Question

If two pipes function simultaneously the reservoir is filled in 12 hrs. One pipe fills the reservoir 10 hours faster than the other. How many hours does the faster pipe take to fill the reservoir

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

the answer will be 30 hours