Difference between revisions of "2014 AMC 10B Problems/Problem 11"
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Assume WLOG that the original price was <math>\$100</math>. Then, option 1 would cost <math>100 \cdot \frac{17}{20} \cdot \frac{17}{20} = \$ 72.25</math>. Option 2 would cost <math>100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = \$72.90</math>. Option 3 ould cost <math>100 \cdot \frac{3}{4} \cdot \frac{19}{20} = \$71.25</math>. Therefore, the largest integer smaller than all of these three is <math>71</math>, so <math>100-71= \boxed{\textbf{(C) } 29}</math>. | Assume WLOG that the original price was <math>\$100</math>. Then, option 1 would cost <math>100 \cdot \frac{17}{20} \cdot \frac{17}{20} = \$ 72.25</math>. Option 2 would cost <math>100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = \$72.90</math>. Option 3 ould cost <math>100 \cdot \frac{3}{4} \cdot \frac{19}{20} = \$71.25</math>. Therefore, the largest integer smaller than all of these three is <math>71</math>, so <math>100-71= \boxed{\textbf{(C) } 29}</math>. | ||
Revision as of 17:03, 11 August 2023
Contents
Problem
For the consumer, a single discount of 
is more advantageous than any of the following discounts:
(1) two successive 
discounts
(2) three successive 
discounts
(3) a 
discount followed by a 
discount
What is the smallest possible positive integer value of 
?
Solution 1
Let the original price be 
. Then, for option 
, the discounted price is 
. For option 
, the discounted price is 
. Finally, for option 
, the discounted price is 
. Therefore, must be greater than 
. It follows 
must be greater than 
. We multiply this by 
to get the percent value, and then round up because is the smallest integer that provides a greater discount than 
, leaving us with the answer of 
Solution 2 (a bit easier)
Assume the original price was dollars. Thus, after a discount of , the price will be 
dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: , 
, 
, and 
. Simplify these to get , 
(rounded down), 
, and 
(rounded down). To have the greatest discount, we need the least price, which is dollars. Now we get the original fractional value of this, and the discount of this is 
, and we round this down to 
. Now, it's pretty easy. The integer value greater than this is
~solution by sakshamsethi
~edited by sosiaops
Solution 3 (Straightforward)
Assume WLOG that the original price was 
. Then, option 1 would cost 
. Option 2 would cost 
. Option 3 ould cost 
. Therefore, the largest integer smaller than all of these three is , so 
.
~MrThinker
Video Solution
~savannahsolver
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
