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Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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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>
 
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
 
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== Solution 4 (Algebra) ==
 
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A muffin costed <math>2</math> quarters, and a bagel costed <math>3</math> quarters.
 
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Suppose that Jane bought <math>m</math> muffins and <math>b</math> bagels, where <math>m</math> and <math>b</math> are nonnegative integers. We need:
 
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<ol style="margin-left: 1.5em;">
 
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  <li><math>m+b=5.</math></li><p>
 
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  <li><math>2m+3b</math> is divisible by <math>4.</math></li><p>
 
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</ol>
 
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From Condition 2, it is clear that <math>b</math> must be even, so we narrow down the choices to either <math>\textbf{(B)}</math> or <math>\textbf{(D)}.</math> By a quick inspection, the only solution is <math>(m,b)=(3,2),</math> from which the answer is <math>b=\boxed{\textbf{(B) } 2}.</math>
 
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~MRENTHUSIASM
 
~MRENTHUSIASM
  

Latest revision as of 21:41, 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.

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?

$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution 1 (Observations: Replacements)

If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.

If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{\textbf{(B) } 2}$ bagels.

~MRENTHUSIASM

Solution 2 (Observations: Answer Choices)

  • If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. So, $\textbf{(A)}$ is incorrect.
  • If Jane bought $2$ bagels, then she bought $3$ muffins. Her total cost for the week would be $75\cdot2+50\cdot3=300$ cents, or $3.00$ dollars. So, $\boxed{\textbf{(B) } 2}$ is correct.

For completeness, we will check $\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)}$ 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 $3$ bagels, then she bought $2$ muffins. Her total cost for the week would be $75\cdot3+50\cdot2=325$ cents, or $3.25$ dollars. So, $\textbf{(C)}$ is incorrect.
  • If Jane bought $4$ bagels, then she bought $1$ muffin. Her total cost for the week would be $75\cdot4+50\cdot1=350$ cents, or $3.50$ dollars. So, $\textbf{(D)}$ is incorrect.
  • If Jane bought $5$ bagels, then she bought $0$ muffins. Her total cost for the week would be $75\cdot5+50\cdot0=375$ cents, or $3.75$ dollars. So, $\textbf{(E)}$ 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 $00$ or $50,$ and the amount spent on bagels must end in one of $00,25,50,$ or $75.$ Furthermore, the number of muffins bought and the number of bagels bought must sum to $5.$

We have two possible cases:

  1. The amounts spent on muffins and bagels both end in $00.$

    2. The number of muffins bought must be even, and the number of bagels bought must be a multiple of $4.$

    In this case, there are no solutions.

  2. The amounts spent on muffins and bagels both end in $50.$

    3. The number of muffins bought must be odd, and the number of bagels bought must be $2$ more than a multiple of $4.$

    In this case, the only solution is $3$ muffins and $2$ bagels, from which the answer is $\boxed{\textbf{(B) } 2}.$

~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

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