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## Comment on

Mixture Questions## Alcohol and water question we

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!!

Beautiful!!

## 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%

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

## 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)

## Looks good!

Looks good!

## Hi Brent,

Please help me with the solution for this problem:

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

Answer: B

## 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...

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