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?

• A. 12
• B. 20
• C. 15
• D. 18

Answer 1

Answer:

B. 20

Solution: 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