2009 AMC 10B Problems/Problem 1

Revision as of 20:04, 2 July 2021 by MRENTHUSIASM (talk | contribs)
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, so we narrow down the choices to either $\text{(B)}$ or $\text{(D)}.$ By a quick inspection, the only solution is $(m,b)=(3,2),$ from which the answer is $b=\boxed{\text{(B) } 2}.$

~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.

  3. The amounts spent on muffins and bagels both end in $50.$
  4. 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 $\boxed{\text{(B) } 2}.$

~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, $\text{(A)}$ 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 $\boxed{\text{(B) } 2}$ 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, $\boxed{\text{(B) } 2}$ 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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