Difference between revisions of "2005 AMC 8 Problems/Problem 5"
Math Kirby (talk | contribs) (Created page with "==Problem 5== If <math> 20\% </math> of a number is <math>12</math>, what is <math> 30\% </math> of the same number? <math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(...") |
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| − | ==Problem | + | ==Problem== |
| − | + | Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda? | |
| − | <math>\ | + | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15 </math> |
==Solution== | ==Solution== | ||
| − | + | Start by buying the largest packs first. After three <math>24</math>-packs, <math>90-3(24)=18</math> cans are left. After one <math>12</math>-pack, <math>18-12=6</math> cans are left. Then buy one more <math>6</math>-pack. The total number of packs is <math>3+1+1=\boxed{\textbf{(B)}\ 5}</math>. | |
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2005|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:08, 5 July 2013
Problem
Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?
Solution
Start by buying the largest packs first. After three 
-packs, 
cans are left. After one 
-pack, 
cans are left. Then buy one more 
-pack. The total number of packs is 
.
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
