# Financial Superannuation Investigation Help?

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 $47,000 per annum. Realising that she needs to provide for her retirement, which she plans to do at age 65 years old. She wants to be able to live comfortably for 20 years. Predict how much she will have to need to have to in her superannuation at retirement to live equivalent living standards that her current income allows and give reasons for the prediction without calculations. #### jonah Beer soaked opinion follows Without calculations? Good luck! #### HallsofIvy MHF Helper A couple of questions- At what interest rate can she save money? What income level is required to "live comfortably"? #### SlipEternal MHF Helper A couple of questions- At what interest rate can she save money? What income level is required to "live comfortably"? My guess would be she invests$\$$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: 1 person #### Vinod 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. #### Vinod 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.