Question

Two pipes together can fill a reservoir in 12hours .if one pipe can fill the reservoir 10hours faster than the other how many hours will the second pipe take to fill the reservoir??

Answer 1

Let the faster pipe fill the reservoir in 'X' hours.
The slower pipe fills it in 'X+10' hours.
Both the pipes together fill in 12 hours.
In 1 hour:
Faster pipe fills: 1/X part
Slower pipe fills: 1/X+10 part
Both pipes fill: 1/12 part

1/12 = 1/X + 1/X+10
1/12 = X+10+X/[X*(X+10)]
1/12 = 2X+10/(X^2+10X)
X^2 + 10X = 24X + 120
X^2 - 14X - 120 = 0
X^2 + 6X - 20X - 120 = 0
X(X+6) - 20(X+6) = 0
(X-20)*(X+6) = 0
Hence X=20, -6.... X=20=faster pipe's time
Hence slower pipe fills in X+10 = 20+10 = 30 hours
Slower pipe(2nd pipe) can fill alone in 30 hours.
Hope it helps.

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.

HOPE THIS ANSWER WILL HELP YOU…...