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Posts tagged ‘Mental Math’

Conquering Proportions, Part 2


In my first “Conquering Proportions” post, I showed how to save time by canceling terms horizontally as well as vertically. In this post you’ll learn how to save even more time with another shortcut. Let’s look at an example to refresh our memory.

Given a proportion such as this:

15   =   5  
 a         3

most people would do the traditional “cross-multiplying” step, to get:

5 x a = 15 x 3  (the x here is a true times sign; that’s why I’m using ‘a‘ as the variable, not ‘x.’)

If you follow the usual steps, the next thing would be to ÷ both sides by 5, to get:

a  =  (15 x 3) ÷ 5

But let’s look more closely at this answer expression:  (15 x 3) ÷ 5

We can conceptualize this expression better if we think of the original proportion:

15   =  5   
 a        3

as containing two DIAGONALS.

One diagonal holds the 15 and the 3; the other diagonal holds the ‘a’ and the 5.

Let’s call the diagonal with the ‘a’ the ‘first diagonal.’ And since ‘5’ accompanies ‘a’ in that diagonal, we’ll call 5 the “variable’s partner.”

We’ll call the other diagonal just that, the “other diagonal.”

Now I know you’re getting ‘antsy’ for the shortcut, so just know it’s right around “the bend.”

Using our new terms, we can better understand the expression we got up above:

a = (15 x 3) ÷ 5

The (15 x 3) is the product (result of multiplication) of the “other diagonal,”
and ‘5’ is the “variable’s partner.

So the answer,

                                      (15 x 3)                     ÷              5

is simply (and here’s the shortcut):

         (product of other diagonal) ÷ by  (“variable’s partner.”)

We’ll call this the Proportion Shortcut Formula, or the PSF, for short.

The PSF saves a BIG STEP; using it, we no longer need to write out the cross-multiplication product the usual way, as:

5 x a = 15 x 3

Instead, using the PSF, we can go straight from the proportion to an expression for ‘a‘:

a  =  (15 x 3) ÷ 5

Let’s see how the PSF works in another proportion, such as:

 9    =   45  
13         a

What’s the “variable’s partner”?  9.
What’s in the “other diagonal”? 13 and 45.

So using PSF, the answer is this:

a  =  (13 x 45) ÷ 9

This simplifies to 65, of course. Isn’t it nice not to have to “cross-multiply” any more?

Another nice thing: the PSF works no matter where the variable is located in the original proportion. All you need to do is identify the “variable’s partner,” and the “other diagonal,” and then you’re all good go with the PSF.

Try a few of these to see how easy and convenient the PSF makes it to solve proportions.

PROBLEMS:

1)   a   =      15  
     12          36

2)   18   =    a  
      24         4

3)   21   =   75  
      14          a

ANSWERS (using the PSF first):

1)   a  =  (12 x 15) ÷ 36
  a  =  5

2)   a  =  (18 x 4) ÷ 24
      a  =  3

3)   a  =  (14 x 75) ÷ 21
      a  =  50

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How to Find the LCM for 3+ Numbers — FAST!


Is there a quick-and-easy way to find the LCM for three or more numbers … WITHOUT prime factorizing?

Of course! We’ll demonstrate the technique by finding the LCM for 10, 14, 20.

High-Octane Boost for Math

High-Octane Boost for Math Ed

To begin, use the technique for finding the GCF for 10, 14, 20 that’s shown in my post:  How to Find GCF for 3+ Numbers — FAST … no prime factorizing. If you don’t want to go to that post, no worries. I’ll re-show the technique here.

1st)   Write the numbers from left to right:

……….   10     14     20

[The periods: …… are just to indent the lines. They have no mathematical meaning.]

2nd)  If possible, rip out a factor common to all numbers. The factor 2 is common. So divide the three numbers by 2 [10 ÷ 2 = 5 and 20 ÷ 2 = 10] and show the result below:

2   |       10     14     20
……….   5      7      10

3rd)  At this point, notice there’s no number that goes into the remaining numbers: 5, 7, 10. That means you’ve found that the GCF is the number pulled out, 2. At this point we’re at a crossroads. We’re done finding the GCF, but now we’re at the start of a new process, finding the LCM.
To proceed toward getting the LCM, see if there’s any number that goes into any pair of remaining numbers. Well, 5 goes into 5 and 10. So divide both those numbers by 5 [5 ÷ 5 = 1 and 10 ÷ 5 = 2] , and show the results below:

2   |       10     14     20
5   |         5      7      10
………..  1       7       2

Notice that if there’s a number 5 doesn’t go into, you leave that number as is. So leave the 7 as 7.

4th)  Repeat. See if there’s a number that goes into two of the remaining numbers. Since nothing goes into 1, 7, and 2, we’re done. To get the LCM, multiply all of the outer numbers. That means you multiply the numbers you pulled out on the left (2 and 5), and also multiply the numbers at the bottom (1, 7 and 2). Ignoring the meaningless 1, you have:  2 x 5 x 7 x 2 = 140, and that’s the LCM.

To see the process in more depth, let’s find the LCM for … not three, not four … but five numbers:
6, 12, 18, 30, 36.

1st)   Write the numbers left to right:

………  6     12     18     30     36

2nd)  If possible, rip out a common factor.  2 is common, so divide all by 2 and show the results below:

2     |    6     12     18     30     36
………. 3      6       9     15     18

3rd)  Repeat. See if there’s a number that goes into the five remaining numbers. 3 goes into all, so divide all by 3 and show the results below:

2     |    6     12     18     30     36
3     |    3       6       9     15     18
……..   1       2       3       5       6

4th)  Repeat. See if any number goes into the last remaining numbers. Nothing goes into all of them, so now you get the GCF by multiplying the left-hand column numbers. GCF = 2 x 3 = 6.
Proceeding to find the LCM, look for any number that goes into two or more of the remaining numbers. One such number is 3, which goes into the remaining 3 and 6. Divide those numbers by 3 and leave the other numbers as they are.

2     |    6     12     18     30     36
3     |    3       6       9     15     18
3     |    1       2       3       5       6
……… 1       2       1       5       2

5th)  Interesting! Notice that 2 goes into the two remaining 2s, so pull out a 2 and show the results below:

2     |    6     12     18     30     36
3     |    3       6       9     15     18
3     |    1       2       3       5       6
2     |    1       2       1       5       2
……..   1      1        1       5       1

6th)  We’ve whittled the bottom row’s numbers so far down that finally there’s no number that goes into two or more of them (except 1, which doesn’t help). So we have all the numbers we need to find the LCM. Multiply them together. The left column gives us:  2 x 3 x 3 x 2. The bottom row gives us 1 x 1 x 1 x 5 x 1. Multiply all of those (non-1) numbers together, you get:
2 x 2 x 3 x 3 x 5 = 180, and that is the LCM! Pretty amazing, huh? And no prime factorizing, to boot.

Some people find that this process takes a bit of practice to get used to it. So here are a few problems to help you become an LCM-finding expert!

a)  12, 18, 30
b)   8, 18, 24
c)  15, 20, 30, 35
d)  16, 24, 40, 56
e)   16, 48, 64, 80, 112

And the answers. LCM for each set is:

a)   180
b)   72
c)   420
d)   1680
e)   6720

How to Find the LCM for Three Numbers


Several readers have said they like my trick for finding the LCM described in the post “How to Find the LCM — FAST!” but wonder how to use the trick for finding the LCM for THREE numbers. Here is how you do that.

Essentially it involves using the same LCM trick three separate times. Here’s how it’s done.

Suppose the numbers for which you need to find the LCM are 6, 8, and 14.

Step 1)  Find the LCM for the any two of those. Using 6 and 8, we find that their LCM = 24.

Step 2)  Find the LCM for another pair from the three numbers. Using 8 and 14, we find that their LCM = 56.

Step 3)  Find the LCM of the two LCMs, meaning that we find the LCM for 24 and 56. The LCM for those two numbers = 168.

And that, my good friends, is the LCM for the three original numbers.

So, to summarize. Find the LCM for two different pairs. Then find the LCM of the two LCMs. The answer you get is the LCM for the three numbers.

Here are a few problems that give you a chance to practice this technique.

Find the LCM for each trio of numbers.

a)  10, 25, 30

b)  16, 28, 40

c)  14, 32, 40

Answers:

The LCMs for each trio are:

a)  150

b)  560

c)  1,120

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which comprise an award-winning program that makes algebra do-able! The books break algebraic concepts down into manageable chunks and provide instruction through a captivating Q&A format. Josh also is the author of PreAlgebra Blastoff!, which presents an engaging, hands-on approach (plus 16-page color comic book) for learning the rules of integers. Josh’s line of unique, student-centered math-help books is published by Singing Turtle Press and can be found on Amazon.com

Find the LCM (aka LCD) in Two Easy Steps


This is really the “Week of the LCM” for me.

Just as I was finishing my last post, on a new way to find the LCM for a pair of numbers, I discovered another way to do the same thing.

Coffee, Pi and More

Coffee, Pi and More

I was looking at the problems at the end of my last post, these problems:

b)   15 and 20;  LCM  =  60

c)   18 and 20;  LCM  =  180

d)   24 and 28;  LCM  =  168, ….

… when I noticed something.

(more…)

Find the LCM in a way that makes sense! (Part 2)


In yesterday’s post on the LCM, I wrote about 375 pages on the topic, and then I said that I left out an idea. Hahaha, you probably thought. Very funny, Josh.

But never fear. I am not going to write another 375 pages on the topic.

What I do need to bring to your attention, though, is that there are two LCM situations that I did not take into account yesterday. So to present a complete picture, I need to explain (for those who have not already figured this out by themselves) how to use my new technique in those two situations.

Coffee, Pi and More

Coffee, Pi and More

You will notice that in my write-up yesterday — and in the practice problems I provided — the gap always divided evenly into the smaller number. How convenient, right? In the first example, we had a gap of 3 dividing into 12; in the next, a gap of 4 going into 20. Of course this does not always happen. Consider a situation in which we want to find the LCM for 10 and 16. The gap of 6 (16 – 10 = 6) does NOT divide evenly into the smaller number, 10. So what would we do here? (more…)

Find the LCM in a way that makes sense! (Part 1)


I don’t know about you folks, but I’ve always been a bit disappointed by the various techniques for finding the Least Common Multiple (LCM) for a pair of numbers.

While there are several techniques that “work” — by which I mean techniques we can teach to students and have them learn quickly — I’ve known of no technique that makes good intuitive sense. In other words, I’ve known no technique whose underlying principle felt obvious.

Feeling frustrated, I started looking for a technique that would have that undeniable “ring of truth.”

Coffee, Pi and More

Coffee, Pi and More

And so, after playing around in my “sandbox of numbers” for quite a while,  I’m happy to report that I’ve finally found what I had been looking for.

In today’s post I will show you a way to find the least common multiple that makes sense, at least to me. I hope it will make sense to you as well.

(more…)

(Divisibility) Practice Makes Perfect


As the saying goes, practice makes perfect.

And boy is that true in math! Of the standard school subjects, math requires the most practice, if you want to excel at it.

That being the case, this strikes me as a great time to practice the divisibility tricks we’ve just learned.

There are many skill areas where divisibility tricks are useful — solving proportions, factoring polynomials, multiplying fractions — but one of the most obvious is the critical skill of reducing fractions.

So now I’m offering you a chance to practice your divisibility skills for 2, 3, 4, 5 and 6. We will save the trick for 7 till we have a few more tricks “up our sleeves.”

For the following problems, answer these four questions:

1)  Which of these numbers — 2, 3, 4, 5 or 6 — divides evenly into the numerator (NM)?

2)  Do the same for the denominator (DNM).

3)  Then choose the largest number that divides into both NM and DNM. For these problems, this number will be the GCF.

4)  Finally, reduce the fraction by dividing both NM and DNM by this number.

Here’s an example that shows what you’d write:

ex)  24/42

1)  NM:  2, 3, 4, 6
2)  DNM:  2, 3, 6
3)  GCF = 6
4)  Answer:   4/7

NOW TRY THESE PROBLEMS:

a)  20/24
b)  25/40
c)   18/48
d)  26/60
e)  21/72
f)  30/85
g)  36/66
h)  56/92
i)  84/102
j)  99/141

ANSWERS:

a)  20/24
1)   NM:  2, 4, 5
2)  DNM:  2, 3, 4, 6
3)  GCF =  4
4)  Answer:   5/6

b)  25/40
1)   NM:  5
2)  DNM:  2, 4, 5
3)  GCF =   5
4)  Answer:  5/8

c)   18/48
1)   NM:  2, 3, 6
2)  DNM:  2, 3, 4, 6
3)  GCF =  6
4)  Answer:  3/8

d)  26/60
1)   NM:  2
2)  DNM:  2, 3, 4, 5, 6
3)  GCF =  2
4)  Answer:  13/30

e)  21/72
1)   NM:  3
2)  DNM:   2, 3, 4, 6
3)  GCF =   3
4)  Answer:   7/24

f)  30/85
1)   NM:  2, 3, 5, 6
2)  DNM:  5
3)  GCF =  5
4)  Answer:  6/17

g)  36/66
1)   NM:  2, 3, 4, 6
2)  DNM:  2, 3, 6
3)  GCF =  6
4)  Answer:  6/11

h)  56/92
1)   NM:  2, 4
2)  DNM:  2, 4
3)  GCF =  4
4)  Answer:  14/23

i)  84/102
1)   NM:  2, 3, 4, 6
2)  DNM:  2, 3, 6
3)  GCF =   6
4)  Answer:   14/17

j)  99/141
1)   NM:   3
2)  DNM:  3
3)  GCF =  3
4)  Answer:   33/47