Difference between revisions of "2006 AMC 12B Problems/Problem 4"
Armalite46 (talk | contribs) (→Problem) |
|||
(10 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Mary is about to pay for five items at the grocery store. The prices of the items are < | + | 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> | ||
Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
− | The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18. | + | 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. | + | <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> | ||
− | |||
== 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 , , , , and . Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the that she will receive in change?
Solution
The total price of the items is
Alternative Solution
We can round the prices to , , , , and .
So
We can make an equation:
If we simplify the equation to "x", we get
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.