If two pipes function simultaneously, the reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir? 1 point
Answers
Answer:
Second Pipe fill reservoir in 30 Hrs
Step-by-step explanation:
two pipes simultaneously fill the reservoir in 12 hours
two pipes simultaneously in 1hr fill the reservoir = 1/12
Let say one pipe fill the reservoir in x hr
Second pipe fills reservoir in x + 10 hr ( as pipe one is 10 hr faster)
Pipe one fill reservoir in 1 hr = 1/x
Second pipe fill reservoir in 1 hr = 1/(x+10)
1/x + 1/(x+10) = 1/12
Multiplying both sides by 12x(x+10)
=> 12(x + 10) + 12x = x(x+10)
=> 12x + 120 + 12x = x² + 10x
=> 0 = x² - 14x -120
=> x² - 14x -120 = 0
=> x² - 20x + 6x - 120 = 0
=> x(x-20) + 6(x-20) =0
=> (x+6)(x-20) = 0
x = 20
pipe one fill reservoir in 20 Hrs
Second Pipe fill reservoir in 20 + 10 = 30 Hrs
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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