A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in $8$ hours. If the outlet pipe is open then the inlet pipe fills the empty tank in $10$ hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

We have a tank with two pipes:

Inlet pipe: Fills the tank

Outlet pipe: Empties the tank

Think of rates like speed - just as a car travels a certain distance per hour, pipes fill or empty a certain fraction of the tank per hour.

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Key Information Given:

Inlet pipe alone: Fills empty tank in 8 hours

Inlet pipe + Outlet pipe: Fills empty tank in 10 hours

Find: Time for outlet pipe alone to make full tank half-full

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Method 1: Using Work Rates

Inlet pipe rate = $\tfrac{1}{8}$ tank per hour (fills the tank)

Combined rate = $\tfrac{1}{10}$ tank per hour (both pipes working)

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When both pipes work together:

Inlet pipe adds water at rate $\tfrac{1}{8}$ per hour

Outlet pipe removes water at some rate (let's call it $\tfrac{1}{x}$ per hour)

Net result: $\tfrac{1}{10}$ tank filled per hour

So: Inlet rate - Outlet rate = Combined rate

$\tfrac{1}{8} - \tfrac{1}{x} = \tfrac{1}{10}$

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$\tfrac{1}{x} = \tfrac{1}{8} - \tfrac{1}{10}$

To subtract fractions, we find common denominator (LCM of 8 and 10 = 40):

$\tfrac{1}{x} = \tfrac{5}{40} - \tfrac{4}{40} = \tfrac{1}{40}$

Therefore: Outlet pipe rate = $\tfrac{1}{40}$ tank per hour

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If outlet pipe empties $\tfrac{1}{40}$ of tank per hour, then to empty $\tfrac{1}{2}$ of tank:

$\text{Time} = \frac{\text{Work to be done}}{\text{Rate}} = \frac{1/2}{1/40} = \frac{1}{2} \times 40 = 20 \text{ hours}$

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Method 2: Using Concrete Numbers (LCM Method)

Instead of working with fractions, we can assume the tank has a capacity equal to LCM(8,10) = 40 units. This makes calculations easier!

Tank capacity = 40 units

Inlet pipe rate = 40 ÷ 8 = 5 units per hour

Combined rate = 40 ÷ 10 = 4 units per hour

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Since: Inlet rate - Outlet rate = Combined rate

$5 - \text{Outlet rate} = 4$

$\text{Outlet rate} = 1 \text{ unit per hour}$

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To empty half tank (20 units) at rate of 1 unit per hour:

$\text{Time} = \frac{20}{1} = 20 \text{ hours}$

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Alternative Thinking:

Normal filling time: 8 hours

With outlet open: 10 hours (2 extra hours needed)

In those 2 extra hours, inlet pipe fills $\tfrac{2}{8} = \tfrac{1}{4}$ of tank

But this $\tfrac{1}{4}$ tank was drained by outlet pipe in 10 hours

So outlet pipe drains $\tfrac{1}{4}$ tank in 10 hours

To drain $\tfrac{1}{2}$ tank: $10 \times \frac{(1/2)}{(1/4)} = 10 \times 2 = 20$ hours

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The outlet pipe alone will take 20 hours to make the full tank half-full.

When dealing with pipes:

Filling rate = positive (adds to tank)

Emptying rate = negative (subtracts from tank)

Combined rate = Sum of individual rates (with proper signs)

This concept applies to all work-rate problems, not just pipes!