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# The concentration of a certain chemical in a full water tank

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Director
Joined: 17 Oct 2005
Posts: 833
The concentration of a certain chemical in a full water tank  [#permalink]

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Updated on: 02 Mar 2014, 04:36
1
4
00:00

Difficulty:

35% (medium)

Question Stats:

76% (02:14) correct 24% (02:57) wrong based on 278 sessions

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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

Originally posted by joemama142000 on 16 Nov 2005, 18:58.
Last edited by Bunuel on 02 Mar 2014, 04:36, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Director
Joined: 21 Aug 2005
Posts: 702

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16 Nov 2005, 19:39
3 + 4/sqrt(5-x) = 6

4/sqrt(5-x) = 3

16/(5-x) = 9

9x = 29 -> x = 3.2
Current Student
Joined: 28 Dec 2004
Posts: 2919
Location: New York City
Schools: Wharton'11 HBS'12

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16 Nov 2005, 19:48
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...
Director
Joined: 06 Jun 2004
Posts: 871
Location: CA

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16 Nov 2005, 19:49
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
Director
Joined: 21 Aug 2005
Posts: 702

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16 Nov 2005, 19:49
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
Current Student
Joined: 28 Dec 2004
Posts: 2919
Location: New York City
Schools: Wharton'11 HBS'12

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16 Nov 2005, 19:51
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
Director
Joined: 21 Aug 2005
Posts: 702

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16 Nov 2005, 19:57
Thumbs up! for a 'go' to gym!
Current Student
Joined: 28 Dec 2004
Posts: 2919
Location: New York City
Schools: Wharton'11 HBS'12

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16 Nov 2005, 20:00
1
gym it is...see you in an hr...

gsr wrote:
Thumbs up! for a 'go' to gym!
Manager
Joined: 08 Nov 2013
Posts: 166
GMAT 1: 710 Q43 V44
GMAT 2: 770 Q49 V46
GPA: 3
Re: The concentration of a certain chemical in a full water tank  [#permalink]

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01 Mar 2014, 13:11
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]

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02 Mar 2014, 04:40
2
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?

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

Hope it's clear.
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Re: The concentration of a certain chemical in a full water tank  [#permalink]

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29 Sep 2015, 03:32
I've been trying to wrap my head around how to set up that equation, but am not able to. Could someone please help me out with why I should be equating 6ppm with the concentration at x feet?
Intern
Joined: 16 Oct 2017
Posts: 37
Re: The concentration of a certain chemical in a full water tank  [#permalink]

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10 Feb 2018, 15:46
Can you show me step-by-step how to solve this one? I'm missing something...
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The concentration of a certain chemical in a full water tank  [#permalink]

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10 Feb 2018, 20:37
1
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 $$3 + \frac{4}{\sqrt{5-x}}$$ 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

OCDianaOC wrote:
Can you show me step-by-step how to solve this one? I'm missing something...

OCDianaOC - This question's wording is not easy. I rephrased it.

Given: a chemical concentration of 6
Given: a formula that will tell how deep the water is at a particular concentration, IF we have the concentration (we do)
Formula: $$3 + \frac{4}{\sqrt{5-x}}$$

Set the formula equal to concentration. The concentration of 6, in tandem with the formula, will yield depth.

$$3 + \frac{4}{\sqrt{5-x}} = 6$$
Subtract 3 from both sides:

$$\frac{4}{\sqrt{5-x}} = 3$$
Square both sides:

$$(\frac{4^2}{(\sqrt{5-x})^2}) = 3^2$$

$$(\frac{16}{(5-x)}) = 9$$
Multiply both sides by denominator and solve:

$$16 = 9(5 - x)$$
$$16 = 45 - 9x$$
$$9x = 29$$

$$x = \frac{29}{9} = 3.2$$ feet

Hope that helps.

**The language "where 0 < x < 4," is mostly irrelevant. It's to keep the denominator in the formula positive. (Can't divide by zero, can't take the square root of a negative number.)
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Posts: 37
Re: The concentration of a certain chemical in a full water tank  [#permalink]

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10 Feb 2018, 20:49
1
generis wrote:
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 $$3 + \frac{4}{\sqrt{5-x}}$$ 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

OCDianaOC wrote:
Can you show me step-by-step how to solve this one? I'm missing something...

OCDianaOC - This question's wording is not easy. I rephrased it.

Given: a chemical concentration of 6
Given: a formula that will tell how deep the water is at a particular concentration, IF we have the concentration (we do)
Formula: $$3 + \frac{4}{\sqrt{5-x}}$$

Set the formula equal to concentration. The concentration of 6, in tandem with the formula, will yield depth.

$$3 + \frac{4}{\sqrt{5-x}} = 6$$
Subtract 3 from both sides:

$$\frac{4}{\sqrt{5-x}} = 3$$
Square both sides:

$$(\frac{4^2}{(\sqrt{5-x})^2}) = 3^2$$

$$(\frac{16}{(5-x)}) = 9$$
Multiply both sides by denominator and solve:

$$16 = 9(5 - x)$$
$$16 = 45 - 9x$$
$$9x = 29$$

$$x = \frac{29}{9} = 3.2$$ feet

Hope that helps.

**The language "where 0 < x < 4," is mostly irrelevant. It's to keep the denominator in the formula positive. (Can't divide by zero, can't take the square root of a negative number.)

YAAAAY! Your explanation is perfect. I finally understand how to solve this now!

Thank you!
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Joined: 03 Jun 2019
Posts: 1500
Location: India
Re: The concentration of a certain chemical in a full water tank  [#permalink]

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21 Aug 2019, 05:07
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 $$3 + \frac{4}{\sqrt{5-x}}$$ 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

Given: 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.

Asked: To the nearest 0.1 foot, at what depth is the concentration equal to 6 parts per million?

6 = $$3 + \frac{4}{\sqrt{5-x}}$$
$$\sqrt{4}{\sqrt{5-x}} = 3$$
5-x = 16/9 = 1.8 to the nearest .1 foot
x = 5 -1.8 = 3.2 to the nearest .1 foot

IMO E
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Re: The concentration of a certain chemical in a full water tank   [#permalink] 21 Aug 2019, 05:07
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