Water enters a distribution network at the source S and leaves only through three outlets T1, T2 and T3. The network has four junctions A, B, C and D, and ten one-way pipes:
S→A, S→B, A→B, A→C, B→C, B→D, C→T1, C→T2, D→T2 and D→T3.
At every junction the total inflow equals the total outflow (no water is stored or lost). All flows are measured in litres per minute (L/min). The following facts are known about the network in its normal state:
1. 100 L/min of water enters the network at S.
2. The flow in S→A is 1.5 times the flow in S→B.
3. The flow in A→C is three times the flow in A→B.
4. Outlet T1 receives 40 L/min and outlet T3 receives 25 L/min.
5. The flow in C→T1 is 15 L/min more than the flow in C→T2.
S→A, S→B, A→B, A→C, B→C, B→D, C→T1, C→T2, D→T2 and D→T3.
At every junction the total inflow equals the total outflow (no water is stored or lost). All flows are measured in litres per minute (L/min). The following facts are known about the network in its normal state:
1. 100 L/min of water enters the network at S.
2. The flow in S→A is 1.5 times the flow in S→B.
3. The flow in A→C is three times the flow in A→B.
4. Outlet T1 receives 40 L/min and outlet T3 receives 25 L/min.
5. The flow in C→T1 is 15 L/min more than the flow in C→T2.
Question 1
Numerical answer
What is the flow (in L/min) in pipe B→D?
Show answer and solution
Answer: 35
Work junction by junction.
- From facts 1 and 2: S→A = 60, S→B = 40.
- At A: A→B + A→C = 60 and A→C = 3·A→B → A→B = 15, A→C = 45.
- T1 is fed only by C, so C→T1 = 40; then C→T2 = 25 (fact 5).
- At C: inflow = A→C + B→C = C→T1 + C→T2 = 65 → B→C = 20.
- At B: inflow = S→B + A→B = 55 = B→C + B→D → B→D = 35.
- At D: D→T3 = 25 (only D feeds T3), so D→T2 = 10.
- From facts 1 and 2: S→A = 60, S→B = 40.
- At A: A→B + A→C = 60 and A→C = 3·A→B → A→B = 15, A→C = 45.
- T1 is fed only by C, so C→T1 = 40; then C→T2 = 25 (fact 5).
- At C: inflow = A→C + B→C = C→T1 + C→T2 = 65 → B→C = 20.
- At B: inflow = S→B + A→B = 55 = B→C + B→D → B→D = 35.
- At D: D→T3 = 25 (only D feeds T3), so D→T2 = 10.
Question 2
Single correct option
How much water (in L/min) does outlet T2 receive?
- A. 25
- B. 30
- C. 35
- D. 40
Show answer and solution
Answer: (C) 35
T2 is fed by C→T2 and D→T2.
C→T2 = 40 − 15 = 25 (fact 5 with C→T1 = 40).
D→T2 = B→D − D→T3 = 35 − 25 = 10.
T2 receives 25 + 10 = 35 L/min.
Quick check: Total outflow must be 100 → T2 = 100 − 40 − 25 = 35 ✓. This check alone answers the question without any junction work.
C→T2 = 40 − 15 = 25 (fact 5 with C→T1 = 40).
D→T2 = B→D − D→T3 = 35 − 25 = 10.
T2 receives 25 + 10 = 35 L/min.
Quick check: Total outflow must be 100 → T2 = 100 − 40 − 25 = 35 ✓. This check alone answers the question without any junction work.
Question 3
Single correct option
Which pipe carries the second-highest flow in the network?
- A. S→B
- B. A→C
- C. B→D
- D. C→T1
Show answer and solution
Answer: (B) A→C
The complete flow table (L/min):
S→A 60, S→B 40, A→B 15, A→C 45, B→C 20, B→D 35, C→T1 40, C→T2 25, D→T2 10, D→T3 25.
In descending order: S→A (60), A→C (45), S→B = C→T1 (40), B→D (35), …
So A→C carries the second-highest flow. S→B and C→T1 tie for third.
S→A 60, S→B 40, A→B 15, A→C 45, B→C 20, B→D 35, C→T1 40, C→T2 25, D→T2 10, D→T3 25.
In descending order: S→A (60), A→C (45), S→B = C→T1 (40), B→D (35), …
So A→C carries the second-highest flow. S→B and C→T1 tie for third.
Question 4
Numerical answer
Suppose the inflow at S is increased by 20%, while facts 2, 3 and 5 continue to hold and the outlets T1 and T3 continue to receive exactly 40 L/min and 25 L/min respectively. If every pipe can safely carry at most 50 L/min, how many pipes would carry more than their safe limit?
Show answer and solution
Answer: 3
New inflow = 120 L/min.
- S→A = 72, S→B = 48 (ratio 3 : 2).
- A→B = 18, A→C = 54 (ratio 1 : 3).
- C→T1 = 40, C→T2 = 25 (unchanged), so C's inflow is 65 → B→C = 65 − 54 = 11.
- At B: 48 + 18 = 66 = 11 + B→D → B→D = 55.
- At D: D→T3 = 25 → D→T2 = 30.
Pipes above 50 L/min: S→A (72), A→C (54), B→D (55) → 3 pipes.
All the extra 20 L/min ends up at T2 (now 55), routed through B→D.
- S→A = 72, S→B = 48 (ratio 3 : 2).
- A→B = 18, A→C = 54 (ratio 1 : 3).
- C→T1 = 40, C→T2 = 25 (unchanged), so C's inflow is 65 → B→C = 65 − 54 = 11.
- At B: 48 + 18 = 66 = 11 + B→D → B→D = 55.
- At D: D→T3 = 25 → D→T2 = 30.
Pipes above 50 L/min: S→A (72), A→C (54), B→D (55) → 3 pipes.
All the extra 20 L/min ends up at T2 (now 55), routed through B→D.
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