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 | + | 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?
Solution 1 (Algebra)
A muffin costed quarters, and a bagel costed
quarters.
Suppose that Jane bought muffins and
bagels, where
and
are nonnegative integers. We need:
is divisible by
From Condition 2, it is clear that must be even. By a quick inspection, the only solution is
so the answer is
~MRENTHUSIASM
Solution 2 (Answer Choices)
- If Jane bought
bagel, then she bought
muffins. Her total cost for the week would be
cents, or
dollars. So,
is incorrect.
- If Jane bought
bagels, then she bought
muffins. Her total cost for the week would be
cents, or
dollars. So,
is correct.
For completeness, we will check and
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
bagels, then she bought
muffins. Her total cost for the week would be
cents, or
dollars. So,
is incorrect.
- If Jane bought
bagels, then she bought
muffin. Her total cost for the week would be
cents, or
dollars. So,
is incorrect.
- If Jane bought
bagels, then she bought
muffins. Her total cost for the week would be
cents, or
dollars. So,
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.