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 |