¶ Mixture
>Try using the spider method.
In this method, you need to subtract Resultant Concentration from Concentration 1 and Concentration 2 (take the positive value after the subtraction). Then you will find the ratio of Concentration 1 - Resultant and Concentration 2 - Resultant.
A Special Kind of Math and the Trick of Solving it:
MIXING TWO MIXTURES:
An example will make it more clear.
1)Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X?
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %
The question asks for the ryegrass so your table should look like this:
| X- mixture 40% | Y- mixture 25% | |
| 30% | ||
| 25 - 30= -5 |-5|= 5 |
40 - 30= 10 |10|= 10 |
So, the ratio of Concentration 1 - Resultant and Concentration 2 - Resultant will be 5:10 or, 1:2. So, for every 1 part of X 2 parts of Y will be in the final mixture.
So, for a 3 kg mixture (for example)=> 1X and 2Y => X=33% of the total. [Answer: B]
This table can be used in other ways also, and this question is an example:
2)How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol to produce a solution that is 25 percent alcohol?
(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4
Your table:
| X- pure alcohol 100% | Y- alcohol 20% | |
| 25% | ||
| 20- 25= -5 |-5|= 5 |
100- 25= 75 |75|= 75 |
So, the ratio of Concentration 1 - Resultant and Concentration 2 - Resultant will be 5:75.
Final ratio: X/Y=5/75
We know that Y is 100 liter so X/100=5x/75=20/3. [Answer: C]
Some important concentrations:
Concentration of Water in any Solution = 0%
Concentration of anything pure = 100%
***Video on “Concept and Practice Problems of Mixture”
MIXTURE PROBLEMS: