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HSC 2001 Question 8c and 10a urgent help needed (1 Viewer)
- Thread starter Twickel
- Start date
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/mathemat_01.pdf
Is that what you mean?
Could you also do question 8c from the 2005 paper
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_05.pdf
Is that what you mean?
Could you also do question 8c from the 2005 paper
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_05.pdf
gurmies
Drover
For 10 (a)
(i) At the beginning of the 2nd year, and the end of the first year are the same thing, so ultimately there will be only 1 years worth of accumulation. To find the formula of what is happening:
1000(1 + 0.06) - 72. [basically, 1000 bux at 0.06 compounding, take away annual payment]
That = $988
(ii) We observe the pattern:
$B1 = 1000(1.06) - 72
$B2 = [1000(1.06) - 72] (1.06) - 72
= 1000(1.06)^2 - 72(1.06) - 72
= 1000(1.06)^2 - 72 (1 + 1.06)
$Bn = 1000(1.06)^n - 72 (1 + 1.06 + ... 1.06^(n-1))
= 1000(1.06)^n - 72 ( (1.06^n -1)/(0.06))
= 1000(1.06)^n - 1200(1.06^n -1)
= 1000(1.06)^n - 1200(1.06^n) + 1200
= 1200 - 200 (1.06)^n
(iii) This is slightly trickier, but I think I have it.
Let's firstly find out how much she can still afford at the end of the 10th year. Using the above formula:
1200 - 200 (1.06)^10 = $841.83
Now, we find a new formula for her account using subtractions of 90 now.
Using the similar method as I did above for $B1, $B2 and thus $Bn, I find the new $Bn to be:
841.83 (1.06)^n - 90 ((1.06^n - 1)/(0.06))
= 1500 - 658.17 (1.06)^n
Now, to find how many more years she can pay, we make the above expression = 0 in that at that amount, she can no longer make any payments using that account.
658.17(1.06)^n = 1500
(1.06)^n = 1500/(658.17)
n log 1.06 = log [(1500)/(658.17)]
n = log [(1500)/(658.17)] / log (1.06)
n = 14.14
Therefore, she can only make an additional 14 FULL payments (14 years).
For question 8 (c) of the 2005 paper, a similar concept is applied.
(i) A1 = 3,000,000(1.12) - 480,000
= (3 x 10^6)(1.12) - (4.8 x 10^5)
A2 = [(3 x 10^6)(1.12) - (4.8 x 10^5)] (1.12) - (4.8 x 10^5)
= (3 x 10^6)(1.12)^2 - (4.8 x 10^5)(1.12) - (4.8 x 10^5)
= (3 x 10^6)(1.12)^2 - (4.8 x 10^5) (1.12 + 1)
(ii) An is found, by the pattern observed in the above 2 expressions, or a 3rd if you have trouble visualising what will occur.
An = (3 x 10^6)(1.12)^n - (4.8 x 10^5)[(1 + 1.12 + ... + (1.12)^(n-1)]
= (3 x 10^6)(1.12)^n - (4.8 x 10^5)[(1.12^n -1)/(0.12)]
= (3 x 10^6)(1.12)^n - (4 x 10^6)(1.12^n -1 )
= (3 x 10^6)(1.12)^n - (4 x 10^6)(1.12)^n + (4 x 10^6)
= (4 x 10^6) - (1 x 10^6)(1.12)^n
= 10^6 [4 - (1.12)^n]
(iii) An = 0 (there will be no amount owing at this point)
Therefore, 10^6 [4 - (1.12)^n] = 0
4 - 1.12^n = 0
1.12^n = 4
n log 1.12 = log 4
n = log 4 / log 1.12
= 12.23
BUT, to clear the debt, the 13th year repayment will have to be last.
(i) At the beginning of the 2nd year, and the end of the first year are the same thing, so ultimately there will be only 1 years worth of accumulation. To find the formula of what is happening:
1000(1 + 0.06) - 72. [basically, 1000 bux at 0.06 compounding, take away annual payment]
That = $988
(ii) We observe the pattern:
$B1 = 1000(1.06) - 72
$B2 = [1000(1.06) - 72] (1.06) - 72
= 1000(1.06)^2 - 72(1.06) - 72
= 1000(1.06)^2 - 72 (1 + 1.06)
$Bn = 1000(1.06)^n - 72 (1 + 1.06 + ... 1.06^(n-1))
= 1000(1.06)^n - 72 ( (1.06^n -1)/(0.06))
= 1000(1.06)^n - 1200(1.06^n -1)
= 1000(1.06)^n - 1200(1.06^n) + 1200
= 1200 - 200 (1.06)^n
(iii) This is slightly trickier, but I think I have it.
Let's firstly find out how much she can still afford at the end of the 10th year. Using the above formula:
1200 - 200 (1.06)^10 = $841.83
Now, we find a new formula for her account using subtractions of 90 now.
Using the similar method as I did above for $B1, $B2 and thus $Bn, I find the new $Bn to be:
841.83 (1.06)^n - 90 ((1.06^n - 1)/(0.06))
= 1500 - 658.17 (1.06)^n
Now, to find how many more years she can pay, we make the above expression = 0 in that at that amount, she can no longer make any payments using that account.
658.17(1.06)^n = 1500
(1.06)^n = 1500/(658.17)
n log 1.06 = log [(1500)/(658.17)]
n = log [(1500)/(658.17)] / log (1.06)
n = 14.14
Therefore, she can only make an additional 14 FULL payments (14 years).
For question 8 (c) of the 2005 paper, a similar concept is applied.
(i) A1 = 3,000,000(1.12) - 480,000
= (3 x 10^6)(1.12) - (4.8 x 10^5)
A2 = [(3 x 10^6)(1.12) - (4.8 x 10^5)] (1.12) - (4.8 x 10^5)
= (3 x 10^6)(1.12)^2 - (4.8 x 10^5)(1.12) - (4.8 x 10^5)
= (3 x 10^6)(1.12)^2 - (4.8 x 10^5) (1.12 + 1)
(ii) An is found, by the pattern observed in the above 2 expressions, or a 3rd if you have trouble visualising what will occur.
An = (3 x 10^6)(1.12)^n - (4.8 x 10^5)[(1 + 1.12 + ... + (1.12)^(n-1)]
= (3 x 10^6)(1.12)^n - (4.8 x 10^5)[(1.12^n -1)/(0.12)]
= (3 x 10^6)(1.12)^n - (4 x 10^6)(1.12^n -1 )
= (3 x 10^6)(1.12)^n - (4 x 10^6)(1.12)^n + (4 x 10^6)
= (4 x 10^6) - (1 x 10^6)(1.12)^n
= 10^6 [4 - (1.12)^n]
(iii) An = 0 (there will be no amount owing at this point)
Therefore, 10^6 [4 - (1.12)^n] = 0
4 - 1.12^n = 0
1.12^n = 4
n log 1.12 = log 4
n = log 4 / log 1.12
= 12.23
BUT, to clear the debt, the 13th year repayment will have to be last.
Last edited:
conics2008
Active Member
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- Mar 26, 2008
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- HSC
- 2005
dam 2unit is becoming demanding lol and im tutoring them haha
gurmies
Drover
^ can you check my answers?
gurmies
Drover
Thank you =)