November 2020
Problem of the Month

Filling A Cistern
Submitted by Brijesh Dave

Pipe S can fill a cistern in 2 hours by itself and pipe K can fill a cistern in 3 hours.

Pipe Y can empty it in 5 hours.

Suppose that all the pipes are turned on at the same time when the cistern is completely empty.

How long will it take to fill?

Solution to the Problem:

It will take 30/19 hours   or
  1   11/19 hours   or
  1 hour 34 minutes 44 seconds.

If Pipe S can fill the cistern in 2 hours, then its rate would be 1/2 work per hour.

If Pipe K can fill the cistern in 3 hours, then its rate would be 1/3 work per hour.

If Pipe Y can empty the cistern in 5 hours, then its rate would be -1/5 work per hour.

Let x = number of hours for to fill the cistern if all three pipes are turned on.

Then (1/2)x + (1/3)x - (1/5)x = 1.

1 represents the whole job or the work to be done.

Multiplying by a common denominator of 30:

15x + 10x - 6x = 30

So, 19x = 30

x = 30/19 hours.

Correctly solved by:

1. Brijesh Dave	Mumbai City, Maharashtra, India
2. Colin (Yowie) Bowey	Beechworth, Victoria, Australia
3. Veena Mg	Bangalore, Karnataka, India
4. Ivy Joseph	Pune, Maharashtra, India
5. Jayda Kenison	Mountain View High School, Mountain View, Wyoming
6. Mishan Kasiparsad	Johannesburg, South Africa
7. James Alarie	Flint, Michigan
8. John Simmons	Memphis, Tennessee

Send any comments or questions to: David Pleacher