Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank
Answers
Answers
Let the time taken by the smaller diameter tap = x
larger = x-10
total time taken = 75 /8
portion filled in one hour by smaller diameter tap = 1/x
and by larger diamter tap = 1/x-10
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
2x+10/x²-10x = 8/75
8(x²-10x) = 75 ×2 (x-5)
8/2 (x²-10x) = 75 (x-5)
4x²-40x = 75x-375
4x² -40x-75x +375 = 0
4x²-115x + 375 = 0
4x²-100x-15x +375 = 0
4x(x-25)-15(x-25)=0
(x-25)(4x-15)
x= 25
x= 15/4
If x= 25
the x-10 = 25-10 = 15
if x = 15/4
x-10 = 15/4 - 10 = 15-40/4 = -25/4
since time cannot be negative therefore x = 25
SOLUTION
Let the tap with smaller diameter fills the tank alone in x hours.
Let the tap with smaller diameter fills the tank alone in x hours.Let the tap with larger diameter fills the tank alone in (x-10)hours.
In 1 hour, the tap with smaller diameter can fill 1/x part of the tank.
In 1 hour, the tap with larger diameter can fill 1/(x-10) part of the tank.
The tank is filled up in 75/8 hours.
Thus, in 1 hour the taps fill 8/75 part of the tank.
=) 1/x+ 1/x-10= 8/75
=) (x-10)+x/x(x-10)= 8/75
=) 2(x-5)/x^2-10x= 8/75
=) x-5/x^2 -10x= 4/75
=) 4x^2-40x= 75x-375
=) 4x^2 -40x-75x+375= 0
=) 4x^2 -115x+375=0
=) 4x^2 -100x-15x+375=0
=) 4x(x-25)-15(x-25)=0
=) (x-25) (4x-15)= 0
=) x= 25 x= 15/4
If x= 15/4,
x-10 = 15/4-10
=) -25/4 (which is not possible)
If x= 25, then (x-10)
=) 25-10 = 15.
Thus, the tap with smaller diameter fills the tank alone in 25 hours where as the larger diameter fills tank alone in 15 hours.