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

## 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 $20 - (8+5+3+2+1) = 1$ $20* \frac{x}{100}=1$ By simplifying to "x", we get $\box{5}$ (Error compiling LaTeX. ! Missing number, treated as zero.)

## See also

 2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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