Difference between revisions of "2009 AMC 10B Problems/Problem 1"
MRENTHUSIASM (talk | contribs) m (→Problem) |
MRENTHUSIASM (talk | contribs) m (→Solution 3.1 (Replacements)) |
||
Line 40: | Line 40: | ||
=== Solution 3.1 (Replacements) === | === Solution 3.1 (Replacements) === | ||
− | If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by <math>25</math> cents. Clearly, she bought one more bagel but one fewer muffin, | + | If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by <math>25</math> cents. Clearly, she bought one more bagel but one fewer muffin at a total cost of <math>3.00</math> dollars. Therefore, she bought <math>\boxed{\text{(B) } 2}</math> bagels. |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 01:59, 18 June 2021
- 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 -cent muffin or a -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
Solution 1 (Algebra)
A muffin costed quarters, and a bagel costed quarters.
Suppose that Jane bought muffins and bagels, where and are nonnegative integers. We need:
- is divisible by
From Condition 2, it is clear that must be even, so we narrow down the choices to either or By a quick inspection, the only solution is from which the answer is
~MRENTHUSIASM
Solution 2 (Arithmetic)
In this solution, all amounts are in the unit of cents.
Note that the amount spent on muffins must end in either or and the amount spent on bagels must end in one of or Furthermore, the number of muffins bought and the number of bagels bought must sum to
We have two possible cases:
- The amounts spent on muffins and bagels both end in
- The amounts spent on muffins and bagels both end in
The number of muffins bought must be even, and the number of bagels bought must be a multiple of
In this case, there are no solutions.
The number of muffins bought must be odd, and the number of bagels bought must be more than a multiple of
In this case, the only solution is muffins and bagels, from which the answer is
~MRENTHUSIASM
Solution 3 (Observations)
If Jane bought bagel, then she bought muffins. Her total cost for the week would be cents, or dollars. So, is incorrect.
Two solutions follow from here:
Solution 3.1 (Replacements)
If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by cents. Clearly, she bought one more bagel but one fewer muffin at a total cost of dollars. Therefore, she bought bagels.
~MRENTHUSIASM
Solution 3.2 (Answer Choices)
- If Jane bought bagels, then she bought muffins. Her total cost for the week would be cents, or dollars. So, is correct.
For completeness, we will check and too. If you decide to use this solution on the real test, then you will not need to do that, as you want to save more time.
- If Jane bought bagels, then she bought muffins. Her total cost for the week would be cents, or dollars. So, is incorrect.
- If Jane bought bagels, then she bought muffin. Her total cost for the week would be cents, or dollars. So, is incorrect.
- If Jane bought bagels, then she bought muffins. Her total cost for the week would be cents, or dollars. So, is incorrect.
~MRENTHUSIASM
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.