Difference between revisions of "2006 AMC 12B Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | Mary is about to pay for five items at the grocery store. The prices of the items are < | + | Mary is about to pay for five items at the grocery store. The prices of the items are <math>7.99</math>, <math>4.99</math>, <math>2.99</math>, <math>1.99</math>, and <math>0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <math>$ </math>20.00<math> that she will receive in change? |
| + | </math> | ||
| + | \text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25 | ||
<math> | <math> | ||
| − | |||
| − | |||
== Solution == | == Solution == | ||
| − | The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95< | + | The total price of the items is </math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95<math> |
| − | <math>20-18.95=1.05< | + | </math>20-18.95=1.05<math> |
| − | <math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)} | + | </math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}$ |
== 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 19:58, 26 July 2013
Problem
Mary is about to pay for five items at the grocery store. The prices of the items are 
, 
, 
, 
, and 
. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the 
20.00
\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25
(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95$$ (Error compiling LaTeX. Unknown error_msg)20-18.95=1.05$$ (Error compiling LaTeX. Unknown error_msg)\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}$
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
