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. 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 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.

Video Series: How to Solve Number Problems Mentally

Last year, I have written a tutorial on How to Solve Number Problems Mentally. This article has been turned into a math tutorial video in Youtube under the Sipnayan channel. Sipnayan is a Youtube Channel which contains math tutorial videos in Tagalog.

Embedded below are the three videos in the series.

Part 1 solves the following problem:

One number is 3 more than the other. Their sum is 45. What are the numbers?

Part 2 solves the following problem:

The sum of two numbers is 53. One number is 7 less than the other. What are the numbers?

Part 3 solves the following problem:

One number is twice the other number. Their sum is 45. What are the numbers?

Enjoy learning. If you have questions or comments, please type them below.

How to Solve Digit Problems Part III

This is the third part of the tutorial series on Solving Digit Problems. In Part 1 and Part 2, we used one variable to solve digit problems. In this post, we learn how to use two variables to solve digit problems. We still use the problem in Part 2.

Problem

The sum of the digits of a 2-digit number is 9. If the digits are reversed, the new number is 45 more than the original number. Find the numbers.

Solution and Explanation

Let t = tens digit and u = units digit.

From the first sentence in the problem, we know that

t + u = 9 (1).

Also, as we have learned in the first two parts of this series, 2-digit numbers with tens digit t and ones digit u can be represented (or has value) 10t + u. For example, the number 25 with t = 2 and u = 5 has value 10(2) + 5.

So, the we can represent the original number as

10t + u.

If we reverse the digit, the specific example which is 25 becomes 52. This becomes 10(5) + 2. Hence, we can represent the reverse number as

10u + t.

Therefore, we our representation is as follows:

original number: 10t + u
new number (with digits reversed): 10u + t.

In the second sentence in the problem, it says when the digits are reversed, the new number is 45 more than the original number. That means that if we add 45 to the original number, they will be equal. That is,

original number + 45 = new number.

Substituting the representations above, we have

10t + u + 45 = 10u + t.

We can simplify the equation by putting the variables on the right.

45 = 10u + t – (10t + u)
45 = 10u + t – 10t – u
45 = 9u – 9t (2).

Thus, we have 2 systems of equations.

t + u = 9 (1)
9u – 9t = 45 (2).

Note: We just change reverse the position of the expressions in equation (2).

We can solve this using elimination or substitution. In this solution, we use substitution.

First, we find the value of t in (1)

t + u = 9.

Subtracting u from both sides, we have
t = 9 – u.

Next, we substitute 9 – u to the value of t in (2)

9u – 9t = 45
9u – 9(9-u) = 45
9u – 81 + 9u = 45
18u – 81 = 45
18u = 45 + 81
18u = 126.

Dividing both sides by 18,

u = 7

So, the units digit is 7.

To find t, we substitute in one of the equations in (1) and (2). We substitute in (1),

t + u = 9
t + 7 = 9
t = 2

So, our number has tens digit 2 and ones digit 7. Therefore, the number is 27.

If we check, if the number is reversed, it becomes 72. Let’s see if the number with reversed digit is 45 more than the original number.

72 – 27 = 45.

Therefore, we are correct.

Having two equations in two variables is an example of systems of equation. In the process above, we solved for the value of one of the variables in (1) and substituted it in (2). We will discuss systems of equations, particularly linear equations in two variables in details in the next posts.

Understanding Conditionals IV: Mixed Conditionals

There are two mixed types of sentences of unreal condition:

1.) If – clause refers to the present and the main clause refers to the past.
e.g. If he were a fast runner, he would have won the race.

If – clause refers to the past and the main clause refers to the present.
e.g. If he had found a job, he wouldn’t be searching for one now.

Sometimes we make sentences which mix Second and Third Conditionals, especially when a past event has an effect in the present.

Example:
a.) If you hadn’t invited me, I wouldn’t have gone to the party. (=I did go to the party – Third Conditional).
If you hadn’t invited me, I wouldn’t be here now. (=I’m at the party now. – Third + Second Conditionals)

b.) If you had planned things properly, you wouldn’t have got into a mess. (=You didn’t plan – Third Conditional)

If you had planned things at the start, we wouldn’t be in this mess now (=We are in trouble now – Third + Second Conditionals)
All types of conditionals can be mixed. Any tense combination is possible if the context permits it.

Conditional clause main clause
If nobody phoned him he won’t come to the meeting.
If he knew her, he would have spoken to her.
If he had found a job, he wouldn’t be searching for one now.

Exercises: Put the verbs in brackets into the correct form.

1. If you (not spend) so much money, I (not be) angry now.
2. If they (post) the parcel yesterday, it (get) here before Friday.
3. If you (not wake) me up in the middle of the night, I (not feel) so tired now.
4. If Tom (be) a bit more ambitious, he (find) himself a better job years ago.
5. If you (know) me better, you (say) that.

1. If you hadn’t spent so much money, I wouldn’t be angry now.
2. If they posted the parcel yesterday, it won’t get here before Friday.
3. If you hadn’t woken me up in the middle of the night, I wouldn’t feel so tired now.
4. If Tom was a bit more ambitious, he would have found himself a better job years ago.
5. If you knew me better, you wouldn’t have said that.

How to Solve Digit Problems Part II

In the previous post, we have discussed the basics of digit problems. We have learned the decimal number system or the number system that we use everyday. In this system, each digit is multiplied by powers of 10. For instance, 871 means $(8 \times 10^2) + (7 \times 10^1) + (1 \times 10^0)$.

Recall that $10^0 = 1$.

In this post, we continue this series by providing another detailed example.

Problem

The sum of the digits of a 2-digit number is $9$. If the digits are reversed, the new number is $45$ more than the original number. What are the numbers?

Solution and Discussion

If the tens digit of the number is $x$, then the ones digit is $9 - x$ (can you see why?).

Since the tens digit is multiplied by $10$, the original number can be represented as $10x + (9 - x)$.

Simplifying the previous expression, we have 10x – x + 9 = 9x + 9.

Now, if we reverse the number, then $9 - x$ becomes the tens digit and the ones digit becomes $x$. So, multiplying the tens digit by 10, we have $10(9 - x) + x$.

Simplifying the expression we have 10 – 10x + x =  90 – 9x.

As shown in the problem, the new number (the reversed number) is $45$ more than the original number. Therefore,

reversed numberoriginal number = 45.

Substituting the expressions above, we have

90 – 9x – (9x + 9) = 45.

Simplifying, we have $90 - 9x - 9x - 9 = 45$ $81 - 18x = 45$ $18x = 81 - 45$ $18x = 36$ $x = 2$.

Therefore, the tens digit of the original number is 2 and the ones digit is $9 - 2 = 7$.

So, the original number is $27$ and the reversed number is $72$.

Now, the problem says that the new number is $45$ more than the original number. And this is correct since $72 - 27 = 45$.

How to Solve Digit Problems Part I

Digit Problems is one of the word problems in Algebra. To be able to solve this problem, you must understand how our number system works. Our number system is called the decimal number system because the numbers in each place value is multiplied by powers of 10 (deci means 10). For instance, the number 284 has digits 2, 8, and 4 but has a value of 200 + 80 + 4. That is, $(100 times 2) + (10 times 8) + (4 times 1) = 284$.

As you can observe, when our number system is expanded, the hundreds digit is multiplied by 100, the tens digit is multiplied by 10, and the units digit (or the ones digit) is multiplied by 1. Then, all those numbers are added. The numbers 100, 10, and 1 are powers of 10: $10^2 = 100$, $10^1 = 10$, and $10^0 = 1$. So, numbers with $h$, $t$, and $u$ as hundreds, tens, units digits respectively has value $100h + 10t + u$.

It is clear that this is also true for higher number of digits such as thousands, ten thousands, hundred thousands, and so on.

Many of the given numbers in this type of problem have their digits reversed. As we can see, if 10t + u is reversed, then it becomes $10u + t$. For instance, $32 = 10(3) + 1(2)$ when reversed is $23 = 10(2)+ 1(3)$. Now, that we have already learned the basics, we proceed to our sample problem.

Worked Example

The tens digit of a number is twice the units digit. If the digits are reversed, the new number is 18 less than the original. What are the numbers?

Solution and Explanation

The tens digit of a number is twice the unit digit. This means that if we let the units digit be $x$, then the tens digit is $2x$. As we have mentioned above, we multiply the tens digit with 10 and the units digit with 1. So, the number is $(10)(2x) + x$.

Now, when the digits are reversed, then x becomes the tens digit and $2x$ becomes the ones digit. So, the value of the number is $(10)(x) + 2x$.

From the problem above, the number with reversed digit is 18 less than the original number. That means, that if we subtract 18 from original number, it will equal the new number. That is, $(10)(2x) + x - 18 = 10(x) + 2x$ $20x + x - 18 = 12x$ $21x - 18 = 12x$ $9x = 18$ $x = 2$ $2x = 4$

So, the number is 42 and the reversed number is 24.

Check: 42 – 24 = 18.

The Fraction-Decimal-Percent Conversion Series

If you are having a hard time converting fractions to decimals and vice versa, converting fraction to percent and vice versa, and converting decimal to percent and vice versa, you can read the posts below. I will be creating practice tests with complete solutions later, so stay posted.

Enjoy learning and good luck for the May 3 exam. 1 2 3 4 5 17