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?
Correct Answer: 6/11
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:
Total rate = 1/30 + 1/20 + 1/10 = (2 + 3 + 6)/60 = 11/60Rate(C) = 1/10 = 6/60Fraction of R = (6/60) / (11/60) = 6/11Verification / 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