Toughest GCSE question ever?

Sky_JP · June 2015

I was reading an article today about a GCSE question. A student's mum took it to work and apparently it took 4 accounts with Maths degrees 2 hours to come up with an answer.

Here's the question;

"There are "N" sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow.

Hannah randomly takes a sweet from the bag and eats it. Hannah then takes another sweet at random and eats it.

The probability that Hannah eats two orange sweets is 1/3.

Prove that n^2-n-90=0."

(Note: N^2 is "N Squared")

I'm going to give it a go. Can anyone beat me to an answer?

Sky_JP · June 2015

To make this a little more relevant, imagine you need runner runner and have a total of 6 outs from a total of "N" cards.

Your chances of getting there are 1/3.

Sky_JP · June 2015

Think I've got there... no one having a go?

tomgoodun · June 2015

Defo fold pre

Benchmark · June 2015

I guess the chances for getting two orange sweets, broken down is:

6/N * 5/N-1= 1/3, with N being the number of sweets in the bag.

or 30/(N^2-N) = 1/3

Now to prove N^2-N-90=0

Here goes

We have 30/N^2-N = 1/3

Which is 30 = (N^2-N)/3

Which is 90 = N^2-N

Which is N^2 - N - 90 = 0

Bench.

bbMike · June 2015

Probably more an indictment on the accountants(?) than the standard of the question.

Probability of selecting first orange sweet is 6/n second time it's 5/(n-1).

Multiply them together gives 30/(n^2-n) which we are told = 1/3.

To express in terms of n you can invert then multiply by 30 to give n^2-n = 90

Therefore n^2-n-90 = 0

Which can be written as (n-10)(n+9) = 0

So the problem has only one true answer since number of sweets is non negative, n = 10 (4 yellows)

Sky_JP · June 2015

Nice efforts Mike and Benchmark!

These are way more simple than mine!

Sky_Dan · June 2015

I felt incredibly stupid reading this post.

NoseyBonk · June 2015

This is going to completely change the way I think when I'm scoffing my way through a bag of Starburst

SkyKirsty · June 2015

In Response to Re: Toughest GCSE question ever?:

I felt incredibly stupid reading this post.
Posted by Sky_Dan

My. Brain. Is. Fried.

GaryQQQ · June 2015

I'd prove it by eating the rest of the sweets, keeping count as I go.

Then insert my count into the equation as n.

jonnyrkd · June 2015

Tree Diagrams innit?

unknown_username · June 2015

I get

N^2-N-90=0

N^2-N=90

N^2=90+N

N=Sqr(90+N)

And just looking at that, I can see 10 works.

gerardirl · June 2015

Ok well N must be 10 as its the only value that gives you zero in that formula.....30secs job to get that!

Whats the relevance of the previous part if you trying to prove the forumula is correct. Am I missing something?

Ger

unknown_username · June 2015

In Response to Re: Toughest GCSE question ever?:

Ok well N must be 10 as its the only value that gives you zero in that formula.....30secs job to get that! Whats the relevance of the previous part if you trying to prove the forumula is correct. Am I missing something? Ger
Posted by gerardirl

Im with you on this, i dont see the connection with the sweets, the question could just be, whats N ?

IDONKCALLU · June 2015

just let me have all the sweets then we dnt have a question

VespaPX · June 2015

The answer to the ultimate question of life, the universe, and everything is... 42

FCHD · June 2015

In Response to Re: Toughest GCSE question ever?:

In Response to Re: Toughest GCSE question ever? : Im with you on this, i dont see the connection with the sweets, the question could just be, whats N ?
Posted by mumsie

Yes, that could be the question, but it isn't. The question isn't asking for values of N, it's testing the candidate's ability to prove a formula.

Sky_JP · June 2015

In Response to Re: Toughest GCSE question ever?:

In Response to Re: Toughest GCSE question ever? : Im with you on this, i dont see the connection with the sweets, the question could just be, whats N ?
Posted by mumsie

I think the idea is to prove that n=10 with the sweets, rather than in the equation.

Benchmark · June 2015

It's proving the formula is correct. Not just that it works for this particular instance.

Benchmark · June 2015

To further explain, bearing in mind I'm just a grunt..

There are six orange sweets in the bag, so the odds of picking the first orange sweet is 6 divided by the number of sweets. We don't know that figure, so it's called 'N' (or x or y or z or anything).

The chances of picking a second sweet are now five (one orange sweet less) divided by the number of sweets left in the bag, which is the total minus the one taken out (i.e. N-1)

To get the chances for both sweets being orange, which are co-dependent, the chances are multiplied together. The top and bottom lines are multiplied separately.

This gives 6*5 divided by N*(N-1). Worked out is 30 divided by N^2-N

We are told the probability is 1/3.

Therefore 30/N^2-N = 1/3

Multiply both sides of the equation by N^2 - N , to simplify.

This changes to 30 = N^2-N/3

Multiply both sides of the equation by 3, to simplify further.

This gives 90 = N^2-N

To get one side of the equation to equal zero, subtract 90 from both sides.

This gives 0 = N^2-N - 90

Hope that clears it up.

gerardirl · June 2015

Awh very good benchmark...i just used excel and changed the value til I got =0 lol.

But I guess you dont have excel in the exams eh!

RLT16 · June 2015

the whole sweet siuation is irrelevant as is the 1/3 odds of her picking out 2 of the same colour. just take the equasion and reverse it.
you know u have to subtract the 90 at the end to = 0
so u know that N^2 - N = 90 so its simply a case of finding which number fits, even if it isnt obvious straight away u could quite simply just work ur way through each number until u find the one that works.

i think if they had asked the question without the silly girl and her sweets this wouldnt of been brought up. simple question IMO, but then again i always did like maths

btw the answer is 10

FCHD · June 2015

The answer isn't 10 because "how many sweets are there?" isn't the question.

RLT16 · June 2015

In Response to Re: Toughest GCSE question ever?:

The answer isn't 10 because "how many sweets are there?" isn't the question.
Posted by FCHD

ah u are correct my bad

gerardirl · June 2015

A question is always very simple if you give the answer you want...instead of answering the actual questions lol.

The real question for me is ....what happened the rest of the sweets!

FCHD · June 2015

In Response to Re: Toughest GCSE question ever?:

A question is always very simple if you give the answer you want...instead of answering the actual questions lol. The real question for me is ....what happened the rest of the sweets!
Posted by gerardirl

Anna Fowler had them.

CHILLIE · June 2015

I hope the kids that got this question right was able to use this in every day life,but somehow i doubt it.

HENDRIK62 · June 2015

In Response to Re: Toughest GCSE question ever?:

To further explain, bearing in mind I'm just a grunt.. There are six orange sweets in the bag, so the odds of picking the first orange sweet is 6 divided by the number of sweets. We don't know that figure, so it's called 'N' (or x or y or z or anything). The chances of picking a second sweet are now five (one orange sweet less) divided by the number of sweets left in the bag, which is the total minus the one taken out (i.e. N-1) To get the chances for both sweets being orange, which are co-dependent, the chances are multiplied together. The top and bottom lines are multiplied separately. This gives 6*5 divided by N*(N-1). Worked out is 30 divided by N^2-N We are told the probability is 1/3. Therefore 30/N^2-N = 1/3 Multiply both sides of the equation by N^2 - N , to simplify. This changes to 30 = N^2-N/3 Multiply both sides of the equation by 3, to simplify further. This gives 90 = N^2-N To get one side of the equation to equal zero, subtract 90 from both sides. This gives 0 = N^2-N - 90 Hope that clears it up.
Posted by Benchmark

One of my friends posted this question on FB and I have to admit I plagiarised your work to look smarter than I am.

Many thanks :-)))

Benchmark · June 2015

The way I play poker, I'm used to people taking all that I have.

;o)

aussie09 · June 2015

In Response to Re: Toughest GCSE question ever?:

To further explain, bearing in mind I'm just a grunt.. There are six orange sweets in the bag, so the odds of picking the first orange sweet is 6 divided by the number of sweets. We don't know that figure, so it's called 'N' (or x or y or z or anything). The chances of picking a second sweet are now five (one orange sweet less) divided by the number of sweets left in the bag, which is the total minus the one taken out (i.e. N-1) To get the chances for both sweets being orange, which are co-dependent, the chances are multiplied together. The top and bottom lines are multiplied separately. This gives 6*5 divided by N*(N-1). Worked out is 30 divided by N^2-N We are told the probability is 1/3. Therefore 30/N^2-N = 1/3 Multiply both sides of the equation by N^2 - N , to simplify. This changes to 30 = N^2-N/3 Multiply both sides of the equation by 3, to simplify further. This gives 90 = N^2-N To get one side of the equation to equal zero, subtract 90 from both sides. This gives 0 = N^2-N - 90 Hope that clears it up.
Posted by Benchmark

brilliant. best post award.

Toughest GCSE question ever?

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