Difference between revisions of "2009 AMC 10B Problems/Problem 1"
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== Problem == | == Problem == | ||
− | Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. | + | 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? |
<math>\text{(A) } 1\qquad\text{(B) } 2\qquad\text{(C) } 3\qquad\text{(D) } 4\qquad\text{(E) } 5</math> | <math>\text{(A) } 1\qquad\text{(B) } 2\qquad\text{(C) } 3\qquad\text{(D) } 4\qquad\text{(E) } 5</math> | ||
+ | |||
+ | == Solution 1 (Algebra) == | ||
+ | A muffin costed <math>2</math> quarters, and a bagel costed <math>3</math> quarters. | ||
+ | |||
+ | Suppose that Jane bought <math>m</math> muffins and <math>b</math> bagels, where <math>m</math> and <math>b</math> are nonnegative integers. We need: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>m+b=5.</math></li><p> | ||
+ | <li><math>2m+3b</math> is divisible by <math>4.</math></li><p> | ||
+ | </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> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
== See also == | == See also == |
Revision as of 15:31, 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 the second condition, it is clear that must be even. By a quick inspection, the only solution is so the answer is
~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.