Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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\mathrm{(E)}\ 5</math> | \mathrm{(E)}\ 5</math> | ||
| − | == Solution == | + | == Solution 1 == |
The only combination of five items with total cost a whole number of dollars is 3 muffins and <math>\boxed {2}</math> bagels. The answer is <math>\mathrm{(B)}</math>. | The only combination of five items with total cost a whole number of dollars is 3 muffins and <math>\boxed {2}</math> bagels. The answer is <math>\mathrm{(B)}</math>. | ||
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| + | == Solution 2 == | ||
| + | Because <math>75</math> ends in a 5, and we want a whole number of dollars, we know that there must be an even number of bagels. Furthermore, this tells us that the number of muffins is odd. Now, because it is a whole number of dollars, and <math>50</math> cents multiplied by an odd number means that it will end in a <math>50</math> , we know that the result of the even number multiplied by <math>75</math>,must end in a 50. Note that the only result that gives this result is when 75 is multiplied by 2. Thus, our answer is <math>\mathrm{(B)}</math>. | ||
== See also == | == See also == | ||
Revision as of 20:00, 27 April 2020
- The following problem is from both the 2009 AMC 10B #1 and 2009 AMC 12B #1, so both problems redirect to this page.
Contents
Problem
Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?
Solution 1
The only combination of five items with total cost a whole number of dollars is 3 muffins and 
bagels. The answer is 
.
Solution 2
Because 
ends in a 5, and we want a whole number of dollars, we know that there must be an even number of bagels. Furthermore, this tells us that the number of muffins is odd. Now, because it is a whole number of dollars, and 
cents multiplied by an odd number means that it will end in a , we know that the result of the even number multiplied by ,must end in a 50. Note that the only result that gives this result is when 75 is multiplied by 2. Thus, our answer is .
See also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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 | ||
| 2009 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Question |
Followed by Problem 2 |
| 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. 
