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Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

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08 Nov 2005, 19:14

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Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution?

Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution? 6:4 6:14 4:4 4:6 3:7

What is faster way to solve this

I am not good at PS but here is my take on this

6+4/5*y*60 = 1/2 (30 +y*60) Solve: 12+96y=30+60y 36 y = 18 y = 1/2

Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution? 6:4 6:14 4:4 4:6 3:7

What is faster way to solve this

Forget the volumes for the time being. You have to mix 20% and 80% solutions to get 50%. This is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities. If this doesn't strike, use w1/w2 = (A2 - Aavg)/(Aavg - A1) w1/w2 = (80 - 50)/(50 - 20) = 1/1 So the volume of the two solutions will be equal. Answer has to be 4:4.

Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution? 6:4 6:14 4:4 4:6 3:7

What is faster way to solve this

Fastest way to solve this Alligations. You can solve any Alligation using this method. Use the diagram, 30:30 = 1:1 hence its C

Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

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16 Dec 2013, 06:57

desiguy wrote:

Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution?

Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

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21 Dec 2014, 08:25

Hello from the GMAT Club BumpBot!

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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

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01 Mar 2015, 04:30

hi karishma why have you considered "percentages" in the scale method...can`t we use the weights of the solutions in the formula as in w1/w2 = (A2 - Aavg)/(Aavg - A1) w1/w2 = (60 - 50)/(50 - 30) = 1/2 please help me with the confusion..

Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

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09 Mar 2015, 16:30

VeritasPrepKarishma wrote:

prashantbacchewar wrote:

Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution? 6:4 6:14 4:4 4:6 3:7

What is faster way to solve this

Forget the volumes for the time being. You have to mix 20% and 80% solutions to get 50%. This is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities. If this doesn't strike, use w1/w2 = (A2 - Aavg)/(Aavg - A1) w1/w2 = (80 - 50)/(50 - 20) = 1/1 So the volume of the two solutions will be equal. Answer has to be 4:4.

Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

Show Tags

03 Apr 2016, 00:11

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Solution A is 20% salt and Solution B is 80% salt. If you have 30 ounces of solution A and 60 ounces of solution B, in what ratio could you mix Solution A with Solution B to produce 50 ounces of a 50% salt solution? 6:4 6:14 4:4 4:6 3:7

What is faster way to solve this

Forget the volumes for the time being. You have to mix 20% and 80% solutions to get 50%. This is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities. If this doesn't strike, use w1/w2 = (A2 - Aavg)/(Aavg - A1) w1/w2 = (80 - 50)/(50 - 20) = 1/1 So the volume of the two solutions will be equal. Answer has to be 4:4.

But as per question asked ,why we cannot take respective weights of salt( i.e if 30 ounces of sol.A has 20% salt) ,then take salt weight as 6 ounces(20% of 30ounces) then isnt is our question becomes "how much 6 ounces and 8 ounces will be added to get 25 ounces of salt"???

w1/w2 = (25-6) / (48 - 25)

You are not required to mix 30 ounces of solution A with some amount of solution B. You are not given that you have to use the entire 30 ounces of solution A. In fact, the volumes of the solution are not required at all since the question asks for the ratio in which A and B should be mixed. We know the concentration of salt in A, concentration of salt in B and average required concentration. This will simply give us the ratio in which the two solutions should be mixed (using the formula). We find that both solutions should be mixed in equal quantities (ratio of 1:1 or 2:2 or 3:3 or 4:4 etc) so to make 50 ounces of mix, we will put 25 ounces of each solution.
_________________

Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]

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23 Apr 2016, 03:25

10 sec solution: always compare the simple average to the weighted average first. Here simple average is 20+80 divided by 2 = 50. And thats our answer. we have taken 50% of A and 50% of be to create the new solution.
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