Percentages with mental math

(This is an older post that I have revised plus added a video to it.)
In this article I want to explore some ideas for using MENTAL math in calculating percents or percentages.

I have also made a video about this topic:



And here are the ideas:

1. Find 10% of some example numbers (by dividing by 10).
2. Find 1% of some example numbers (by dividing by 100).
3. Find 20%, 30%, 40% etc. of these numbers.
   FIRST find 10% of the number, then multiply by 2, 3, 4, etc.
   For example, find 20% of 18. Find 40% of $44. Find 80% of 120.
   
   I know you can teach the student to go 0.2 × 18, 0.4 × 0.44, and 0.8 × 120 - however when using mental math, the above method seems to me to be more natural.
4. Find 3%, 4%, 6% etc. of these numbers.
   FIRST find 1% of the number, then multiply.
5. Find 15% of some numbers.
   First find 10%, halve that to find 5%, and add the two results.
6. Find 25% and then 75% of some numbers. 25% of a number is 1/4 of it, so you find it by dividing by 4. For example, 25% of 16 is 4. To find 75%, first find 25% and multiply that by 3.
7. Calculate some simple discounts. If an item is discounted 20%, 15%, 25%, 75% etc., then find the new price.
   
   For example, a book costs $40 and is discounted by 15%. What is the new price?
   First find 15% of $40 (10% of $40 is $4 and 5% of $40 is $2... so 15% of it is $6). Then subtract $40 - $6. So the new price is $34.
   
8. "40% of a number is 56. What is the number?" - types of problems.
   
   You CAN do this mentally: First FIND 10% of the number, and then multiply that result by 10, and you'll get 100% of the number - which is the number itself.
   
   So if 40% is 56, then 10% is 14 (divide by 4). Then, 100% of the number is 140. This result is reasonable, because 40% of this number was 56, so the actual number (140) needs to be more than double that.
9. "34% of a number is 129. What is the number?" (A calculator will help here.)
   
   You don't need to write an equation. You can just first find 1% of this number, and then find 100% of the number.
   
   If 34% of a number is 129, then 1% of that number is 129/34. Find that, and multiply the result by 100.

Hope you enjoyed these little mental math ideas! They also help students understand the concept of percent where they don't end up relying too much on mechanical calculations or equations.

Dividing decimals

I feel students need to get grounded conceptually in this topic. So many times, all they learn about decimal division are the rules of how to go about decimal division when using long division, and it becomes an "empty" skill - a skill that lacks the conceptual foundation.

So for starters, we can do two different kinds of mental math division problems.

1. Division by a whole number - using mental math
   
   Here it is easy to think, "So much is divided between so many persons".
   
   0.9 ÷ 3 is like "You have nine tenths and you divide it between three people. How much does each one get?" The answer is quite easy; each person "gets" 0.3 or three tenths.
   
   And... remember ALWAYS that you can check division problems by multiplication. Since 3 × 0.3 = 0.9, we know the answer was right.
   
   0.4 ÷ 100 turns out to be an easy problem if you write 0.4 as 0.400:
   0.400 ÷ 100 is like "You have 400 thousandths and you divide it between 100 people; how much does each one get?" The answer is of course 4 thousandths, or 0.004. Check: 100 × 0.004 which is 100 × 4/1000 = 400/1000 or 0.400 = 0.4.
   
   Here are some more similar ones:
   
   0.27 ÷ 9
   
   0.505 ÷ 5
   
   0.99 ÷ 11
   ...and you can make more, just think of the multiplication tables.
   



2. Division where the quotient (answer) is a whole number
   
   This time it helps to think, "How many times does the divisor go into the dividend?" In these types of mental math problems, the answer ends up being a whole number. (Of course the teacher has to plan these problems just right.)
   
   For example, 0.4 ÷ 0.2. Ask, "How many times does 0.2 fit into 0.4?" The answer is, 2 times. So 0.4 ÷ 0.2 = 2. Again, we can check it by multiplying: 2 × 0.2 = 0.4.
   
   Other similar division problems to solve mentally:
   
   1 ÷ 0.5
   
   3 ÷ 0.5
   
   0.09 ÷ 0.03
   
   0.9 ÷ 0.1
   
   2 ÷ 0.4
   
   1 ÷ 0.01
   
   ...and so on.
   

This decimal division lesson taken from my Decimals 2 book illustrates these two kinds of mental division problems.

Towards the general case

After the student is familiar with the two special cases above, we can go forward and study decimal division problems in general. Even here, we will divide the problems into two classes, depending on whether the divisor is a whole number or not.

1. The divisor is a whole number.
   
   For example, 3.589 ÷ 4 or 0.1938 ÷ 83. These can simply be solved by long division as they are. Just put the decimal point in the same place in the quotient as where it is in the dividend.
   
   The "stumbling block" may come when the division is not even (this also leads into the study of repeating decimals). Generally, you can continue the division indefinitely by tagging zeros to the dividend, such as making 3.589 to be 3.589000. Then when you've continued the division as long as you wish (or as long as the book tells you to do it), cut the decimal off at a desired accuracy and round it.
   
   Typical problem in a textbook would say, "Do 2.494 ÷ 3 and give your answer with 3 decimal digits." For this, you need to do the long division until the fourth decimal digit - so as to be able to round to 3 decimal digits. Since 2.494 does not have four decimal digits, you tag a zero to it to make it have so (2.4940).
   
   Fortunately, this process is not generally difficult. It's the second case that's more of a problem.
   
   



2. The divisor is not a whole number.
   
   Here, we do something quite special before dividing, and turn the problem into one where the divisor is a whole number. Then, the actual division is done like explained above.
   
   I say this is special, because this special thing that we do is based on a very important general principle of arithmetic:
   
   If you multiply both the dividend and the divisor by some same number, the quotient won't change.
   
   Let's see it in action with some easy numbers:
   
   1000 ÷ 200 = 5
   
   100 ÷ 20 = 5
   
   10 ÷ 2 = 5
   
   Each time both the dividend and the divisor change by a factor of ten, but the quotient does not change.
   
   We can also try it using a factor of 3 (or any other number):
   
   8 ÷ 2 = 4
   24 ÷ 6 = 4
   72 ÷ 18 = 4
   
   Let's try one more time, with a factor of 2:
   
   30 ÷ 6 = 5
   
   15 ÷ 3 = 5
   
   7.5 ÷ 1.5 = 5
   
   3.75 ÷ 0.75 = 5
   
   H hopefully by now you have convinced the student(s) of this principle. Now we can apply it to those pesky decimal division problems.
   
   
   
   
   This image shows how the decimal division problem 0.644 ÷ 0.023 can be changed into another problem, with a whole number divisor, and with the same answer.
   
   In each step, we multiply both the dividend and the divisor by 10. This, of course, is the same process as moving the decimal point.
   
   Many textbooks only show the student the "trick" of moving the decimal point... but don't show him what that idea is based on.
   
   An example
   
   To solve 13.29 ÷ 5.19, we need to first change the problem so that the divisor 5.19 is a whole number. We multiply both the dividend and the divisor by 10 as many times as needful to accomplish that:
   
   13.29 ÷ 5.19
   = 132.9 ÷ 51.9
   = 1329 ÷ 519, and now off you go to do long division... I'm not saying it's the easiest long division problem in the world, since the divisor is 519. Let's try an easier one.
   
   
   2,916 ÷ 0.02
   = 29,160 ÷ 0.2
   = 291,600 ÷ 2 and now you can do the long division.
   
   Of course, in reality you can also multiply by 100 instead of taking two steps of multiplying by 10. But students can start out by multiplying by 10 as many times as needed.
   

Please also see the lesson on dividing decimals by decimals, from my Math Mammoth Decimals 2 book.

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

Squares that aren't squares?

Updated with solutions!

Today I want to highlight a square problem I saw at MathNotations. I hope Dave Marain doesn't mind me showing this picture and problem on my blog... I have no problem acknowledging it's from his blog. I COULD just tell you all to "go read it at Dave's blog....

BUT I don't feel that's the best way, IF I want you to think about this. I can just guess that most of the folks would feel too lazy to click on the link and go read it there (would you?). So I want to show it here.


Figures not drawn to scale! And this is important!

Now here's the question:

Does the given information in each diagram guarantee that each is a square?

If you don't think so, your mission is to draw a quadrilateral with the given information but that clearly does NOT look like a square.

The IDEA is to make our students THINK LOGICALLY, or practice their deductive reasoning skills. A great little problem.

The answers:


Figure 1 is not necessarily a square. The upper left corner angle can be of any size. The upper side can be of any length. And so on. See here two examples.

Figure two is not necessarily a square either since the "top" side can be of any length. But it is a rectangle.

Figure 3 in the original problem IS always a square!

Now, I'll write another post where we'll extend this idea to some parallelograms.

A bar diagram to solve a ratio problem

Dave at MathNotations had an interesting ratio problem:

In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?

He asked if it can be solved using "Singapore" style bar model.

I'm not sure if this is exactly how they'd do it, but this is how I'd do it... so here goes.

After I made the diagrams, I soon saw that Dave's numbers are two awkward; the bar diagram drawing would get too messy because we'd need to divide it into too tiny parts to see anything.

BUT... you probably know about the PROBLEM SOLVING STRATEGY called "solve an easier problem". My agenda is therefore:

• show how to solve a few related simple ratio problems using the bar diagram


• solve a variant of the original problem (with friendly numbers)


• solve the original problem.

1. Here's a bar diagram representing the fact that the ratio of the number of juniors to seniors is 7:5.

As you can see, the "whole" ends up divided into 12 parts (7 + 5).

We can use this diagram to solve problems such as:

If there are 768 juniors and seniors in total, how many juniors are in the school?
(Divide that into 12 parts, and then take 7 of those parts.)

Or...

If there are 95 seniors, how many juniors are there?
(Divide the number of seniors by 5, then multiply by 7.)

Or..
If there are 49 Juniors, how many students are there in all?
(Divide the number of juniors by 7, then multiply by 12.)

The diagram makes all this dividing/multiplying by 5/7/12 all crystal clear.

---

2. Let's change the numbers from the original problem to make them "friendly" for this approach. Let's solve this instead:

In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 4:3.
The ratio of senior males to senior females is 7:3.
What is the ratio of junior males to senior females?

Our diagram becomes, first of all, like before:

But then we add the additional information into it. The ratio of junior males to females is 4:3, which means juniors as a whole are divided into 7 parts (4 + 3). Similarly, seniors need divided into 10 parts.

Voila! It just so happens (wonder why?) that our original division into 7 and 5 parts works beautifully to give us juniors divided into 7 parts (they already are!) and seniors into 10 parts if we only split each part into 2.

So, what is the ratio of junior males to senior females?

In the diagram, the WHOLE is now divided into 24 parts. Junior males are 8 of those parts, and senior females are 3 of those parts. Their ratio is therefore 8:3.

---

3. Solving the original problem, with those unfriendly numbers.... the diagram looks sort of like this:

It gets messy... I'd prefer using the bar model as a stepping stone and to illustrate the basic situation, but eventually using the "least common denominator" or algebraic methods.

Shorter-wider table problem

I failed to publish this the other day since Blogger was acting up. I'll try it now.

A rectangular kitchen table is three times as long as it is wide.
If it were 3 m shorter and 3 m wider it would be a square.
What are the dimensions of the rectangular table?

1) You can solve this using algebra: set width to be x, the length is then 3x. We know these two will be equal if the former is increased by 3, and the latter is decreased by 3:

x + 3 = 3x - 3

2x = 6

x = 3.
The dimensions are 3 and 9.

But since this is supposed to be a 6th grade problem, surely we can find another way to solve it, as well.

2) Think of the two quantities length and width as just numbers. If you reduce one by 3, and increase the other by 3, they will "meet" or be the same. Below, I've drawn the two numbers as lines; you could use bars.


length |---------------------------------|

width |----------|


If I decrease the length by 3 and increase the width by 3, they'll be equal:

length |---------------------|<------------|

width |----------|--------->|

So... the DIFFERENCE between these two is 6. The difference of 6 is also TWO TIMES the original width, because remember the length is 3 times the width.

So the original width is 3 m and the original length is 9 m (a giant table!).

3) Solving it geometrically is also quite easy. First draw a table that looks about 3 times as long as it is wide. Then trace the sides if it was a square, making the long side shorter, and short side longer:

It sure looks like that the side of the square should be double 3, or 6. The original sides would then be 3 and 9. And you can easily check out that those work.




--------------------------------------------------------------

I feel like the numbers in this problem are almost too easy. But maybe this problem could be used as a stepping stone for a harder version of the same.

Tips for teaching integers

The main struggle with integers comes, not with the numbers themselves, but with some of the operations. There seem to be so many little rules to remember (though less than with fractions).

Some good real-life MODELS for integers are:

- temperature in a thermometer

- altitude vs. sea depth

- earning money vs. being in debt.

When first teaching integer operations, tie them in with one of these models.

I'll take for example the temperature.

Assuming n is a positive integer, the simple rules governing this situation are:

* x + n   means the temperature is x° and RISES by n degrees.
* x − n   means the temperature is x° and DROPS by n degrees.

It's all about MOVEMENT — moving either "up" or "down" the thermometer n degrees.

For example:

• 6 − 7 means: temperature is first 6° and drops 7 degrees.


• (-6) − 7 means: temperature is first -6° and drops 7 degrees (it's even colder!).


• (-2) + 5 means: temperature is first -2° and rises 5 degrees.


• 4 + 5 means: temperature is first 4° and rises 5 degrees.

These simple situations handle adding or subtracting a positive integer. Practice those first, until kids are familiar with these cases.

The remaining cases to handle ar adding or subtracting a negative integer:

• (-2) + (-5) would mean: temperature is first -2° and you "add" more negatives so it gets even colder.

The last case is least intuitive one:

• 1 − (-5) or subtracting a negative integer. I personally just remember the little rule of "two negatives turns into a positive".
  
  Some people explain it this way. In (-7) − (-3) you can think that you have 7 negatives at first, and you "take away" three of those negatives, leaving -4.
  
  This rule of "two negatives makes a positive" might seem counterintuitive at first, but it is needful so that many principles of mathematics can continue to apply (for example distribuitive property).

See also an excellent treatise of integers vs. submarine depth at Text Savvy. Excerpt:

"When you add or subtract with integers, you are NOT combining collections or extracting from collections; you are moving in certain directions."

The "collections" idea does work nicely, for ADDITION:

• 7 + (-4) means you have a collection of 7 red balls and 4 blue balls. A pair of one red and one blue ball "cancels" or becomes zero. Total therefore will be 3 red balls.


• (-3) + (-9) means you have 3 blue balls and 9 blue balls more. Total 12 blue balls, or -12.

However, this "moving" idea is exactly how I have always intuitively done simple integer problems — except adding (negative) + (negative) and subtracting a negative, which I change to adding a positive.

Some books might present the rules for adding integers this way:

• If you add two integers that have the same sign, add the absolute values and put the same sign as what the numbers had.


• If you add two integers that have a different sign, subtract their absolute values and the answer will have the same sign as the number wiht greater absolute value.

Then they instruct to change the subtraction to addition; for example 5 − 7 becomes 5 + (-7) and (-4) − 2 becomes (-4) + (-2), and 5 − (-3) becomes 5 + 3.

While these are technically totally correct, I find it SO much easier to use the "moving" idea for most quick integer calculations. It is easier to start that way, and then learn these other rules to be used with more complex expressions, such as when adding many integers, or with negative and positive decimals.

Please feel free to download a simple integer addition and subtraction worksheet that practices some of these ideas (from Math Mammoth Grade 6-B Worksheets collection).

You will also benefit greatly from reading my previous article on teaching integers, which goes through all four integer operations.