how long will it take each pipe alone to fill the tank?

Question

2 pipes, A and B, are used to fill a water tank. The empty tank is filled in 100 hours if the 2 pipes are used together/ If pipe A alone is used for 6 hours and then turned off, pipe B will take over and finish filling the tank in 18 hours. IF it takes A alone x hours and B alone y hours, A fills 1/x of the tank per hour, and B fills 1/y per hour.

Answer 1

Since it takes 100 hours to fill the tank when both pipes are used together, in 1 hour, 1/100 of the tank will be filled. Pipe A contributes 1/x of the tank in one hour and pipe B contributes 1/y of the tank. So the two fractions added together fill 1/100 of the tank in an hour: 1/x+1/y=1/100.

Since pipe A fills 1/x tank in an hour, it would take it x hours to fill the tank by itself. Similarly, pipe B would take y hours.

Pipe A fills 6/x of the tank in 6 hours, leaving 1-6/x of the tank to be filled. Since pipe B takes y hours to fill the tank, it would take fewer hours if the tank was already partly filled. So, by proportion, it would take y(1-6/x) hours to top up the tank by itself. We know this time is 18 hours, so y(1-6/x)=18.

We have two equations in x and y, so we can work out the two variables. But there's something wrong! If A and B working together take 100 hours, it surely takes longer to fill the tank working in relay. So I think that 100 hours should be 10 hours. If we proceed as the question indicates, we'll see that we have a pipe removing rather than adding water to the tank. 

Multiply the first equation through by 100xy: 100y+100x=xy. Multiply the second equation through by x: xy-6y=18x. Therefore xy=18x+6y=100y+100x, so 94y+82x=0 and x or y would be negative.

Let's replace 100 with 10: 18x+6y=10x+10y, 8x=4y so y=2x. 10x+20x=2x^2 (substituting for y in 10x+10y=xy). So x^2=15x, and x=15 (we disregard x=0). Therefore y=30.

Check: 1/15+1/30=3/30=1/10 and it takes pipe A 6 hours to fill 6/15=2/5 of a tank, leaving 3/5 left to fill. This will take pipe B 3/5 of 30 hours=18 hours to complete.

So with the adjustment, x=15 hours and y=30 hours.

Answer 2

It takes pipe A x hours alone to fill the water tank, or it fills 1/x per hour.
It takes pipe B y hours along to fill the water tank, or it iflls 1/y per hour.

Working togeather, they fill the water tank in 100 hours. This means that pipe A fills 100/x of the water tank and pipe B fills 100/y of the water tank.
We get equation
1): 100/x+100/y=1

If pipe A is working alone for 6hours, it fills 6/x of the water tank. Then pipe B takes over and for 18 hours it fills 18/y of the tank.We have equation 2): 6/x+18/y=1.

Now set the two equations equal and we have:
100/x+100/y=6/x+18/y
100y+100x=6y+18x
82x+94y=0
y =-41x/47

Now substituting back into equation 1)
100/x+100/(-41x/47)=1
x = -600/41

and y = (-41(-600/41))/47
y= 600/47

if the question is "2 pipes, A & B, are used to fill a water tank. The empty tank is filled in 10 hrs if the two pipes are used together. If pipe A alone is used for 6 hrs and then turned off, pipe B will take over and finish the tank in 18 hrs. How long will it take each pipe alone to fill the tank.if it takes A alone x hrs and B alone y hrs, A fills 1/x of the tank per hour, and B fills 1/y per hour "


Then the answer is
It takes pipe A x hours alone to fill the water tank, or it fills 1/x per hour.
It takes pipe B y hours along to fill the water tank, or it iflls 1/y per hour.

Working togeather, they fill the water tank in ten hours. This means that pipe A fills 10/x of the water tank and pipe B fills 10/y of the water tank.
We get equation 1): 10/x+10/y=1

If pipe A is working alone for 6hours, it fills 6/x of the water tank. Then pipe B takes over and for 18 hours it fills 18/y of the tank.We have equation 2): 6/x+18/y=1.

Now set the two equations equal and we have:
10/x+10/y=6/x+18/y
10y+10x=6y+18x
4y=8x
y=2x. Now substituting back into equation 1)
10/x+10/2x=1
10+5=x, so x=15 hrs and y=2x=30 hrs.

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