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Trava [24]
3 years ago
14

It takes the first pipe 9 more hours to fill the pool than the first and the second pipes together and 7 less hours than it woul

d take the second pipe if it was working alone. How long would it take to fill up the pool if both pipes were working together?
Mathematics
1 answer:
solniwko [45]3 years ago
6 0
If Pipe 1 (P1) takes x hours to fill the pool, Pipe 2 (P1) and pipe 2 (P2) takes (x-9) hours to fill the pool, and pipe 2 (P2) takes (x+7) hours to fill the pool.

That is,
P1 = x hrs
P1+P2 = (x-9) hrs
P3 = (x+7) hrs

In 1 hour, P1 fills 1/x of the pool, P1+P2 fills 1/(x-9) of the pool and P2 fills 1/(1+7) of the pool.
Therefore,
1/x+1/(1+7) = 1/(x-9) => ((x+7)+x)/(x)(x+7)=1/(x-9) => (2x+7)/x^2+7x = 1/(x-9) => (2x+7)(x-9)=x^2+7x => x^2-18x-63 =0 
Solving for x
x= (-b+/- sqrt (b^2-4ac)/2a, where a=1, b=18, and c=63
Substituting;
x1=21 and x2=-3 (the  negative x is ignored as it does not make sense).
Therefore, x = 21
This means,
P1 takes 21 hours to fill the pool
P1+P2 takes (21-9) hours = 12 hours to fill the pool while P3 takes (21+7) hours = 28 hours
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