to fill a swimming for two pipes are used if the pipe of the larger diameter used for 4 hours and the pipe of the smaller diameter for 9 hours only half of the full can be filled find how long it would take for each pipe to fill the pool separately if the pipe of the smaller diameter takes 10 hours more than the pipe of the larger diameter to fill the pool
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Let x hour be the time taken by larger pipe to fill the tank
So in one hour it would fill 1/x part of the tank
Same way as y hour is needed for smaller pipe then in one hour it would fill 1/y part
Now it is given y - x = 10
==> y = x + 10
Also 4 *1/x + 9 *1/y = 1/2
Now plug y = x + 10 in this
We get 4/x + 9/(x+10) = 1/2
Rearranging we get x^2 - 16x - 80 = 0
OR (x -20)(x+4) = 0
Hence x = 20 h
And y = 30 h
So larger dia pipe would fill in 20 h and smaller dia pipe would fill the pool in 30 h
priyasingh123:
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Let the bigger pipe fill the tank in x hours.the smaller pipe fills the tanks in (x+10) hours.
∴(4/x) + (9/(x+10)) = 1/2
(4x+40+9x) / (x^2+10x) = 1/2
2(13x+40)= x^2+10x
26x+80= x^2+10x
x^2-16x-80 =0
x^2-20x+4x-80 =0
x(x-20) + 4(x-20) =0
(x+4)(x-20) = 0 ∴ x = 20 and x = -4 ( dont consider)
Hence, the pipe with larger diameter fills the tank in 20 hours. and the pipe with smaller diameter fills the tank in 30 hours.
∴(4/x) + (9/(x+10)) = 1/2
(4x+40+9x) / (x^2+10x) = 1/2
2(13x+40)= x^2+10x
26x+80= x^2+10x
x^2-16x-80 =0
x^2-20x+4x-80 =0
x(x-20) + 4(x-20) =0
(x+4)(x-20) = 0 ∴ x = 20 and x = -4 ( dont consider)
Hence, the pipe with larger diameter fills the tank in 20 hours. and the pipe with smaller diameter fills the tank in 30 hours.
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