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math question,help please! (1 Viewer)
- Thread starter luzq1025
- Start date
I think I feel quite confident with Q2 here and please bear in mind the fact that I'm only a year 9/10 student but I gave it a good try. I might not be right, so please tell me if my answer is correct and perhaps I could ask another person I know (carrotsticks or spiralflex or largarithmic) to clarify my answer and the process I have used.
Here, you just have to find the original loan the person took out. Let the original amount be P, rate is 1 + (0.115/12) and time period be 300 months.
At the beginning of the loan period:
 - 800\\A2 = A1(1+23/2400) - 800 \\= P(1+23/2400)2 - 800((1+23/2400) + 1)\\A3 = P(1+23/2400)3 - 800((1+23/2400)2 + (1+23/2400) + 1)\\Continued...\\A300 = P(1+23/2400)300-800((1+23/2400)299 (1+23/2400)298+ ... + (1+23/2400) + 1)\\\text{but } A300 = 0, \text{ as the loan is repayed, and} ((1+23/2400)299 (1+23/2400)298+ ... + (1+23/2400) + 1) \text {is a geometric series with 300 terms, first term 1 and ratio }(1+23/2400), \text{ so }\\0 = P(1+23/2400)300 - 800((1+23/2400)300 - 1)/(23/2400) - 1\\P(1+23/2400)300 = 800((1+23/2400)300 - 1)/((23/2400)-1)\\P = { 800((1+23/2400)300 - 1)/((23/2400)-1) } / { (1+23/2400)300 }\\= $78,703.83 \text{ So now we must round to the nearest hundred }\\= $78,700 )
Here, you just have to find the original loan the person took out. Let the original amount be P, rate is 1 + (0.115/12) and time period be 300 months.
At the beginning of the loan period:
Last edited:
michaeljennings
Active Member
This is the correct answerI think I feel quite confident with Q2 here and please bear in mind the fact that I'm only a year 9/10 student but I gave it a good try. I might not be right, so please tell me if my answer is correct and perhaps I could ask another person I know (carrotsticks or spiralflex or largarithmic) to clarify my answer and the process I have used.
Here, you just have to find the original loan the person took out. Let the original amount be P, rate is 1 + (0.115/12) and time period be 300 months.
At the beginning of the loan period:
Perhaps one of the 'better' bosers here who are more capable in maths will probably give us a simpler and faster approach to the problem.
lydia1992
scientificfield
there is this formula but i'm not sure if it works with all loan problems :
M= (Pr (1+r )^n)/ (1+r)^n -1
800=( P(23/2400)(2423/2400)^300)/(2423/2400)^300 -1
13187.55202 = P(23/2400) (2423/2400)^300
P= 13187.55202/(23/2400)(2423/2400)^300
P=78700
YaY it works
M= (Pr (1+r )^n)/ (1+r)^n -1
800=( P(23/2400)(2423/2400)^300)/(2423/2400)^300 -1
13187.55202 = P(23/2400) (2423/2400)^300
P= 13187.55202/(23/2400)(2423/2400)^300
P=78700
YaY it works
Last edited:
Carrotsticks
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The formula works because it is just a general result acquired by assigning letters (having a mindblank right now, forgot the proper term for it) to things like interest, time etc , then continuing a question as usual, then simplifying the expression, hence acquiring a 'general formula'there is this formula but i'm not sure if it works with all loan problems :
M= (Pr (1+r )^n)/ (1+r)^n -1
800=( P(23/2400)(2423/2400)^300)/(2423/2400)^300 -1
13187.55202 = P(23/2400) (2423/2400)^300
P= 13187.55202/(23/2400)(2423/2400)^300
P=78700
YaY it works
This is similar to the 'shortcut' that some people use for the 'Greatest Coefficient' topic, which is in fact not accepted by HSC markers.
Nevertheless, it is a very good way to verify your answer.