1
$\begingroup$

This is from the controversial GCSE question in the UKs recent exams. The orginal question is thus:

There are $n$ sweets in a bag. $6$ of the sweets are orange, the rest are yellow.

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

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

a) show that $n^2 - n - 90 = 0$

This in itself isn't too difficult:

$$ \frac{6}{n}\frac{5}{n-1} = \frac{30}{n^2 - n} = \frac{1}{3} \Rightarrow \frac{90}{n^2 -n} = 1 $$ so that $$ n^2 - n -90 = 0 $$ But, what I'm trying to figure out is the value of n, the only way to do this I have thought of is trial and error, quickly finding the answer to be 10. Is there a different option here?

$\endgroup$
5
  • 1
    $\begingroup$ Factorise the equation to give $(n-10)(n+9)=0$, resulting in $n=10$ as $n$ is positive. $\endgroup$
    – Hypergeometricx
    Jun 9, 2015 at 15:06
  • 2
    $\begingroup$ What about the question is controversial? $\endgroup$
    – Daniel Fischer
    Jun 9, 2015 at 15:07
  • $\begingroup$ If you assume what you are asked to show you can use the quadratic formula to find the roots of the equation or factorise as suggested in the comment above then test the conditions to check it works. $\endgroup$
    – Warren Hill
    Jun 9, 2015 at 15:08
  • $\begingroup$ Also: math.stackexchange.com/questions/1313076/… $\endgroup$
    – MattAllegro
    Jun 9, 2015 at 19:28
  • $\begingroup$ @DanielFischer It was a GCSE level question, there was no explanation based on the facts. It is somewhat advanced and was a bit here are some random facts, a b and c, and here's a question Z. Here is the news article: bbc.co.uk/news/education-33017299 $\endgroup$
    – MrDobilina
    Jun 10, 2015 at 9:37

3 Answers 3

Reset to default
3
$\begingroup$

You could factor $-90$. You need $a,b$ such that $ab = -90$ and $a+b = -1$. There aren't that many factors, so you can quickly find $a,b = -10,9$, so that $n^2 - n - 90 = (n-10)(n+9)$, which gets you to the solutions $n = 10, -9$, the second of which is obviously invalid.

Alternatively, if you don't notice the factoring, there's always the good old standby: the quadratic formula (assuming you know your squares!)

$$\begin{split} n &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{1 \pm \sqrt{361}}{2} \\ &= \frac{1 \pm 19}{2} \\ &= 10, -9 \end{split}$$

$\endgroup$
1
  • $\begingroup$ I do like this, had forgotten about the quadratic formula, been a while! $\endgroup$
    – MrDobilina
    Jun 9, 2015 at 15:18
0
$\begingroup$

I suppose use of quadratic formula can help(http://en.wikipedia.org/wiki/Quadratic_formula). Just substitute coefficients to expression $x=(-b-sqrt(b^2-4ac))/2a, y=(-b+sqrt(b^2-4ac))/2a$ where $a=1, b=-1, c=-90.$ Obviously, only positive solution makes sense, so you see n=10.

$\endgroup$
0
$\begingroup$

Ok write the quadratic expression as $$n^2-10n+9n-90=0$$ then factorise it to get $(n+9)(n-10)=0$.

$n$ can't be negative so the answer is only $10$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.