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Math Help - simple word problems

  1. #1
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    Post simple word problems

    1. A and B working together can finish painting the house in 6 days. A, working alone can finish it in 5 days less than B. How long will it take each of them to finish the work alone?

    2. A 400mm diameter pipe can fill the tank alone in 5 hours and another 600mm diameter pipe can fill the tank in 4 hours. A drain pipe 300 mm in diameter can empty the tank in 20 hours. With all the pipes open, how long will it take to fill the tank?

    3. A merchant has 3 items on sale: namely, a radio for $50, a clock for $30, and a flashlight for $1. At the end of the day, she has sold a total of 100 of the 3 items and has taken exactly $1000 on the total sales. How many radios did he sell?

    I don't have equations to actually get started. Help please!
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  2. #2
    Eater of Worlds
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    #1:

    Both workers can paint 1/6th of the house in one day.

    Therefore, \frac{1}{t}+\frac{1}{t+5}=\frac{1}{6}

    Solve for t.


    #2:

    The 400mm pipe can fill 1/5th of the tank in one hour and the 600mm can fill 1/4th of it in one hour. But another empties 1/20th of it in one hour.

    \frac{1}{4}+\frac{1}{5}-\frac{1}{20}=\frac{1}{t}

    Solve for t.


    For #3, did you forget a portion of the statement?. I may be mistaken, but it appears there's not enough info. The way it is you have two equations and 3 unknowns.
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  3. #3
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    Hello, signorinadulce!

    Obviously, setting up the equations is your main task . . .


    1. A and B working together can finish painting the house in 6 days.
    A, working alone can finish it in 5 days less than B.
    How long will it take each of them to finish the work alone?

    Let b = number of days for B to do the job alone.
    In one day, he can do \frac{1}{b} of the job.
    In 6 days, he can do \frac{6}{b} of the job.

    Then b-5 = number of days for A to do the job alone.
    In one, he can do \frac{1}{b-5} of the job.
    In 6 days, he can do \frac{6}{b-5} of the job.

    Together, in 6 days, they can do: . \frac{6}{b} + \frac{6}{b-5} of the job.

    But we are told that, in 6 days, they can complete the job (1 job).

    There is our equation! . . . . \frac{6}{b} + \frac{6}{b-5} \;=\;1

    Solve for b.



    2. One pipe can fill the tank alone in 5 hours
    and another pipe can fill the tank in 4 hours.
    A drain pipe empty the tank in 20 hours.
    With all the pipes open, how long will it take to fill the tank?

    The first pipe takes 5 hours to fill the tank.
    In one hour, it can fill \frac{1}{5} of the tank.
    In x hours, it can fill \frac{x}{5} of the tank.

    The second pipe takes 4 hours to fill the tank.
    In one hour, it can fill \frac{1}{4} of the tank.
    In x hours, it can fill \frac{x}{4} of the tank.

    The drain pipe takes 20 hours to drain the tank.
    In one hour, it empties \frac{1}{20} of the tank.
    In x hours, it empties \frac{x}{20} of the tank.

    Together, in x hours, they fill: . \frac{x}{5} + \frac{x}{4} - \frac{x}{20} of the tank.

    But in x hours, we expect them to fill the entire tank (1 tank).

    There is our equation!. . . . \frac{x}{5} + \frac{x}{4} - \frac{x}{20} \;=\;1

    Solve for x.



    3. A merchant has 3 items on sale: namely, a radio for $50, a clock for $30,
    and a flashlight for $1. At the end of the day, she has sold a total of 100 of the 3 items
    and has taken exactly $1000 on the total sales.
    How many radios did she sell?

    Let: . \begin{array}{ccc}R & = & \text{number of radios sold} \\ C & = & \text{number of clocks sold} \\F & = & \text{number of flashlights sold} \end{array}

    100 items were sold: . R + C + F \:=\:100 .[1]

    R radios cost 50R dollars.
    C clocks cost 30C dollars.
    F flashlights cost 1F dollars.

    Their total cost was $1000: . 50R + 30C + F \:=\:1000 .[2]

    Subtract [1] from [2]: . 49R + 29C \:=\:900 \quad\Rightarrow\quad R \:=\:\frac{900-29C}{49}

    Since R is a whole number, 900-29C must be divisible by 49.

    We can try values of C until we find one that works
    . . . . . or we can try a little algebra.
    We can rewrite the equation: . R \;=\;16 + \frac{116-29C}{49} \;=\;16 + \frac{29(4 - C)}{40}

    And we see that C = 4 gives us: . \boxed{R = 16}

    Therefore, 16 radios were sold.

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