IGCSE MATHEMATICS Topic : Venn Diagrams – A Summary

Some of the Subsets of a Venn Diagram are shown below:

AB={The set of elements of A that are also in B}
A/B (also written A-B or AB') ={The set of elements of A that are not in B}
B/A (also written B-A or BA') ={The set of elements of B that are not in A}
/A/B (also written -A-B or A'B') ={The set of elements that are not in A or in B}
AB={The set of elements that are in either A or B}. This is shaded in the Venn diagram above.
The meaning of some other symbols is shown below.

 

IGCSE MATHEMATICS Topic : Plotting Lines From Exponential Equations

Given a straight line plotted on a graph we can estimate the values of the gradient and either using a point on the line or by estimating the y – intercept, we can find the equation of the line in the form The most convenient form in which to analyse the relationship between two variables x and y is the linear form, and we must see if, given a suspected relationship between two variables, we can construct a linear graph from the suspected relationship.

Suppose we suspect an exponential relationship between two variables. We also have the following data.

Time, t in s	10	20	30	40	50
Mass m in g	40.3	27	18	12.2	8.1

Our relationship will take the formTo transform this into a straight line we take logs of both sides obtaining,We recalculate the table.

Time, t in s	10	20	30	40	50
log (m)	3.696	3.296	2.890	2.501	2.092

And sketch the graph:

Comparingwithwe seeso andsothen

IGCSE MATHEMATICS Topic : Formulae

The equation is solved using

You can choose r from n possibilities in possible ways.

You can line up r from n possibilities in different ways

Binomial Theorem :

For any triangle ∆ABC ,

The straight line passing through has gradient and equation

The distance between two points, is

A straight linepassing though points A and B is given by

The angle between two vectors is given by The angle between two lines and is given by

To change to a straight line equation, take logs

and

IGCSE MATHEMATICS Topic : Combinations and Permutations

This topic deals with the numbers of ways ways we can pick a selection from a number of possible combinations. For instance, suppose we have 10 people lined up and we have to pick a team of 4.

The number of ways we can pick 4 from 10 is written or

Working from first principles we can pick the first from 10, the second from 9, the third from 8, the fourth from 7, hence 10*9*8*7=5040. But the order of the picking will not matter here. The four people can be picked in any order and we have not taken account of this. To take account of this objection we notice that 4 people can be arranged in 4*3*2*1=4!=24 ways, so now we divide 5040 by 24 to get 210.

The order did not matter for the above question, but sometimes the order does matter, For example 10 runners in a race will obviously differentiate between first, second and third place. In this case we finddifferent possibilities.

Sometimes we have combinations of combinations. Suppose we have 6 men and 5 women. We have to form from these a team of 4 men and 3 women. We can pick the four men indifferent ways and the 3 women in different ways. The choices of men and women are completely independent. INDEPENDENT! That should ring a bell. If probabilities are independent we multiply, and so with combinations. Hence the number of ways in which we can pick four men and three women from 6 men and five women is

Sometimes though, we have to write down list of possible arrangements because not every arrangement is acceptable.

Suppose a committee of 5 people is to be selected from 6 men and 4 women. We are required to find the number of selections which has more men than women.

We could have 5 men and no women:possible choices.

We could have 4 men and 1 woman:possible choices.

We could have 3 men and 2 woman:possible choices.

Hence there are 6+60+120 possibilities.