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21 Mar 2025, 00:30
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Difficulty:
45%
(medium)
Question Stats:
66% (01:49) correct
34%
(02:12)
wrong
based on 181
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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%
A. 4%
B. 6%
C. 8%
D. 10%
E. 15%
21 Mar 2025, 01:48
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Here the amount of flourine in 1000 galloons is 0.06 * 1000 = 60
amount of water is 1000 - 60 = 940
in 25 days, 25 * 10 gallons of pure water will be gone.
So Amount of water is 940 - 250 = 690
Amount of flourine remains the same = 60
% of flourine in the resulting solution is 60/(690+60) * 100 = 60/750 * 100 = 8%
amount of water is 1000 - 60 = 940
in 25 days, 25 * 10 gallons of pure water will be gone.
So Amount of water is 940 - 250 = 690
Amount of flourine remains the same = 60
% of flourine in the resulting solution is 60/(690+60) * 100 = 60/750 * 100 = 8%
21 Mar 2025, 02:21
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(6/100) *1000 =60 - fluorine amount
Water given vol. 1000 gallons
25*10=250 ( evaporates in 10days)
Leftover amount of water after 10 days =1000-250
=750 gallons
Keeping fluorine volume same as only water gets evaporated
Con%( fluorine) = (60/750)*1000
=8%
Water given vol. 1000 gallons
25*10=250 ( evaporates in 10days)
Leftover amount of water after 10 days =1000-250
=750 gallons
Keeping fluorine volume same as only water gets evaporated
Con%( fluorine) = (60/750)*1000
=8%
