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=0
^ is “to the power of”
n2-n-90=0
- Thread starter Chinman3000
- Start date
Got something to say or just want fewer pesky ads? Join us... 😊
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=0
^ is “to the power of”
The Tactician
Well-known member
- Feb 18, 2013
- 1,062
I will admit-I was actually in the exam yesterday and got this question-and answered it without any problems. I'm rubbish at Maths as well-C/B grade sadly. My worst subject by far but I don't see what everyone is complaining about...
daveinprague
New member
Thought the answer was 'A fish'
brightn'ove
cringe
Aren't these type of questions supposed to test the top 10%?
If so it makes sense that the bottom 90% are complaining about it (i.e. a lot of annoying children with twitter).
If so it makes sense that the bottom 90% are complaining about it (i.e. a lot of annoying children with twitter).
SPOILER BELOW
The chance of getting an orange sweet first time = 6/n where n is the total number of sweets.
Having already taken an orange sweet, the chance of getting an orange sweet next time = 5/(n-1)
The probability of both being orange sweets is one third, so 6/n x 5/(n-1) = 1/3
Or (6x5)/n(n-1) = 1/3
Then 30/(n^2-n) = 1/3
(3x30)/(n^2-n) = 1
90/(n^2-n) = 1
90=n^2-n
n^2-n-90=0
QED
n is obviously 10, but that wasn't the question.
I passed my O level maths in 1985.
The chance of getting an orange sweet first time = 6/n where n is the total number of sweets.
Having already taken an orange sweet, the chance of getting an orange sweet next time = 5/(n-1)
The probability of both being orange sweets is one third, so 6/n x 5/(n-1) = 1/3
Or (6x5)/n(n-1) = 1/3
Then 30/(n^2-n) = 1/3
(3x30)/(n^2-n) = 1
90/(n^2-n) = 1
90=n^2-n
n^2-n-90=0
QED
n is obviously 10, but that wasn't the question.
I passed my O level maths in 1985.
Buzzer
Languidly Clinical
- Oct 1, 2006
- 26,121
Actually, I think most people on here, myself included have got the answer wrong. We weren't asked to solve the equation but to prove it. And the pre-amble is vital in proving the formula. Here's the answer:
There are 6 orange sweets and n sweets overall. So, if Hannah takes one, there is 6/n chance of getting an orange sweet. When she takes one,, there is one less orange sweet and one less overall meaning that the probability is now (6-1)/(n-1)=5/n-1.
To find the probability of getting the orange sweet both times, multiply the two fractions: 6/n* 5/n-1 =30/n^2-n.
It shows the probability of taking two orange sweets (1/3) is: 1/3=30/n^2-n.
The denominators then need to be the same, so multiply 1/3 by 30 which would then make 30/90=30/n^2-n.
Discounting the 30 on both sides of the equation makes n^2-n=90. By moving 90 onto the other side of the equation, it will equal zero.
Not so easy, was it?
EDIT - MAC04 beat me to it. Well done matey! Mr Quittenton would be proud of you.
There are 6 orange sweets and n sweets overall. So, if Hannah takes one, there is 6/n chance of getting an orange sweet. When she takes one,, there is one less orange sweet and one less overall meaning that the probability is now (6-1)/(n-1)=5/n-1.
To find the probability of getting the orange sweet both times, multiply the two fractions: 6/n* 5/n-1 =30/n^2-n.
It shows the probability of taking two orange sweets (1/3) is: 1/3=30/n^2-n.
The denominators then need to be the same, so multiply 1/3 by 30 which would then make 30/90=30/n^2-n.
Discounting the 30 on both sides of the equation makes n^2-n=90. By moving 90 onto the other side of the equation, it will equal zero.
Not so easy, was it?
EDIT - MAC04 beat me to it. Well done matey! Mr Quittenton would be proud of you.
Last edited:
Quite.
n doesn't have to equal 10 (in the raw quadratic equation).
My mind is clearly ODD, because when I said I'd answered it in 15 seconds, I hadn't even spotted the obvious answer, but had answered -9.
-9 x -9 = 81
81 - -9 = 90
n = -9
My mind is clearly ODD, because when I said I'd answered it in 15 seconds, I hadn't even spotted the obvious answer, but had answered -9.
-9 x -9 = 81
81 - -9 = 90
n = -9
Prince Monolulu
Everything in Moderation
Just goes to prove the point that was emphasised over and over at Ashford PTC by the training skippers......R.F.Q.
They even wrote it on the board at the beginning of each exam.
They even wrote it on the board at the beginning of each exam.
Pavilionaire
Well-known member
- Jul 7, 2003
- 31,694
Just when you think NSC is full of CJTCs you come along - good work, this took me back 30+ years!
happypig
Staring at the rude boys
Tony Meolas Loan Spell
Slut Faced Whores
Well, I think that the exam board are being arses with this question.
_LRG.jpg)
Yes, the quadratic is easy but why deliberately confuse students with the rest of the guff? The kids are stressed enough already without playing silly buggers.
Thank you! Thats what I was trying to say.
n doesn't have to equal 10 (in the raw quadratic equation).
My mind is clearly ODD, because when I said I'd answered it in 15 seconds, I hadn't even spotted the obvious answer, but had answered -9.
-9 x -9 = 81
81 - -9 = 90
n = -9
If I went to a shop and paid good money for -9 sweets I'd come back and shoot the f***ers
Prince Monolulu
Everything in Moderation
This is pretty basic probability theory. If you have been taught this, it should be reasonably straight forward? The knack is realising it is a probability question, but the clues are there.