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

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Video:
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https://www.youtube.com/watch?v=6hRHZxSieKc
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By MathTalks
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Revision as of 12:57, 14 January 2023

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 half of the blue gumdrops with brown gumdrops, then $15\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$

See Also

Video:

https://www.youtube.com/watch?v=6hRHZxSieKc

By MathTalks


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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