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?
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<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>
  
==Solution==
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==Solution 1==
We do <math>100-30-20-15-10</math> to find the percent of gumdrops that are green. We find that <math>25%</math> of the gumdrops are green. That means there are <math>120</math> gumdrops. If we replace all blue gumdrops with green gumdrops, then <math>35%</math> of the jar's gumdrops are brown. <math>.35 \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}</math>
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We do <math>100-30-20-15-10</math> to find the percent of gumdrops that are green. We find that <math>25\%</math> of the gumdrops are green. That means there are <math>120</math> gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then <math>15\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}</math>
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==Solution 2==
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We first calculate the percentage of gumdrops that are green. Subtracting all the other percentages from \(100\%\), we have:
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<math>100-30-20-15-10</math> = <math>25\%</math>.
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Thus, <math>25\%</math> of the gumdrops are green.
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Next, we set up a proportion: \(\frac{30}{25\%} = \frac{x}{30\%}\) to find the number of blue gumdrops. Cross-multiplying, we solve for \(x\): 36
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So there are 36 blue gumdrops.
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We divide the number of blue gumdrops by 2 to get 18
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Now, we calculate the number of brown marbles using the proportion:
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\(\frac{30}{25\%} = \frac{y}{20\%}\),
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where \(y\) is the number of brown marbles. Solving, we find y = 24.
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Adding these together gives:
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24 + 18 = 42.
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Thus, the answer is <math>\boxed{\textbf{(C)}\ 42}</math>
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~ algebraic_algorithmic
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==Video by MathTalks==
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https://www.youtube.com/watch?v=6hRHZxSieKc
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==See Also==
 
==See Also==
{{AMC8 box|year=2011|num-b=14|num-a=16}}
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{{AMC8 box|year=2010|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 20:59, 31 December 2024

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 1

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. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$

Solution 2

We first calculate the percentage of gumdrops that are green. Subtracting all the other percentages from \(100\%\), we have: $100-30-20-15-10$ = $25\%$.

Thus, $25\%$ of the gumdrops are green.

Next, we set up a proportion: \(\frac{30}{25\%} = \frac{x}{30\%}\) to find the number of blue gumdrops. Cross-multiplying, we solve for \(x\): 36

So there are 36 blue gumdrops.

We divide the number of blue gumdrops by 2 to get 18


Now, we calculate the number of brown marbles using the proportion:

\(\frac{30}{25\%} = \frac{y}{20\%}\),

where \(y\) is the number of brown marbles. Solving, we find y = 24.

Adding these together gives:

24 + 18 = 42.

Thus, the answer is $\boxed{\textbf{(C)}\ 42}$

~ algebraic_algorithmic

Video by MathTalks

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



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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