asked 174k views
3 votes
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. 192B. 195C. 200D. 205E. 208

asked
User Harika
by
8.0k points

1 Answer

7 votes

Answer: D 205

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

answered
User Mmcorrelo
by
7.6k points
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