Percentages

After an 8% pay rise Mr Brown's salary was £15 714.
What was his salary before the increase?

What do I do here?
Thanks

Okay, so you need to adopt this sort of approach.

100% -----> original salary
108% -----> 15714
1% ------> 15714/108
100% ------> 15714/180 * 100

Thanks I'll do it like that now.
But why is it wrong when I do 15 714 * 0.92 ?
Isn't 0.92 the decimal multiplier to reduce by 8% (100-8=92) ? Hmmm

Nopes, you've taken the original amount and increased it by a factor of 1.08, multiplying it by 0.92 doesn't reverse this, unfortunately.

Try it out.

We take 100, increase it by 1.08 to get 108. Now decrease this by 0.92 to get 99.36 =/= 100.

Instead, in questions like this, always use the method I showed you; assign 100% to the original price/value/whatever, find what information you know, find 1% and then multiply 1% by whatever number to get whatever percentage required.

Thanks!!

After an 8% pay rise Mr Brown's salary was £15 714.
What was his salary before the increase?

Is it £14456.88 ?

Instead, in questions like this, always use the method I showed you;

Just a little too dogmatic there, I'd suggest, Mr Zacken. There are other approaches..

Just a little too dogmatic there, I'd suggest, Mr Zacken. There are other approaches..

I teach both approaches and the students who have strong algebra tend to prefer the algebraic method.

I like the algebra approach because all you need to do is perform an easy non-reverse percentage operation e.g. 'Increase by 8%' to a variable instead of a number.

I find that the students who prefer and utilise the algebraic approach rarely make mistakes.

I don't really see a difference between approaches 🙈

I teach both approaches and the students who have strong algebra tend to prefer the algebraic method.

I like the algebra approach because all you need to do is perform an easy non-reverse percentage operation e.g. 'Increase by 8%' to a variable instead of a number.

I find that the students who prefer and utilise the algebraic approach rarely make mistakes.

I teach both approaches and the students who have strong algebra tend to prefer the algebraic method.

I like the algebra approach because all you need to do is perform an easy non-reverse percentage operation e.g. 'Increase by 8%' to a variable instead of a number.

I find that the students who prefer and utilise the algebraic approach rarely make mistakes.