# Lesson: Mixture Questions

## Comment on Mixture Questions

### Alcohol and water question we

Alcohol and water question we can also solve as under.
Initially, there 40 ml alcohol and 120 ml water. We are adding pure alcohol so volume of water is fixed. To make alcohol to water ratio 60:40 (=120:80 = 180:120) we need total 180 ml alcohol and 120 ml water (water is already there with 40 ml alcohol. so we need 180-40 = 140 ml alcohol.

Beautiful!!

### Hello there

Hello there
I did not get the way that Sandy175 solve this question
how did he get the ratio numbers?
thanks so much

### Sandy is referring to the

Sandy is referring to the question that starts at 2:45.
In the beginning, the 160 ml of solution is 25% alcohol.
In other words, the ratio of alcohol to non-alcohol is 1:3, which is the same as 40:120.

Our goal is to create a mixture that is 60% alcohol. In other words, we want an alcohol to non-alcohol ratio of 60:40, which is the SAME as an alcohol to non-alcohol ratio of 180:120

Notice that the original alcohol to non-alcohol ratio is 40:120 and the TARGET alcohol to non-alcohol ratio is 180:120. In BOTH ratios, the value in the non-alcohol position is 120.

Since we're adding only alcohol to the original mix, the only part of the ratio that changes is the first value (the amount of alcohol).

We need to increase that value from 40 to 180. This is an increase of 140. So, we need to add 140 ml of alcohol.

### How many liters of a 40%

How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140

After making equation I don't know what to do

### You're referring to https:/

You're referring to https://gmatclub.com/forum/how-many-liters-of-a-40-iodine-solution-need-...

What's your equation? Also, can you tell me the steps you took to get that equation?

### 0.4 + 7 for iodine then 0.6

0.4 + 7 for iodine then 0.6 +28 for water then total x + 35 I made three tanks and gave information to them which was available to me from the question

### Those algebraic expressions

Those algebraic expressions are missing some variables.
I have posted a solution to the question here: https://gmatclub.com/forum/how-many-liters-of-a-40-iodine-solution-need-...

### 40+x = 0.6(160+x)

40+x = 0.6(160+x)

Looks good!

### Hi Brent,

Hi Brent,

A rabbit on a controlled diet is fed daily 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 % percent protein and Food Y contains 15 % protein. If the rabbit’s diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture ?

(A) 100
(B) 140
(C) 150
(D) 160
(E) 200

### One option is to keep track

One option is to keep track of the PROTEIN

Let x = the number of grams of Food X needed
So, 300 - x = the number of grams of Food Y needed (since the total weight is 300 grams)

Food X is 10% protein, so the number of grams of protein from Food X = 10% of x = 0.10x

Food Y is 15% protein, so the number of grams of protein from Food Y = 15% of (300 - x) = 0.15(300 - x)

The total protein from the mixture is 38 grams

So, we can write: (protein from Food X) + (protein from Food Y) = 38 grams

Or....0.10x + 0.15(300 - x) = 38

Expand: 0.10x + 45 - 0.15x = 38

Simplify: -0.05x + 45 = 38

Subtract 45 from both sides: -0.05x = -7

Divide both sides by -0.05 to get: x = 140

### Dear Brent,

Dear Brent,
can you please help solve this question without using the weighted average formula that everyone is suggesting?
Thanks!

https://gmatclub.com/forum/jackie-has-two-solutions-that-are-2-percent-sulfuric-acid-204112.html?kudos=1

Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42

### You bet.

You bet.
My solution to the above question can be found here: https://gmatclub.com/forum/jackie-has-two-solutions-that-are-2-percent-s...

### Dear Brent,

Dear Brent,

Could you explain the difference between the second question in the video and this question below
"A 48-gallon solution of salt and water is 10% salt. How many gallons of water must be added to the solution in order to decrease the salt to 8% of the volume?"

As I observed there is one difference in the solution which the question in the video has been solved in this way (40%)(160)+(x)=(60%)(160+x) while the question I refer to was solved in that way (10%)(48)=(8%)(48+x)

thank you in advance

### The second question in the

The second question in the video (which starts at 2:40) takes a 25% solution and adds pure alcohol to make it MORE concentrated (60%).

In the question you've posted, we take a 10% solution and add water to make it LESS concentrated (8%).

As such, the solutions will be a little different. However, the logic we apply remains the same.

Does that help?

Cheers,
Brent

### Hi Brent can you provide me

Hi Brent can you provide me with a solution for OG 2018 ques no 227

### You bet.

You bet.
Here's my step-by-step solution: https://gmatclub.com/forum/seed-mixture-x-is-40-percent-ryegrass-and-60-...

Cheers,
Brent

### https://gmatclub.com/forum

https://gmatclub.com/forum/two-mixtures-a-and-b-contain-milk-and-water-in-the-ratios-57862.html

### I'm happy to help.

I'm happy to help.
Here's my full solution: https://gmatclub.com/forum/two-mixtures-a-and-b-contain-milk-and-water-i...

Cheers,
Brent

### Hi Brent,

Hi Brent,

Are following Q types pertaining to Mixtures required for GMAT?
1. A portion of the solution containing x% liquid X is replaced and filled with liquid Y. The process is repeated multiple times. Find the resultant concentration of liquid X.

2. Mixture of 3 quantities.

Thanks & Regards,
Abhirup

### 1) I've seen questions from

1) I've seen questions from test-prep companies that feature the process you describe, but I don't believe I've seen an official question that does. IF there exists such an official question, it's definitely 720+

2) Rare, but this would be a reasonable question.

Cheers,
Brent

### Hi Brent,

Hi Brent,

Need your help with this question:

https://gmatclub.com/forum/if-a-milkman-wishes-to-earn-a-profit-of-16-66-by-selling-milk-at-cost-285347.html

### https://gmatclub.com/forum

https://gmatclub.com/forum/m03-183589.html

Cheers,
Brent

### Two mixtures A and B contain

Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

I approached this using the following:

Mixture A = A
Milk mixture in A = 2/7A
Mixture B = 90 gallons
Milk in new mixture 40% (or 2/5)

2/7(A) + 50 / A + 90 = 2 / 5
Cross multiply
4/7 A = 70
A = 122.5

Is this right? Is this a longer and more tedious way of doing the calculations?

### That solution is perfect.

That solution is perfect.
Your calculations are pretty much the same as mine in this solution: https://gmatclub.com/forum/two-mixtures-a-and-b-contain-milk-and-water-i... (except you're not using a diagram)

Cheers,
Brent

### I tried using the same

I tried using the same approach for this question:

https://gmatclub.com/forum/m03-183589.html

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

The wording is a bit irritating but I'm assuming that when they say replace they mean mixed.

0.5X + 0.25Y / X + Y = 3 /10
Y=4X

So the ratio of X:Y = 4:1
And seeing that the only answer choice that matched was 80% I'm guessing my approach worked?

### Question link: https:/

Interesting. I (incorrectly) read question as asking "What percent of the original 50% solution was removed (and then replaced with the 25% solution)? But, I got the correct answer anyway (since those two scenarios are very similar)

The same applies to your solution. You skipped the part about removing some of the 50% solution, but you also got the correct answer.

Cheers,
Brent

### Hi Brent!

Hi Brent!

I tried to solve this question ( https://gmatclub.com/forum/two-containers-of-equal-capacity-are-full-of-a-mixture-of-oil-and-wat-113800.html) the same way jamifahad (the first answer) did, but I don't understand why it isn't 7/18 and 11/18 in the second container?

Thank you!
BR Pia

### jamifahad's solution: https:

You are correct; jamifahad's solution SHOULD have 7/18 oil and 11/18 water (NOT 7/11) in the second container.

Cheers,
Brent

### Could you help solve this

Could you help solve this please?
https://gmatclub.com/forum/m21-184282.html
Aman

### Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/m21-184282.html#p2285599

Cheers,
Brent

### Hi Brent,

Hi Brent,

In the below DS problem

https://gmatclub.com/forum/material-a-costs-3-per-kilogram-and-material-b-costs-5-per-kilogram-109070.html

I understood the solution provided by Bunnel. However, I am confused with the case where x=3 and y=7. The cost of 10 kg of Mixture will still be less than 40. But in this case y will be greater than x.

I think I am missing something.

Thanks for your help.

Regards,
Sushant

### Hi Sushant,

Hi Sushant,

Sorry, but I don's see where Bunuel uses x=3 and y=7. Can you copy and paste that part of his solution?

In the meantime, I've also answered the question here: https://gmatclub.com/forum/material-a-costs-3-per-kilogram-and-material-...

### Hi Brent,

Hi Brent,

The ratio of a cup of flour to cups of milk is 3:2. 6 cups of milk are added to the cake, and the new ratio is 3:4, how many cups of flour are there in the cake ?
> A.6, B.8, C.9, D. 10, E.12

### Oof, that's a poorly-worded

Oof, that's a poorly-worded question! All of a sudden, "cake" somehow enters the picture :-)

Presumably the question is intended to read:
A CAKE recipe requires a flour to milk ratio of 3:2. If 6 cups of milk are added to the cake, and the new ratio is 3:4, how many cups of flour are there in the cake ?

Let F = the cups of flour required
Let M = the cups of milk required

Given: The recipe requires a flour to milk ratio of 3:2
We can write: F : M = 3 : 2
In other words: F/M = 3/2
Cross multiply to get: 2F = 3M

Given: If 6 cups of milk are added to the cake, and the new ratio is 3:4
We can write: F : M + 6 = 3/4
In other words: F/(M + 6) = 3/4
Cross multiply to get: 4F = 3(M + 6)
Simplify to get: 4F = 3M + 18
Subtract 18 from both sides to get: 4F - 18 = 3M

We now have two equations: 2F = 3M and 4F - 18 = 3M
Since both equations are set equal to 3M, we can write: 2F = 4F - 18
Solve to get: F = 9

### Hi Brent,

Hi Brent,
Could you please help on this one ? thank you

A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%. What's the ratio of the amount of the solution that was replaced to the amount of solution that was not replaced ?
> A. 1:3, B. 1:2, C. 2:3, D. 2:1, E. 3:1

### Hi Brent I have this question

Hi Brent I have this question about solutions

A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%. What’s the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?
A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

### I've answered the question

I've answered the question here: https://gmatclub.com/forum/a-certain-quantity-of-40-solution-is-replaced...

The only difference is that the linked question asks, "What is the FRACTION of the solution that was replaced?"
The answer is 1/3 of the solution was replaced

Your question asks, "What’s the RATIO of the amount of solution that was replaced to the amount of solution that was not replaced?"
If 1/3 of the solution was replaced, then 2/3 of the solution was NOT replaced.
So, the ratio is 1/3 : 2/3
This ratio isn't among the answer choices, so we need to find an equivalent ratio.
Take the ratio 1/3 : 2/3
And multiply both parts by 3 to get the equivalent ratio 1 : 2