C1 and C2 DOUBTS HERE!!!!!

[elemis]

http://answers.yahoo.com/question/index?qid=20101106111046AAZVRlB

[Ivo]

http://answers.yahoo.com/question/index?qid=20101106111046AAZVRlB

Thanks for the link, but I still don't really understand what they are trying to say.  Hope you can help.

[astarmathsandphysics]

see attach

[**RoRo**]

I have 2 questions:

1. Can someone please explain to me how do you come to a conclusion about Question 9 (e) in June 2005 paper?

2. The answer to question 7 (d) in the marking scheme is 26 years old? How is that? [January 2006]

Thanks in advance!!

[Alpha]

I have 2 questions:

1. Can someone please explain to me how do you come to a conclusion about Question 9 (e) in June 2005 paper?

2. The answer to question 7 (d) in the marking scheme is 26 years old? How is that? [January 2006]

Thanks in advance!!

Can you post the link to those papers please? Or upload them?

I think it's Edexcel, right? Don't find the papers on Xtreme. 

[elemis]

For part 2 :

3000 = n/2 (2*500 + (n-1)200)

64000 = 800n +200n²

2n²+8n-640=0

n² +4n -320 = 0

(n+20)(n-16)= 0

n= -20 or 16

Disallow -20 and consider only 16.

The above formula INCLUDES her 11th birthday. So we must add 16 to 10 to give 26.

[Alpha]

For part 2 :

3000 = n/2 (2*500 + (n-1)200)

64000 = 800n +200n²

2n²+8n-640=0

n² +4n -320 = 0

(n+20)(n-16)= 0

n= -20 or 16

Disallow -20 and consider only 16.

The above formula INCLUDES her 11th birthday. So we must add 16 to 10 to give 26.

Where did you find the papers? I'm still looking for them. 

And thank you Lion.

[elemis]

For the 1st part :

If you input 100 into the equation a + (n-1)d to find the amount he would be paying in tthe 100 th month you will see that you get an answer of -49.

However, a person cant pay $-49 to a bank hence making it a nonsensical answer.

[elemis]

Where did you find the papers? I'm still looking for them. 

And thank you Lion.

I downloaded them off the net. However, the site I got it off is now dead.

[Alpha]

Okie, Xtremepapers doesn't have them right now either.

Thank you.

[**RoRo**]

Can you post the link to those papers please? Or upload them?

I think it's Edexcel, right? Don't find the papers on Xtreme. 

I have the past papers in a booklet here!

I tried to find the link on www.freeexampapers.com but it just won't open! :/

Check this link, it has some of the past papers: https://eiewebvip.edexcel.org.uk/PastPapers/Default.aspx

[elemis]

I have the past papers in a booklet here!

I tried to find the link on www.freeexampapers.com but it just won't open! :/

Check this link, it has some of the past papers: https://eiewebvip.edexcel.org.uk/PastPapers/Default.aspx

I've answered your doubts already.

[**RoRo**]

For the 1st part :

If you input 100 into the equation a + (n-1)d to find the amount he would be paying in tthe 100 th month you will see that you get an answer of -49.

However, a person cant pay $-49 to a bank hence making it a nonsensical answer.

Thank you very much!

& I apologize for the late reply, my Internet got dc for a while!

[elemis]

Thank you very much!

& I apologize for the late reply, my Internet got dc for a while!

No worries. The explanation for the 2nd question is here :

For part 2 :

3000 = n/2 (2*500 + (n-1)200)

64000 = 800n +200n²

2n²+8n-640=0

n² +4n -320 = 0

(n+20)(n-16)= 0

n= -20 or 16

Disallow -20 and consider only 16.

The above formula INCLUDES her 11th birthday. So we must add 16 to 10 to give 26.

[**RoRo**]

No worries. The explanation for the 2nd question is here :

I read your explanation, but see, in part (c) we found that the total of the allowances up to and including her 18^th birthday was 39600...how come her allowance would stop at the age of 26 at 32000? It's supposed to increase, right?

& one more question please:

In the specimen paper, [it has no particular date info, so I'll type the question here]:

The curve C has equation y=f(x) and the point P (3,5) lies on C.
Given that f'(x) = 3x²-8x+6

(a) Find f(x)
The answer is f(x)=x³-4x²+6x-4

(b) Not needed now

The point Q also lies on C, and the tangent to C at Q is parallel to the tangent to C at P.
(c) Find the x-coordinate of Q.

I found the gradient to be 9 [using the formula for f'(x)]
& it's correct according to the marking scheme.

And then I equated the formula:
3x²-8x+6=9

I got 2 solutions: x=3, x=-1/3

Which one would be the correct one? The MS says it's -1/3 but I don't understand why!