{(E)}\ \textdollar{25}$

Solution

Since the worker receives a $20$% cut in wages, his present wage would be $\frac{4}{5}$ of the original wage before the cut. To regain his original pay he would have to obtain a raise of $1$ $\div$ $\frac{4}{5}$ = $1$$\frac{1}{4}$.

Therefore, the worker would have to get a $\fbox{{\bf(B)} 25\%}$ raise to acquire his original pay.

Solution by awesomechoco

See Also

1955 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 26 27 28 29 30
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png