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: Let Time Taken by Larger Pipe
Let larger pipe take x hours.
Then smaller pipe takes x + 10 hours.

Rate of work:
Large pipe = 1/x
Small pipe = 1/(x + 10)

Given condition:
4/x + 9/(x + 10) = 1/2

Step 2: Take LCM
[4(x + 10) + 9x] / [x(x + 10)] = 1/2

Simplify numerator:
4x + 40 + 9x = 13x + 40

So,
(13x + 40) / (x² + 10x) = 1/2

Step 3: Cross Multiply
2(13x + 40) = x² + 10x

26x + 80 = x² + 10x

Bring all terms to one side:
x² − 16x − 80 = 0

Factorise:
x² − 20x + 4x − 80 = 0
(x − 20)(x + 4) = 0

x = 20 or x = −4

Since time cannot be negative,
x = 20

Step 4: Find Time for Smaller Pipe
Smaller pipe = x + 10
= 20 + 10
= 30 hours

Final Answer:
Larger pipe takes 20 hours.
Smaller pipe takes 30 hours.