We started the lesson today by trying to recall the first few prime numbers that we discuss a couple of lessons ago.
We then reminded ourselves of the work we were doing last lesson on estimation of answers to calculations. We talked about how dividing by a number less than 1 gives an answer that is bigger than the number we started with.
Dividing by a number less than 1 is the same as multipling by its reciprocal.
For example: to find the answer to 
. We must remember that 0.1 is the same as 
and so the reciprocal of 0.1 is 10.
So = 75 x 10 = 750
We identified some more equivalent fraction and decimal that we need to remember:
0.1 = (reciprocal = 10)
0.2 = 
(reciprocal = 4)
0.125 = 
(reciprocal = 8 )
The first part of Module 3 chapter 2 is concerned with estimating answers to calculations by rounding the numbers to one significant figure (1sf).
If you are not sure how to round to significant figures you can look at the Mymaths lesson on the topic.
Example:
Estimate the answer to 3.86 x 2.14
3.86 is 4 to 1sf and 2.14 is 2 to 1sf
so 3.86 x 2.14 ≈ 4 x 2 = 8
It is difficult to calculate with decimals so it is a good idea to estimate answers to calculations using this method to help you check that you are correct if you do the exact calculation.
There is a Mymaths lesson on estimating calculations that covers the work from this lesson and the next couple of lessons.
The last part of Module 3 chapter 1 was reciprocals.
The reciprocal of a number is found by inverting (or turning the number upside down)
For example the reciprocal of 
is 
.
The reciprocal of a whole number is found by thinking of the number as a fraction over 1.
So 5 can be written as 
and the reciprocal is 
To find the reciprocal of a decimal is is often easiest to change it to a fraction.
For example 0.3 is the same as 
and the reciprocal is 
We used the sieve of Eratosthenes to identify all of the prime numbers below 200.
This involves crossing out all the numbers that have factors other than 1 or themselves, by crossing out multiples of 2,3,5,7,9, and 11. All of the remaining numbers are prime.
The presentation below shows the result.
We looked at how to find the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of two numbers by finding the prime factors of both numbers and using a Venn diagram.
For example to find the HCF and LCM of 84 and 140:
1. Use factor trees to find the prime factors of both numbers
2. Put the prime factors of both numbers in a Venn diagram. Any prime factors in common go in the centre and others go to the sides.
3. To work out the Highest Common Factor, multiple the numbers in the overlapping area together
HCF = 2 x 2 x 7 = 28
4. To work out the Lowest Common Multiple, Multiple all the numbers in the Venn diagram together.
LCM = 3 x 2 x 2 x 7 x 5 = 420
You can find more details and other examples on the Mymaths.co.uk lessons for HCF and LCM.
(you will need to log in to Mymaths to access the lessons)
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Welcome to the class blog for Mr Dolan’s Year 10 top set in 2008.
This blog is a place to record what we do in class, collect resources to support the learning that we do and to discuss how we can improve our understanding of Maths, achieve excellent results and enjoy the process of doing it.
