At a certain industrial manufacturer, an open tank contains

BTGmoderatorDC wrote:At a certain industrial manufacturer, an open tank contains 6% fluorine dissolved in water. If the tank holds 1000 gallons, and 10 gallons of the total volume evaporate as water each day, what is the concentration of fluorine after 25 days

A. 4%
B. 6%
C. 8%
D. 10%
E. 15%

Since 6% of the 1000-gallon tank is fluorine, the amount of fluorine = (6/100)(1000) = 60 gallons.
Since the total volume decreases at a rate of 10 gallons per day for 25 days, the decrease in the total volume = 10*25 = 250 gallons.
Remaining total volume = 1000-250 = 750 gallons.
Since 60 of the remaining 750 gallons will be fluorine, we get:
Resulting fluorine percentage = 60/750 * 100 = 8%.

The correct answer is C.

Alternate approach:
After 250 of the 1000 gallons evaporate, 3/4 of the total volume remains.
Since the total volume is multiplied by a factor of 3/4, the fluorine percentage is multiplied by a factor of 4/3:
(4/3)(6) = 8%.

$$Quality\ of\ fluorine=6\%\ of\ 1000gallons\ =60gallons$$
$$10gallons\ of\ water\ evaporates\ everyday$$
$$in\ 25days=25\cdot10=250gallons\ of\ water\ evaporates$$
$$the\ remaining\ quantity\ will\ be\ =1000-250=750gallons$$
$$concentration\ of\ the\ fluorine\ after\ 25days\ is\ $$
$$\frac{60}{750}\cdot100=0.08\cdot100=8\%$$
$$Answer=\ option\ C$$