Mixture composition under simultaneous inflow Three pipes A, B and C fill a tank in 30 min, 20 min and 10 min respectively. When all are opened together from empty, A brings solution P, B brings Q, and C brings R. What fraction of R is present in the tank after 3 minutes of inflow?

Introduction / Context:
When multiple inlets add different solutions simultaneously, the instantaneous composition is proportional to the inlets’ rates. For a short interval where all run throughout, the mixture’s fraction of each solution equals that solution’s rate divided by the total rate.

Given Data / Assumptions:

• A: 30 min ⇒ 1/30 per min (solution P).
• B: 20 min ⇒ 1/20 per min (solution Q).
• C: 10 min ⇒ 1/10 per min (solution R).
• All three run for the entire 3 min.

Concept / Approach:
Fraction of R = rate(C) / (rate(A) + rate(B) + rate(C)). Time cancels out because all run for the same duration.

Step-by-Step Solution:

Verification / Alternative check:
In 3 min the tank has 3*(11/60) = 11/20 full. Of this, portion from C is 3*(1/10) = 3/10. (3/10) / (11/20) = 6/11.

Why Other Options Are Wrong:
8/11, 5/11, 7/11 are not the ratio of C’s rate to the total rate here.

Common Pitfalls:
Computing fraction using absolute volumes without dividing by the total mixed volume.

Final Answer:
6/11