2010 AMC 8 Problems/Problem 15: Difference between revisions

Revision as of 19:57, 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 $\boxed{\textbf{(C)}\ 42}$

Video by MathTalks

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

2010 AMC 8 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 8: / Line 8: @@
 ==Solution 2==
-=== Solution 2 ===
 We first calculate the percentage of gumdrops that are green. Subtracting all the other percentages from \(100\%\), we have:
-\[
+<math>100-30-20-15-10</math> = <math>25\%</math>.
-- 30 - 20 - 15 - 10 = 25.
-\]
+Thus, <math>25\%</math> of the gumdrops are green.
-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
-Next, we set up a proportion: \(\frac{30}{25\%} = \frac{x}{30\%}\) to find the number of blue gumdrops. Cross-multiplying, we solve for \(x\):
-\[
-x = 36.
-\]
 So there are 36 blue gumdrops.
-We divide the number of blue gumdrops by 2 to get:
+We divide the number of blue gumdrops by 2 to get 18
-\[
-\frac{36}{2} = 18.
-\]
 Now, we calculate the number of brown marbles using the proportion:
-\[
-\frac{600}{25\%} = y,
+\(\frac{30}{25\%} = \frac{y}{20\%}\),
-\]
-where \(y\) is the number of brown marbles. Solving, we find:
+where \(y\) is the number of brown marbles. Solving, we find y = 24.
-\[
-y = 24.
-\]
 Adding these together gives:
-\[
 + 18 = 42.
-\]
-Thus, the answer is \(\boxed{42}\).
+Thus, the answer is <math>\boxed{\textbf{(C)}\ 42}</math>
 ==Video by MathTalks==

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