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

(Undo revision 56739 by Weegee13 (talk))
(Solution)
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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==
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We can round the prices to <math>8</math>, <math>5</math>, <math>3</math>, <math>2</math>, and <math>1</math>.
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So <math> 20 - (8+5+3+2+1) = 1 </math>
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<math> 20* \frac{x}{100}=1 </math>
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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 21:04, 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 $(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95$

$20-18.95=1.05$

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

Alternative Solution

We can round the prices to $8$, $5$, $3$, $2$, and $1$. So $20 - (8+5+3+2+1) = 1$ $20* \frac{x}{100}=1$ By simplifying to "x", we get $\box{5}$ (Error compiling LaTeX. Unknown error_msg)

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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