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At a certain industrial manufacturer, an open tank contains 6% fluorin

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At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 30 Jun 2018, 06:50
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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%
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Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 30 Jun 2018, 09:13
Quantity of fluorine - 6% of 1000= 60 gallons

10 gallons of water evaporate everyday
So in 25 days - 250 gallons of water evaporates
remaining quantity -1000-250=750
concentration of fluorine after 25 days - (60/750 )*100=8% Ans C

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At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 30 Jun 2018, 18:46
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wishalsp333 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%

Track on fluorine.

1) Not much algebra
1,000 gallons liquid
6% fluorine = (.06 * 1,000) = 60 gals fluorine

Water evaporates. Water is 0.0% fluorine
NO EFFECT on quantity of fluorine

10 gals of water/day evaporate for 25 days: 250 gallons of liquid are gone

(1,000 - 250) = 750 gals liquid remain

Percent fluorine after =\(\frac{60}{750}=\)
\(\frac{6}{75}=.08*100\)= 8 percent

Answer C

2) Algebra
Let x/100 = percent fluorine

Water is 0.0% fluorine

Water evaporation goes on the RHS in this equation. Water decrease = fluorine concentration increase

\(.06(1,000)=\frac{x}{100}*(1,000 - 250)\)
\(60=\frac{750}{100}x\)
\(60=7.5x\)
\(x=\frac{60}{7.5}=8\) percent

(Percent is already accounted for)

Answer C
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Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 30 Jun 2018, 19:03
60/1000
60/750
Answer C
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Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 04 Apr 2019, 18:39
wishalsp333 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%


At the beginning, there are 0.06 x 1000 = 60 gallons of fluorine. After 25 days, there are 1000 - 10 x 25 = 1000 - 250 = 750 gallons of liquid left, but there are still 60 gallons of fluorine since only water evaporates. Therefore, the concentration of fluorine after 25 days is:

60/750 = 6/75 = 2/25 = 8/100 = 8%

Answer: C
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Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 08 Apr 2019, 03:37
If the question asked was as
10% of water evaporates everyday then after 25 days concentration?

If this type of question can be asked in GMAT?

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Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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New post 10 Apr 2019, 03:22
guys , what if they say that 10 gallons evaporate every day (and didn't say as water )
just curious
appreciate any explanation
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Re: At a certain industrial manufacturer, an open tank contains 6% fluorin   [#permalink] 10 Apr 2019, 03:22
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