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

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== Problem ==
 
== Problem ==
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Mary is about to pay for five items at the grocery store.  The prices of the items are <math>\</math> 7.99, <math>\</math> 4.99, <math>\</math> 2.99, <math>\</math> 1.99, and <math>\</math> 0.99.  Mary will pay with a twenty-dollar bill.  Which of the following is closest to the percentage of the <math>\</math>20.00 that she will receive in change?
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<math>
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\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25
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</math>
  
 
== Solution ==
 
== Solution ==
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The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.05</math>
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<math>20-18.05=1.95</math>
  
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<math>\frac{1.95}{20}=.0975 \Rightarrow \text{(C)}</math>
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12B Problems]]
 
* [[2006 AMC 12B Problems]]

Revision as of 08:28, 14 November 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 $(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.05$

$20-18.05=1.95$

$\frac{1.95}{20}=.0975 \Rightarrow \text{(C)}$

See also

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