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

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A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   19 Nov 2018, 09:24
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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) \(15\)%
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BrentGMATPrepNow
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   19 Nov 2018, 10:30
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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


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-----------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.

Answer: A

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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   19 Nov 2018, 11:47
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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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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   19 Nov 2018, 09:54
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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.
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   19 Nov 2018, 11:29
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


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-----------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.

Answer: A

RELATED VIDEO FROM OUR COURSE





nice explanation Brent, :) but what if volumes of mixtures arent equal ? any advice is appreciated :)
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   23 Nov 2018, 02:54
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


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS


-----------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.

Answer: A

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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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   23 Nov 2018, 09:16
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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


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS


-----------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.

Answer: A

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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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   07 Sep 2023, 22:41
Hi Experts, Bunuel, KarishmaB,

Can you please share your approach to this question? I usually find myself a bit lost in the details in Mixture questions like these. I would like to know your approach to this.

Thanks in advance! :)
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   07 Sep 2023, 23:46
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ChandlerBong wrote:
Hi Experts, Bunuel, KarishmaB,

Can you please share your approach to this question? I usually find myself a bit lost in the details in Mixture questions like these. I would like to know your approach to this.

Thanks in advance! :)


Imagine it to make it easier for yourself.

Say you have a 5 ltr jug with 4 ltr 15% salt solution. You remove 1.5 ltr out of it. So you will be left with 2.5 ltr 15% salt solution.
Since the jug has a capacity of 5 ltr, you can put it 2.5 ltr of water.

Essentially, you mixed 2.5 ltr of 15% salt solution with 2.5 ltr of water (0% salt solution)
Since you are mixing equal quantities of both, the weights will be the same so the average concentration will be the simple average i.e. average of 15% and 0% which is 7.5%.

Or you can use the formula: 1/1 = 0 - Avg/Avg - 15
Avg = 7.5

Answer (A)

Check out this video on mixtures: https://www.youtube.com/watch?v=VdBl9Hw0HBg
and this post: https://anaprep.com/arithmetic-mixtures/
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   08 Sep 2023, 00:35
KarishmaB wrote:
ChandlerBong wrote:
Hi Experts, Bunuel, KarishmaB,

Can you please share your approach to this question? I usually find myself a bit lost in the details in Mixture questions like these. I would like to know your approach to this.

Thanks in advance! :)


Imagine it to make it easier for yourself.

Say you have a 5 ltr jug with 4 ltr 15% salt solution. You remove 1.5 ltr out of it. So you will be left with 2.5 ltr 15% salt solution.
Since the jug has a capacity of 5 ltr, you can put it 2.5 ltr of water.

Essentially, you mixed 2.5 ltr of 15% salt solution with 2.5 ltr of water (0% salt solution)
Since you are mixing equal quantities of both, the weights will be the same so the average concentration will be the simple average i.e. an average of 15% and 0% which is 7.5%.

Or you can use the formula: 1/1 = 0 - Avg/Avg - 15
Avg = 7.5

Answer (A)

Check out this video on mixtures: https://www.youtube.com/watch?v=VdBl9Hw0HBg
and this post: https://anaprep.com/arithmetic-mixtures/



Thanks, KarishmaB for such a clear explanation!

I had a small doubt, let's say instead of both the weights being 2.5 ltr, one is 2.5 ltr and the other is 5 ltr, how would we approach the question then?

Just wanted to strentghen the concept. :)
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   08 Sep 2023, 00:51
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ChandlerBong wrote:
KarishmaB wrote:
ChandlerBong wrote:
Hi Experts, Bunuel, KarishmaB,

Can you please share your approach to this question? I usually find myself a bit lost in the details in Mixture questions like these. I would like to know your approach to this.

Thanks in advance! :)


Imagine it to make it easier for yourself.

Say you have a 5 ltr jug with 4 ltr 15% salt solution. You remove 1.5 ltr out of it. So you will be left with 2.5 ltr 15% salt solution.
Since the jug has a capacity of 5 ltr, you can put it 2.5 ltr of water.

Essentially, you mixed 2.5 ltr of 15% salt solution with 2.5 ltr of water (0% salt solution)
Since you are mixing equal quantities of both, the weights will be the same so the average concentration will be the simple average i.e. an average of 15% and 0% which is 7.5%.

Or you can use the formula: 1/1 = 0 - Avg/Avg - 15
Avg = 7.5

Answer (A)

Check out this video on mixtures: https://www.youtube.com/watch?v=VdBl9Hw0HBg
and this post: https://anaprep.com/arithmetic-mixtures/



Thanks, KarishmaB for such a clear explanation!

I had a small doubt, let's say instead of both the weights being 2.5 ltr, one is 2.5 ltr and the other is 5 ltr, how would we approach the question then?

Just wanted to strentghen the concept. :)


It doesn't matter what the actual weights are. We can use the formula or the scale in all cases. If 5 ltr of water were added to 2.5 ltr of 15% solution,

we would get 2.5/5 = (0% - Avg)/(Avg - 15%)
Avg = 5%

or on the scale, the distance between 0% to 15% will be divided in the ratio 1:2 because the weights are 5 and 2.5 i.e. a ratio of 2 : 1.

You should check out both the methods in this video: https://anaprep.com/arithmetic-weighted-averages/
They are quite helpful.
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   08 Sep 2023, 00:57
Thanks again for explaining so clearly KarishmaB! Will check out the methods! :)
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Re: A 5-liter jug contains 4 liters of a saltwater solution that is 15 per [#permalink] New post   08 Sep 2023, 13:42
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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) \(15\)%


The percentage of salt in the resulting mixture can be expressed as the weighted average of the percentages of salt in the components of the resulting mixture.

Component A:

4 – 1.5 = 2.5 liters [This is the amount that remains of the original solution.]

15% [This is the percentage of salt in the original solution.]

Component B:

5 – 2.5 = 2.5 liters [This is the amount of water used to fill the jug to capacity.]

0% [This is the percentage of salt in the added water. Water is just water and nothing else.]

So, the percentage of salt in the resulting solution is:

[2.5(15) + 2.5(0)]/5 = 2.5(3) = 7.5 => 7.5%

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