Question

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

Answer 1

Let the 1st papi takes x to fill alone
2nd takes x-10 ------------—------------
1st. in x hrs the water filled is 1
in 1hr 1st tap fills=(1/x)
similarly in 1hr 2nd tap fills 1/x-10
(1/x +1/x-10)12=1
(x-10+x)/x²-10x=1/12
2x-10=(x²-10x)/12
(2x-10)12=x²-10x
24x-120=x²-10x
x²-34x+120=0
x²-30x-4x+120=0
x(x-30)-4(x-30)=0
x=30
x-10=20

Answer 2

let there are 2 pipes =x &y
time taken by x = x hour
" by y = (x-10) h
l/x + 1/x-10 = 1/12
2x - 10/x²- 10x = 1/12
24x -120 = x² - 10x
x²- 34x -120 = 0
x(x-30)-4(x-30) = 0
x=30 h ans.