Thread: Fractions & ratios when items are removed

1. Fractions & ratios when items are removed

I have a word equation i cannot figure out! anyone have any ideas?

There is a large bag of red and blue counters. One red and one blue counter are removed from the bag. This process is repeated until 1/4 of the red counter and 3/5 of the blue counters have been removed. What fraction of the original counters remain in the bag?

Im stumped!
I've written that:

Red - n = 3/4 Red
blue - n = 2/5 blue
n is the number of counters removed as it the same number removed from both.

therefore n = 1/4red and n = 3/5 blue. therefore 1/4 red = 3/5 blue. if you cross multiply that you get that 5 red = 12 blue.... I dont know how that helps!

2. Re: Fractions & ratios when items are removed

Hint:
r = 24, b = 10, n = 6

r: 24 - 6 = 18 : 3/4 left
b: 10 - 6 = 04 : 2/5 left

3. Re: Fractions & ratios when items are removed

Originally Posted by Jessic17
I have a word equation i cannot figure out! anyone have any ideas?

There is a large bag of red and blue counters. One red and one blue counter are removed from the bag. This process is repeated until 1/4 of the red counter and 3/5 of the blue counters have been removed. What fraction of the original counters remain in the bag?

Im stumped!
I've written that:

Red - n = 3/4 Red
blue - n = 2/5 blue
n is the number of counters removed as it the same number removed from both.

therefore n = 1/4red and n = 3/5 blue. therefore 1/4 red = 3/5 blue. if you cross multiply that you get that 5 red = 12 blue.... I dont know how that helps!

we have

$\displaystyle r=4n$ and $\displaystyle b=\frac{5}{3}n$

the fraction we are looking for is

$\displaystyle \frac{r+b-2n}{r+b}$

4. Re: Fractions & ratios when items are removed

$n = \dfrac{r}{4} = \dfrac{3b}{5}$

$\dfrac{3b}{5} \div \dfrac{r}{4} = \dfrac{n}{n} \implies \dfrac{12}{5} \cdot \dfrac{b}{r} = 1 \implies \dfrac{b}{r} = \dfrac{5}{12}$

$b = 5k$ and $r=12k$ for some positive integer $k$

ratio of counters remaining

$\dfrac{\frac{3}{4} \cdot 12k + \frac{2}{5} \cdot 5k}{12k+5k}= \dfrac{11}{17}$