Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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Each morning of her five-day workweek, Jane bought either a <math>50</math>-cent muffin or a <math>75</math>-cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy? | Each morning of her five-day workweek, Jane bought either a <math>50</math>-cent muffin or a <math>75</math>-cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy? | ||
− | <math>\ | + | <math>\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5</math> |
− | == Solution 1 ( | + | == Solution 1 (Observations: 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. | |
− | + | 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. 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{\textbf{(B) } 2}</math> bagels. | |
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− | + | ~MRENTHUSIASM | |
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− | </ | + | == Solution 2 (Observations: 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>\textbf{(A)}</math> is incorrect. | |
+ | |||
+ | * If Jane bought <math>2</math> bagels, then she bought <math>3</math> muffins. Her total cost for the week would be <math>75\cdot2+50\cdot3=300</math> cents, or <math>3.00</math> dollars. So, <math>\boxed{\textbf{(B) } 2}</math> is correct. | ||
+ | |||
+ | For completeness, we will check <math>\textbf{(C)},\textbf{(D)},</math> and <math>\textbf{(E)}</math> 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 <math>3</math> bagels, then she bought <math>2</math> muffins. Her total cost for the week would be <math>75\cdot3+50\cdot2=325</math> cents, or <math>3.25</math> dollars. So, <math>\textbf{(C)}</math> is incorrect. | ||
+ | |||
+ | * If Jane bought <math>4</math> bagels, then she bought <math>1</math> muffin. Her total cost for the week would be <math>75\cdot4+50\cdot1=350</math> cents, or <math>3.50</math> dollars. So, <math>\textbf{(D)}</math> is incorrect. | ||
+ | |||
+ | * If Jane bought <math>5</math> bagels, then she bought <math>0</math> muffins. Her total cost for the week would be <math>75\cdot5+50\cdot0=375</math> cents, or <math>3.75</math> dollars. So, <math>\textbf{(E)}</math> is incorrect. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution | + | == Solution 3 (Arithmetic) == |
In this solution, all amounts are in the unit of cents. | In this solution, all amounts are in the unit of cents. | ||
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<li>The amounts spent on muffins and bagels both end in <math>50.</math></li><p> | <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> | 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>\boxed{\ | + | In this case, the only solution is <math>3</math> muffins and <math>2</math> bagels, from which the answer is <math>\boxed{\textbf{(B) } 2}.</math> |
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
Latest revision as of 22:41, 14 September 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 (Observations: Replacements)
If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by cents.
If Jane bought bagel, then she bought muffins. Her total cost for the week would be cents, or dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of dollars. Therefore, she bought bagels.
~MRENTHUSIASM
Solution 2 (Observations: 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
Solution 3 (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
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.