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

Is it £14456.88 ?
2. (Original post by _Xenon_)
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
3. (Original post by Zacken)
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
4. (Original post by _Xenon_)
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.
5. (Original post by Zacken)
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!!
6. (Original post by _Xenon_)
Thanks!!
No problem.
7. (Original post by _Xenon_)
After an 8% pay rise Mr Brown's salary was £15 714.
What was his salary before the increase?

Is it £14456.88 ?
You can make an equation then solve it:

1. You want the salary before the increase, but you don't know it - call it .
2. You know that was increased by 8% of , and that the total was 15714, and 8% of so:

Can you complete this?
8. (Original post by Zacken)
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..
9. (Original post by atsruser)
Just a little too dogmatic there, I'd suggest, Mr Zacken. There are other approaches..
10. (Original post by atsruser)
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.
11. (Original post by notnek)
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 🙈
12. (Original post by Kvothe the arcane)
I don't really see a difference between approaches 🙈
Mathematically they are similar but the thought process is quite different.

E.g. if you require the initial amount that has increased by 5% to 1200.

With the algebraic approach you assign a variable and increase it by 5%.

With the other approach you start with the final amount and aim to get from 105% to 100%.
13. (Original post by notnek)
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.
Ah, yeah - fair point. I tend to completely ignore the algebraic approach because when I teach this topic to my tutor group who learn about this sort of stuff before being introduced to algebra and variables it's really the only method I can show, so I've gotten stuck in a rut, seemingly.
14. (Original post by notnek)
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.
Algebaraic method is 1000000 times better, thank you so much

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