Question

If two pipes function simultaneously the tank will be filled in 12 hours. one pipe fills the tank 10 hours faster than the other. how many hours does the faster pipe take to fill the tank?

Answer 1

let the reservoir be filled by first pipe in x hours.

Then ,second pipe fill it in (x+10)hrs.

Therefore (1/x)+(1/x+10)=(1/12) <=>(x+10+x)/(x(x+10))=(1/12).

x^2 –14x-120=0 <=> (x-20)(x+6)=0

<=> x=20 [neglecting the negative value of x]

so, the second pipe will take (20+10)hrs. (i.e) 30 hours to fill the reservoir

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…...