Independent solution

How to solve this Loan Amortization question

Setup

Setup

Interest-only payments during the first decade do not change principal.

B10=100000B_{10}=100000

Model

Model

During the second decade, paying 150% of interest applies 100% to interest and the remaining 50% of interest, or 5% of balance, to principal.

B20=100000(1−0.05)10=59873.69B_{20}=100000(1-0.05)^{10}=59873.69

Compute

Compute

Ten successive 5% balance reductions leave 59,873.69; this becomes the present value of the final ten level payments.

B20=Xa10‾∣ 0.10B_{20}=Xa_{\overline{10}|\,0.10}

Answer

Answer

The resulting level amount is about 9,744, which is choice D.

X=59873.69/6.14457=9744(D)\boxed{X=59873.69/6.14457=9744\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; 10 N; 10 I/Y; 59873.69 PV; 0 FV; CPT PMTPMT = -9744.17END mode; I/Y is the annual effective rate. The positive balance and negative payment signs are opposite cash-flow directions.