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

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.

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

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

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

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

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!!

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.

weighted average

20x+80y/x+y=50

20x+80y=50x+50y

30y=30x

y/x=1/1 or 4/4

C

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.

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?

Responding to a pm:



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.

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.

let A and B=ounces of A and B to be mixed
.2A+.8B=.5(A+B)→
A=B
A:B=1:1=4:4
C

Take the middle 25 ounces each, luckily it has 50% salt, the average. then it should be k:k for any integer k.


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We see that we need 25 ounces of salt. We can go through the choices:

A) 6:4

6x + 4x = 50

10x = 50

x = 5

So we have 6(5) = 30 ounces of A and 4(5) = 20 ounces of B. Does this produce 25 ounces of salt?

30 x 0.2 + 20 x 0.8 = 6 + 16 = 22 → No

B) 6:14

6x + 14x = 50

20x = 50

x = 2.5

So we have 6(2.5) = 15 ounces of A and 14(2.5) = 35 ounces of B. Does this produce 25 ounces of salt?

15 x 0.2 + 35 x 0.8 = 3 + 28 = 31 → No

C) 4:4

This means we have 25 ounces of A and 25 ounces of B. Does this produce 25 ounces of salt?

25 x 0.2 + 25 x 0.8 = 5 + 20 = 25 → Yes!

Alternate Solution:

Let A be the amount of Solution A and B be the amount of Solution B that produces a 50% salt solution when mixed. Since we want 50 ounces of 50% salt solution, there must be 50 x 0.5 = 25 ounces of salt in our solution.

We have the following:

A + B = 50

0.2A + 0.8B = 25

Let’s rewrite these two equations:

2A + 2B = 100

2A + 8B = 250

Let’s subtract the first equation from the second:

6B = 150

B = 25

Since A + B = 50, A = 25 also. Since there are A = 25 ounces of solution A and B = 25 ounces of solution B in our mixture, we need to mix A and B in the ratio of 1:1; or equivalently, 4:4.

Answer: C

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

You don't require any quantities, as weighted average method will give you the answer directly..
A....average....B
20%...50%...80%
so A to B = \(\frac{80-50}{50-20}=\frac{30}{30}\)=1:1
4:4 is same as 1:1
C