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Solution A is 20% salt and Solution B is 80% salt. If you

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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
[Reveal] Spoiler: OA

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(1-x) = 0.5(a+b)
xa - 0.5a = 0.5b - b(1-x)
a(x-0.5) = b (x-0.5)
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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09 Nov 2005, 11:13
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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
let in the final solution : contribution of A = x. B's = 50 -x.
0.2x + 0.8(50-x) = 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 ... ge-brutes/
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 26 May 2005 Posts: 565 Followers: 18 Kudos [?]: 207 [2] , given: 13 Re: Mixture Problem [#permalink] Show Tags 06 Jul 2011, 04:19 2 This post received KUDOS 4 This post was BOOKMARKED 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 Attachments Allegations.jpg [ 17.06 KiB | Viewed 6052 times ] Intern Joined: 11 Jan 2013 Posts: 16 Location: United States Followers: 0 Kudos [?]: 31 [0], given: 12 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 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 = (0.8-0.5)/(0.5-0.2) = 0.3/0.3 = 1/1 => The only multiple of 1/1 in the answer choices is 4/4 Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 64 Kudos [?]: 604 [0], given: 355 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 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!! GMAT Club Legend Joined: 09 Sep 2013 Posts: 13938 Followers: 590 Kudos [?]: 167 [0], given: 0 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 21 Dec 2014, 07:25 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. _________________ Manager Joined: 27 May 2014 Posts: 85 Followers: 0 Kudos [?]: 19 [0], given: 21 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 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. Director Joined: 23 Jan 2013 Posts: 582 Schools: Cambridge'16 Followers: 1 Kudos [?]: 21 [1] , given: 40 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 21 Dec 2014, 23:17 1 This post received KUDOS weighted average 20x+80y/x+y=50 20x+80y=50x+50y 30y=30x y/x=1/1 or 4/4 C Intern Joined: 08 Dec 2013 Posts: 47 Followers: 0 Kudos [?]: 0 [0], given: 23 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 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. Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 443 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) Followers: 2 Kudos [?]: 112 [0], given: 169 Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 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 ... ge-brutes/ 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? GMAT Club Legend Joined: 09 Sep 2013 Posts: 13938 Followers: 590 Kudos [?]: 167 [0], given: 0 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 02 Apr 2016, 23:11 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. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7185 Location: Pune, India Followers: 2167 Kudos [?]: 14018 [0], given: 222 Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] Show Tags 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 ... ge-brutes/ 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 = (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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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] 23 Apr 2016, 02:25
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