Difference between revisions of "2010 AMC 8 Problems/Problem 15"

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==Problem==
 
==Problem==
A jar contains <math>5</math> different colors of gumdrops. <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> are red, <math>10\%</math> are yellow, and other <math>30</math> gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how gumdrops will be brown?
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A jar contains <math>5</math> different colors of gumdrops. <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> are red, <math>10\%</math> are yellow, and other <math>30</math> gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
  
 
<math> \textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64 </math>
 
<math> \textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64 </math>

Revision as of 09:16, 8 October 2015

Problem

A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?

$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$

Solution

We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace all blue gumdrops with green gumdrops, then $35\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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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