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?

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

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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

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

The 2 pipes together take 2.7 hr, or 2 hrs and 42 min

to fill 3/4 of the pool

Step-by-step explanation:

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

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.