Problem Solving

Whenever you have a problem that presents two either-or categories (male or female, rainbow or speckled), the best way to approach the problem is to set up a matrix as follows:



Use M and F to represent the total number of males and females, respectively. We know that together they add up to the total number, so:

M + F = 645

If the number of males is 45 more than twice the number of females, then:

M = 45 + 2F

Now we have a system of equations - 2 equations and 2 variables - so we can substitute for M:

(45 + 2F) + F = 645
3F = 600
F = 200
M = 445

If the ratio of female speckled to male rainbow is 4/3, we know that the number of female speckled is 200, so:

(200/x) = (4/3)
4x = 600
x = 150

There are 150 male rainbow trout, so put that in the the grid. If the ratio of male rainbow to all trout is 3/20, then:

(150/y) = (3/20)
3Y = 3000
Y = 1000



There are 1,000 total trout. Once we add this to our grid, we can easily solve for the number of female rainbow trout. There are 645 speckled trout and 1000 total trout, so:

1000 - 645 = 355 There are 355 total rainbow trout.

Total rainbow trout - male rainbow trout = female rainbow trout, so:

355 - 150 = 205



There are 205 female rainbow trout. The answer is D.