Difference between revisions of "2018 AMC 12B Problems/Problem 6"
MRENTHUSIASM (talk | contribs) (Reconstructed the original solution so it is clearer in terms of unit and logic.) |
MRENTHUSIASM (talk | contribs) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | |||
Suppose <math>S</math> cans of soda can be purchased from a vending machine for <math>Q</math> quarters. Which of the following expressions describes the number of cans of soda that can be purchased for <math>D</math> dollars, where <math>1</math> dollar is worth <math>4</math> quarters? | Suppose <math>S</math> cans of soda can be purchased from a vending machine for <math>Q</math> quarters. Which of the following expressions describes the number of cans of soda that can be purchased for <math>D</math> dollars, where <math>1</math> dollar is worth <math>4</math> quarters? | ||
<math>\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}</math> | <math>\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | Each can of soda costs <math>\frac QS</math> quarters, or <math>\frac{Q}{4S}</math> dollars. Therefore, <math>D</math> dollars can purchase <math>\frac{D}{\left(\tfrac{Q}{4S}\right)}=\boxed{\textbf{(B) } \frac{4DS}{Q}}</math> cans of soda. | ||
− | + | ~MRENTHUSIASM | |
+ | |||
+ | ==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 units 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, from which we eliminate <math>\textbf{(E)}.</math> Therefore, the answer is <math>\boxed{\textbf{(B) } \frac{4DS}{Q}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 21:13, 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 units 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, from which we eliminate
Therefore, 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.