[Maths Class Notes] on Mix Slider Pdf for Exam

The mixture problem in Maths combines two or more things and determines some characteristics of either the ingredients used or the resulting mixture. For example, we might need to determine how much water should be added to dilute a saline solution, or might want to know the percentage of concentrate in a jug of apple juice. Solving such types of mix slider or mixture problems generally involves solving systems of equations. In this, we will discuss how to solve mixture problems in Maths that involve concentrations through examples for a better understanding.

Mixtures Introduction

A mixture is a combination of two or more two substances. We generally use ratio, percentage, or fractions to describe quantities in a mixture.

Example:

What percentage of the solution is salt, if 500 gms of a saline solution has 50 gms of Salt?

Solutions:

[frac{50}{500}] = 0.10 = 10%

Solving Mixture Problems

Here are the steps on how to solve mixture problems in Maths:

Step 1: Read the problem carefully. Find the known or unknown quantity. Let x or another variable to represent the unknown quantity.

Step 2: If required, using another variable, write an expression to represent any other unknown quantity.

Step 3:  Write an algebraic equation that needs to be solved.

Step 4: Solve the equations

Step 5: Check the answer and ask yourself “ Is the answer reasonable?”

Step 6: Write a sentence that defines what was asked in the problem, and ensure to include a unit as part of the solution.

Mix Slider Examples With Solution

1. How many Gallons of 15% sugar solution do we need to mix with 5 Gallons of 50% percent Solution to get 30% Sugar Solution?

Solution:

Step 1: We will choose the Variable to represent the unknown Quantity. 

Let x be the quantity of 15% sugar solution to be mixed with 5 gallons of 50% percent solution. 

Step 2: We use different Variables to Determine the Other Unknown Quantity.

Let y be the quantity of the final 30% sugar solution.

Accordingly,

x + 5 = y  (1)

Step 3: We will now express Mathematically that the quantity of sugar solution in x gallons plus the quantity of sugar solutions in 5 gallons  is equal to the quantity of sugar solution in y gallons. But remember the sugar solution is measured in percentage terms. According, the algebraic equation will be

15% * x + 50% * 5 = 30%* y  

Step 4: Solving the Equation

Substituting the y values in the equation, 

15% * x + 40% * 5 = 30%* ( x + 5)

Changing Percentage Into Fraction

[frac{15}{100}]x + 5 *
[frac{40}{100}]  =
[frac{30}{100}]  ( x + 5)

Multiplying all the teams  by 100, we get

15x + 5 * 40 = 30x + 150

Solving the variable x

15x + 200 = 30x + 150

15 x – 30 x = 150 – 200

-15x = -50

x = [frac{50}{15}]

x = [frac{10}{3}]

Step 5: Checking the answer by substituting the value of x , we get

15% *
[frac{10}{3}]  + 40% * 5 = 30%* (
[frac{10}{3}] + 5)

Multiply all the terms by 100 we get,

15(
[frac{10}{3}]) + 40(5) = 30(
[frac{10}{3}] + 5)

50 + 200 = 100 + 150

250 = 250

Step 6: Hence, we will mix
[frac{10}{3}]  Gallons of 15% sugar solution to 5 Gallons of 40% Sugar Solution to get 30% Sugar Solution.

2. How much 40% Ethyl Alcohol do we need to add to 10 liters of 90% Ethyl Alcohol to obtain a 50% solution of Ethyl Alcohol?

Solution: 

We can also solve the given mix slider problem by creating the table as shown below:

The rows of the table describes the type of mixture that you have. The columns of the table describe the amount of each compound you have and the concentration of that amount in each mixture (represented in decimal).

As some information is unknown, we use some variables to represent unknown quantities.

The amount of 40% Ethyl Alcohol is also unknown. Let it represent it with x.  The amount of 90% Ethyl Alcohol is also unknown but should be 10 -x liters to get the 10 liters of the final solution.

The quantity of the alcohol that each part of the mixture adds to the final solution is equal to the quantity of each solution that is mixed, times the fraction of alcohol that solution is obtained from.

Now, we will use the fourth column of the table as an equation to solve the variable x.

The 10 liters of the final solution must have a total volume of 5 liters of alcohol in order to get 50% alcohol. This 5 liter of alcohol is a combination of the 40% of the solution we mix and the amount of 90% solution. If the 40% of the solution that we mix is x then 0.4x (in litres) will be the amount of the alcohol that is contributed. Similarly, 0.9 ( 10 -x) will be the amount of alcohol contributed by the 10-x litres of 90% of alcohol solution that we add.  Hence, in total 0.4 + 0.9(10 – x) should be equivalent to 5 litres we require in the final solution to get 50% alcohol. 

Solving the equation, we get

0.4 + (9 – 0.9 x) =  5

 -0.5 x + 9 = 5

0.5 x = 4

x =
[frac{4}{0.5}]  = 8

Hence, we need 8 liters of the 40% solution.