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FE Magazine - Issue 37: It's a Post-16 Thing


Created on 14 August 2014 by ncetm_administrator
Updated on 27 August 2014 by ncetm_administrator

 

FE Magazine - Issue 37part of equation from article
 

It's a Post-16 Thing

One of the most popular and well-used features of the NCETM’s website has always been the section devoted to helping teachers audit their own mathematical knowledge, and, in parallel, think about the associated pedagogy. These are the self-evaluation tools, and the section has recently been enhanced with new material for post-16 GCSE teachers, particularly designed with the unique characteristics of teaching GCSE re-sit students in mind.

This section is slightly different from the other Self-evaluation Tools.

There is a Mathematical Knowledge section and a Mathematics-specific Pedagogy Post-16 GCSE section.

In the Mathematical Knowledge section you are asked to complete a series of problems to find out how confident you are about a topic.

Here’s an example in the Algebra: expressing relations section:

3. How confident are you that you can understand and use mathematical arguments, relying on rearrangement of equivalent expressions?

Problem 1

Problem: Show how 3(x + 10) + 9(x - 2) simplifies to a(x + b) and find the values of a and b.

Prompt 1

First you need to remove the two sets of brackets. How will you do that?

Prompt 2

When you have removed both pairs of brackets, you should collect like terms. What does ‘collect like terms’ mean? What different kinds of term are there in this expression?

Prompt 3

When you have collected like terms you should have an expression of the form Qx + P or P + Qx where P and Q are numbers.

Now factorise Qx + P. That is, find the largest number that will divide into both P and Q and write it before brackets. That is, if Qx + P = (A × Bx) + (A × C), then Qx + P = A(Bx+ C).

Solution

3(x + 10) + 9(x - 2) = 3x + 30 + 9x - 18
  = 12x + 12
  = 12(x + 1)

a is 12 and b is 1.

Problem Two

Problem: A number is increased by P%.

The number obtained is then decreased by P%. The result is N.

What is the original number when expressed in terms of P and N as a fraction in its simplest form?

Prompt 1

If you use a new letter to represent the original number, the task is to express that new letter in terms of P and N.

Prompt 2

If you use x to represent the original number,

  • how can you write P% of x as an expression?
  • how can you write (100% plus P%) of x as an expression?

Prompt 3

How can you write P% of ((100% plus P%) of x) as an expression?

How can you write (100% minus P%) of ((100% plus P%) of x) as an expression?

Prompt 4

Simplify your expression of (100% minus P%) of ((100% plus P%) of x) as far as possible by factorising. (Could you use the difference of two squares?) Then write it as a single fraction over a common denominator.

Prompt 5

If x is the original number, then your simplified expression of (100% minus P%) of ((100% plus P%) of x) is N.

Prompt 6

Put your simplified expression of (100% minus P%) of ((100% plus P%) of x) equal to N, then rearrange the formula to make x the subject of the equation.

Solution

Suppose that the original number is x. We want to find an expression for x in terms of P and N. The result of increasing x by P% is:

solution: equation 1

The result of now decreasing this previous result by P% is

solution: equation 2

...which is N.

Therefore:

solution: equation 3

The original number expressed in terms of P and N is:

solution: equation 4

 

 

 
 
 
 
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