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Math Help - Logic help.

  1. #1
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    Logic help.

    2 men and 7 boys can do a piece of work in 14 days,
    3 men and 8 boys can do it in 11 days.
    8 men and 6 boys can do this 3 times of work in how many days?


    Logic: I found either the first or the second sentence is redundant, but when I ignored one and tried solving, it gives me wrong answers, Kindly help how to approach?
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  2. #2
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    Re: Logic help.

    Let x represent the amount of work that can be done by a man, and let y represent the amount of work that can be done by a boy. Then \displaystyle \begin{align*} 2x + 7y = 14 \end{align*} and \displaystyle \begin{align*} 3x + 8y = 11 \end{align*}. Solve these two equations simultaneously for x and y, then use this to answer the remainder of your question.
    Thanks from hisajesh
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  3. #3
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    Re: Logic help.

    Hello, hisajesh!

    I assume that all the men work at the same constant rate
    and that all the boys work at the same constant rate.


    2 men and 7 boys can do a piece of work in 14 days,
    3 men and 8 boys can do it in 11 days.
    8 men and 6 boys can do this 3 times of work in how many days?

    Let m = the number of days it takes one man to do the job..
    In one day, a man can do \tfrac{1}{m} of the job.
    In one day, 2 men can do \tfrac{2}{m} of the job.

    Let b = the number of days it takes one boy to do the job.
    In one day, a boy can do \tfrac{1}{b} of the job.
    In one day, 7 boys can do \tfrac{7}{b} of the job.

    So, in one day, 2 men and 7 boys can do \tfrac{2}{m} + \tfrac{7}{b} of the job.
    . . But this equals \tfrac{1}{14} of the job.
    Hence: . \tfrac{2}{m} + \tfrac{7}{b} \:=\:\tfrac{1}{14}

    In one day, 3 men can do \tfrac{3}{m} of the job.
    In one day, 8 boys can do \tfrac{8}{b} of the job.

    So in one day, 3 men and 8 boys can do \tfrac{3}{m} + \tfrac{8}{b} of the job.
    . . But this equals \tfrac{1}{11} of the job.
    Hence: . \tfrac{3}{m} + \tfrac{8}{b} \:=\:\tfrac{1}{11}

    We have a system of equations: . \begin{Bmatrix}\frac{2}{m} + \frac{7}{b} \:=\:\frac{1}{14} & [1] \\ \\[-4mm] \frac{3}{m} + \frac{8}{b} \:=\:\frac{1}{11} & [2] \end{Bmatrix}

    \begin{array}{ccccccc}\text{Multiply [1] by 3:} & \frac{6}{m} + \frac{21}{b} &=& \frac{3}{14} \\ \\[-4mm] \text{Multiply [2] by -}2\!: & \text{-}\frac{6}{m} - \frac{16}{b} &=& \text{-}\frac{2}{11} \end{array}

    Add: . \tfrac{21}{b} - \tfrac{16}{b} \:=\:\tfrac{3}{14} - \tfrac{2}{11} \quad\Rightarrow\quad \tfrac{5}{b} \:=\:\tfrac{5}{154} \quad\Rightarrow\quad b \:=\:154

    Substitute into [1]: . \tfrac{2}{m} + \tfrac{7}{154} \:=\:\tfrac{1}{14} \quad\Rightarrow\quad \tfrac{2}{m} \:=\:\tfrac{2}{77} \quad\Rightarrow\quad m \:=\:77


    Now we know the following:

    In one day, a man can do \tfrac{1}{77} of the jpb.
    In one day, 8 men can do \tfrac{8}{77} of the job.
    In x days, 8 men can do \tfrac{8x}{77} of the job.

    In one day, a boy can do \tfrac{1}{154} of the job.
    In one day, 6 boys can do \tfrac{6}{154} of the job.
    In x days, 6 boys can do \tfrac{6x}{154} of the job.

    In x days, 8 men and 6 boys can do: \tfrac{8x}{77} + \tfrac{6x}{154} of the job.
    But this equals 3 jobs.

    Hence: . \frac{8x}{77} + \frac{6x}{154} \:=\:3

    Multiply by 77: . 8x + 3x \:=\:231 \quad\Rightarrow\quad 11x \:=\:231 \quad\Rightarrow\quad x \:=\:21

    Answer: 21 days.
    Thanks from hisajesh
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