Independent solution

How to solve this Loan Amortization question

Setup

Setup

Payments begin at year five, so first determine how many full 600 payments the loan value can support with a four-year deferral.

4000=600v4an‾∣ 0.064000=600v^4a_{\overline n|\,0.06}

Model

Model

The implied noninteger term is 12.07 payment periods, which means 11 full payments precede the final smaller payment.

an‾∣ 0.06=(4000/600)(1.06)4=8.4165,n=12.07a_{\overline n|\,0.06}=(4000/600)(1.06)^4=8.4165,\qquad n=12.07

Compute

Compute

Value those 11 payments and the balloon at the loan date; solving the residual equation gives 639.43.

4000=600v4a11‾∣+Xv164000=600v^4a_{\overline{11}|}+Xv^{16}

Answer

Answer

The final balloon payment is about 639, matching choice B.

X=639.43(B)\boxed{X=639.43\quad\text{(B)}}

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; 4 N; 6 I/Y; 4000 +/- PV; 0 PMT; CPT FVFV = 5049.91END mode; I/Y is annual. This accumulates the advance to the first-payment comparison date.
  2. 2nd CLR TVM; 2nd I/Y; 1 ENTER; ↓; 1 ENTER; 2nd CPT; 2nd PMT; if BGN is displayed, 2nd ENTER; 2nd CPT; 6 I/Y; 5049.91 PV; 600 +/- PMT; 0 FV; CPT NN = 12.07END mode; I/Y is annual. The positive balance and negative repayments make the signs explicit.
  3. 2nd CLR TVM; 2nd I/Y; 1 ENTER; ↓; 1 ENTER; 2nd CPT; 2nd PMT; if BGN is displayed, 2nd ENTER; 2nd CPT; 11 N; 6 I/Y; 5049.91 PV; 600 +/- PMT; CPT FVFV = -603.25; next-year payoff = 639.44END mode; I/Y is annual. Accumulate the residual magnitude for one more year to obtain the balloon.