Difference between revisions of "2010 AMC 8 Problems/Problem 15"
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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== | + | ==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 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> | 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> | ||
| + | |||
| + | ==Solution 2== | ||
| + | === 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 to find the number of blue gumdrops. Let \(x\) represent the number of blue gumdrops. Using the proportion: | ||
| + | \[ | ||
| + | \frac{30}{25\%} = \frac{x}{30\%}. | ||
| + | \] | ||
| + | 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: | ||
| + | \[ | ||
| + | \frac{36}{2} = 18. | ||
| + | \] | ||
| + | |||
| + | Now, we calculate the number of brown marbles using the proportion: | ||
| + | \[ | ||
| + | \frac{600}{25\%} = y, | ||
| + | \] | ||
| + | where \(y\) is the number of brown marbles. Solving, we find: | ||
| + | \[ | ||
| + | y = 24. | ||
| + | \] | ||
| + | |||
| + | Adding these together gives: | ||
| + | \[ | ||
| + | 24 + 18 = 42. | ||
| + | \] | ||
| + | Thus, the answer is \(\boxed{42}\). | ||
| + | |||
| + | |||
==Video by MathTalks== | ==Video by MathTalks== | ||
Revision as of 19:43, 31 December 2024
Problem
A jar contains 
different colors of gumdrops. 
are blue, 
are brown, 
are red, 
are yellow, and other 
gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
Solution 1
We do 
to find the percent of gumdrops that are green. We find that 
of the gumdrops are green. That means there are 
gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then of the jar's gumdrops are brown. 
Solution 2
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 to find the number of blue gumdrops. Let \(x\) represent the number of blue gumdrops. Using the proportion: \[ \frac{30}{25\%} = \frac{x}{30\%}. \] 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: \[ \frac{36}{2} = 18. \]
Now, we calculate the number of brown marbles using the proportion: \[ \frac{600}{25\%} = y, \] where \(y\) is the number of brown marbles. Solving, we find: \[ y = 24. \]
Adding these together gives: \[ 24 + 18 = 42. \] Thus, the answer is \(\boxed{42}\).
Video by MathTalks
https://www.youtube.com/watch?v=6hRHZxSieKc
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
