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Hannah's Sweets

What's this question about "Hannah's sweets" which was on the Edexcel GCSE paper? It's all over the news, I'm dying to know what the question was.

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It had something to do with a girl having sweets and like a 1/3 chance of getting an orange sweet then it said show that n^2-n-90 = 0... Can't remember even though I just did the exam...
Reply 2
Avatar for Kieran S
Kieran S
2
Original post by lizard54142
What's this question about "Hannah's sweets" which was on the Edexcel GCSE paper? It's all over the news, I'm dying to know what the question was.

Im bored so i might as well break it down.

Key
/ means a fraction. * means multiplication. ^ means power

Question
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=o

Answer
There are 6 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 1 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 2 orange sweet 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. If you are confused, i'll help you understand :smile:
(edited 8 years ago)
Reply 3
Avatar for lizard54142
lizard54142
OP
6
Original post by Kieran S
Im bored so i might as well break it down.

Key
/ means a fraction. * means multiplication. ^ means power

Question
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=o

Answer
There are 6 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 1 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 2 orange sweet 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. If you are confused, i'll help you understand :smile:


I just wanted to know the question, I am doing A levels! But cheers :wink: that's quite hard for GCSE
Original post by lizard54142
I just wanted to know the question, I am doing A levels! But cheers :wink: that's quite hard for GCSE


Are you sure you would be able to answer that question, its super hard for someone of your level. :colondollar:
This was a GCSE question?!
Holy smokes, I didn't even understand the lovely, detailed explanation up there by Kieran S. :eek:
Reply 6
Avatar for lizard54142
lizard54142
OP
6
Original post by Jai Sandhu
Are you sure you would be able to answer that question, its super hard for someone of your level. :colondollar:


Actually Jai I am still confused :frown: can you explain please...
So pleased I might have got the marks for that one - I probably ended up spending the best part of an hour on that one + the cone/cyclinder one!
Original post by lizard54142
Actually Jai I am still confused :frown: can you explain please...


I would if I could, but the question is impossibly hard :angry:
Reply 9
Avatar for dezaiyr
dezaiyr
9
lmao I fully flopped that paper! Didnt even understand Hanna's ******* sweets. IM PRAYING THEY MAKE THE BOUNDARIES ALOT LOWER THIS YEAR *fingerscrossed*
Original post by Jai Sandhu
I would if I could, but the question is impossibly hard :angry:


Like how can you have "n" sweets, n isn't even a number... definitely a typo
Reply 11
Avatar for hamiel
hamiel
3
https://www.dropbox.com/sh/ijwhfhd81n67614/AADBoJB30HJHmkyKJa2tDhCOa?dl=0 :smile:

**side note** hannah's sweets is right but last q and 16d is not. this is done by a teacher in a rush
Here is the solutionDSC_0024.jpg
After reading about it online that is so much easier than everything made it out to be!
guys, any assumption about the grade boundaries
????
Honestly the question is worded to be more difficult than it actually is:

There's n sweets originally, 6 of them orange and you're told the probability of getting two oranges in a row is 1/3

Initially there's a 6/n chance of getting an orange, then once one orange is removed the chance changes to 5/n-1

Therefore:
6/n * 5/n-1 = 1/3

= 30/n^2-n = 1/3
Times both sides by n^2 -n: 30 = 1/3n^2 - 1/3n
Times both sides by 3: 90 = n^2 - n
Subtract 90 from both sides: n^2 - n - 90 = 0
Is this really meant to be a super hard question? I remember having similar questions in my GCSE. It's not even bad.

Original post by itshuda
lmao I fully flopped that paper! Didnt even understand Hanna's f*cking sweets. I honestly hope she chokes on them and dies .. greedy ass b*tch UGH IM PRAYING THEY MAKE THE BOUNDARIES ALOT LOWER THIS YEAR *fingerscrossed*

Funniest thing I've read all day. :biggrin:
(edited 8 years ago)
Reply 17
Avatar for Notnek
Notnek
21
I've seen a similar question appear a few times in IGCSE - not sure if a question like this has been in a recent GCSE exam.

But there's nothing wrong with the wording of the question. For anyone struggling with questions like this, I recommend starting by using a number instead of n. Then it becomes a standard conditional probability question.

Then follow the same method but replace the number with n.
OMg i think i may have put all that,. Actually most, i mucked up and times by 3 after 30/n2-n but i still managed the the rest :^) maybe half marks?
Original post by Kieran S
Im bored so i might as well break it down.

Key
/ means a fraction. * means multiplication. ^ means power

Question
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=o

Answer
There are 6 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 1 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 2 orange sweet 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. If you are confused, i'll help you understand :smile:
(edited 8 years ago)
Original post by Magicaltoaster
OMg i think i may have put all that,. Actually most, i mucked up and times by 3 after 30/n2-n but i still managed the the rest :^) maybe half marks?


To be honest, when the most trending news has the headline of "Students traumatized as they show frustration towards today's maths exam", i wouldn't really worry about 1 question as grade boundaries will be up to your knees.

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