Independent solution

How to solve this Piecewise Interest Accumulation question

Setup

Setup

Move all cash flows to time 10. The initial deposit and the later unknown deposit first accumulate at 6%, then pass through the stated varying-force interval.

75000=[10000(1.06)10+X(1.06)7]exp⁡(∫510dtt+1)75000=\left[10000(1.06)^{10}+X(1.06)^7\right]\exp\left(\int_5^{10}\frac{dt}{t+1}\right)

Model

Model

The force integral from 5 to 10 is ln(11/6), so its accumulation factor is 11/6. Insert that factor in the terminal-value equation.

exp⁡(∫510dtt+1)=11/6\exp\left(\int_5^{10}\frac{dt}{t+1}\right)=11/6

Compute

Compute

Solving the resulting linear equation gives X = 24,498.78, which rounds to 24,500.

X=24498.78X=24498.78

Answer

Answer

The calculation gives 24,500 for piecewise interest accumulation, matching published choice C.

X=24500(C)\boxed{X=24500\quad\text{(C)}}