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A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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Question Stats: 64% (02:11) correct 36% (02:47) wrong based on 162 sessions

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A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) $$7\frac{1}{2}$$%
(B) $$9\frac{3}{8}$$%
(C) $$10\frac{1}{2}$$%
(D) $$12$$%
(E) $$15$$%

Project PS Butler : Question #28

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Posts: 3777
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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HKD1710 wrote:
A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) $$7\frac{1}{2}$$%
(B) $$9\frac{3}{8}$$%
(C) $$10\frac{1}{2}$$%
(D) $$12$$%
(E) $$15$$%

Project PS Butler : Question #28

-----------ASIDE------------
IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.
-------------------------------

Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution
Since the jug holds 5 liters, we then add 2.5 liters of water (aka a 0% salt solution)

Since we're combining EQUAL VOLUMES of 15% and 0% solutions, the salt content of the resulting solution will equal the average of 15% and 0%
(15 + 0)/2 = 7.5

So, the resulting solution will be 7.5% salt.

RELATED VIDEO FROM OUR COURSE

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General Discussion
Intern  Joined: 15 Nov 2018
Posts: 13
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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As a person that is bad at math I did this:

Original mixture
Total = 40ml
Salt = 40 * 0.15 = 6ml
Water = 40 - 6 = 34ml

After spill
Total = 40 - 15 = 25
Salt = 6ml - (15 * 0.15) = 3.75ml

After refill
Total = 50ml
Salt = 3.75 (unchanged)

3.75/50 = 7.5% [A]

Let me know if I did this wrong.
VP  D
Joined: 09 Mar 2016
Posts: 1279
A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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GMATPrepNow wrote:
HKD1710 wrote:
A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) $$7\frac{1}{2}$$%
(B) $$9\frac{3}{8}$$%
(C) $$10\frac{1}{2}$$%
(D) $$12$$%
(E) $$15$$%

Project PS Butler : Question #28

-----------ASIDE------------
IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.
-------------------------------

Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution
Since the jug holds 5 liters, we then add 2.5 liters of water (aka a 0% salt solution)

Since we're combining EQUAL VOLUMES of 15% and 0% solutions, the salt content of the resulting solution will equal the average of 15% and 0%
(15 + 0)/2 = 7.5

So, the resulting solution will be 7.5% salt.

RELATED VIDEO FROM OUR COURSE

nice explanation Brent, but what if volumes of mixtures arent equal ? any advice is appreciated VP  P
Joined: 07 Dec 2014
Posts: 1198
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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HKD1710 wrote:
A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) $$7\frac{1}{2}$$%
(B) $$9\frac{3}{8}$$%
(C) $$10\frac{1}{2}$$%
(D) $$12$$%
(E) $$15$$%

let x=% salt in final solution
2.5*.15=5x
x=.075=7 1/2%
A
CEO  V
Joined: 12 Sep 2015
Posts: 3777
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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dave13 wrote:
nice explanation Brent, but what if volumes of mixtures aren't equal ? any advice is appreciated Thanks, Dave!
If the volumes aren't equal, you can use the Weighted Averages formula (discussed in the video), or you might be able to use the answer choices to your advantage.

For example, let's say that we're adding 3 liters of 0% salt solution to 2.5 liters of 22% salt solution, and the answer choices are:
A) 10%
B) 11%
C) 13%
D) 14%
E) 17%

We know that, if we were adding EQUAL volumes of the two solutions, then the resulting solution would be 11% (since 11% is the AVERAGE of 0% and 22%)
Since we're actually add more 0% solution, then the resulting solution will be LESS THAN 11% salt.
Since answer choice A is the only answer that is less than 11%, the correct answer must be A

Cheers,
Brent
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Posts: 1279
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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keatross wrote:
As a person that is bad at math I did this:

Original mixture
Total = 40ml
Salt = 40 * 0.15 = 6ml
Water = 40 - 6 = 34ml

After spill
Total = 40 - 15 = 25
Salt = 6ml - (15 * 0.15) = 3.75ml

After refill
Total = 50ml
Salt = 3.75 (unchanged)

3.75/50 = 7.5% [A]

Let me know if I did this wrong.

hi there keatross you are very good at math i think I wonder where from did you get 15. see below

After spill / Total = 40 - 15= 25

thanks Intern  Joined: 15 Nov 2018
Posts: 13
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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dave13 wrote:
keatross wrote:
As a person that is bad at math I did this:

Original mixture
Total = 40ml
Salt = 40 * 0.15 = 6ml
Water = 40 - 6 = 34ml

After spill
Total = 40 - 15 = 25
Salt = 6ml - (15 * 0.15) = 3.75ml

After refill
Total = 50ml
Salt = 3.75 (unchanged)

3.75/50 = 7.5% [A]

Let me know if I did this wrong.

hi there you are very good at math i think I wonder where from did you get 15. see below

After spill / Total = 40 - 15= 25

thanks I just multiplied the units in the question by 10 to help with the math a bit. 4L= 40, 1.5L= 15. Helps to create whole units for the "salt" portion in case the second part of the question got more confusing with ratios, etc, too.
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Joined: 13 Nov 2018
Posts: 63
Location: India
GMAT 1: 700 Q51 V32 Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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keatross wrote:
As a person that is bad at math I did this:

Original mixture
Total = 40ml
Salt = 40 * 0.15 = 6ml
Water = 40 - 6 = 34ml

After spill
Total = 40 - 15 = 25
Salt = 6ml - (15 * 0.15) = 3.75ml

After refill
Total = 50ml
Salt = 3.75 (unchanged)

3.75/50 = 7.5% [A]

Let me know if I did this wrong.

Did u assumed total as 40 ml

i am very bad in mixtures, i always mess up with them!!!
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Anyways...Thanks for trying Intern  Joined: 15 Nov 2018
Posts: 13
A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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gmatdordie wrote:
keatross wrote:
As a person that is bad at math I did this:

Original mixture
Total = 40ml
Salt = 40 * 0.15 = 6ml
Water = 40 - 6 = 34ml

After spill
Total = 40 - 15 = 25
Salt = 6ml - (15 * 0.15) = 3.75ml

After refill
Total = 50ml
Salt = 3.75 (unchanged)

3.75/50 = 7.5% [A]

Let me know if I did this wrong.

Did u assumed total as 40 ml

i am very bad in mixtures, i always mess up with them!!!

Sorry, I just assigned arbitrary unit measurements for those (again to help my brain visualize). For this question a conversion is not required so it doesn't really matter if you label them litres, millilitres, etc as long as you remember what the question is asking imo. You could theoretically multiply everything by 1000 and use mL as your common unit measurement, though, and I've seen plenty of questions like that.
VP  D
Joined: 09 Mar 2016
Posts: 1279
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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GMATPrepNow wrote:
HKD1710 wrote:
A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) $$7\frac{1}{2}$$%
(B) $$9\frac{3}{8}$$%
(C) $$10\frac{1}{2}$$%
(D) $$12$$%
(E) $$15$$%

Project PS Butler : Question #28

-----------ASIDE------------
IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.
-------------------------------

Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution
Since the jug holds 5 liters, we then add 2.5 liters of water (aka a 0% salt solution)

Since we're combining EQUAL VOLUMES of 15% and 0% solutions, the salt content of the resulting solution will equal the average of 15% and 0%
(15 + 0)/2 = 7.5

So, the resulting solution will be 7.5% salt.

RELATED VIDEO FROM OUR COURSE

hello Brent, GMATPrepNow many thanks for detailed explanation.

i have one question you write "Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution" But when we pour out 1.5 litres and we are left with 2.5 , shouldnt salt percentage reduce as well because you still write 15% before poring out we had 4 litres of solution that is 15 % of salt, and after pouring out logically 15% of salt shoulb be reduced as well, no ? thanks!
D
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Joined: 12 Sep 2015
Posts: 3777
Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per  [#permalink]

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Top Contributor
dave13 wrote:
GMATPrepNow wrote:
HKD1710 wrote:
A 5-liter jug contains 4 liters of a saltwater solution that is 15 percent salt. If 1.5 liters of the solution spills out of the jug, and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?

(A) $$7\frac{1}{2}$$%
(B) $$9\frac{3}{8}$$%
(C) $$10\frac{1}{2}$$%
(D) $$12$$%
(E) $$15$$%

Project PS Butler : Question #28

-----------ASIDE------------
IMPORTANT CONCEPT: If EQUAL VOLUMES of 2 solutions are combined, the concentration of the resulting mixture will be the AVERAGE of the 2 solutions.
Example: If 3 liters of a 10% alcohol solution are combined with 3 liters of a 30% alcohol solution, the resulting solution will be 20% alcohol.
Likewise, if 11 liters of 6% alcohol solution are combined with 11 liters of a 7% alcohol solution, the resulting solution will be 6.5% alcohol.
-------------------------------

Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution
Since the jug holds 5 liters, we then add 2.5 liters of water (aka a 0% salt solution)

Since we're combining EQUAL VOLUMES of 15% and 0% solutions, the salt content of the resulting solution will equal the average of 15% and 0%
(15 + 0)/2 = 7.5

So, the resulting solution will be 7.5% salt.

RELATED VIDEO FROM OUR COURSE

hello Brent, GMATPrepNow many thanks for detailed explanation.

i have one question you write "Once we pour out 1.5 liters, we're left with 2.5 liters of 15% salt solution" But when we pour out 1.5 litres and we are left with 2.5 , shouldnt salt percentage reduce as well because you still write 15% before poring out we had 4 litres of solution that is 15 % of salt, and after pouring out logically 15% of salt shoulb be reduced as well, no ? thanks!
D

The salt is totally dissolved into the water.
So, no matter how much solution you examine, it will contain 15% salt.
So, 1 ml of solution will be 15% salt, 10 ml of solution will be 15% salt, etc.

Here's another example:
Let's say a beer is 5% alcohol.
If you take a sip of beer, the amount you drink will contain 5% alcohol, and the beer that's remaining in your glass is also 5% alcohol.

Does that help?

Cheers,
Brent
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