Hannah's Sweets

Took me 30 seconds. To those people saying grade boundaries should be lowered, it's all done automatically so a certain number of people get a set grade. It's not like they're going to artificially lower it because some people didn't like one of the questions

Sweets are taken without replacement so it becomes conditional.

The probability that Hannah takes an orange sweet on the second pick is dependent on whether she takes/doesn't take an orange sweet on the first pick.

If a question like this can make the news then I hope the exam boards are ready for the outcry when students take the new GCSE in 2017.

For anyone interested, read through the 3 specimen papers from p.101 of this document.

Hannah's sweets will look easy compared to e.g. Q15 on Paper 3.

Took A Level maths myself, still found his question very difficult. The wording of it isn't particularly clear for what they're trying to get you to calculate.

But then equally I appreciate that they need to have some very tough questions in there just to separate out the ultra high achievers.

The decimal one?

Yes. Parts b) and c) are nasty.

So many words for part 3 . Well he hasn't shown it to be true because he hasn't actually found the reciprocal and he hasn't verified that 3 is the least value for n. It could be 2 or 1.

Edit: Cool. I can't tell whether someone at GCSE level would find it hard.

Can you not simply factorise the equation?

N^2-n-90 = (n-10)(n+9)

So n must be 10, then show that 6/10 x 5/9 = 1/3??

Can you not simply factorise the equation?

N^2-n-90 = (n-10)(n+9)

So n must be 10, then show that 6/10 x 5/9 = 1/3??

I'm pretty sure most would. I find the theorem and the stuff before it are badly written. I would have written all the information in the theorem in italics.

I also think most students would struggle to see what to do in part b).

while you are correct you are not answering the question, it wants you to prove the equation n^2-n-90=0 not find out what n is or prove the probability of two orange sweets to be 1/3.

Can you not simply factorise the equation?

N^2-n-90 = (n-10)(n+9)

So n must be 10, then show that 6/10 x 5/9 = 1/3??

It is proving it, just in the reverse order. If n^2-n-90 didn't equal zero then this method wouldn't work.

If you can show that n is 10 then it follows that $n^2-n-90=0$

i.e. $n=10 \Rightarrow n^2-n-90=0$


But how do you know that n=10 is the only solution? Your method is to assume the equation $n^2-n-90=0$ without proving it.

It could be the case that n=10 works and so does n=5 then

$n=10$ or $n=5$ $\Rightarrow n^2-n-90=0$

This is incorrect.

but you're using the thing they've given you to prove, to prove it. If you use an equation to prove itself then you're not actually proving anything, except maybe 1=1

You might be lucky with the markscheme but unfortunately it's more than likely they won't accept that..

You're correct, the other solution is -9 but as you can't have a negative number of sweets then it must be 10, there is no other solution to making that equation equal to zero and have all integer values.