Question

two pipes together fill a reservoir in 12 hours. if one pipe can fill the reservoir10 hours faster than the other, how many hours will thesecond pipe take to fill the reservoir

Answer 1

if one fill ten time faster tjen the other pipe will take 2 hours

Answer 2

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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