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Difference between revisions of "2005 AMC 8 Problems/Problem 5"

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==Problem==
 
==Problem==
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If <math> 20\% </math> of a number is <math>12</math>, what is <math> 30\% </math> of the same number?
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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?
  
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<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30</math>
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<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15 </math>
  
 
==Solution==
 
==Solution==
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Since <math> 20 \% </math> of a number is <math> 12 </math>, then <math> 10 \% </math> of the number is <math> 6 </math>, so <math> 30\% </math> of the number is <math> \boxed{\mathrm{(B)}\ 18} </math>.
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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>.
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==See Also==
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{{AMC8 box|year=2005|num-b=4|num-a=6}}

Revision as of 16:14, 24 December 2012

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?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$

Solution

Start by buying the largest packs first. After three $24$-packs, $90-3(24)=18$ cans are left. After one $12$-pack, $18-12=6$ cans are left. Then buy one more $6$-pack. The total number of packs is $3+1+1=\boxed{\textbf{(B)}\ 5}$.

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
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