2006 AMC 10B Problems/Problem 22: Difference between revisions
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The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math> | The only integer solutions for <math>B</math> and <math>J</math> are <math>B=2</math> and <math>J=3</math> | ||
Therefore the cost of the jam Elmo uses to make the sandwiches is <math>3\cdot5\cdot11=165</math>¢ <math> = </math> | Therefore the cost of the jam Elmo uses to make the sandwiches is <math>3\cdot5\cdot11=165</math>¢ <math> = </math> \$1.65 \Rightarrow D $ | ||
== See Also == | == See Also == | ||
Revision as of 14:00, 25 October 2015
Problem
Elmo makes 
sandwiches for a fundraiser. For each sandwich he uses 
globs of peanut butter at 
¢ per glob and 
blobs of jam at 
¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is 
. Assume that , , and are positive integers with 
. What is the cost of the jam Elmo uses to make the sandwiches?
Solution
The peanut butter and jam for each sandwich costs 
¢, so the peanut butter and jam for sandwiches costs 
¢.
Setting this equal to 
¢:
The only possible positive integer pairs 
whose product is are: 
The first pair violates and the third and fourth pair have no positive integer solutions for and .
So, 
and 
The only integer solutions for and are 
and 
Therefore the cost of the jam Elmo uses to make the sandwiches is 
¢ 
\$1.65 \Rightarrow D $
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 10 Problems and Solutions | ||
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