Can you help please?

Here is a question and answer with the solution.

A large fresh water reservoir has two types of drainage system. Small pipes and large pipes. 6 large pipes, on their own, can drain the reservoir in 12 hours. 3 large pipes and 9 small pipes, at the same time, can drain the reservoir in 8 hours.
How long will 5 small pipes, on their own, take to drain the reservoir?

21 hours and 36 minutes. The exact size of the pipes and reservoir does not matter. If we label the large pipes L, the small pipes S and if the reservoir has a total of T litres,

then:
T / 6L = 12 which means L = T / 72
and
T / (3L + 9S) = 8 which means that S = T / 108
we want
T / 5S which gives 108 / 5 hours = 21.6 hours = 21 hours and 36 minutes.

How do we get S = T / 108 from T / (3L + 9S) = 8?

Thanks
geecee

Hi geecee,

Since T / (3L + 9S) = 8 you have 3L + 9S = T/8. But L = T/72 and hence

3L + 9S = T/8
3(T/72) + 9S = T/8
T/24 + 9S = T/8
9S = T/8 - T/24 = 3T/24 - T/24 = T/12 and thus
S = T/(9 12) = T/108

I solved this problem in a different way. L is the number of litres per hour that flow through a large pipe and S is the number of litres per hour that flow through a small pipe. You are given

T/6L = 12 or T = 72 L, and
T/(3L + 9S) = 8 or 3L + 9S = T/8

Substitute L = T/72 into the later expression to get

3L + 9S = 72L/8 = 9L so
9S = 6L

Thus 9 small pipes will drain the water at the same rate as 6 large pipes. But you want 5 small pipes.

5S = 5/9 (9S) = 5/9 (6L) = 10/3 L

Thus the time required for 5 small pipes to drain the tank is

T/5S = T/(10/3 L) = 72 L/(10/3 L) = 216/10 = 21.6 hours.

Penny