Difference between revisions of "2010 AMC 10B Problems/Problem 11"
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<math>A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150</math> | <math>A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150</math> | ||
− | We see here that the greatest possible value for <math>p</math> is <math> 150 </math>, thus <math> y = 100 + 150 = 250 </math> and the smallest value for p is <math> 100 </math> so <math> x = 100 + 100 = 200 </math>. | + | We see here that the greatest possible value for <math>p</math> is <math> 150 </math>, thus <math> y = 100 + 150 = 250 </math> and the smallest value for <math>p</math> is <math> 100 </math> so <math> x = 100 + 100 = 200 </math>. |
The difference between <math>y</math> and <math>x</math> is <math>y - x = 250 - 200 = \boxed{\textbf{(A)}\ 50}</math> | The difference between <math>y</math> and <math>x</math> is <math>y - x = 250 - 200 = \boxed{\textbf{(A)}\ 50}</math> |
Revision as of 20:46, 12 January 2020
Problem
A shopper plans to purchase an item that has a listed price greater than and can use any one of the three coupons. Coupon A gives off the listed price, Coupon B gives off the listed price, and Coupon C gives off the amount by which the listed price exceeds
.
Let and be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is ?
Solution
Let the listed price be , where
Coupon A saves us:
Coupon B saves us:
Coupon C saves us:
Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities:
We see here that the greatest possible value for is , thus and the smallest value for is so .
The difference between and is
See Also
2010 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 |
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