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 <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>&#036; </math>20.00<math> that she will receive in change?
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Mary is about to pay for five items at the grocery store.  The prices of the items are <math>&#036; </math>7.99<math>, </math>&#036; <math>4.99</math>, <math>&#036; </math>2.99<math>, </math>&#036; <math>1.99</math>, and <math>&#036; </math>0.99<math>.  Mary will pay with a twenty-dollar bill.  Which of the following is closest to the percentage of the </math>&#036; <math>20.00</math> 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
 
</math>
 
</math>
\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25
 
<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<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>20-18.95=1.05<math>
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<math>20-18.95=1.05</math>
  
</math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}$
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<math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}</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 20:58, 26 July 2013

Problem

Mary is about to pay for five items at the grocery store. The prices of the items are $&#036;$7.99$,$$ $4.99$, $&#036;$2.99$,$$ $1.99$, and $&#036;$0.99$.  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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