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

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==Problem==
 
==Problem==
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Susie pays for <math>4</math> muffins and <math>3</math> bananas. Calvin spends twice as much paying for <math>2</math> muffins and <math>16</math> bananas. A muffin is how many times as expensive as a banana?
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<math> \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}</math>
  
 
==Solution==
 
==Solution==
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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=\boxed{\frac{5}{3}\rightarrow \text{(B)}}</cmath>.
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/_AvJzq4QiUg
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/7_-ZvunSC8w
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2014|ab=B|before=num-b=3|num-a=5}}
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{{AMC10 box|year=2014|ab=B|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:36, 26 September 2023

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, \[2m+16b=2(4m+3b)=8m+6b\Rightarrow 10b=6m\Rightarrow m=\frac{10}{6}b=\boxed{\frac{5}{3}\rightarrow \text{(B)}}\].

Video Solution (CREATIVE THINKING)

https://youtu.be/_AvJzq4QiUg

~Education, the Study of Everything



Video Solution

https://youtu.be/7_-ZvunSC8w

~savannahsolver

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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