Independent solution

How to solve this Annuities and Perpetuities question

Setup

Setup

Use the date of the last deposit, six months before the first tuition payment, as the valuation date.

j=0.06/12=0.005,P=25000(1.005)−6a¨4‾∣ ((1.005)12−1)=88917.16j=0.06/12=0.005,\qquad P=25000(1.005)^{-6}\ddot a_{\overline{4}|\,((1.005)^{12}-1)}=88917.16

Model

Model

The four annual tuition payments form a due annuity shifted six months; deposits accumulate monthly to the same date.

1000sn‾∣ 0.005=88917.161000s_{\overline{n}|\,0.005}=88917.16

Compute

Compute

Solving the monthly accumulation equation gives 73.75 deposits before enforcing the integer requirement.

(1.005)n=1.444586,n=73.75(1.005)^n=1.444586,\qquad n=73.75

Answer

Answer

Because a fractional deposit count cannot fund the full obligation, at least 74 deposits are needed, choice D.

nmin⁡=74(D)\boxed{n_{\min}=74\quad\text{(D)}}

Calculator reproduction

BA II Plus keystrokes

Check END/BGN, period, sign, TVM, and cash-flow setup

  1. 2nd CLR TVM; 2nd I/Y; 1 ENTER; ↓; 1 ENTER; 2nd CPT; 2nd PMT; if BGN is displayed, 2nd ENTER; 2nd CPT; 0.5 I/Y; 0 PV; 1000 +/- PMT; 88917.16 FV; CPT NN = 73.75END mode; I/Y is the monthly effective rate. Round up to 74 deposits.