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ElenaW [278]
3 years ago
7

Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more tha

n 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. 192B. 195C. 200D. 205E. 208
Mathematics
1 answer:
Kryger [21]3 years ago
7 0

Answer: D 205

Step-by-step explanation:

Let,

Number of all trouts = N

Number of speckled trouts = Ns = 645

Number of rainbow trouts = Nr

Number of male speckled trouts = Ms

Number of female speckled trouts = Fs

Number of male rainbow trouts = Mr

Number of female rainbow trouts = Fr

Since, Ms = 2Fs + 45

Also, Ms + Fs = 645

Therefore, 2Fs + 45 + Fs = 645

Fs = (645-45)/3 = 200

Female speckled trouts = 200

Since

Ms + Fs = 645

Ms = 645- 200 = 445

Since, Fs/Mr = 4/3

Mr = 3x200/4 = 150

Since,

Mr/N = 3/20

N = 20x 150/3 = 1000

Recall that,

N = Ms+Fs+Mr+Fr

Fr = N-Ms-Fs-Mr

Fr = 1000-445-200-150

Fr = 205

Therefore, the number of rainbow female trouts = 205

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