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The concentration of a certain chemical in a full water tank

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Director
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The concentration of a certain chemical in a full water tank [#permalink] New post 16 Nov 2005, 17:58
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83% (02:23) correct 16% (02:55) wrong based on 31 sessions
The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is 3 + \frac{4}{\sqrt{5-x}} parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?

(A) 2.4 ft
(B) 2.5 ft
(C) 2.8 ft
(D) 3.0 ft
(E) 3.2 ft
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Mar 2014, 03:36, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Director
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 [#permalink] New post 16 Nov 2005, 18:39
3 + 4/sqrt(5-x) = 6

4/sqrt(5-x) = 3

16/(5-x) = 9

9x = 29 -> x = 3.2
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 [#permalink] New post 16 Nov 2005, 18:48
I was going to go with 3.0...but then the question says witin 0.1 feet...and with 3.0..we get within 0.2 feet...so I would have just picked 3.2 at that point....

anyway

4/sqrt(5-x)=3

16/(5-x)=9

16=45-9x

solve for X...
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Re: Problem Solving Proportion [#permalink] New post 16 Nov 2005, 18:49
joemama142000 wrote:
The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is 3 + 4/sqrt(5-x) parts per million, where 0< x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?

A) 2.4 ft
B) 2.5 ft
C) 2.8 ft
D) 3.0 ft
E) 3.2 ft


3 + 4/sqrt(5-x) = 6

solve for x, x = 29/9 = 3.2 ft

E
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 [#permalink] New post 16 Nov 2005, 18:49
fresinha12 wrote:
I was going to go with 3.0...but then the question says witin 0.1 feet...and with 3.0..we get within 0.2 feet...so I would have just picked 3.2 at that point....
anyway
4/sqrt(5-x)=3
16/(5-x)=9
16=45-9x
solve for X...


You are tired. Go to bed :sleeping:
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 [#permalink] New post 16 Nov 2005, 18:51
not yet...its raining here...and I want to go to the gym

to go or not to go to gym is the question...what would GSR do?hmm

gsr wrote:
fresinha12 wrote:
I was going to go with 3.0...but then the question says witin 0.1 feet...and with 3.0..we get within 0.2 feet...so I would have just picked 3.2 at that point....
anyway
4/sqrt(5-x)=3
16/(5-x)=9
16=45-9x
solve for X...


You are tired. Go to bed :sleeping:
Director
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 [#permalink] New post 16 Nov 2005, 18:57
Thumbs up! for a 'go' to gym!
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 [#permalink] New post 16 Nov 2005, 19:00
gym it is...see you in an hr...

gsr wrote:
Thumbs up! for a 'go' to gym!
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Re: The concentration of a certain chemical in a full water tank [#permalink] New post 01 Mar 2014, 12:11
My brain is cloudy, going to gym to clear it up....but how did you guys know to set the equation equal to 6? Once we get to the equation the math is easy but I did not know where to begin. Can someone explain?
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Re: The concentration of a certain chemical in a full water tank [#permalink] New post 02 Mar 2014, 03:40
Expert's post
PeterHAllen wrote:
The concentration of a certain chemical in a full water tank depends on the depth of the water. At a depth that is x feet below the top of the tank, the concentration is 3 + \frac{4}{\sqrt{5-x}} parts per million, where 0 < x < 4. To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?

(A) 2.4 ft
(B) 2.5 ft
(C) 2.8 ft
(D) 3.0 ft
(E) 3.2 ft

My brain is cloudy, going to gym to clear it up....but how did you guys know to set the equation equal to 6? Once we get to the equation the math is easy but I did not know where to begin. Can someone explain?


Given: at a depth that is x feet below the top of the tank, the concentration is 3 + \frac{4}{\sqrt{5-x}} parts per million.

Question: at what depth, for which x, is the concentration equal to 6 parts per million? So, for which x, is 3 + \frac{4}{\sqrt{5-x}} equal to 6?

3 + \frac{4}{\sqrt{5-x}}=6 --> x\approx{3.2}.

Answer: E.

Hope it's clear.
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Re: The concentration of a certain chemical in a full water tank   [#permalink] 02 Mar 2014, 03:40
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