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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?
  
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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 1 dollar is worth 4 quarters?
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<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>
  
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<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>
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==Solution 1==
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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.
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~MRENTHUSIASM
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==Solution 2==
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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>
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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>
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Finally, the answer is <math>\boxed{\textbf{(B) } \frac{4DS}{Q}}.</math>
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~MRENTHUSIASM
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 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/Yednk_TWgBI
  
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==Solution 1==
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~Education, the Study of Everything
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The unit price for a can of soda (in quarters) is <math>\frac{S}{Q}</math>. Thus, the number of cans which can be bought for <math>D</math> dollars (<math>4D</math> quarters) is<math> \boxed {\textbf{(B)} \frac{4DS}{Q}}</math>
 
  
 
==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

Problem

Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?

$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$

Solution 1

Each can of soda costs $\frac QS$ quarters, or $\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\frac{D}{\left(\tfrac{Q}{4S}\right)}=\boxed{\textbf{(B) } \frac{4DS}{Q}}$ cans of soda.

~MRENTHUSIASM

Solution 2

Note that $S$ is in the unit of $\text{can}.$ On the other hand, $Q$ and $D$ are both in the unit of $\text{cost}.$

The units of $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)}$ are $\frac{\text{cost}^2}{\text{can}},\text{can},\frac{1}{\text{can}},\frac{\text{cost}^2}{\text{can}},$ and $\text{can},$ respectively. Since the answer is in the unit of $\text{can},$ we eliminate $\textbf{(A)},\textbf{(C)},$ and $\textbf{(D)}.$ Moreover, it is clear that $D$ dollars can purchase more than $S=\frac{DS}{4Q}$ cans of soda, so we eliminate $\textbf{(E)}.$

Finally, the answer is $\boxed{\textbf{(B) } \frac{4DS}{Q}}.$

~MRENTHUSIASM

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/Yednk_TWgBI

~Education, the Study of Everything

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
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Problem 5 		Followed by
Problem 7
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All AMC 12 Problems and Solutions

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