Difference between revisions of "2018 AMC 12B Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
− | Note that <math>S</math> is in the unit of <math>\text{can}.</math> On the other hand, <math>Q</math> and <math>D</math> are both in the | + | Note that <math>S</math> is in the unit of <math>\text{can}.</math> On the other hand, <math>Q</math> and <math>D</math> are both in the unit of <math>\text{cost}.</math> |
− | The units of <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},</math> and <math>\textbf{(E)}</math> are <math>\frac{\text{cost}^2}{\text{can}},\text{can},\frac{1}{\text{can}},\frac{\text{cost}^2}{\text{can}},</math> and <math>\text{can},</math> respectively. Since the answer is in the unit of <math>\text{can},</math> we eliminate <math>\textbf{(A)},\textbf{(C)},</math> and <math>\textbf{(D)}.</math> Moreover, it is clear that <math>D</math> dollars can purchase more than <math>S=\frac{DS}{4Q}</math> cans of soda, | + | The units of <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},</math> and <math>\textbf{(E)}</math> are <math>\frac{\text{cost}^2}{\text{can}},\text{can},\frac{1}{\text{can}},\frac{\text{cost}^2}{\text{can}},</math> and <math>\text{can},</math> respectively. Since the answer is in the unit of <math>\text{can},</math> we eliminate <math>\textbf{(A)},\textbf{(C)},</math> and <math>\textbf{(D)}.</math> Moreover, it is clear that <math>D</math> dollars can purchase more than <math>S=\frac{DS}{4Q}</math> cans of soda, so we eliminate <math>\textbf{(E)}.</math> |
+ | |||
+ | Finally, the answer is <math>\boxed{\textbf{(B) } \frac{4DS}{Q}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Latest revision as of 21:18, 18 September 2021
Contents
Problem
Suppose cans of soda can be purchased from a vending machine for quarters. Which of the following expressions describes the number of cans of soda that can be purchased for dollars, where dollar is worth quarters?
Solution 1
Each can of soda costs quarters, or dollars. Therefore, dollars can purchase cans of soda.
~MRENTHUSIASM
Solution 2
Note that is in the unit of On the other hand, and are both in the unit of
The units of and are and respectively. Since the answer is in the unit of we eliminate and Moreover, it is clear that dollars can purchase more than cans of soda, so we eliminate
Finally, the answer is
~MRENTHUSIASM
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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.