Toughest GCSE question ever?

Sky_JP

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

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

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

tomgoodun

Defo fold pre

Benchmark

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

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

Nice efforts Mike and Benchmark!

These are way more simple than mine!

Sky_Dan

I felt incredibly stupid reading this post.

NoseyBonk

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

SkyKirsty

In Response to Re: Toughest GCSE question ever?:

I felt incredibly stupid reading this post.
Posted by Sky_Dan

My. Brain. Is. Fried.

GaryQQQ

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

Tree Diagrams innit?

mumsie

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

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

mumsie

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 ?

VespaPX

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

FCHD

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

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

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

Benchmark

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

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!