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

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Manager
Joined: 18 Jul 2017
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At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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30 Jun 2018, 06:50
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Difficulty:

35% (medium)

Question Stats:

68% (01:46) correct 32% (02:10) wrong based on 102 sessions

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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%
Manager
Status: Student
Joined: 07 Jun 2018
Posts: 61
Location: India
Concentration: Finance, Marketing
GPA: 4
Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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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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30 Jun 2018, 18:46
1
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

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

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Manager
Joined: 07 Feb 2017
Posts: 183
Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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

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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%

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Intern
Joined: 06 Apr 2019
Posts: 12
Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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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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Manager
Joined: 09 Jun 2017
Posts: 107
GMAT 1: 640 Q44 V35
Re: At a certain industrial manufacturer, an open tank contains 6% fluorin  [#permalink]

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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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