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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, 18:14
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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? A. 6:4 B. 6:14 C. 4:4 D. 4:6 E. 3:7
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Last edited by Bunuel on 08 Sep 2013, 06:47, edited 1 time in total.
Edited the question.



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Re: Solutions [#permalink]
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08 Nov 2005, 20:13
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the ACs are quite strange.
weingt of a = x
xa+b(1x) = 0.5(a+b)
xa  0.5a = 0.5b  b(1x)
a(x0.5) = b (x0.5)
a = b =25.
total = a+b = 25+25 = 50
salt = 50% 0f 50 = 25%
salt from a = 20% of 25 = 5
salt from b = 80% of 25 = 20
so total = 25 which is 50% of 50.
so it should be 4:4.



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let:
x = ounces taken from solution A (20% salt)
y = ounces taken from solution B (80% salt)
to prepare 50 ounce 50% salt.
first equation is simple:
x + y = 50
to get another equation so as to be able to solve, compute salt contents.
20% of x + 80% of y = 50% of 50 or
x/5 + 4/5 * y = 25 or
x+4y = 125
solve two equations to get:
x = 25
y = 25
so solutions has to mix in
1:1 oops 4:4



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Re: Mixture Problem [#permalink]
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05 Jul 2011, 06:00
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 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 So y of 60 = 30 Ratio is 1:1 which is 4:4. C



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Re: Mixture Problem [#permalink]
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05 Jul 2011, 07:12
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let in the final solution : contribution of A = x. B's = 50 x. 0.2x + 0.8(50x) = 50*0.5 solving x =25 = A's, B's = 25.
hence, 4:4, C



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Re: Mixture Problem [#permalink]
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05 Jul 2011, 08:47
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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. For details of this formula, see http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2011/04 ... gebrutes/http://www.veritasprep.com/blog/2011/04 ... mixtures/
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Re: Mixture Problem [#permalink]
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06 Jul 2011, 04:19
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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 Fastest way to solve this Alligations. You can solve any Alligation using this method. Use the diagram, 30:30 = 1:1 hence its C
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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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08 Sep 2013, 06:32
If you have to look for the weights of different ingredients to come up with a target mix, there is an easy formula that you can apply:
W = weight C = concentration
Wa/Wb = (Cb  Cavg)/(Cavg  Ca)
=> Wa/Wb = (0.80.5)/(0.50.2) = 0.3/0.3 = 1/1
=> The only multiple of 1/1 in the answer choices is 4/4



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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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16 Dec 2013, 05:57
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?
A. 6:4 B. 6:14 C. 4:4 D. 4:6 E. 3:7 Applying differentials. Forget the volumes 3x+3y=0 3x=3y x=y So they have to be the same weight both So 1:1 = 4:4 C is the correct answer Cheers! J Let the Kudos rain begin!!



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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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21 Dec 2014, 14:55
How come everyone is ignoring the announces. If I take 50% of the 30g and 50% of the 60g I only have 45 g not 50.



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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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21 Dec 2014, 23:17
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weighted average
20x+80y/x+y=50
20x+80y=50x+50y
30y=30x
y/x=1/1 or 4/4
C



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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, 03:30
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..
thanks a lot.



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Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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09 Mar 2015, 15:30
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. For details of this formula, see http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2011/04 ... gebrutes/http://www.veritasprep.com/blog/2011/04 ... mixtures/Hi Karishma, My question is also why are we ignoring the quantities. The way I started thinking about it was like this: (0.20) * (30) * (x) + (0.80) * (60) * (y) = (0.50) * (50) * (x+y) However, this ends up in y/x = 9/23. I then noticed that we are ignoring the actual quantities. Why is this so? Are we ignoring them because they anyway have to do with salt?



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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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21 Apr 2016, 00:21
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. For details of this formula, see http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2011/04 ... gebrutes/http://www.veritasprep.com/blog/2011/04 ... mixtures/Responding to a pm: Quote: 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 = (256) / (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.
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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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23 Apr 2016, 02:25
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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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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11 Aug 2017, 14:34
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?
A. 6:4 B. 6:14 C. 4:4 D. 4:6 E. 3:7 .2A+.8B=.5(A+B) A/B=1 4:4 C



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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink]
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11 Aug 2017, 17:00
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?
A. 6:4 B. 6:14 C. 4:4 D. 4:6 E. 3:7 Take the middle 25 ounces each, luckily it has 50% salt, the average. then it should be k:k for any integer k. Sent from my iPhone using GMAT Club Forum mobile app




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