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Manager  Joined: 02 Dec 2012
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A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%
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A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

"The remaining solution will be approximately what percent sodium chloride?" means: what percent of the remaining solution is sodium chloride. Now, since the remaining solution is 10,000-2,500=7,500 gallons and sodium chloride is 500 gallons (5% of initial solution of 10,000 gallons) then sodium chloride is 500/7,500*100=~6.66% of the remaining solution of 7,500 gallons.

Hope it's clear.
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

5% of 10,000 = 500.

required %age = 500 * 100/7500 = 6.67%
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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I have a question.

Are we supposed to assume none of the sodium chloride was evaporated from the same original 10000 Gallons.

Hence of the 500 Gallons of original Sodium Chloride wouldn't the amount of sodium chloride in the solution also decrease by the same rate as the solution without Sodium Chloride?
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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hfbamafan wrote:
I have a question.

Are we supposed to assume none of the sodium chloride was evaporated from the same original 10000 Gallons.

Hence of the 500 Gallons of original Sodium Chloride wouldn't the amount of sodium chloride in the solution also decrease by the same rate as the solution without Sodium Chloride?

We are told that from 10,000 gallons of a solution, evaporated 2,500 gallons of water. So, salt did not evaporate.

If it were the way you suggest (if salt evaporated at the same rate as the water) then the answer would simply be 5%.
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Bunuel wrote:
hfbamafan wrote:
I have a question.

Are we supposed to assume none of the sodium chloride was evaporated from the same original 10000 Gallons.

Hence of the 500 Gallons of original Sodium Chloride wouldn't the amount of sodium chloride in the solution also decrease by the same rate as the solution without Sodium Chloride?

We are told that from 10,000 gallons of a solution, evaporated 2,500 gallons of water. So, salt did not evaporate.

If it were the way you suggest (if salt evaporated at the same rate as the water) then the answer would simply be 5%.

Ok, I assumed that the two solutions were mixed in the same container.

Thank you Bunuel
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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This is a inverse proportional problem.

With the salt quantity remaining constant, if the water decreases, it means the percentage of salt increases.

$$\frac{10000}{7500} * 5$$

$$= \frac{20}{3}$$

= 6.67%

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GMAT 1: 620 Q42 V33 Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Bunuel, even though the answer doesn't change, I still don't grasp how WATER = SOLUTION evaporated.

I have 10,000 gallons of both: NaCl and Water.
5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind it does.

I don´t know if the problem stem were to change a bit, maybe my explanation might help.
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Enael wrote:
Bunuel, even though the answer doesn't change, I still don't grasp how WATER = SOLUTION evaporated.

I have 10,000 gallons of both: NaCl and Water.
5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind it does.

I don´t know if the problem stem were to change a bit, maybe my explanation might help.

What contradiction do you see between you solution and mine?
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GMAT 1: 620 Q42 V33 Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Bunuel wrote:
Enael wrote:
Bunuel, even though the answer doesn't change, I still don't grasp how WATER = SOLUTION evaporated.

I have 10,000 gallons of both: NaCl and Water.
5% of NaCl = 500 gallons so water HAS to be 9,500 gallons.

It explicitly states that 2,500 gallons of water are evaporated, thus our base reduced should be 9,500 - 2,500 = 7,000 gallons of water.

So: NaCl / (NaCl+Water) = 500/(7000+500) = 1/15 = almost 7%.

As said, the answer doesn't change but the concept behind it does.

I don´t know if the problem stem were to change a bit, maybe my explanation might help.

What contradiction do you see between you solution and mine?

Just read it again, and indeed it is the same. The difference is that I separate both components first, and then subtract 2,500 from Water. You do it the other way around. It should work both ways.

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GMAT 1: 710 Q49 V38 Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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How do you convert 1/15 to decimals quickly during the test? Posted from GMAT ToolKit
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Sammyh21,

Have you tried doing the math by hand? There's a pattern in the calculation that shouldn't take you too long to find; by extension, the math probably won't take you as long as you might think.

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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Initial amount of water must be 95% of total quantity (9500 gallons); as it's given that 5% (500 gallons) is sodium chloride. lf 2500 gallons of water evaporates, remaining solution should have 7000 gallons of water and 500 gallons of sodium chloride. Then we should have something like: 500/7000, which is clearly wrong.

Could someone please correct where l'm wrong? Regards.
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Initial amount of water must be 95% of total quantity (9500 gallons); as it's given that 5% (500 gallons) is sodium chloride. lf 2500 gallons of water evaporates, remaining solution should have 7000 gallons of water and 500 gallons of sodium chloride. Then we should have something like: 500/7000, which is clearly wrong.

Could someone please correct where l'm wrong? Regards.

The remaining solution will be approximately what percent sodium chloride?

The remaining solution = 7000 gallons of water + 500 gallons of sodium chloride = 7500 gallons.

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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Initial amount of water must be 95% of total quantity (9500 gallons); as it's given that 5% (500 gallons) is sodium chloride. lf 2500 gallons of water evaporates, remaining solution should have 7000 gallons of water and 500 gallons of sodium chloride. Then we should have something like: 500/7000, which is clearly wrong.

Could someone please correct where l'm wrong? Regards.

Hope this helps -

Attachment: Capture.PNG [ 3.75 KiB | Viewed 9117 times ]

Thus the required percentage = $$\frac{500}{7500}$$x$$100$$ => 6.67% _________________
Thanks and Regards

Abhishek....

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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

We start with 10,000 gallons of a solution that is 5% sodium chloride by volume. This means that there are 0.05 x 10,000 = 500 gallons of sodium chloride.

When 2,500 gallons of water evaporate we are left with 7,500 gallons of solution. From here we can determine what percent of the 7,500 gallon solution is sodium chloride.

(sodium chloride/total solution) x 100 = ?

(500/7,500) x 100 = ?

5/75 x 100 = ?

1/15 x 100 = ?

100/15 = 20/3 = 6 2/3 = 6.67%

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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Sammyh21 wrote:
How do you convert 1/15 to decimals quickly during the test? Posted from GMAT ToolKit

$$\frac{1}{15} = \frac{1}{3}*\frac{1}{5}$$ = 0.333~ * 0.2 = 0.6667~
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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Bunuel wrote:
A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

"The remaining solution will be approximately what percent sodium chloride?" means: what percent of the remaining solution is sodium chloride. Now, since the remaining solution is 10,000-2,500=7,500 gallons and sodium chloride is 500 gallons (5% of initial solution of 10,000 gallons) then sodium chloride is 500/7,500*100=~6.66% of the remaining solution of 7,500 gallons.

Hope it's clear.

Why did you take out 2500 from the 10000 gallons? The question said 2500 gallons of water, so why not 9500-2500, since 500g were sodium?
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Re: A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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guilherme28 wrote:
Bunuel wrote:
A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

"The remaining solution will be approximately what percent sodium chloride?" means: what percent of the remaining solution is sodium chloride. Now, since the remaining solution is 10,000-2,500=7,500 gallons and sodium chloride is 500 gallons (5% of initial solution of 10,000 gallons) then sodium chloride is 500/7,500*100=~6.66% of the remaining solution of 7,500 gallons.

Hope it's clear.

Why did you take out 2500 from the 10000 gallons? The question said 2500 gallons of water, so why not 9500-2500, since 500g were sodium?

The solution is 10,000 gallons (9,500 gallons of water and 500 gallons of sodium chloride) out of which 2,500 gallons of water evaporates and we are left with 7,000 gallons of water and 500 gallons of sodium chloride, so total of 7,500 gallons.
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A tank contains 10,000 gallons of a solution that is 5  [#permalink]

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A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

Conceptually, it helps to remember that unless told otherwise, water is assumed to have zero percent sodium chloride (or zero percent alcohol, etc.).

The prompt refers to a solution that is 5 percent sodium chloride by volume. It's part water, part sodium chloride.

To make these numbers easier to work with, divide each original volume total by 100.

Let A = 100 gallons of 5 percent (=.05) sodium chloride solution

Let B = 25 gallons of water with zero percent (=0.0) sodium chloride

Resultant solution is A - B

(Concentration A)(Volume A) - (Concentration B)(Volume B) = (Resultant concentration of A - B)(Volume of resultant solution A - B)

In the equation below, arithmetically, LHS doesn't need the second term. It's just there to show that, in weighted average terms, it is being subtracted (it evaporates).

(.05)(100) - (0.0)(25) = $$\frac{x}{100}$$(100 - 25)

5 - 0 = $$\frac{75}{100}$$x

5 = $$\frac{3}{4}$$x

$$\frac{4}{3}$$*(5) = x

x = $$\frac{20}{3}$$ = 6.67

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