college word problems

Question

swimming pool 19.7 feet wide and 29.5 feet long It is an average of 5 feet deep , some water evaporates and Jill needs to add 10 inches of water to the depth of the pool using her garden hose Garden hose runs 10 gallons per min How many liters or gallons is this how long does hose need to run ?

Answer 1

Angela:

 

Since the pool is rectangular in shape, we know that the Volume=Length*Width*Depth.

 

In this problem, we know that 10 inches needs to be added to the depth of the pool (note that 10 inches is equal to 10/12 ft or 5/6 ft).  Hence, the total volume of water which needs to be added using the hose is:

 

Volume to be added = 5/6 ft * 19.7 ft * 29.5 ft = 484.29 cubic feet

 

We also know that there are 7.48 gallons in a cubic foot.  We can therefore convert the answer to gallons, as follows:

 

Volume to be added =484.29 cubic feet * 7.48 gallons per cubit foot = 3622.5 gallons of water to be added

 

Since the hose spews water  at 10 gallons per minute, the total time to generate the 3622.5 gallons needed to fill the pool is:

 

3622.5 gallons / 10 gallons per minute = 362.25 minutes

 

This can be converted to hours, knowing that there are 60 minutes in an hour, as follows:

 

362.25 minutes / 60 minutes per hour = 6.0 hours (rounding to nearest tenth of an hour).

 

Hope this helps!

 

George T.

Answer 2

Volume of the pool when full = l * w * h (or depth)

 

V = 29.5 * 19.7 * 5 = 2905.75 ft³

 

1 Gallon = 3.78541 L

 

 

Flow rate = 10 g/min * 3.78541 L/g = 37.8541 L/min  * 1000mL/1L = 37854.1 mL/min = 37854.1 cm³/min

 

 

1 cm³ = 3.53146667 × 10^-5 cubic foot

 

37854.1 cm3/min * 0.0000353146667 ft³/1cm³ =  1.33680 ft³/min

 

 

How much water was lost in ft³?

 

10 inches = 5/6 ft.

 

V to replace = L*W* (depth lost) = 29.5 * 19.7 * (5/6) = 484.2917 ft³

 

 

How many minutes will it take at 1.33680 ft³/min to replace the 484.2917 ft³?

 

484.2917ft³/(1.33680 ft³/min)  = 362.28 min

 

6 hours and 2.28 minutes.

 

Answer 3

Information that you need to look up:

 

1 cubic foot = 7.5 gallons

12 inches = 1 foot

 

Amount of water needed to be added to increase depth by 10 inches:

 

First in cubic feet:

 

19.7 ft  X  29.5 ft  X  (10 in / (12 in / ft) )  =  484.3 ft³

 

Then convert to gallons:

 

484.3 ft³ / (1 ft³ / 7.5 gallons)  =  3632 gallons

 

Then how long does the hose need to run?

 

3632 gallons / (10 gallons / minute)  =  363.2 minutes  =  6 hours

 

Significant Figures: I did not follow sig. fig. rules. With calculators these are not as important as when calculations were done on slide rulers or by long hand.

 

 

 

 

Answer 4

Hi,

 

I agree with George T.  If you calculate this with 4 digit precision all the way, and do not round, then you get 6 hours and 2 minutes.  I doubt if your teacher cares about such detail but if you were filling a tank of crude oil, it would be important :-)