Difference between revisions of "2005 AMC 8 Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
Each folder can also be <math> 5/2</math> dollars, and <math>20\%</math> can be shown as <math>(1/5)</math>. We can multiply <math>(5/2) \cdot (1/5) = (1/2)</math>. <math>(1/2)</math> is also <math>50</math> cents or the amount of money that is saved after the <math>20\%</math> discount. So each folder is <math>2.50-0.5 = \textdollar2</math>.Since Karl bought 5 folders all of the folders after the discount is <math>(5)(2) = 10</math>, and the money bought before the discount is <math>(5)(2.50) = \textdollar12.50</math>. To find the money Karl saves all we have to do is subtract <math>12.50 - 10 = 2.50</math>. Thus the answer is <math>\boxed{\textbf{(C)}\ \textdollar 2.50}</math>. | Each folder can also be <math> 5/2</math> dollars, and <math>20\%</math> can be shown as <math>(1/5)</math>. We can multiply <math>(5/2) \cdot (1/5) = (1/2)</math>. <math>(1/2)</math> is also <math>50</math> cents or the amount of money that is saved after the <math>20\%</math> discount. So each folder is <math>2.50-0.5 = \textdollar2</math>.Since Karl bought 5 folders all of the folders after the discount is <math>(5)(2) = 10</math>, and the money bought before the discount is <math>(5)(2.50) = \textdollar12.50</math>. To find the money Karl saves all we have to do is subtract <math>12.50 - 10 = 2.50</math>. Thus the answer is <math>\boxed{\textbf{(C)}\ \textdollar 2.50}</math>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | Having a <math>20\%</math> off sale is equivalent to a "Buy 4 get 1 free" sale. Thus, he would have saved the price of 1 folder, which is <math>\boxed{\textbf{(C)}\ \textdollar 2.50}</math>. | ||
| + | |||
| + | -JeffersonJ | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=1|num-a=3}} | {{AMC8 box|year=2005|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:26, 19 June 2022
Problem
Karl bought five folders from Pay-A-Lot at a cost of 
each.
Pay-A-Lot had a 20%-off sale the following day. How much could
Karl have saved on the purchase by waiting a day?
Solution
Karl paid 
. 
of this cost that he saved is 
.
Solution 2
Each folder can also be 
dollars, and 
can be shown as 
. We can multiply 
. 
is also 
cents or the amount of money that is saved after the discount. So each folder is 
.Since Karl bought 5 folders all of the folders after the discount is 
, and the money bought before the discount is 
. To find the money Karl saves all we have to do is subtract 
. Thus the answer is 
.
Solution 3
Having a off sale is equivalent to a "Buy 4 get 1 free" sale. Thus, he would have saved the price of 1 folder, which is .
-JeffersonJ
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
