Independent solution

How to solve this Periodic Deposit Accumulation question

Setup

Setup

Let x be the accumulation factor over one four-year deposit interval. The time-40 fund contains ten beginning-of-interval deposits, whereas the time-20 fund contains the first five.

x=(1+i)4x=(1+i)^4

Model

Model

Value both funds on their respective measurement dates and divide the two geometric sums. Clearing the common factors reduces the fivefold-fund condition to the displayed polynomial in x.

x−x11x−x6=5⟹(x5−1)(x5−4)=0\frac{x-x^{11}}{x-x^6}=5\Longrightarrow(x^5-1)(x^5-4)=0

Compute

Compute

The admissible root is x = 4 to the power 1/5 = 1.319508; x = 1 would contradict the fivefold growth condition. Accumulating all ten deposits then gives 6,195.1, which rounds to the listed whole-dollar amount.

x=41/5=1.319508,X=100∑k=110xk=6195x=4^{1/5}=1.319508,\quad X=100\sum_{k=1}^{10}x^k=6195

Answer

Answer

The calculation gives 6195 for periodic deposit accumulation, matching published choice E.

X=6195(E)\boxed{X=6195\quad\text{(E)}}