# Ratio word problem: What fraction of the original counters remain in the bag

#### Jessica15

##### New member
I have a word equation i cannot figure out! any one 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!

#### skeeter

##### Well-known member
MHB Math Helper
n = (3/5)B
----------- $\implies$ 1 = (12B)/(5R) $\implies$ B/R = 5/12 ... 5 blue, 12 red, 17 total
n = (1/4)R

if n = 3, 3 blue counters are removed = 3/5 ofthe original 5 ... 3 red counters are removed = 1/4 of the original 12

2 blue and 9 red counters remain ... 11/17 of the original counters remain

#### Wilmer

##### In Memoriam
Hint:
r = 24, b = 10, n = 6

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

#### skeeter

##### Well-known member
MHB Math Helper
B/r = 5/12 = 10/24 = 15/36 = ...