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Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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   <li><math>2m+3b</math> is divisible by <math>4.</math></li><p>
 
   <li><math>2m+3b</math> is divisible by <math>4.</math></li><p>
 
</ol>
 
</ol>
−
From the second condition, it is clear that <math>b</math> must be even. By a quick inspection, the only solution is <math>(m,b)=(3,2),</math> so the answer is <math>b=\boxed{\text{(B) } 2}.</math>
+
From Condition 2, it is clear that <math>b</math> must be even. By a quick inspection, the only solution is <math>(m,b)=(3,2),</math> so the answer is <math>b=\boxed{\text{(B) } 2}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
== Solution 2 (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>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{\text{(B) } 2}</math> is correct.
 +
 
 +
For completeness, we will check <math>\text{(C)},\text{(D)},</math> and <math>\text{(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>\text{(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>\text{(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>\text{(E)}</math> is incorrect.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 16:53, 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.

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?

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

Solution 1 (Algebra)

A muffin costed $2$ quarters, and a bagel costed $3$ quarters.

Suppose that Jane bought $m$ muffins and $b$ bagels, where $m$ and $b$ are nonnegative integers. We need:

  1. $m+b=5.$

  2. $2m+3b$ is divisible by $4.$

From Condition 2, it is clear that $b$ must be even. By a quick inspection, the only solution is $(m,b)=(3,2),$ so the answer is $b=\boxed{\text{(B) } 2}.$

~MRENTHUSIASM

Solution 2 (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, $\text{(A)}$ 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, $\boxed{\text{(B) } 2}$ is correct.

For completeness, we will check $\text{(C)},\text{(D)},$ and $\text{(E)}$ 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, $\text{(C)}$ 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, $\text{(D)}$ 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, $\text{(E)}$ 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

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