Difference between revisions of "2014 AMC 10B Problems/Problem 4"

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==Solution==
 
==Solution==
Let <math>m</math> be the cost of a muffin and <math>b</math> be the cost of a banana. From the given information, <cmath>2m+16b=2(4m+3b)=8m+6b\Rightarrow 10b=6m\Rightarrow m=\frac{10}{6}b=\text{(B)} \boxed{\frac{5}{3}b}</cmath>.
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Let <math>m</math> be the cost of a muffin and <math>b</math> be the cost of a banana. From the given information, <cmath>2m+16b=2(4m+3b)=8m+6b\Rightarrow 10b=6m\Rightarrow m=\frac{10}{6}b=\text{(B) } \boxed{\frac{5}{3}b}</cmath>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=3|num-a=5}}
 
{{AMC10 box|year=2014|ab=B|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:10, 26 January 2016

Problem

Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana?

$\textbf {(A) } \frac{3}{2} \qquad \textbf {(B) } \frac{5}{3} \qquad \textbf {(C) } \frac{7}{4} \qquad \textbf {(D) } 2 \qquad \textbf {(E) } \frac{13}{4}$

Solution

Let $m$ be the cost of a muffin and $b$ be the cost of a banana. From the given information, \[2m+16b=2(4m+3b)=8m+6b\Rightarrow 10b=6m\Rightarrow m=\frac{10}{6}b=\text{(B) } \boxed{\frac{5}{3}b}\].

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 10 Problems and Solutions

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