if two pipes function simultaneously a Reservoir will be filled in 12 hours 15 will take the Reservoir 10 hours faster than the other how many hours will the second pipe take to fill the Reservoir?
Answers
faster pipe alone can fill the reservoir in (x−10)(x−10) hours.
Part filled by slower pipe in 1 hr =1x=1x
Part filled by faster pipe in 1 hr =1x−10=1x−10
If pipes functions simultaneously, the reservoir will be filled in 12 hours.
=> Part filled by both the popes in 1hr = 112112
Thus we have,
1x+1x−10=11212(x−10)+12x=x(x−10)12x−120+12x=x2−10xx2−34x+120=0(x−30)(x−4)=0x=30 or 41x+1x−10=11212(x−10)+12x=x(x−10)12x−120+12x=x2−10xx2−34x+120=0(x−30)(x−4)=0x=30 or 4
xx cannot be 4 because (x−10)(x−10) will be negative.
Therefore, x=30x=30
Part filled by second pipe (slower pipe) in 1 hr = 130130
=> Second pipe (slower pipe) alone can fill the reservoir in 30 hours.
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SOLUTION :
Let the faster pipe fill the reservoir in x h.
Then, the slower pipe the reservoir in (x + 10) h
In 1 hour the faster pipe fills the portion of the reservoir : 1/x
In 12 hour the faster pipe fills the portion of the reservoir : 12 × 1/x = 12/x
In 1 hour the slower pipe fills the portion of the reservoir : 1/(x + 10)
In 12 hour the faster pipe fills the portion of the reservoir : 12 × 1/(x + 10) = 12/(x +10)
A.T.Q
12/x + 12/(x +10) = 1
12(1/x + 1/(x + 10) ) = 1
1/x + 1/(x + 10) = 1/12
(x + 10 + x ) / [x(x + 10)] = 1/12
[By taking LCM]
2x +10 /(x² + 10x) = 1/12
x² + 10x = 12(2x +10)
x² + 10x = 24x + 120
x² + 10x - 24x - 120 = 0
x² - 14x - 120 = 0
x² + 6x - 20x - 120 = 0
[By splitting middle term]
x(x + 6) - 20(x + 6) = 0
(x - 20) (x + 6) = 0
(x - 20) or (x + 6) = 0
x = 20 or x = - 6
Since, time cannot be negative, so x ≠ - 6
Therefore, x = 20
The faster pipe takes 20 hours to fill the reservoir
Hence, the slower pipe (second) takes (x + 20) = 30 hours to fill the reservoir.
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