Difference between revisions of "2018 AMC 12B Problems/Problem 6"
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==Problem== | ==Problem== | ||
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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 21:05, 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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