Difference between revisions of "2006 AMC 12B Problems/Problem 4"

(Problem)
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== Solution ==
 
== Solution ==
The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.05</math>
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The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95</math>
  
<math>20-18.05=1.95</math>
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<math>20-18.95=1.05</math>
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<math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}</math>
  
<math>\frac{1.95}{20}=.0975 \Rightarrow \text{(C)}</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}}

Revision as of 10:03, 3 December 2007

Problem

Mary is about to pay for five items at the grocery store. The prices of the items are $$ $7.99$, $$ $4.99$, $$ $2.99$, $$ $1.99$, and $$ $0.99$. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the $$ $20.00$ that she will receive in change?

$\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25$

Solution

The total price of the items is $(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95$

$20-18.95=1.05$

$\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}$

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
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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