Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 2 (Answer Choices) == | + | == Solution 2 (Arithmetic) == |
+ | In this solution, all amounts are in the unit of cents. | ||
+ | |||
+ | The amount spent on muffins must end in either <math>00</math> or <math>50,</math> and the amount spent on bagels must end in one of <math>00,25,50,</math> or <math>75.</math> | ||
+ | |||
+ | We have two possible cases: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>The amounts spent on muffins and bagels both end in <math>00.</math></li><p> | ||
+ | The number of muffins bought must be even, and the number of bagels bought must be a multiple of <math>4.</math> <p> | ||
+ | In this case, there are no solutions. | ||
+ | <li>The amounts spent on muffins and bagels both end in <math>50.</math></li><p> | ||
+ | The number of muffins bought must be odd, and the number of bagels bought must be <math>2</math> more than a multiple of <math>4.</math> <p> | ||
+ | In this case, the only solution is <math>3</math> muffins and <math>2</math> bagels, from which the answer is <math>b=\boxed{\text{(B) } 2}.</math> | ||
+ | </ol> | ||
+ | |||
+ | == Solution 3 (Answer Choices) == | ||
* If Jane bought <math>1</math> bagel, then she bought <math>4</math> muffins. Her total cost for the week would be <math>75\cdot1+50\cdot4=275</math> cents, or <math>2.75</math> dollars. So, <math>\text{(A)}</math> is incorrect. | * If Jane bought <math>1</math> bagel, then she bought <math>4</math> muffins. Her total cost for the week would be <math>75\cdot1+50\cdot4=275</math> cents, or <math>2.75</math> dollars. So, <math>\text{(A)}</math> is incorrect. | ||
Revision as of 21:43, 9 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 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 (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.
The amount spent on muffins must end in either or and the amount spent on bagels must end in one of or
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
Solution 3 (Answer Choices)
- If Jane bought bagel, 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 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.