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Mixture Problems

Algebra mixture problems are the problems which derive at a final solution by adding solution of two or more results of the same problem. A given algebra word problem consists of data which are arranged to solve part by part and then to derive the solution for end results.

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Value Mixture Problems

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A value mixture problems are involved combining two ingredients that have different price into a single blend. A solution of these types of problems based on the equation AC = V, where A is the amount of the integrent, C is the cost per unit of the integrent and V is the value of the integrent.

How to Solve Mixture Problems


Given below are the steps for solving mixture problems:
Steps for Solving Mixture Problems:

Step 1:
Write a expression for the amount of the ingredient used, the unit cost of ingredient and the value of the amount used.

Step 2: Write numerical or variable expression for the amount, the unit cost of the mixture and the value of the amount.

Step 3: Draw all the values in a table and solve the expression.

Given below are the examples in solving mixture problems.

Solved Example

Question: How many ounces of a gold alloy that costs 300 dollars an ounce mixed with 120 oz of an alloy that costs 150 dollars an ounce to make a mixture that costs 200 dollars an ounce?
Solution:
Let ounces of the 300 dollars gold alloy = x



Amount
Cost
Value
300 dollar alloy
x
300 300x
100 dollar alloy 120
150
120 * 150
Mixture x + 120
200
200(x + 120)


=> 300x + 120 * 150 = 200(x + 120)

Solve for x,

=> 300x + 18000 = 200x + 24000

=> 300x - 200x = 24000 - 18000

=> 100x = 6000

=> x = 60

The mixture contain 60 oz of the 300 dollars gold alloy.

Mixture Word Problems

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Given below are some of the mixture problems with solution.

Solved Examples

Question 1: How many ounces of a solution that is 10% alcohol should be mixed with 12 ounces of a solution that is is 24% alcohol to create a solution that is 15% alcohol?
Solution:
 

Let ounces of the 10 % of alcohol = x

PERCENT AMOUNT OF SOLUTION AMOUNT OF TARGET
1ST CONTAINER 0.10 X 0.10X
2ND CONTAINER 0.24 12 0.24(12)
MIXTURE 0.15 X + 12 0.15(X + 12)

=> 10 X + 0.24 (12) = 0.15 (X + 12)

=> 0.10 X + 2.88 = 0.15X + 1.80

=>10 X + 288 = 15 X + 180 (Multiply by 100)

=> 10 X - 15 X = 180 - 288

=> -5X = - 108

X = $\frac{108}{5}$ = 21.6

Therefore, 21.6 OUNCES @ 10 %.


 

Question 2: Finding how many liters of a solution that is 20% alcohol should be combined with 10 liters of a solution that is 50% alcohol to create a solution that is 30% alcohol?
Solution:

The primary technique that we will develop here is to trace the target ingredient in the problem. The equation will state that the amount of alcohol in the first container and the amount of alcohol in the second will equal the amount of alcohol in the final mixture.

Table Showing Final Mixture


PERCENT AMOUNT OF SOLUTION AMOUNT OF TARGET
1ST CONTAINER 0.50 10 5
2ND CONTAINER 0.20 x
0.20 x
MIXTURE 0.30 x + 10 0.30 (x + 10)

The amount of alcohol in the first container plus the amount of alcohol in the second container equals the amount of alcohol in the final mixture.

Now, we solve the equation

5 + 0.20x = 0.30(x + 10)

5 + 0.20x = 0.30x + 3

Subtract 0.20x from both the sides

5 + 0.20x - 0.20x = 0.30x + 3 - 0.20x

5 = 0.10x + 3

5 - 3 = 0.10x + 3 - 3

$\frac{2}{0.10}$ = $\frac{0.10x}{0.10}$

X = 20.



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