You are Here: Home >< Maths

# Easy GCSE question watch

1. Hey I read this question and I want to figure out how to work it out
basically it says that in 2001 there was 2 rabbits and in 2002 there was 4 rabbits and in 2003 there was 8 rabbits

which means that it doubles every year but what year will it be when there is 1 million rabbits?

I know the answer is 2020 but how do I get there?
2. (Original post by high)
Hey I read this question and I want to figure out how to work it out
basically it says that in 2001 there was 2 rabbits and in 2002 there was 4 rabbits and in 2003 there was 8 rabbits

which means that it doubles every year but what year will it be when there is 1 million rabbits?

I know the answer is 2020 but how do I get there?
show it in calculations
3. Keep doubling until you get to a million. (Takes a long time, worth the marks though)
4. Lol I know I could do that but isn't there a method?
5. if you havent done sequences or logs, then i cant think of any other way. i dont think sequences or logarithms are on the gcse syllabus, are they?
6. Doable with logs. Without that just do it longhand. It's GCSE. If you put the answer down without writing anything you'd probably still get the marks.
7. (Original post by RK92)
if you havent done sequences or logs, then i cant think of any other way. i dont think sequences or logarithms are on the gcse syllabus, are they?
Nope they're not but I can still do them lol...
so would it be log2 1000000?
Thanks a lot never thought it would come in handy....
8. the bunnies would all be inbred and genetically broken
9. (Original post by sparrow-legs)
the bunnies would all be inbred and genetically broken
10. The sequence is 2 to the power n.

You want 1000000 rabbits so
2^n = 1000000

Taking logs of both sides

nlog2 = log1000000

therefore n= (log1000000)/(log2)

Add the answer to 2000 to get the year since you started in 2001

EDIT: or use log to the base 2 as you stated. Makes it a bit simpler.
11. (Original post by high)
Nope they're not but I can still do them lol...
so would it be log2 1000000?
Thanks a lot never thought it would come in handy....
yeah, it would be log base 2 of 1,000,000 which is approx 19.9years so youd say 2020
12. If you let 2001, 2002, 2003 etc = 1, 2, 3 etc. and call this number (set) as n, then x (the answer) is 2 to the power of n. (2^n)

so 1mil = 2 ^ y y is an unknown value of n
log both sides to get log(1mil)=ylog(2)

log(1mil) / log (2) =y
y- 19.93 = 20
so 2000+20 = 2020
13. 2001 - 2^1
2002 - 2^2
2003 - 2^3....

X - 2^20 ( about 1 million)

so x=2020
14. 2^n = 1000000

Just work it out. They won't ask for more in GCSE.
15. Assuming that there were 2 rabbits at the beginning of 2001 and 4 at the beginning of 2002 etc. then there will 1,000,000 rabbits by midway through December in 2019, not 2020.

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: May 7, 2011
Today on TSR

Poll
Useful resources

### 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