Financial Superannuation Investigation Help?

May 2018
1
0
Nowhere
Hi guys, I am really confused for this one and need help. Thank you guys in advanced. A 23 year old woman has just finished her final year of studying in university and is now working full time at a job that pays $47,000 per annum. Realising that she needs to provide for her retirement, which she plans to do at age 65 years old. She wants to be able to live comfortably for 20 years. Predict how much she will have to need to have to in her superannuation at retirement to live equivalent living standards that her current income allows and give reasons for the prediction without calculations.
 
Apr 2008
230
48
Beer soaked opinion follows
Without calculations?
Good luck!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
A couple of questions- At what interest rate can she save money? What income level is required to "live comfortably"?
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
A couple of questions- At what interest rate can she save money? What income level is required to "live comfortably"?
My guess would be she invests $\$$x per year, so her living wage would be $\$$47,000-x. So, my guess would be, at an annual percentage yield of $n$%, she would want to save

\(\displaystyle 47000\dfrac{\left(1+\dfrac{n}{100}\right)^{20}-1}{\left(1+\dfrac{n}{100}\right)^{20}+\left( 1 + \dfrac{n}{100} \right)^{62}-2}\)

per year. If there is no interest, this becomes a simpler calculation:

$42x = 20(47000-x)$

which gives

$x \approx 15161$

So, her "comfortable living" would be $\$31839$.
 
Last edited:
  • Like
Reactions: 1 person
Sep 2011
393
8
Mumbai (Bombay),Maharashtra State,India
My guess would be she invests $\$$x per year, so her living wage would be $\$$47,000-x. So, my guess would be, at an annual percentage yield of $n$%, she would want to save

\(\displaystyle 47000\dfrac{\left(1+\dfrac{n}{100}\right)^{20}-1}{\left(1+\dfrac{n}{100}\right)^{20}+\left( 1 + \dfrac{n}{100} \right)^{62}-2}\)

per year. If there is no interest, this becomes a simpler calculation:

$42x = 20(47000-x)$

which gives

$x \approx 15161$

So, her "comfortable living" would be $\$31839$.
Hello,
Brilliant answer.
 
Sep 2011
393
8
Mumbai (Bombay),Maharashtra State,India
My guess would be she invests $\$$x per year, so her living wage would be $\$$47,000-x. So, my guess would be, at an annual percentage yield of $n$%, she would want to save

\(\displaystyle 47000\dfrac{\left(1+\dfrac{n}{100}\right)^{20}-1}{\left(1+\dfrac{n}{100}\right)^{20}+\left( 1 + \dfrac{n}{100} \right)^{62}-2}\)

per year. If there is no interest, this becomes a simpler calculation:

$42x = 20(47000-x)$

which gives

$x \approx 15161$

So, her "comfortable living" would be $\$31839$.
Hello,
My answer to this question is $X *(\displaystyle\sum_{k=1}^{42} (1+n\%)^k )=20(47000-X)$ If we assume interest rate to be 5%, the amount to be saved each year would be $5802.7077.