BTGmoderatorDC wrote:Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more than twice the number of females. If the ratio of female speckled trout to male rainbow trout is 4:3 and the ratio of male rainbow trout to all trout is 3:20, how many female rainbow trout are there?

A. 192
B. 195
C. 200
D. 205
E. 208

OA D

Source: Manhattan Prep

We can let m = male speckled trout, n = male rainbow trout, f = female speckled trout and g = female rainbow trout. Therefore, we have:

m + f = 645

m = 2f + 45

f/n = 4/3

And

n/(645 + n + g) = 3/20

Substitute 2f + 45 for m in the first equation, we have:

2f + 45 + f = 645

3f = 600

f = 200

Substitute 200 for f in the third equation, we have:

200/n = 4/3

4n = 600

n = 150

Finally, substitute 150 for n in the fourth equation, we have:

150/(645 + 150 + g) = 3/20

150/(795 + g) = 3/20

3000 = 2385 + 3g

615 = 3g

205 = g

Answer: D