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10 November, 16:26

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

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  1. 10 November, 20:00
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    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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