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Difference between revisions of "2009 AMC 10B Problems/Problem 1"

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{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #1]] and [[2009 AMC 12B Problems|2009 AMC 12B #1]]}}
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{{duplicate|[[2009 AMC 10B Problems#Problem 1|2009 AMC 10B #1]] and [[2009 AMC 12B Problems#Problem 1|2009 AMC 12B #1]]}}
  
 
== Problem ==
 
== Problem ==
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Each morning of her five-day workweek, Jane bought either a 100000-cent muffin or a 10000000000000000000000000000000000000000000000000000-cent bagel. Her total cost for the week was a whole number of dollars,  How many bagels did she buy?
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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?
  
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<math>\mathrm{(A)}\ 1\qquad
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<math>\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5</math>
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\mathrm{(B)}\ 2\qquad
 
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\mathrm{(C)}\ 3\qquad
 
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\mathrm{(D)}\ 4\qquad
 
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\mathrm{(E)}\ 5</math>
 
  
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== Solution 1 ==
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== Solution 1 (Observations: Replacements) ==
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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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If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by <math>25</math> cents.
  
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== Solution 2 ==
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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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Because <math>75</math> ends in a <math>5</math>, 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 <math>50</math>. Note that the only result that gives this result is when <math>75</math> is multiplied by <math>2</math>. Thus, our answer is <math>\mathrm{(c )}</math> you dumb idoit.
 
  
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~coolmath2017
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~MRENTHUSIASM
 +
 
 +
== 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.
 +
 
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* 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.
 +
 
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* 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
 +
 
 +
== 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 <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> Furthermore, the number of muffins bought and the number of bagels bought must sum to <math>5.</math>
 +
 
 +
We have two possible cases:
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<ol style="margin-left: 1.5em;">
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  <li>The amounts spent on muffins and bagels both end in <math>00.</math></li><p>
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The number of muffins bought must be even, and the number of bagels bought must be a multiple of <math>4.</math> <p>
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In this case, there are no solutions.
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  <li>The amounts spent on muffins and bagels both end in <math>50.</math></li><p>
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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>
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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>
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</ol>
 +
~MRENTHUSIASM
  
 
== See also ==
 
== See also ==

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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