Compound Interest (1 Viewer)

[mystify]

Sorry to be a real pest but maths is my weak point and i have no idea how wrong or right my answers to the following question are. For those who have any time to spare hope you wouldnt mind doing them and posting up the answers just so that I can have some feedback since i cant find the solutions to them. Thanks

(Btw its question10 of the 2001 HSC paper)

Helen sets up a prize fund with a single investment of $1000 to provide her school with an annual prize valued at $72. The fund accrues interest at a rate of 6% p.a. compounded annually. The first prize is awarded one year after the investment is set up.

(i) Calculate the balance in the fund at the beginning of the 2nd year.

(ii) Let $B'n' be the balance in the fund at the end of 'n' yrs (and after the nth prize has been awarded). Show that B'n' = 1200 - 200 x (1.06^n).

(iii) At the end of the 10th yr (and after the 10th prize has been awarded) its decided to increase the prize value to $90. For how many more yrs can the prize fund be used to award the prize.

 

[Jago]

i)
B1 = 1000x1.06-72
B2 = (1000x1.06-72)1.06-72
= 1000x1.06²-72(1 + 1.06)
= $1050.54

Edit: oops you only need to find B1 i.e.
B1 = $988

ii)
Bn = 1000x1.06^n - 72[(1.06^n - 1)/.06]
= 1000x1.06^n - 72[(1.06^n/.06) - (1/.06)]
= 1000x1.06^n - 1200x1.06^n + 1200
= -200x1.06^n + 1200
= 1200 - 200 x 1.06^n

iii)
B10 = 1200 - 200x1.06^n

Bm = 0 = B10 x 1.06^m - 90[(1.06^m)/.06) - (1/0.06)]
= B10 x 1.06^m - 1500x1.06^m + 1500
1500 = (1500 - B10) x 1.06^m
1500/(1500-B10) = 1.06^m

then use logs to find m which should be 14.14...... therefore the prize fund can be used for 14 more years.

I apologise for the = signs not being directly under each other and any mistakes i have made, i haven't revised compound interest for......3 weeks now.

Goodluck mystify

 

[mystify]

Thanks heaps *Jago*

 

[Jago]

np, i spent a good 8 minutes on those questions. Took me a while to get part 2. Good luck with your studies.

 

[mojako]

do u think investment and loan payment shouldnt be in 2U?
coz they seem to be out of standard with the other 2U stuff...

 

[Jago]

they're just applications of GPs. Just like how dynamics is an application of calculus...

wait, dynamics is that displacement, velocity and acceleration stuff right?

 

[mojako]

but they are too hard!!
who else disagree with me??

dynamics is a study of the effect of forces on bodies which, if initially at rest, are caused to move, or, if initially moving, have their state of motion altered. some aspects of it is discusses in HSC extension 2 maths.

2U maths and ext1 study kinematics
which is the study of the motion of a body without reference to the causes of its motion.

now, why am I wasting my time copying from a handout I got (most likely from New Senior Maths)? I really dont know haha...

 

[Estel]

heh 2U GP's always hog q9 and 10 anyway, and they're pretty easy after you do 10 or so of them.

 

[mojako]

still.....

ive seen that in an extension1 paper.. from Neap
and its not harder than the one here..
I think.. coz I havent really read this one

 

[Jago]

post the one from the extension 1 paper then

 

[mojako]

post the one from the extension 1 paper then

what for?
I mean, I can certainly post it but I dont wanna turn this into an argument.

 

[Jago]

so i can decide for myself whether it is indeed more difficult

 

[mojako]

here u go
is it harder?

EDIT: attachment removed becoz... its a copyrighted paper

 

[Jago]

looks exactly the same. I'll do it later. Maybe reducible interest doesn't get any harder at high school.

Is there only 7 questions for the 3u trial?

Edit: for the trig question, was the answer 3pi/4 and 5pi/4?

 

[mojako]

there are always 7 questions for all extension 1 trials

for the trig, its (2/3)pi and pi
whats wrong with you
BTW for some reason I thought u were at uni... until I looked at ur profile.

 

[Jago]

i knew i had it wrong...gimme a second while i try the question again. Me in uni? Hah!

Edit: nope, can't do it. Can you set out the method?

 

[mojako]

i knew i had it wrong...gimme a second while i try the question again. Me in uni? Hah!

Edit: nope, can't do it. Can you set out the method?

lucky I actually click on the link on the email
(I usually do anyway haha)

you can use the t-formula (dont forget.. if the quadratics in t disappears, pi will also be a solution)
or you can use the method of expressing sin(x) + cos(x) as a single sin or cos function (using auxiliary / subsidiary angle)

 

[Jago]

how does the t method work again? 1 + t², 1 - t², t²?

 

[Xayma]

t method is as follows.

Let t=tan θ/2
Then
sin θ=2t/(1+t²)
tan θ=2t/(1-t²)
cos θ=(1-t²)/(1+t²)

 

[mojako]

hint for memorising:
for tan@ its just the double angle
for sin@, use the numerator of the one for tan@, on 1+t^2
for cos@, use the denominator on 1+t^2
sin@ is numerator becoz tan@=sin@/cos@.. sin@ at the top