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Pipe A can fill the tank in 3hrs and pipe B can fill the tank in 4hrs. An outlet pipe C can empty the tank in 6 hrs. If all pipes are kept open simultaneously, in how many hours will the tank be half full?
- a)6/5
- b)12/5
- c)17/5
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Pipe A can fill the tank in 3hrs and pipe B can fill the tank in 4hrs....
Tank capacity = 12 L
A + B → 7L in 1 min
C → -2L in 1 min
So Net result = 5L in 1min
So, time taken = 12/5
A + B → 7L in 1 min
C → -2L in 1 min
So Net result = 5L in 1min
So, time taken = 12/5
Most Upvoted Answer
Pipe A can fill the tank in 3hrs and pipe B can fill the tank in 4hrs....
Understanding the Problem
To solve the problem, we first need to determine the rates at which each pipe fills or empties the tank.
Rates of Pipes
- Pipe A fills the tank in 3 hours:
- Rate of A = 1 tank / 3 hours = 1/3 tank per hour.
- Pipe B fills the tank in 4 hours:
- Rate of B = 1 tank / 4 hours = 1/4 tank per hour.
- Pipe C empties the tank in 6 hours:
- Rate of C = 1 tank / 6 hours = 1/6 tank per hour (negative rate since it empties).
Combined Rate of All Pipes
When all pipes are open, we can combine their rates:
- Combined Rate = Rate of A + Rate of B - Rate of C
- Combined Rate = (1/3 + 1/4 - 1/6)
Finding a Common Denominator
To add these fractions, we need a common denominator. The least common multiple of 3, 4, and 6 is 12.
- Rate of A in terms of 12 = 4/12
- Rate of B in terms of 12 = 3/12
- Rate of C in terms of 12 = 2/12 (negative)
Combining these gives:
- Combined Rate = (4/12 + 3/12 - 2/12) = (5/12) tank per hour.
Time to Fill Half the Tank
To find the time taken to fill half the tank:
- Time = Amount of tank / Rate = (1/2) / (5/12) = (1/2) * (12/5) = 12/10 = 6/5 hours.
Thus, the tank will be half full in 12/5 hours, which corresponds to option 'B'.
To solve the problem, we first need to determine the rates at which each pipe fills or empties the tank.
Rates of Pipes
- Pipe A fills the tank in 3 hours:
- Rate of A = 1 tank / 3 hours = 1/3 tank per hour.
- Pipe B fills the tank in 4 hours:
- Rate of B = 1 tank / 4 hours = 1/4 tank per hour.
- Pipe C empties the tank in 6 hours:
- Rate of C = 1 tank / 6 hours = 1/6 tank per hour (negative rate since it empties).
Combined Rate of All Pipes
When all pipes are open, we can combine their rates:
- Combined Rate = Rate of A + Rate of B - Rate of C
- Combined Rate = (1/3 + 1/4 - 1/6)
Finding a Common Denominator
To add these fractions, we need a common denominator. The least common multiple of 3, 4, and 6 is 12.
- Rate of A in terms of 12 = 4/12
- Rate of B in terms of 12 = 3/12
- Rate of C in terms of 12 = 2/12 (negative)
Combining these gives:
- Combined Rate = (4/12 + 3/12 - 2/12) = (5/12) tank per hour.
Time to Fill Half the Tank
To find the time taken to fill half the tank:
- Time = Amount of tank / Rate = (1/2) / (5/12) = (1/2) * (12/5) = 12/10 = 6/5 hours.
Thus, the tank will be half full in 12/5 hours, which corresponds to option 'B'.
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