Difference between revisions of "2018 AMC 12B Problems/Problem 6"
MRENTHUSIASM (talk | contribs) m (→Problem: Reformatted a little.) |
MRENTHUSIASM (talk | contribs) (Reconstructed the original solution so it is clearer in terms of unit and logic.) |
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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== |
− | + | ||
+ | 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 | ||
==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 20:32, 18 September 2021
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
Each can of soda costs quarters, or dollars. Therefore, dollars can purchase cans of soda.
~MRENTHUSIASM
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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