Question

The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to

• A. \(76\)
• B. \(78\)
• C. \(73\)
• D. \(71\)

Hint:

In pipes and cisterns problems:
Convert ratios into actual rates using any one known filling time.
Add rates (not times) when pipes work together.
Do all calculations in hours first, and convert to minutes only at the end to avoid mistakes.

Answer 1

The Correct Option is C

Approach: Work in pipe-A units instead of fractions. Since the rates are $4:9:36$, the three pipes together act like $\dfrac{49}{4}$ copies of pipe A $-$ so just divide A's solo time by that factor.

Step 1: A alone takes $15$ hours $=900$ minutes.

Step 2: Combined rate $=4+9+36=49$ "rate-units", and pipe A is $4$ rate-units. So all three together are $\dfrac{49}{4}$ times as fast as A alone.

Step 3: Combined time $=\dfrac{900}{49/4}=\dfrac{900\times4}{49}=\dfrac{3600}{49}\approx73.47$ minutes.

Step 4: Rounding to the nearest whole minute gives $73.$

Final answer: Nearest to $73$ minutes.