2010 AMC 8 Problems/Problem 15

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: \(\frac{30}{25\%} = \frac{x}{30\%}\) 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: \[ \frac{36}{2} = 18. \]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.