Angles, Parallel Lines and Bearings

Posted on February 27, 2013 by admin

 

Angles In Par­al­lel Lines

We use arrow­heads to indi­cate that 2 lines are parallel:-

Arrowheads to indicate parallel lines

 

 

 

 

 

When a line crosses 2 par­al­lel lines it cre­ates pairs of equal angles

Inte­rior Angles

In the dia­gram below x and y are inte­rior angles.

Inte­rior angles add up to 180°

so x + y = 180°

Interior angles add up to 180 degrees

 

 

 

 

 

Here’s a video that explains inte­rior (or co-interior angles) very clearly:-

Cor­re­spond­ing Angles

What are cor­re­spond­ing angles? That’s best explained by a cou­ple of exam­ples. In the fol­low­ing dia­grams; a and b and c and d are cor­re­spond­ing angles. Note that the lines that join up cor­re­spond­ing angles make an ‘F’ shape:-

Corresponding angles are equal

 

 

 

 

 

Here’s a video that explains cor­re­spond­ing angles clearly and in detail:-

Alter­nate Angles

This dia­grams shows 2 pairs of alter­nate angles.

Alter­nate angles are equal. Notice how the lines make a ‘Z’ shape.

Alternate angles on parallel lines are equal

Here’s a video to explain alter­nate angles in par­al­lel lines:-

Bear­ings

Bear­ings are mea­sured clock­wise from the North:-

Bearings- example of 60 degrees

 

 

 

 

 

 

To get a fuller under­stand­ing of all bear­ings, con­sider the com­pass. Can you work out each direc­tion shown here as a bearing?

Compass bearings

 

 

 

 

 

 

 

 

NE = 045

E = 090

SE = 135

S = 180

SW = 225

W = 270

NW = 315

Here’s a video that explains bear­ings in some detail:-

Here’s another video that walks through some Bear­ings exam questions:-

Par­el­lel Lines and Bearings

You can use the fact that North lines are always par­al­lel and knowl­edge of angle facts to work out bearings.

This is best illus­trated with an exam­ple. In the fol­low­ing dia­gram, work out the bear­ing of B from A.

Bearings question using North , parallel lines and angle rules

 

 

 

 

 

 

 

 

There are 2 pos­si­ble approaches to answer­ing this ques­tion. Both start with draw­ing the North line from A so that we have 2 par­al­lel lines point­ing North.

Draw the parallel North line

 

 

 

 

 

 

 

 

The first method we can use is based on the fact that co-interior angles add up to 180°

There­fore we know the angle below marked in red must be 180° — 135° = 45°

Interior Angles add up to 180 degrees

 

 

 

 

 

 

 

 

But remem­ber bear­ings are mea­sured clock­wise from the North. So the bear­ing of B from A = 360 — 45 = 315 see the dia­gram below:-

Co-Interior Angle Solution to Bearing Question

 

 

 

 

 

 

 

 

 

Here’s an alter­nate solu­tion based on.….alternate angles. Again we start by draw­ing the North line from A:-

Draw the parallel North line

 

 

 

 

 

 

 

 

Then we find the alter­nate angles (it helps to extend the line below A and high­light the ‘Z’ shape we’re look­ing for):-

Deduce Bearings using alternate angles

 

 

 

 

 

 

 

 

 

 

 

Finally, remem­ber­ing that bear­ings are always mea­sured clock­wise from the North we just have to cal­cu­late the answer:-

Use alternate angles to deduce bearing- answer

Posted in 22. Angles and Bearings | Leave a comment

Further Algebra

Posted on February 5, 2013 by admin

This another sec­tion where all the ques­tions are in the B to A* range. I think that algebra’s rep­u­ta­tion as being very dif­fi­cult is exag­ger­ated. Like all maths top­ics if you really get to grips with the basics, you’ll be able to tackle the more chal­leng­ing stuff. If you’re strug­gling step back to a lower level and when you’re more con­fi­dent come back to the higher level. A lit­tle rep­e­ti­tion and prac­tice goes a long way. You might not get it the first time but if you keep com­ing back to it, you’re more likely to succeed.

Fur­ther Simul­ta­ne­ous Equations

First some key facts:

When 2 straight lines cross or inter­sect they only do so in one point.

When you plot a qua­dratic and lin­ear func­tion on the same graph, there are poten­tial outcomes:-

  • The lin­ear graph crosses or inter­sects the qua­dratic graph at 2 points
  • The lin­ear graph touches the qua­dratic graph at one point
  • The lin­ear graph and the qua­dratic graph do not intersect

Here’s the graph of y = x² — 4x + 1 and y = x + 1

Graph of quadratic function y = x² -4x + 1 and linear function y = x + 1

 

 

 

 

 

 

 

 

 

 

The lines cross or inter­sect at 2 points, (0, 1) and (5, 6)

So you can solve these simul­ta­ne­ous equa­tions by plot­ting the graphs but you also use algebra:-

1. Start with the lin­ear equa­tion and, if you need to, make x or y the subject.

So we have:-

y = x + 1

and

y = x² — 4x + 1

2. Next sub­sti­tute the expres­sion for “y” from the lin­ear equa­tion into the qua­dratic equation:-

x + 1 = x² — 4x + 1

3. Adjust so that you move all the terms to the left hand side:-

x + 1 = x² — 4x + 1

x² — 4x + 1 = x + 1

x² — 5x = 0

x(x — 5) = 0

So x = 0 or 5 

i.e. when x is 0 or 5, then x(x — 5) does equal zero. 

to be continued…

Posted in 21. Further Algebra | Leave a comment

20. Quadratic Equations

Posted on February 1, 2013 by admin

The whole of this topic is in the range grade B to grade A*. I sus­pect that the mere men­tion of “qua­dratic equa­tions” is enough to put off many peo­ple. There is no doubt that some of the equa­tions look scary but if you work through them and prac­tice them, you’ll see that there’s no need to be afraid.

A qua­dratic expres­sion is where the high­est power of x is x².

Exam­ples of qua­dratic expressions:-

4x² + 3x + 7             x² — 4x — 3                 6x² +9             16x² — 4x

There are cer­tain types of expres­sion which are one square num­ber sub­tracted from another square num­ber. Note in the fol­low­ing exam­ples all the num­bers are square numbers:-

x² — 9                    a² — 25                  16c² — 36

Where you have this type of expres­sion in the form a² — b², where a and b are either num­bers or alge­braic terms, it’s known as the dif­fer­ence of two squares.

You need to remem­ber that a² — b² = (a — b)(a + b)

Check:  (a-b)(a + b) = a² + ab — ba — b² = a² — b²

Fac­tor­ize x² — 16

x² — 16  =  x² — 4²  = (x — 4)(x + 4)

Fac­toris­ing qua­drat­ics x² + bx + c

to be continued…

Posted in 20. Quadratic Equations | Leave a comment

Maths News Headlines

Posted on January 31, 2013 by admin

Links to any maths news that catches my eye. Could be zero or twenty sto­ries on any par­tic­u­lar day.

Fri­day Feb­ru­ary 1st 2013

1. Elite Maths School Pro­posed for Exeter

New Free school to be estab­lished in Exeter will be a cen­tre of excel­lence for Maths. This has been sup­ported by local Labour MP Ben Brad­shaw but opposed by teach­ing unions.

In my opin­ion this is mis­sion creep from the orig­i­nal Free School prin­ci­ples. It will be an elit­ist school and only pupils with excep­tional maths skills will be eli­gi­ble. If this was repli­cated across other sub­jects and spread to all parts of the coun­try, wouldn’t this amount to selec­tion (of the type we saw when we had gram­mar and sec­ondary schools)  by the back door? Read more here.

Thurs­day Jan­u­ary 31st 2013

1. OCF joins new ini­tia­tive to improve adults’ maths skills

OCF (Online Cen­tres Foun­da­tion) is join­ing Maths4us along with over 20 other orga­ni­za­tions to boost adults’ maths skills. An action plan will be devel­oped over the com­ing weeks. Read more here.

At the moment the details seem vague but I guess these will become clear as the action plan is devel­oped. In the mean­time Maths4us has a web­site which has some use­ful infor­ma­tion for all adults want­ing to improve their maths skills and for all maths and numer­acy teachers.

2. UK’s Games Indus­try Calls for Improved Stan­dards in Maths and Sci­ence Education

TIGA (The Inde­pen­dent Game Developer’s Asso­ci­a­tion) has responded to Sir Tim’s Berner-Lee’s com­ments about com­puter sci­ence edu­ca­tion. Read full press release here.

3. Maths Should Be Cool Not Scary

Ok this is not really news but it’s an inter­est­ing Guardian blog about the way that maths is per­ceived by chil­dren and soci­ety at large. It’s all the more com­pelling because it’s writ­ten by an Eng­lish teacher! The com­ments are also a good read. Read arti­cle here.

Posted in Maths News Headlines | Tagged | Leave a comment

Linear Graphs

Posted on January 31, 2013 by admin

You could be intim­i­dated by graphs but I don’t think there’s any need. You just need to learn a few def­i­n­i­tions and prac­tice a few tech­niques to pick up some rel­a­tively easy marks.

Line Seg­ments and Mid-points

A graph paints a thou­sand words:

Line Segments and Mid-points

 

 

 

 

 

 

 

 

 

 

In exam­ples like the one above you can just use com­mon sense to find the mid-point of a line seg­ment like the one above but you can cal­cu­late the mid-point from just the co-ordinates of a line seg­ment with this formula:-

Formula to calculate the mid-point of a line segment

 

 

 

So, for exam­ple a line seg­ment AB has coordinates

A: (2, 10)
B: (10, 16)

What is the mid=point?

Using the for­mula from above, we have

Mid­point = ((2+10)÷2), ((10+16)÷2)

= (6, 13)

You can see this on a graph:-

Graph mid-point of line segment

 

Plot­ting Straight-Line Graphs

When you plot a lin­ear func­tion on a graph you get a straight line. A lin­ear func­tion is usu­ally expressed in the for­mat y= mx + b.

You can plot a lin­ear func­tion by feed­ing in val­ues of ‘X’ and cal­cu­lat­ing ‘Y’. In the­ory, because a lin­ear func­tion will result in a straight line you only need two sets of coor­di­nates. Nor­mally you will be asked to cal­cu­late a few sets of coor­di­nates in a table and then plot the line.

For exam­ple; draw the graph of y = 3x + 3 for val­ues of x from –3 to +3

Just 3 sim­ple steps; draw up a table, cal­cu­late the val­ues of y and plot the graph:

Table of Val­ues and Graph of Lin­ear Func­tion y = 3x + 3

Plot x and y values linear function

 

 

Graph of Linear Function y = 3x +3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

To be continued…

Posted in 19. Linear Graphs | Leave a comment

Percentages

Posted on January 24, 2013 by admin

This sec­tion builds on Unit 1 Sta­tis­tics and Num­ber, Sec­tion 2 Frac­tions, Dec­i­mals and Per­cent­ages. It’s my kind of maths, mainly because it’s not really math­e­mat­ics it’s just arith­metic. I think if you learn one or two def­i­n­i­tions and do a few prac­tice ques­tions, you  should find this one of the eas­ier parts of the Maths GCSE syllabus.

Per­cent­age Increase and Decrease

Here you are given a start num­ber and you have to increase or decrease that num­ber by a cer­tain percentage.

There are two ways to solve these type of problems.

The first way is nor­mally how you would do it in your head or on paper. For exam­ple: Fred earns £110 per week and gets a 5% increase, how much will his new weekly wage be? In your head you would work out 5% of £110 and add it to the orig­i­nal sum. 5% of £110 is £5.50, add this to £110 to give the answer of £115.50.

Alter­na­tively, if you were using a cal­cu­la­tor you would add the per­cent­age increase to 100% to give 105%, con­vert this to a dec­i­mal to give 1.05 and then mul­ti­ply it by the orig­i­nal wage, 1.05 x £110 = £115.50.

Per­cent­age Profit or Loss

Is the per­cent­age profit or loss of the cost price.

So if the cost price is £10 and the sales price is £20, then the profit is £10 (£20 — £10) and the per­cent­age profit = £10/£10 x 100% = 100%

In words:-

Per­cent­age Profit (or loss) = actual profit (or loss)/cost price X 100%

Repeated Per­cent­age Change

This is most often used in finance. Usu­ally when you invest money, inter­est is cal­cu­lated using com­pound inter­est. This is where the inter­est is paid on the ini­tial invest­ment plus on any inter­est already earned. Sounds more com­pli­cated than it is! Best to look at an actual example:

John invests £1000 in a savers account which earns 5% inter­est annu­ally com­pound  inter­est. Assum­ing he makes no with­drawals, how much will he have in his account;

a) After 1 year

b) After 2 years

Answer

a) £1000 x 5% = £50, so after one year he’ll have £1,000 + £50 = £1,050

b) After 2 years he’ll have £1,050 (from the first year) + 5% inter­est on £1,050=

£1,050 + (£1,050 x 0.05) = £1,050 + £52.50* = £1,102.50

*1% of £1,050 = £10.50, so 5% of £1,050 = 5 x £10.50 = £52.50

Reverse Per­cent­ages

This is another case where the two dif­fer­ent meth­ods, one with­out a cal­cu­la­tor and one with. Best explained with an example.

A shirt is on sale at £60 and the ticket says that it has been reduced by 20%. What was the orig­i­nal price?

If you don’t have a cal­cu­la­tor, fol­low these two steps:

a) Work out what per­cent­age the new price rep­re­sents of the orig­i­nal price:-

In this case 100% — 20% = 80%

b) Mul­ti­ply the revised price by 100/(your answer from (a))

In this case £60 x 100/80

At this point could divide by 80 and mul­ti­ple by 100 BUT it’s often a lot sim­pler to use your knowl­edge of fractions:-

so £60 x 100/80 = £60 x 10/8 = £60 x 5/4 = £60 x 1.25 (we know that 0.25 = a quarter)

so £60 x 1.25 = £60 + a quar­ter of £60 = £60 + £15 = £75.

If you do have a cal­cu­la­tor fol­low these steps (note the first step is the same as the “with­out a cal­cu­la­tor method” shown above):

a) Work out what per­cent­age the new price rep­re­sents of the orig­i­nal price:-

100% — 20% = 80%

b) Divide the answer by 100 to get a “multiplier”:-

80/100 = 0.8

c) Divide the new price by the mul­ti­plier to get the orig­i­nal price:-

£60/0.8 = £75

 

Posted in 18. Percentages | Leave a comment

Sequences and Proof

Posted on December 6, 2012 by admin

When I started look­ing at “Sequences and Proof”, I fell into the trap of look­ing at some of the most dif­fi­cult ques­tions (A and A*) first. What a sur­prise to find that I couldn’t do them! I now think it’s espe­cially impor­tant with this topic, to mas­ter the basic, sim­ple stuff and then move onto more chal­leng­ing areas such as proof using algebra.

So, the basics:-

A sequence is a list of num­bers in a given order

Con­sec­u­tive num­bers are num­bers next to each other

We call the num­bers in a sequence, terms


This leads us to the most basic sequence:-

A liner sequence increases or decreases in equal sized steps.


For exam­ple:-   2, 4, 6, 8

In this sequence the terms increase by 2 in each step.

We can say:  The nth term = 2n (i.e the 1st term = 1 x 2 = 2, the 2nd term = 2 x 2 = 4, the 3rd term = 3 x 2 = 6 and the 4th term = 4 x 2 = 8).

When you’re try­ing to under­stand a sequence it’s very use­ful to write the dif­fer­ences between each term under the sequence, like this:-

Simple Sequence

 

 

 

 

When you have a lin­ear sequence if you look at dif­fer­ences between the terms to start to define the sequence, if the dif­fer­ences are all 2 then the sequence def­i­n­i­tion will include 2n, if the dif­fer­ences are all 3 then the sequence def­i­n­i­tion will include the term 3n, if the dif­fer­ences are all 4 then the the sequence def­i­n­i­tion will include the term 4n etc., etc.

This is best explained with a cou­ple of examples:-

Find the nth term of the sequence 4, 6, 8, 10,.…

First look for the dif­fer­ence between each term:-

Simple Linear Sequence

 

 

 

 

You now know that the sequence includes 2n. Set the infor­ma­tion in a table includ­ing 2n:-

2n sequence

 

 

 

 

 

 

 

 

 

This method might seem over the top for sim­ple lin­ear sequences, but it shows its worth with more com­pli­cated examples:-

Find the nth term of the sequence 1, 9, 17, 25

So start by find­ing the dif­fer­ence between each term and then com­plete the grid as shown above:-

Linear sequence example 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Most ques­tions ask you to find the nth term but some­times this is turned on its head. For exam­ple what is the 10th term if the nth term = 8n — 7? Just feed the actual num­ber into the statement:

10th term = (8 x 10) — 7 = 73 

Alter­na­tively you might be asked to find out if a par­tic­u­lar num­ber is a term in the sequence, for exam­ple is 155 a term in the above sequence? Again just feed the num­ber into the sequence definition.

If 155 is in the sequence, then 8n — 7 = 155

and 8n = 162

n = 162/8 = 20.25

So “n” is not a whole num­ber and 155 is not in the sequence.

Another exam­ple using the same sequence, is 177 a term in the sequence?

If 177 is in the sequence then 8n — 7 = 177

and 8n = 184

n = 184/8 = 23

So “n” is a whole num­ber and 177 is in the sequence.

Qua­dratic Sequences

The nth term of a qua­dratic includes an “n²”. To find out if a sequence is qua­dratic you check the sec­ond dif­fer­ences. So what is a sec­ond dif­fer­ence? We used the first dif­fer­ences between terms when we looked at lin­ear sequences above. The sec­ond dif­fer­ence is the dif­fer­ence between the dif­fer­ences. This is best under­stood by using an exam­ple. Let’s invent our own qua­dratic sequence, n² + 1. So look­ing at the first few terms and the dif­fer­ences and the sec­ond dif­fer­ences we have:-

quadratic sequence example

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

You may get a ques­tion where the answer is a mul­ti­ple or frac­tion of n². In these cases the sec­ond dif­fer­ence will not be 2, but it will be con­stant. For exam­ple, find the nth term of this sequence:-    0.25, 1, 2.25, 4, 6.25

Quadratic sequence example 0.25n2

 

 

 

 

 

 

 

 

 

 

 

 

 

Proofs and Proof using alge­bra to follow.

Posted in 17. Sequences and Proof | Leave a comment

How to Multiply Using the Lattice Method

Posted on October 29, 2012 by admin

The Lat­tice Method of Multiplication

The lat­tice method has a num­ber of dif­fer­ent names, includ­ing: Gelosia, Sieve, Shabakh, Venet­ian Squares, Hindu Lat­tice. I dis­cov­ered it the other day when my son was strug­gling to mul­ti­ply dec­i­mals using the grid method. My daugh­ter saw him get­ting frus­trated and me strug­gling to explain where he had gone wrong, and said “Why don’t you just use the lat­tice method, it’s really easy”. I was skep­ti­cal but after she showed me how quickly and con­fi­dently she could do long mul­ti­pli­ca­tion, with or with­out dec­i­mals, I’m con­vinced it’s a valid and easy to under­stand method.

Exam­ple of the Lat­tice Method of Multiplication

The method can only really be explained by using an example:

723 x 63

Step 1 Draw the Blank Lattice

This exam­ple involves the mul­ti­pli­ca­tion of a 3 digit num­ber by a 2 digit num­ber. So we need a lat­tice with 3 columns and 2 rows (you will see that you’ll also need space around the lat­tice). This looks like a grid but as we progress it’s trans­formed into a lattice.

Blank lattice to multiply numbers

Step 2 Add the num­bers to be multiplied:

Add Numbers to the Lattice

Step 3 Draw Diag­o­nals to form the Lattice

Notice how the diag­o­nals project beyond the ini­tial columns and rows. This gives space to write the answer.

Lattice Method Add the Diagonals

Step 4 Mul­ti­ply the Lattice

Notice how the mul­ti­pli­ca­tion results are entered into the lat­tice. There are 6 multiplications:-

7 x 6 = 42

2 x 6 = 12

3 x 6 = 18

7 x 3 = 21

2 x 3 = 6

3 x 3 = 9

Where there is a 2 digit answer, the first digit is entered to the left of the diag­o­nal and the sec­ond digit is entered to the right of the diag­o­nal. Where there is a one digit answer, zero is entered to left of the diag­o­nal and the sin­gle digit entered to the right of the diagonal.

Multiply Out the Lattice

 

Step 5 Add up the Diagonals

Work­ing from the right-hand side we have:-

9 = 9

8 + 6 =14 NB when you have a 2 digit answer, write down the sec­ond digit and carry the first digit over to the next diag­o­nal. In this case we write down “4” and carry over “1” (see  the num­ber one writ­ten in green)

1 (the one car­ried over, see above) + 1 + 2 + 1 =5

1 + 2 + 2 = 5

4 = 4

Lattice Method Add the Diagonals

Step 6 Read off the final answer

Lattice Method of Multiplication Final Answer

Final Answer 723 x 63 = 45,549

Advan­tages of the Lat­tice Method of Multiplication

1. You only need to know the times tables from 1 x 1 through to 9 x 9 to be able to com­plete any mul­ti­pli­ca­tion ques­tion.  Check out “How to Learn Your Times Tables Fast”  to become more con­fi­dent with your times tables.

2. The method is step by step and rel­a­tively easy to follow.

3. You mul­ti­ply and then add in 2 dis­tinct sep­a­rate steps.

4. As long as you take care draw­ing the diag­o­nals you should not mud­dle units, tens, hun­dreds, thou­sands etc.

5. You can use this method to mul­ti­ply dec­i­mals (sep­a­rate arti­cle to follow).

Dis­ad­van­tages of the Lat­tice Method of Multiplication

1. If you’re care­less or untidy in prepar­ing the lat­tice you are likely to make mistakes.

2. It’s pos­si­ble to learn this method by rote with­out under­stand­ing why it works.

Posted in 8. How to Multiply Using the Lattice Method | Leave a comment

Multiplication Grid, The Grid Method of Multiplication and Multiplying Decimals using the Grid Method

Posted on October 8, 2012 by admin

I decided to inves­ti­gate the grid method of mul­ti­pli­ca­tion because my son, in his first year of sec­ondary school (Year 7), came home with his first bit of maths home­work. The home­work included the fol­low­ing multiplication:-

7.23 x 6.3

My son is ok at Maths, in fact he’s quite good, he was one of only a few at his pri­mary school that achieved “Level 6″. How­ever, he was really strug­gling with this ques­tion. He was using the grid method of mul­ti­pli­ca­tion. The grid method was never used when I was at school but now it’s com­monly used as a step­ping stone to the tra­di­tional method of long multiplication.

The Grid Method of Multiplication

The grid method can only really be explained by using an example.

So let’s use:-

16 x 23

First you draw up grid. In this exam­ple which is mul­ti­ply­ing a two digit by a two digit num­ber, we need 2 columns and two rows. Next we split the num­bers into tens and dig­its. So 16 becomes 10 and 6 and 23 becomes 20 and 3 and enter as below. Then mul­ti­ply out (refer the grid below) 20 x 10 = 200, 20 x 6 = 120, 3 x 10 = 30 and 3 x 6 =18. Then add up each col­umn 200 + 30 = 230 and 120 + 18 = 138. Finally (see the sum beneath the grid) just add 230 +138 = 368.

Multiplication Grid Example

 

 

 

 

 

 

 

 

 

 

 

 

Here’s a video that takes you through this exam­ple, step-by-step:-

I can see the advan­tages of using the grid method. It is highly visual, con­trast how dif­fi­cult it was to fol­low my writ­ten expla­na­tion com­pared to how easy it was to just look at the actual grid! The other advan­tage is that it clearly sep­a­rates tens and units (and hun­dreds and thou­sands etc. for larger num­bers). In my view this helps chil­dren to under­stand how it works.

Mul­ti­ply­ing Dec­i­mals Using The Grid Method

As my son now realises, you have to be care­ful when you use the grid method to mul­ti­ply dec­i­mals. As I men­tioned above he had to solve this mul­ti­pli­ca­tion question:-

7.23 x 6.3

This was roughly how he set out his grid to answer this ques­tion (THIS IS AN EXAMPLE OF HOW NOT TO DO IT!):-

Multipliaction Grid with Decimals- How NOT to do it

 

 

 

 

 

 

 

 

 

The cells in the grid above are cor­rect. My son, using pen and paper and the typ­i­cally less than neat pre­sen­ta­tion skills of an eleven year old, man­aged to get the dec­i­mal point in the wrong place in more than one of the cells. The trick here is to elim­i­nate the dec­i­mal point when you use the grid and use a sim­ple rule to intro­duce it back after you’ve used the grid. So we have:

7.23 x 6.3

Step 1 — Elim­i­nate the dec­i­mal points;

7.23 x 6.3 becomes 723 x 63

Step 2 — Use the grid method;

Grid Method for Multiplying Decimals

Step 3: Rein­tro­duce the dec­i­mal point using this sim­ple method.

Count the num­ber of dig­its in the orig­i­nal ques­tion that are after the dec­i­mal point and then alter the answer from the grid so that there are that num­ber of dig­its after the dec­i­mal place. Hmmm… that’s not easy to put into words! Best look at our example:-

The orig­i­nal ques­tion was 7.23 (2 dig­its after the dec­i­mal place) x 6.3 (1 digit after the dec­i­mal place), so in this case there are 3 dig­its after the dec­i­mal place. So we need to take our answer from the grid:- 45549 and alter it so that there are 3 dig­its after the dec­i­mal place = 45.549

Grid Method Multiplying Decimals

 

 

 

 

 

 

 

 

 

 

Final Answer 7.23 x 6.3 = 45.549

Posted in 6. How to Multiply Using the Grid Method, 7. How to Multiply with Decimals Using the Grid Method | Leave a comment

How to Divide Fractions, Including Mixed Numbers and Improper Fractions

Posted on September 26, 2012 by admin

Most peo­ple imag­ine that divid­ing by a frac­tion will be very com­pli­cated but it’s really quite straight­for­ward. You just have to “flip” (a tech­ni­cal maths term) the divid­ing frac­tion by swap­ping the top num­ber (the numer­a­tor) with the bot­tom num­ber (the denom­i­na­tor). You then just have to mul­ti­ply, which we’ve already seen is simple.

You don’t need to know this, but the fol­low­ing proof explains why flip­ping the divid­ing frac­tion and then mul­ti­ply­ing works. It’s not dif­fi­cult to fol­low and if you under­stand it, you’ll never for­get how to divide by a frac­tion. It’s also, I think, more sat­is­fy­ing to know why some­thing works rather than just know how to do it. The proof is also quite ele­gant (math­e­mati­cians some­times say that proofs are “beau­ti­ful”, that’s a bit over the top). Any­way, here’s the proof:-

Proof: To Divide by a Frac­tion, Just Flip it and Multiply

Proof to Divide a Fraction, Flip and Multiply

Proof to Divide a Frac­tion, Flip and Multiply

Just as saw when we were mul­ti­ply­ing frac­tions, you will prob­a­bly have to deal with mixed num­bers and improper frac­tions. This video walks through the five sim­ple steps that will cover all pos­si­bilites:-

The video looks at the fol­low­ing division:-

Dividing by Fractions Question

 

 

 

These are the five steps that are shown in the video and that can be used to answer divide by a frac­tion question:-

Step 1:-  Con­vert any Mixed Num­bers to Improper Fractions

A mixed num­ber is a com­bi­na­tion of an inte­ger (or whole num­ber) and a frac­tion. Any mixed num­bers need to be con­verted to an improper frac­tion, a frac­tion where the top num­ber (numer­a­tor) is greater then the bot­tom num­ber (denominator):-

Convert Mixed Number to Improper Fraction

Step 2:- Take the divid­ing num­ber and switch or flip its numer­a­tor and denominator.

The proof shown above shows why this works. Don’t for­get to change the divide sign into the mul­ti­ply sign:-

Flip the dividing fraction

Step 3:- If it’s pos­si­ble sim­plify any of the fractions

Again, this is not always pos­si­ble, but if you can sim­plify it makes the remain­ing steps easier:-

Simplify the Fractions

Step 4:- Multiply

Now we’re ready to just mul­ti­ply the numer­a­tors and the denominators:-

Multiply Numerators and Denominators

Step 5:- If Required Change your answer from an Improper Frac­tion to a Mixed Number

Your ini­tial answer may or may not be an improper frac­tion (numer­a­tor greater than the denom­i­na­tor) but if it is you should con­vert it to a mixed num­ber. You may lose marks (or to be more pos­i­tive gain extra marks if you do) if you don’t com­plete this final step:-

Convert Improper Fraction to Mixed Number

Re-cap: How to Divide by a Frac­tion in Five Sim­ple Steps

Here’s all five steps summarized:-

How to Divide By a Fraction in 5 Simple Ste

Posted in 5. How to Divide by a Fraction in 5 Simple Steps | Leave a comment