Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
m
Line 14: Line 14:
 
   <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 Condition 2, it is clear that <math>b</math> must be even, so we narrow down the choices to either <math>\text{(B)}</math> or <math>\text{(D)}.</math> By a quick inspection, the only solution is <math>(m,b)=(3,2),</math> from which the answer is <math>b=\boxed{\text{(B) } 2}.</math>
+
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>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 30: Line 30:
 
   <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{\text{(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>
 
</ol>
 
</ol>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
 
== Solution 3 (Observations) ==
 
== Solution 3 (Observations) ==
−
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>\textbf{(A)}</math> is incorrect.
  
 
Two solutions follow from here:
 
Two solutions follow from here:
  
 
=== Solution 3.1 (Replacements) ===
 
=== Solution 3.1 (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. 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{\text{(B) } 2}</math> bagels.
+
If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by <math>25</math> cents. 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.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
 
=== Solution 3.2 (Answer Choices) ===
 
=== Solution 3.2 (Answer Choices) ===
−
* 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.
+
* 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>\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.
+
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>\text{(C)}</math> is incorrect.
+
* 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>\text{(D)}</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>\text{(E)}</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

Revision as of 03:08, 5 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 (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, so we narrow down the choices to either $\textbf{(B)}$ or $\textbf{(D)}.$ By a quick inspection, the only solution is $(m,b)=(3,2),$ from which the answer is $b=\boxed{\textbf{(B) } 2}.$

~MRENTHUSIASM

Solution 2 (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

Solution 3 (Observations)

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.

Two solutions follow from here:

Solution 3.1 (Replacements)

If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents. 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 3.2 (Answer Choices)

  • 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_1&oldid=161606"