Calculating deposit based on interest and required withdrawal in future
I am really stuck with example 2-5 and 2-6. I don't really understand example 2-6 and example 2-5 I just can't figure out...I was able to do example 2-4 which was easy...
For example 2-4 I did F=P(1+ interest)^n
6500=P(1+0.03)^4
I solved for P and got the deposit value...Example 2-5 or 2-6 are different and don't work the same way so I am not sure what to do here..
2 Answers
$\begingroup$2-5: $\frac{6500}{1.03^4} + \frac{3000}{1.03^5}$
2-6: $900,000 + 200,000[\frac{1}{1.04} + \frac{1}{1.04^2} + \frac{1}{1.04^3} + \frac{1}{1.04^4} + \frac{1}{1.04^5}]$ assuming costs due at end of each year and interest paid annually
2-7: $500,000 + 300,000[\frac{1}{1.04} + \frac{1}{1.04^2} + \frac{1}{1.04^3} + \frac{1}{1.04^4} + \frac{1}{1.04^5}]$
These arent good questions
$\endgroup$ 5 $\begingroup$For example 2-5: you want
$$ 6500 + Q = P (1.03)^4$$
and
$$3000 = Q (1.03)$$
so that
$$ = 6500 + \frac{3000}{1.03} = P (1.03)^4$$
Solve for $P$.
$\endgroup$ 2
