Difference between revisions of "1951 AHSME Problems/Problem 8"
Yankeesfan (talk | contribs) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 7: | Line 7: | ||
[[Without loss of generality]], let the price of the article be 100. Thus, the new price is <math>.9\cdot 100= 90</math>. Then, to restore it to the original price, we solve <math>90x=100</math>. We find <math>x=1\dfrac{1}{9}</math>, thus, the percent increase is <math>1\dfrac{1}{9}-1=\dfrac{1}{9}=\boxed{\textbf{(C) \ } 11\frac{1}{9}\%}</math>. | [[Without loss of generality]], let the price of the article be 100. Thus, the new price is <math>.9\cdot 100= 90</math>. Then, to restore it to the original price, we solve <math>90x=100</math>. We find <math>x=1\dfrac{1}{9}</math>, thus, the percent increase is <math>1\dfrac{1}{9}-1=\dfrac{1}{9}=\boxed{\textbf{(C) \ } 11\frac{1}{9}\%}</math>. | ||
| − | == See | + | == See Also == |
| − | {{AHSME box|year=1951|num-b=7|num-a=9}} | + | {{AHSME 50p box|year=1951|num-b=7|num-a=9}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:20, 5 July 2013
Problem
The price of an article is cut 
To restore it to its former value, the new price must be increased by:
Solution
Without loss of generality, let the price of the article be 100. Thus, the new price is 
. Then, to restore it to the original price, we solve 
. We find 
, thus, the percent increase is 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
