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 19:32, 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
