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 3

In the previous post, we have discussed three examples on how to solve mixture problems. In this post, we are going to learn how to set up the givens in a mixture problem in a table, so it is easier to solve.  Let’s have the fourth example.

Example 4

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

Solution and Explanation

In the first column, we placed the liquid or solution. We have three kinds: water, the liquid with 20% salt, and liquid with 5% salt. In the second column, we place the volumes of ths liquid. We do not know the volume of water, so we represent it with x. Since we have to combine water with 20% salt solution, we have to add the volumes of the two liquid.  This makes sense since if we add more water, the amount of salt is getting lesser in relation to the total volume of the liquid.

mixture problems

Again, the total amount of salt when water is combined with 20% salt solution should also be equal to the total amount of salt in the 5% salt solution. That means that in the last column, we have to add the first and the second row and then equated to the third row.  That is,

3 = 0.05(x + 15)

3 = 0.05x + 0.75

2.25 = 0.05x

45 = x

That means that we need 45L of water to turn a 20% salt solution to a 5% solution.

Check:

3 = 0.05(x + 15)

3 = 0.05(45 + 15)

3 = 0.05(60)

3 = 3

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.

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.

The Solving Digit Problems in Algebra Tutorial Series

The Solving Digit Problems in Algebra Tutorial Series is a series of tutorials on solving digit problems. Digit Problems is one of the types of Word Problems in Algebra. Below is the list of posts.

  • How to Solve Digit Problems Part I  explains the basics of digit problems including the concepts behind the strategy in solving. The particular concepts discussed are place values and the decimal number system.  One problem involving 2 digit numbers was also discussed including its detailed solution. The solution to the problem involves one variable.
  • How to Solve Digit Problems Part II is the continuation of the first part. This is a review on the number system and one more sample problem involving 2 digit numbers is discussed. The solution to this problem also involves one variable
  • How to Solve Digit Problems Part III introduces the use of two variables in solving digit problems with numbers containing 2 digits.
  • How to Solve Digit Problems Part IV introduces the basic of solving digit problems involving 3-digit numbers. Three variables are used to solve the problem in this part.
  • How to Solve Digit Problems Part V presents another detailed example on how to solve digit problems involving 3-digit numbers. Systems of Equations in 3 variables was used to solve the problem in this part.

If you have comments about the problems, kindly type them below.

If you want to learn more about word problems, visit our Word Problem page.

If you want to learn more about mathematics, please go to the Math page.

To view the complete list of posts, you can visit the Post List page.

How to Solve Digit Problems in Algebra Part V

This is the last part of the series on How to Solve Digit Problems in Algebra. In this post, we are going to solve another digit problem involving 3-digit numbers just like in the previous post.

Problem

The sum of the digits of a 3-digit number is 10. The hundreds digit is 3 more than the tens digits. If 198 is subtracted from the number, then the digits are reversed. What is the number?

Solution and Explanation

Since this is already the fifth part of the series, we won’t be as detailed as the previous parts.

Let

h = hundreds digit of the number
t = tens digit of the number
u = units (or ones) digit of the number.

The first sentence says the sum of the digits of the number is 10. So,
h + t + u = 10 –> (Equation 1).

The hundreds digit is 3 more than the tens digit.
h = t + 3 –> (Equation 2).

If 198 is subtracted from the number, the digits are reversed.

100h + 10t + u - 198 = 100u + 10t + h.

Simplifying the preceding equation, we have

(100h - h) + (10t - 10t) + (u - 100u) = 198
99h - 99u = 198.

Dividing both sides by 99, we have
h - u = 2 –> (Equation 3).

We can eliminate u by adding Equation 1 and Equation 3.
(h + t + u) + (h - u) = 10 + 2
2h + t = 12 (*).

We substitute t + 3 from  Equation 2 to h in (*)
2(t + 3) + t = 12
2t + 6 + t = 12
3t + 6 = 12
3t = 6
t = 2.

Substituting the value of t in Equation 2,
h = t + 3
h = 2 + 3
h = 5.

Substituting the values of h and t in Equation 1,
h + t + u = 10
5 + 2 + u = 10
7 + u = 10
u = 10 - 7
u = 3.

So, the hundreds digit is 5, tens digit is 2, and ones digit is 3. Therefore, the number is 523.

If we check with the conditions above, we have
(1) The sum of the digits is 10. That is, 5 + 2 + 3 = 10
(2) The hundreds digit is 3 more than the tens digit. The hundred digit is 5 is 3 more than 2
(3) If 198 is subtracted from the number, the digits are reversed. That is, 523 – 198 = 325.

How to Solve Digit Problems in Algebra Part IV

In the previous posts, Part 1, Part 2, and Part 3, we have learned how to solve digit problems involving 2-digit numbers. In this post, we are going to discuss digit problems involving 3-digit numbers. Let’s have the following problem.

Problem

The hundreds digit of a 3-digit number is twice its units digit and its tens digit is 1 more than its units digit. If 297 is subtracted from the number, then its digits are reversed. What is the number?

Solution and Explanation

Before we solve the problem, let’s recall that in the first two parts of this series that a number with tens digit t and units digit u can be represented by 10t + u and when we reverse its digits can be represented by 10u + t. For example, the number 52 with t = 5 and u = 2 can be represented as 10(5) + 2 and when we reverse the digits, it becomes 25 = 10(2) + 5.

In the same way, if we let h be the hundreds digit of a number, t be the tens digit, and u be the units digit, then we can represent the number as

100h + 10t + u

and with its digits reversed as

100u + 10t +h.

In the first sentence of the problem above, it says that the hundreds digit is twice the units digit. Therefore,

h = 2t (*).

In the second sentence, it says that if 297 is subtracted from the number, then the digits is reversed. Putting this in equation, we have

number – 297 = number with digits reversed.

That is,

100h + 10t + u - 297 = 100u + 10t +h.

Simplifying, we have

(100h - h) + (10t - 10t) + (u - 100u) = 297
99h - 99u = 297.

Dividing both sides by 99, we have

h - u = 3.

But from (*), h = 2u. So, substituting, we have

2u - u = u = 3.

So, the units digit is equal to 3.

Now, the tens digit is 1 more than the units digit, so it’s 4.

The hundreds digit is twice the units digit, so it’s 6.

Therefore, the number is 643.

Check: Let’s try to subtract 297 and see if the digits are reversed.

643 - 297 = 346.

Related Posts Plugin for WordPress, Blogger...
1 8 9 10 11 12 29