Question

2) One tank can be filled up two taps in 6 hrs the smaller tap alone take 5 hrs more than the
bigger tap alone. Find the time required by each tap to fill the tank separately.​

Answer 1

Answer:

Let the large tap takes x hours to fill up the tank alone.

Then, in 1 hour the large tap fills the  part of the tank.

So, as per the given condition, the small tap will fill the tank alone in (x + 5) hours.

Then, in 1 hour the large tap fills the 1/x+5 part of the tank.

When both of them are open then in 1 hour they will fill 1/x + 1/ x+5  part of the tank.

Given that, both the taps when open they fill the tank in 6 hours.

So, when both the tanks are open they will fill  part of the tank in 1 hour.

Therefore, we can write the equation as

1/x + 1/ x+5 = 1/6

⇒ 6(2x + 5) = x(x + 5)

⇒ x² - 7x - 30 = 0

⇒ x² - 10x + 3x - 30 = 0

⇒ (x - 10)(x + 3) = 0

So, x = 10 hours {As x can not be negative}

So, x + 5 = 15 hours.

Therefore, the small tap fills the tank in 15 hours and the large tap fills the tank in 10 hours.