Difference between revisions of "2018 AMC 12B Problems/Problem 6"

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==Solution 1==
 
==Solution 1==
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> (Giraffefun)
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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}}

Revision as of 02:08, 19 February 2018

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

The unit price for a can of soda (in quarters) is $\frac{S}{Q}$. Thus, the number of cans which can be bought for $D$ dollars ($4D$ quarters) is$\boxed {\textbf{(B)} \frac{4DS}{Q}}$

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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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