Independent solution

How to solve this Loan Amortization question

Setup

Setup

For a level-payment loan, the difference between successive principal portions reveals the monthly rate.

1400=X−60000j,1414=X−58600j1400=X-60000j,\qquad1414=X-58600j

Model

Model

Subtracting the first two principal equations gives j = 1%; substituting back gives a regular payment of 2,000.

14=1400j⟹j=0.01,X=200014=1400j\Longrightarrow j=0.01,\qquad X=2000

Compute

Compute

The exact annuity term is 35.8455, so 35 full payments are followed by a smaller payment at month 36.

60000=2000a35‾∣ 0.01+Pv3660000=2000a_{\overline{35}|\,0.01}+Pv^{36}

Answer

Answer

Valuing the residual at that date gives a drop payment of approximately 1,692, choice E.

P=1692(E)\boxed{P=1692\quad\text{(E)}}

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; 35 N; 1 I/Y; 60000 PV; 2000 +/- PMT; CPT FVFV = -1675.61; next-month payoff = 1692.37END mode; I/Y is the monthly effective rate. Accumulate the residual magnitude for one month to obtain the drop payment.