Difference between revisions of "2018 AMC 12B Problems/Problem 6"
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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> | <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== |
| + | 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 16:06, 16 February 2018
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 1 dollar is worth 4 quarters?
Solution 1
The unit price for a can of soda (in quarters) is 
. Thus, the number of cans which can be bought for dollars (
quarters) is
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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