Question

Two pipes are used to fill a swimming pool. If the pipe of the larger diameter is used for 4 hours and the pipe of the smaller diameter for 9 hours, only half of the pool can be filled. Find how long it would take for each pipe to fill the pool, separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.

Hint:

In work-rate problems, remember that "half filled" means the equation should be set equal to $1/2$, not $1$.

Answer 1

Step 1: Understanding the Concept:
Rate of work is the reciprocal of the time taken to complete the job. Rate = 1 / Time.
Step 2: Key Formula or Approach:
Let large pipe take $x$ hours. Small pipe takes $(x + 10)$ hours.
Work done in 1 hour: Large $= 1/x$, Small $= 1/(x+10)$.
Condition: $4/x + 9/(x+10) = 1/2$.
Step 3: Detailed Explanation:
\[ \frac{4(x + 10) + 9x}{x(x + 10)} = \frac{1}{2} \]
\[ \frac{13x + 40}{x^2 + 10x} = \frac{1}{2} \]
\[ 26x + 80 = x^2 + 10x \implies x^2 - 16x - 80 = 0 \]
Factoring:
\[ x^2 - 20x + 4x - 80 = 0 \implies (x - 20)(x + 4) = 0 \]
Since $x>0$, $x = 20$ hours.
Large pipe $= 20$ hours. Small pipe $= 20 + 10 = 30$ hours.
Step 4: Final Answer:
The larger pipe takes 20 hours and the smaller pipe takes 30 hours.