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2006 AMC 12B Problems/Problem 4: Difference between revisions

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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>\</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?
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>20.00</math> that she will receive in change?


<math>
<math>
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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>
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>
<math>20-18.95=1.05</math>
 
<math>\frac{1.05}{20}=.0525 \Rightarrow \text{(A)}</math>
 
==Alternative Solution==
We can round the prices to <math>8</math>, <math>5</math>, <math>3</math>, <math>2</math>, and <math>1</math>.  
 
So <math> 20 - (8+5+3+2+1) = 1 </math>
 
We can make an equation: <math> 20* \frac{x}{100}=1 </math>
 
If we simplify the equation to "x", we get <math>x=5</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}}
{{MAA Notice}}

Latest revision as of 15:11, 8 December 2013

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 $(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95$

$20-18.95=1.05$

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

Alternative Solution

We can round the prices to $8$, $5$, $3$, $2$, and $1$.

So $20 - (8+5+3+2+1) = 1$

We can make an equation: $20* \frac{x}{100}=1$

If we simplify the equation to "x", we get $x=5$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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