But hang on, what might happen is the first time yes she gets two oraange
ones if she cannot see the colour, howeve she might get 1 of each, so the
next bit is then changed. its not a real world problem, this is what always
made me annoyed.
Brian

--
From the Sofa of Brian Gaff Reply address is active
"Tim Watts" wrote in message
...
On 05/06/15 15:32, Tim Watts wrote:
On 05/06/15 15:23, Muddymike wrote:
GCSE's getting harder causes Paul Dacre's head to explode.

or

"Slightly hard" GCSE Maths Question causes outrage...


http://www.bbc.co.uk/news/education-33017299



The actual question was:

Hannah has 6 orange sweets and some yellow sweets.
Overall, she has n sweets.
The probability of her taking 2 orange sweets is 1/3.

Prove that: n^2-n-90=0




There are six orange sweets and n sweets overall. If she takes one,
there is a 6/n chance of getting and orange sweet. When she takes one,
there is one less orange sweet and one less sweet overall.

If she took another orange sweet, the probability would be
(6-1)/(n-1)=5/n-1. Now, you have to find the probability if she gets two
orange sweets so you simply times the two fractions: 6/n * 5/n-1 =
30/n^2-n.

It tells us the probability of two orange sweets is 1/3 which means
1/3=30/n^2-n.

We need to make the denominators the same so simply times 1/3 by 30/30
which would equal 30/90. if 30/90 = 30/n^2-n, then n^2-n=90. if n^2-n=90
then n^2-n-90 will equal zero.

Mike
(with a little help from Google)

Which being in the real 2015 world is exactly what I would do if this
were a real problem, so who need to pass the exam?

And this was my effort (complete with ****ty handwriting as it was on a
phone screen *cough*):

http://tinypic.com/m/ip86qs/3


I had a look back.

For comparison, here is avery uk.d-i-y oritent Uni London Board O-Level
question from 1988:

http://tinypic.com/m/ip8900/3

And here's a 1957 Cambridge paper, also O-Level:

Paste error:

http://www.cambridgeassessment.org.u...tion-paper.pdf