One pipe can fill a pool in 9 hours. Another pipe can fill the pool in 6 hours. How long would it take

One pipe can fill a pool in 9 hours. Another pipe can fill the pool in 6 hours. How long would it take them to fill the pool if they were working together?
help asap please

Related Posts

This Post Has 4 Comments

  1. 3.6 hours

    Step-by-step explanation:

    Let X be the total job

    Combined speed = X/9 + X/6

    LCM = 18

    (2X + 3X)/18

    5X/18 = X/T

    T = 18/5

    T = 3.6

  2. 3.6 hours

    Step-by-step explanation:

    To determine the time needed if they are working together, we use the formula

    1/Ta + 1/Tb = 1 Tc

    where Ta is the the time for the first item working alone

    Tb is the time for the second item alone

    Tc is the time for them working together

    1/9 + 1/6 = 1/Tc

    I multiply by 54Tc  to clear the fractions

    54Tc (1/9 + 1/6) = 1/Tc *54Tc

    6Tc + 9 Tc = 54

    Combine like terms

    15Tc = 54

    Divide each side by 15

    15 Tc /15 = 54/15

    Tc =3.6 hours

  3. Answer= 4 hours.
    The first pipe can fill one pool in nine hours. So in one hour, it can fill 1/9th of the pool. Same with the other pipe, it can fill 1/6th of the pool in other hour.
    So 1/9 + 1/6 would represent the first hour of filling the pool. Since 1/9 + 1/6 = .277.. , 27% of the pool is filled. Repeat this until you get to 100% (full pool).

    First hour, 1/6 + 1/9 = .27
    Second hour, 2/6 + 2/9 = .5
    Third hour, 3/6 + 3/9 = .83
    Fourth hour, 4/6 + 4/9 = 1.11

    So in 4 hours, the pool will be filled.

Leave a Reply

Your email address will not be published. Required fields are marked *