Your thinking is on the right track, but the conclusion that \(\displaystyle 5r = 12b\) is incorrect. We can see why this cannot possibly be the case by considering a simple example. Suppose we had r = 5 and b = 12. Since we can never remove a fractional number of counters, can you see why this means it's impossible to ever have *exactly* 1/4 of the original number of red counters? In order for there to be exactly 1/4 of the red counters left, we must have \(\displaystyle r = 20k, \: b = 48k\) for some integer *k*.

Using the simplest possible case where there's 20 red marbles, we see that in order for leave exactly 1/4 of the red counters behind, we have to remove 15 of each color counter. This means 33 blue counters are left, but \(\displaystyle 33/48 = 0.6875\) so it's impossible to have 25% of the red counters and 60% of the blue counters at the same time. A little bit of thinking can easily show that it's likewise impossible for any integer value of *k*.

Now, to see what went wrong and how to correct it. It should be obvious why our starting amount of counters must be 100% of the red counters and 100% of the blue counters. After removing some amount (*n*) of counters, we're left with 25% of the red counters and 60% of the blue counters. Can you see why that means the amount we removed must be 75% (3/4) of the red counters and 40% (2/5) of the blue counters? So we have \(\displaystyle n = \frac{3r}{4}\) and \(\displaystyle n = \frac{2b}{5}\).

Perform the same cross-multiplication and now that you've got the correct values here you'll get the correct ratio. Then use this ratio in the method Dr.Peterson described.