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.

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 triangle problem to solve

I have another neat problem for you to solve. This one should be accessible to middle schoolers on up.

The sides of triangles A and B measure 5, 5, 8 and 5, 5, 6 respectively. What is the ratio of the area of triangle A to that of triangle B? Express in simplest a:b form.

This problem is original to John Morse. He is a local mathematics researcher/
author/tutor/computer programmer in Delmar, NY, and has written this problem to help learners use creative and problem-solving skills in various ways. And I think this problem can indeed help in that - it can be solved in many different ways.

[update - solution follows]

I like this problem because there are many ways to solve it and to use it with different grade-level students - such as is already mentioned in the comments.

1) You could use this as a drawing and measuring exercise with 6th or 7th graders who have learned to do compass and ruler constructions.

Once they know how to construct a triangle given its three sides (or see here as well), they can construct these two, then draw & measure the altitudes, and calculate the areas.

Of course the downside is that measurements are bound to be inaccurate, and so they are unlikely to get exactly 12 square units as the area. But then again, you could discuss this topic as well (inaccuracy of measurements).


2) Another way to (clumsily) solve this is to calculate the areas of the two triangles using Heron's formula (if you know it exists!) Just plug the numbers in.

3) The elegant way... accessible to students who know about Pythagorean Theorem. I'll take you through it step by step.

The first step is to NOTICE that the triangles are isosceles ("same-legged"), and to draw a sketch.

(Mine is not a "rough sketch" but as exact as can be done in PhotoImpact, because I started out by drawing the three sides with the given lengths. Then I calculated one angle, and turned one side in that angle to get a "corner".)


In order for the students to easily see the solution, it is fairly important to be able to sketch the two triangles.

When drawing isosceles triangles, think of the capital letter "A" and its two "legs". We just need to draw one that "stands" taller and one that "stands" with its legs wider open.

Notice one of them is close to an equilateral triangle since it's sides are "almost" the same.. 5, 5, and 6. So you would sketch it as almost an equilateral triangle, just make those "legs" to open a little more. Then the 5-5-8 triangle needs to open its "legs" quite wide to make for the wide 8-unit base.



Now, we are concerned about their AREAS, so naturally we need an altitude. Once you draw the two altitudes, the idea is usually to find how many units long the altitudes are, so you could calculate the area.

But in this case, there's sort of a shortcut. You will realize (or remember) that the altitude forms two little RIGHT triangles in each triangle.



In the triangle on the left, the two known side lengths of the little triangle are 5 and 3, and in the other the two known lengths are 5 and 4.

This would HOPEFULLY ring a bell.... DING! REMEMBER the 3-4-5 right triangle? Would it fit here?

Oh yes, it does. In BOTH triangles, the little triangles are 3-4-5 right triangles!

To prove it, you CAN use the Pythagorean theorem: in one little triangle, you'd have x, 4, and 5 (5 being the hypotenuse), so solving x² + 4² = 5² gives you x = 3, and in the other triangle, solving x² + 3² = 5² gives you x = 4.

But there's no need to go through the solving if you remember that 3-4-5 is a right triangle.



OK, now we know the altitudes and could calculate the areas... BUT wait! If the little triangles are identical (congruent), then their areas are the same, and the areas of the large triangles are the same as well. We only need to find the ratio of the two areas, not the actual areas.

So with two triangles with the same area, the ratio is of course 1:1.

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.

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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.

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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.

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.

Percent of change

When some quantity changes, such as a price or the amount of students, we can measure either the absolute change ("The price increased by $5" or "There were 93 less students this year"), or the percent change.

In percent change, we express WHAT PART of the original quantity the change was.

For example, if a gadget costs $44 and the price is increased by $5, we measure the percent change by first considering WHAT PART $5 is of $44. Of course the answer is easy: it is 5/44 or five forty-fourths parts.

To make it percent change, however, we need to express that part using hundredths and not 44th parts. this happens to be easy, too. As seen in my previous post, you COULD make a proportion to find out how many hundredths 5/44 is:

5/44 = x/100

To solve this, you simply go 5/44 x 100, which is easy enough to remember in itself. In fact, this is the rule often given: you compare the PART to the WHOLE using division (5/44), and multiply that by 100.

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There were 568 students one year, and 480 the next year. By how may percent did the student population decrease?

You first calculate the absolute change, which is 568 − 480 = 88. Then we find what part of the original population is 88 (it is 88/568), and express that using hundredth parts (percents):

88/568 x 100 = 15.49%

The student population decreased by 15.49%.

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Often we are given the opposite problem: we know the percent of change and the original situation, and are asked about the new situation.

The price was $4.55 and increased by 14.78%. What is the new price?

Here, we'd need to find the price increase, or the absolute change in price first. We know the percent part of the total (it is 14.78/100) and the total amount, so multiplying those we get the part as a dollar-amount: 14.78/100 x $4.55 = $0.67249. So this is the increase. To find the new price, add the increase to the original: $4.55 + $0.67249 = $5.22249 = $5.22.

Instead of multiplying by 14.78/100, it is far quicker to multiply by 0.1478 — or to change the percent-amount 14.78% to a decimal 0.1478 and multiply by it.

And, since in the end we need to add the original total, the whole calculation looks like this:

0.1478 × $4.55 + $4.55

Here, using distributive property we can make it look like this:

= $4.55 (0.1478 + 1) = $4.55(1.1478) = 1.1478 × $4.55

So it can all be done in one multiplication. Instead of multiplying by the decimal 0.1478, you add 1 to it before multiplying.

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Then one more possible problem type is that you know the percent of change and the actual change amount (absolute change), and are asked the original and/or the new total.

The price increased by 13%, or by $10.14. What was the original price?

Let the original price be p. Then you can build an equation based on the idea that the price increase is 13% of p:

0.13p = $10.14   or   13/100p = $10.14

p = $10.14/0.13 = $78.

The new price would be found by adding.

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Some other lessons to read are below:

Percent Of Change - Lesson and Problems

General Increase and Decrease Examples from Purplemath.com

Percent of change calculator - enter the original and the changed quantities, and it calculates the percent of change.

PERCENT INCREASE OR DECREASE lesson from TheMathPage.com

Math Mammoth Grade 7 (prealgebra) worksheets available

Just as of today, I got online and available the new Math Mammoth Grade 7-A and 7-B worksheets collections.

These pre-algebra sheets contain problems for beginning algebra topics such as expressions, linear equations, and slope - plus your typical 7th grade math topics such as integers, fractions, decimals, geometry, statistics, and probability.

You will find the problems are very varying as these worksheets have been created one by one (not script made).

Go download and enjoy the free sample sheets! - including a Percent Fact Sheet.