Browse Tag: mixture problems with solutions

PCSR Civil Service Exam Review Guide 14

After learning work problems, we now learn how to answer mixture problems.

PART I: MATH

A. Videos
Mixture Problem Part 1
Mixture Problem Part 2
Mixture Problem Part 3
Mixture Problem Part 4

B. Articles

How to Solve Mixture Problem Part 1
How to Solve Mixture Problem Part 2
How to Solve Mixture Problem Part 3
How to Solve Mixture Problem Part 4
How to Solve Mixture Problem Part 5

Part II: ENGLISH
Civil Service Exam Vocabulary Review 13

Part III: For Review Guides 1-23, click here.

The Solving Mixture Problems Series

The Solving Mixture Problems Series is a series of tutorials that explain how to solve mixture problems. Mixture problems can be classified into two, those involved with percents and the other one involved with prices.

How to Solve Mixture Problem Part 1 discusses the basics of base and percentage. This is a preparation of the mixture problems involving percents.

How to Solve Mixture Problem Part 2 discusses the most basic of the mixture problems. A detailed solution is discussed about the following problem.

How many liters of 80% alcohol solution must be added to 60 liters of 40% alcohol solution to produce a 50% alcohol solution.

How to Solve Mixture Problem Part 3 discusses another mixture problems involving percents. A detailed solution is discussed about the following problem.

How many liters of pure water must be added to 15 liters of a 20% salt solution to make a 5% salt solution?

How to Solve Mixture Problem Part 4 discusses one more mixture problems involving percents. A detailed solution is discussed about the following problem.

A chemist creates a mixture with 5% boric acid and combined it with another mixture containing 40% boric acid to obtain a 800 ml of mixture with 12% boric acid. How much of each mixture did he use?

How to Solve Mixture Problem Part 5 discusses the basics of mixture problem involving prices. A detailed solution is discussed about the following problem.

A seller mixes 20 kilograms of candy worth 80 pesos per kilogram to candies worth 50 pesos per kilogram. He sold at 60 pesos per kilogram. After selling all the candies he discovered that he had no gain or loss. How much of the 50-pesos per kilogram candies did he use?

How to Solve Mixture Problem Part 6 discusses one more mixture problem involving prices. A detailed solution is discussed about the following problem.

A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black chocolates worth $10 per pack and packs of ordinary chocolates worth $7 each pack. How many packs of each kind should he use to make 15 packs which he can sell for $8 per pack?

How to Solve Mixture Problems Part 1

If you have followed this blog, then you would know that we have been tackling a lot of math word problems. In this series, we will learn to solve another type of math word problem called mixture problems. Mixture problems are easy if you know how to set up the equation.

Mixture problems can be classified into two, those which deals with percent and the other which deals with price. In any case, the method of solving is almost the same. Of course, in working with percent, you must be able to know the basics of percentage problems. Let’s have our first example.

Example 1

How many ml of alcohol does a 80 ml mixture if it contains 12% alcohol?

Solution and Explanation

In this example,the word mixture means that alcohol is mixed with another liquid (we don’t know what it is and we don’t need to know). The total mixture contains 80 ml and we are looking for the pure alcohol content which is 12% of the entire mixture. Therefore,

pure alcohol content = 12%× 80 ml = 0.12 x 80 ml = 9.6 ml

In the calculation above, we converted percent to decimal (12% to 0.12) and then multiply it with 80. This means that in the 80 ml alcohol, 9.6 ml is pure alcohol.

Example 2

What is the total alcohol content of an 80 ml mixture containing 12% alcohol and a 110 ml mixture containing 8% alcohol?

Solution and Explanation

In this problem, we need to “extract” the pure alcohol content of both mixtures and add them in order to find the volume of the pure alcohol content of both mixtures.

In Example 1, we already know that it contains 9.6 ml of alcohol, therefore, we only need to solve for the second mixture.

Pure alcohol content of Example 2 = 8% × 110 ml = 0.08 x 110 ml = 8.8 ml

In the calculation above, we converted 8% to decimal so it became 0.08. Multiplying 0.08 by 110 gives us 8.8 ml. Therefore,

total alcohol content = 9.6 ml + 8.8 ml = 18.4 ml.

That means that the two mixtures contain 18.4 ml of pure alcohol in total.

The second problem shows that if we have more than one mixture, and we want to find the total amount of pure content, then we need to add the pure contents in the mixtures. Now, does the percentages add up? Will the total mixture contain 20% alcohol? Let’s see.

The total amount of liquid = 110 ml + 80 ml = 190 ml
Total amount of alcohol = 18.4 ml

18.4 ml / 190 ml = 0.0968

As we can see, if we convert 0.0968 to percent, it becomes 9.68% and not 20%. In what case will they add up?

We will use the concepts we have learned in these two problems to solve more complicated problems in the next post. Stay tuned by subscribing in the email subscription box on the right part of the page.