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

m (Problem 5)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
If <math> 20\% </math> of a number is <math>12</math>, what is <math> 30\% </math> of the same number?
+
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>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30</math>
+
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15 </math>
  
 
==Solution==
 
==Solution==
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>.
+
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 01:08, 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?

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

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 (ProblemsAnswer KeyResources)
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. AMC logo.png