Difference between revisions of "2006 AMC 12B Problems/Problem 4"
(→Solution) |
(→Alternative Solution) |
||
| Line 15: | Line 15: | ||
==Alternative Solution== | ==Alternative Solution== | ||
We can round the prices to <math>8</math>, <math>5</math>, <math>3</math>, <math>2</math>, and <math>1</math>. | 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> | So <math> 20 - (8+5+3+2+1) = 1 </math> | ||
<math> 20* \frac{x}{100}=1 </math> | <math> 20* \frac{x}{100}=1 </math> | ||
Revision as of 20:04, 26 July 2013
Problem
Mary is about to pay for five items at the grocery store. The prices of the items are 
7.99
$ 
, 2.99$ 
, and 0.99
$ 
that she will receive in change?
Solution
The total price of the items is 
Alternative Solution
We can round the prices to 
, 
, 
, 
, and 
.
So 
By simplifying to "x", we get $\box{5}$ (Error compiling LaTeX. Unknown error_msg)
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. 
