Question

One tank can be filled up by two taps in 6 hours.The smaller tap alone takes 5 hours more than the bigger tap alone.Find the time required by each tap to fill the tank separately.

Answer 1

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

Step-by-step explanation:

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

Then, in 1 hour the large tap fills the $\frac{1}{x}$ 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 $\frac{1}{x + 5}$ part of the tank.

When both of them are open then in 1 hour they will fill $(\frac{1}{x} + \frac{1}{x + 5})$ 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 $\frac{1}{6}$ part of the tank in 1 hour.

Therefore, we can write the equation as

$\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}$

⇒ 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. (Answer)

Answer 2

Answer:

The larger tank takes 10 hours and the smaller tank takes 15 hours to fill the tank separately.

Step-by-step explanation:

Here we have been given to find the time required by each of the two taps to fill up the tank separately. For this, we assume the larger tank fills the tank in y hours.

Therefore in 1 hour, it will fill $\frac{1}{y}$ part of the tank.

According to the question, the smaller tank will fill the tank in y +5 hours

Hence in 1 hour, it will fill $\frac{1}{y+5}$ part of the tank.

When used together they take 6 hours to fill the tank.

Hence in 1 hour, they will fill $\frac{1}{6}$ part of the tank.

Hence we have,

$\frac{1}{y} +\frac{1}{y+5} = \frac{1}{6}$

⇒ $\frac{y+5 + y}{y(y+5)} = \frac{1}{6}$

⇒ (2y +5) × 6 = y(y+5)

⇒ 12y + 30 = $y^{2}$ + 5y

⇒ 12y + 30 - $y^{2}$ - 5y = 0

⇒ $- y^{2} + 7y + 30 = 0$

⇒ $y^{2} - 7y - 30 = 0$

⇒ $y^{2} - ( 10-3) y - 30 = 0$

⇒ $y^{2} - 10y + 3y -30 =0$

⇒$y(y-10) + 3(y-10) = 0$

⇒ (y+3)(y-10) = 0

Hence y = -3 or 10

Therefore y = 10 hours as time cannot be in negative value.

Larger tap takes 10 hours

Smaller tap takes = y + 5 = 10 +5 = 15 hours

The final answer is that the larger tank takes a total of 10 hours and the smaller tank takes 15 hours to fill the tank separately.