x Turn on thread page Beta
 You are Here: Home >< Maths

# tricky question (ratios?) watch

1. When a number X is increased by 10% its value becomes Y. When anumber Z is decreased by 10% its value becomes Y. By that percentagemust X be increased so its value equals Z?

All i have is Y= 11X/10 and Y=9Z/10
2. 22.22%
3. (Original post by ErniePicks)
When a number X is increased by 10% its value becomes Y. When anumber Z is decreased by 10% its value becomes Y. By that percentagemust X be increased so its value equals Z?

All i have is Y= 11X/10 and Y=9Z/10
You have two simultaneous equations there.. what can you do?
4. (Original post by SeanFM)
You have two simultaneous equations there.. what can you do?
If i substitute it in then it's 11X/10=9Z/10 i have no diea what to do next.
5. (Original post by ErniePicks)
If i substitute it in then it's 11X/10=9Z/10 i have no diea what to do next.
Good job

So if you look at going from X to Y, a 10% increase means you do 11/10 * X = Y (which you've done already), so the number you're going from * the percentage it's increasing/decreasing by = the new number.

So if you wanted to make your substituted equation look like the X to Y one, what would you do?
6. (Original post by SeanFM)
Good job

So if you look at going from X to Y, a 10% increase means you do 11/10 * X = Y (which you've done already), so the number you're going from * the percentage it's increasing/decreasing by = the new number.

So if you wanted to make your substituted equation look like the X to Y one, what would you do?
I'm really sorry but i don't really understand it. Is the equation

(11/10*X)* N%=(9/10X)

That doesn't look right to me, but even if it i have no clue where to go from there
7. (Original post by ErniePicks)
I'm really sorry but i don't really understand it. Is the equation

(11/10*X)* N%=(9/10X)

That doesn't look right to me, but even if it i have no clue where to go from there
Not quite.

For X and Z you have a single fraction (which represents the percantage) multiplied by X or Z to get 1 * Y (or just Y).

So you want there to be a single fraction that you multiply X by to get Z and from that fraction you can work out the answer.
8. (Original post by SeanFM)
Not quite.

For X and Z you have a single fraction (which represents the percantage) multiplied by X or Z to get 1 * Y (or just Y).

So you want there to be a single fraction that you multiply X by to get Z and from that fraction you can work out the answer.
I am so sorry i just can't seem to wrap my head around this. Is there a worked example of a similar question to this explaining the steps that you have a link for or the name of what type of question this is so i could search around?

Once again so sorry, i really appreciated your help
9. (Original post by ErniePicks)
I am so sorry i just can't seem to wrap my head around this. Is there a worked example of a similar question to this explaining the steps that you have a link for or the name of what type of question this is so i could search around?

Once again so sorry, i really appreciated your help
Not to worry. I think if you understand how to do this question you'll be fine if you ever see it again.

If I asked you to make Z the subject of the formula (the substituted one) what would you do? And how does that answer look similar to the two equations that you started with?
10. (Original post by SeanFM)
Not to worry. I think if you understand how to do this question you'll be fine if you ever see it again.

If I asked you to make Z the subject of the formula (the substituted one) what would you do? And how does that answer look similar to the two equations that you started with?
I would just divide both sides by 9/10 and that means it would be Z= 11/10*10/9*X
which is 110/90*X which is equivalent to 1.22*z= x OHHHHHH SO THATS HOW X HAS TO INCREASE BY 22% TO EQUAL Z!!!!!!! jesus sosrry i have no idea how i didn't manage to put that together. Thank you so much for the help, i think if i ahd put it in as ratios (0.9 and 1.1) i would have noticed sooner. Thanks so much mAN

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 22, 2015
Today on TSR

### How much will your degree earn you?

Find out where yours ranks...

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE