The internet has once again been abuzz with a tricky Maths question (following the extremely challenging problem set in a Singaporean exam a few weeks ago). Today's question came from the Edexcel GCSE exam paper. The question and solution are below.

Question:

There are *n *sweets in a bag.

6 of the sweets are orange.

The rest of the sweets are yellow.

Hannah takes a random sweet from the bag.

She eats the sweet.

Hannah then takes at random another sweet from the bag.

She eats the sweet.

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

Show that n^{2} - n - 90 = 0

Click here to view the solution.

#### Solution

The initial thought when seeing this question would be factorise the equation to give:

(n + 9) (n - 10) = 0

This would tell you that n = 10 or -9. As there can't be a negative number of sweets there therefore must be 10 sweets. However, while this solves the equation it doesn't prove the equation in the context of the question. To show that the equation is correct we must prove it from the information given. There are two ways of doing this:

**Option 1:**

As per the above if the equation is true then there must be 10 sweets.

The probability of Hannah taking an orange seet first is:

6/10

As she has now eaten one orange sweet there are 5 orange sweets left and 9 sweets left in total. The probability of Hannah taking an orange seet second is:

5/9

In order to work out the probability of two separate events happening we must multiple the probabilities of each event together:

6/10 × 5/ 9

= 30/90

=1/3

**Option 2:**

The probability of Hannah taking an orange sweet first is:

6/n

As there are now 5 orange sweets left and also one less than n sweets left in the bag, the probability of Hannah taking an orange weet second is:

5/(n-1)

In order to work out the probability of two separate events happening we must multiple the probabilities of each event together allowing as to form the equation:

6/n × 5/(n-1) = 1/3

If we multiply this out we get:

30/(n^{2} - n) = 1/3

Multiply both sides by 3:

90/(n^{2} - n) = 1

Multiply both sides by (n^{2} - n):

90 = n^{2} - n

Subtract 90 from both sides:

0 = n^{2} - n - 90

Which is the same as n^{2} - n - 90 = 0

As ever it's **crucial to show all your working to get all the marks available**.