Question

If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?
1. 12
2. 20
3. 15
4. 18

Answer 1

Correct Answer - Option 2 : 20

Given:

Two pipes working simultaneously can fill a tank in 12 hours.

One pipe fills the tank 10 hours faster than the other.

Calculation:

Let the slower pipe can fill the tank in x hours,

Then faster pipe can fill the tank in (x – 10) hours.

Tank filled by slower pipe in 1 hour = 1/x

Tank filled by faster pipe in 1 hour = 1/(x – 10)

Part filled by both the pipes in 1 hour = 1/12

According to the question

1/x + 1/(x – 10) = 1/12

⇒ [(x – 10) + x]/x(x – 10) = 1/12

⇒ 12(x – 10) + 12x = x(x – 10)

⇒ 12x – 120 + 12x = x² – 10x

⇒ x² – 34x + 120 = 0

⇒ x² – 30x – 4x + 120 = 0

⇒ x(x – 30) – 4(x – 30) = 0

⇒ (x – 30)(x – 4) = 0

⇒ x = 30, 4

But x = 4 is not possible since (x – 10) i.e. (4 – 10) will give a negative value

Hence slower pipe can fill the tank in 30 hours and

Faster pipe can fill the tank  in = 30 – 10 hours = 20 hours

∴ The faster pipe alone take to fill the tank in 20 hours