Browse Tag: mixture problems algebra

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 6

This is the 6th part and last part of the Solving Mixture Problems Series. In the previous 4 parts, we have learned how to solve mixture problems involving percent and in part 5, we have learned how to solve problems involving percents. In this post, we solve another problem involving percent.

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?

Solution

Let x = number of $10 packs
15 – x = number of $7 packs

Multiplying the cost per pack and the number of packs we have

($10)(x) = total cost of $10 packs
($7)(15 – x) = total cost of the $7 packs
($8)(15) = total cost of all the chocolates

Now, we know that

total cost of $10 packs + total cost of the $7 packs = total cost of all the chocolates.

Substituting the values, we have

($10)(x) + ($7)(15 – x) = ($8)(15).

Eliminating the dollar sign and solving for x, we have

(10)(x) + (7)(15 – x) = (8)(15)
10x + 105 – 7x = 120
3x + 105 = 120
3x = 120 – 105
3x = 15
x = 5.
This means that we need 5 packs of $10 and 10 packs of $7 chocolates.

Check:
($10)(5) + ($7)(10) = ($8)(15)
$50 + $70 = $120
$120 = $120
Therefore, we are correct.

How to Solve Mixture Problems Part 5

In the previous posts, we have learned how to solve mixture problems involving percentages and liquid mixture problems. In this post, we are going to solve mixture problems involving prices. Although these two types of problems are different, they are very similar when you set up the equation. Below is our first problem.

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.

candy solve mixture problems

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?

Solution and Explanation

Let x = number of kilograms of candy worth 50-pesos per kilogram.

The total price of 20 kilograms of candy at 80 pesos per kilogram is (20 kg)(80 pesos/kg) = 1600 pesos.

The total price of x kilograms of candy at 50 pesos per kilogram is (x kg)(50 pesos/kg) pesos.

When we add these 2, the number of kilograms of candy is x + 20 (can you see why?) and it is sold at 60 pesos. So, its total price is (x + 20)(60 pesos/kg).

Using these facts, we have the following equation:

total price of 80pesos/kg candy + total price of 50pesos/kg candies = total price of 60pesos/kg candies.

Substituting the expressions above, we have

(20 kg)(80 pesos/kg) + (x kg)(50 pesos/kg) = (x + 20 kg)(60 pesos/kg)

1600 + 50x = 60(x + 20)
1600 + 50x = 60x + 1200
1600 – 1200 = 60x – 50x
400 = 10x
40 = x.

Therefore, he used 40 kilograms of candy worth 50 pesos per kilogram.

Check:

1600 + 50(40) = 60(40 + 20)
1600 + 2000 = 60(60)
3600 = 3600.

This means that we are correct.

How to Solve Mixture Problems Part 4

This is the fifth post of the Solving Mixture Problems Series on PH Civil Service Review. In this post, we are going to solve a problem in which only the total amount of mixture is given.

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?

Solution and Explanation

Let

x = mixture with 5% boric acid

800 – x = mixture with 40% boric acid.

Note that if these two mixtures are combined, we will produce a mixture with 800 ml of solution with 12% boric acid. As we have learned in the previous tutorials, the amount of boric acid in the first mixture added to the amount of boric acid in the second mixture is equal to the amount of boric acid in the combined mixtures. That is,

(5%)(x) + (40%)(800-x) = (12%)(800).

Take note that x, 800 – x, and 800 are amount of the mixture and if these are multiplied by the percentage of boric acid, then we will get the exact amount of pure boric acid. 

Converting percent to decimals, we have

(0.05)(x) + (0.4)(800-x) = (0.12)(800)

0.05x + 320 – 0.4x = 96.

Simplifying, we have

-0.35x = -224

x= 640.

That means that we need 640 ml of mixture with 5% boric acid and 800 – 640 = 160 ml of mixture with 40% boric acid.

Check:

Amount of boric acid in mixture with 5% boric acid: (640 ml)(0.05) = 32ml
Amount of boric acid in mixture with 40% boric acid: (160 ml)(0.4) = 64ml
Amount of boric acid in mixture with 12% boric acid: (800 ml)(0.12) = 96ml

As we can see, 32 ml + 64 ml = 96 ml which means that we are correct.

How to Solve Mixture Problems Part 2

In the previous post, we have learned the basics of mixture problems. We have learned that if solutions are added, then the pure content of the combined solution is equal to the sum of the all the amount of pure content in the added solutions.

In this post, we are going to discuss two more mixture problems. We have already finished two examples in the previous part, so we start with Example 3.

Example 3

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

Solution and Explanation

The first thing that you will notice is that we don’t know the amount of liquid with 80% alcohol solution. So, if we let x = volume of the solution of liquid with 80% alcohol content. So, the amount of pure alcohol content is 80% times x.

We also know that the amount of alcohol in the second solution is 60% times 40.

Now, if we let solution 1 be equal to the solution with 80% alcohol, solution 2 with 40% alcohol, and solution 3 be the combined solutions, we have

amt of alchohol in solution 1 = 0.8x
amt of alcohol in solution 2 = (0.4)(60)
amt of alchohol in solution 3 = 0.50(0.8x + 60)

Note that we have already converted the percentages to decimals in the calculation above: 80% = 0.8, 40% = 0.4, and 50% = 0.5.

amt. of alcohol in solution 1 + amt. of alcohol in solution 2 = amt. of alcohol in solution 3

Solving, we have

0.8x + (0.4)(60) = 0.50(x + 60)

0.8x + 24 = 0.5x + 30

0.8x – 0.5x = 30 – 24

0.3x = 6

To get rid of the decimal, we multiply both sides by 10.

3x = 60

x = 60/3

x = 20.

This means that we need 20 liters of solution 1, the solution containing 80% alcohol. We need to combine this to solution 2, to get solution 3 which has a 50% alcohol content.

In problems like this, you can check your answers by substituting the value of x to the original equation.

0.8x + (0.4)(60) = 0.50(x + 60)

0.8(20)+ (0.4)(60) = 0.50(20 + 60)

16 + 24 = 0.50(80)
40 = 40

Indeed, the amount of alcohol in the left hand side is the same as the amount of alcohol in the right hand side.