Difference between revisions of "2018 AMC 12B Problems/Problem 6"
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==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 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, so we eliminate <math>\textbf{(E)}.</math> | ||
+ | |||
+ | Finally, the answer is <math>\boxed{\textbf{(B) } \frac{4DS}{Q}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/Yednk_TWgBI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}} | {{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:05, 27 May 2023
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
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
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 |
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