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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] New post 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.
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Re: Solutions [#permalink] New post 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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 [#permalink] New post 09 Nov 2005, 11:13
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] New post 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] New post 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] New post 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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Re: Mixture Problem [#permalink] New post 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
Attachments

Allegations.jpg
Allegations.jpg [ 17.06 KiB | Viewed 3257 times ]

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

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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] New post 21 Dec 2014, 07:25
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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] New post 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] New post 21 Dec 2014, 23:17
weighted average

20x+80y/x+y=50

20x+80y=50x+50y

30y=30x

y/x=1/1 or 4/4

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Re: Solution A is 20% salt and Solution B is 80% salt. If you [#permalink] New post 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] New post 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?
Solution A is 20% salt and Solution B is 80% salt. If you   [#permalink] 09 Mar 2015, 15:30
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