Question

the water from a swimming pool evaporates at a rate of 6gal/hr in shade and 19gal/hr in sun.for several weeks in august the amount of water lost in shade is equal to the amount lost in sun.what was the average rate of evaporation in the pool?

Answer 1

Answer: 9.12

Explanation:

r. This is like an uphill and downhill bicycle problems. The same

amount of water is lost in the sun, as in the shade. First, we should find a number that is

easy to work with as the number of gallons of water evaporated. This number should be the

LCM of 6 and 19 which is 114. Then we find how long it took to evaporate in the sun and in

the shade. For the sun, it took 114

19

= 6 hours to evaporate. In the shade it took 114

6

= 19

hours to evaporate. This gives us a total of 25 hours to evaporate 228 gallons of water. The

average rate of evaporation is 228

25

= 9.12 gallons per hour.

Answer 2

Let's denote the average rate of evaporation in the pool as R gallons per hour.

The rate of evaporation in the shade is given as 6 gallons per hour, and the rate in the sun is given as 19 gallons per hour. We are told that the amount of water lost in the shade is equal to the amount lost in the sun during these weeks in August.

So, we can set up the following equation:

Rate of evaporation in shade (6 gal/hr) * Time in shade (hours) = Rate of evaporation in sun (19 gal/hr) * Time in sun (hours)

Let's say the time in the shade is t hours, and the time in the sun is also t hours.

6 * t = 19 * t

Now, we can solve for t:

19t - 6t = 0

13t = 0

t = 0

Wait, something doesn't seem right. The equation implies that t is 0, which means there was no time spent in either the shade or the sun, which is not possible. This implies there may be an error in the provided information or question.

If there was a typo or an omission in the question, please provide the correct or complete information so that I can assist you further.