Question

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can seperately fill the tank

Answer 1

Let x be the time taken by the small tap to fill the tank.

Then,

time taken by the big tap to fill the tank = (x-9) hours

Now,

part of the tank filled by the small tap in 1 hour = 1/x

part of the tank filled by small tap in 6 hours = (1/x) × 6 = 6/x

And,

part of tank filled by big tap in 1 hour = 1/(x-9)

part of tank filled by big tap in 6 hours = 1/(x-9) × 6 = 6/(x-9)

Since both the taps fills the tank in 6 hours

part of tank filled by small tap in 6 hours + part of tank filled by big tap in 6 hours = total part of the tank filled = 1

=> 6/x + 6/(x-9) = 1

=> (6(x-9) + 6x)/(x(x-9)) = 1

=> (6x - 54 + 6x)/(x^2 - 9x) = 1

=> (12x - 54)/(x^2 - 9x) = 1

=> 12x - 54 = x^2 - 9x

=> x^2 - 9x - 12x + 54 = 0

=> x^2 - 21x + 54 = 0

=> x^2 - 3x - 18x + 54 = 0

=> x(x - 3) - 18(x - 3) = 0

=> (x-3)(x-18) = 0

either x-3 = 0

=> x = 3

or x-18 = 0

=> x = 18

here x = 3 hours or 18 hours

now,

taking x = 3 hours

time taken for the big tap to fill the tank = (x-9) hours = (3-9) hours = -6 hours (which is negative therefore rejected)

again,

taking x = 18 hours

time taken for the big tap to fill the tank = (x-9) hours = (18-9) = 9 hours

Hence time taken for the small tap to fill the tank = 18 hours and time taken for the big tap to fill the tank = 9 hours

Answer 2

Step-by-step explanation:

the bigger tap takes 9 hours & the smaller tap takes 18 hours