Difference between revisions of "2006 AMC 12B Problems/Problem 4"
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<math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}</math> | <math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}</math> | ||
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+ | ==Alternative Solution== | ||
+ | We can round the prices to <math>8</math>, <math>5</math>, <math>3</math>, <math>2</math>, and <math>1</math>. | ||
+ | So <math> 20 - (8+5+3+2+1) = 1 </math> | ||
+ | <math> 20* \frac{x}{100}=1 </math> | ||
+ | By simplifying to "x", we get <math>\box{5}</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2006|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:04, 26 July 2013
Problem
Mary is about to pay for five items at the grocery store. The prices of the items are 7.99$ , 2.99$ , and 0.99$ that she will receive in change?
Solution
The total price of the items is
Alternative Solution
We can round the prices to , , , , and . So By simplifying to "x", we get $\box{5}$ (Error compiling LaTeX. Unknown error_msg)
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 12 Problems and Solutions |
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