Interest Formulas and Cash Flow Questions and Answers

Answer: C

Correct answer: present sum whose interest will meet its costs for ever

Capitalised cost is the amount which, invested at the given rate, would generate interest exactly sufficient to meet the asset’s costs indefinitely, so it is \( A/i \) plus the first cost. It is the natural way to compare permanent works, since a finite study period would arbitrarily favour whichever alternative it happened to end on.

Answer: D

Maintenance costs rising by a fixed sum each year are the standard example, and the increment is the same whatever the level already reached. A constant percentage change is a geometric gradient and needs a different factor entirely.

Answer: A

Correct answer: a geometric gradient

A cost rising by a constant percentage forms a geometric series, each term a fixed multiple of the last. An arithmetic gradient rises by a constant amount instead, and the two behave quite differently over a long horizon - the geometric form compounds while the arithmetic one does not.

Answer: B

The gradient is defined as zero in the first period and G, 2G and so on thereafter, so a series that starts at A and rises is treated as a uniform A plus a gradient beginning in period two. Overlooking that decomposition is a frequent error.

Answer: C

Correct answer: a uniform series of 1000 plus a gradient of 200

The series rises by a constant 200 each year, which is an arithmetic gradient. The standard treatment splits it into a uniform series equal to the first payment and a gradient starting at zero in the first year, because the gradient factors are tabulated on exactly that convention. Twenty per cent would only fit the first step, not the later ones.

Answer: D

The general formula becomes indeterminate at that point and the limit gives the simpler expression, each payment's discounted value being the same. Every term contributes equally, which is why n appears as a plain multiplier.

Answer: A

Correct answer: find the effective rate for the payment period

The factors assume that interest is compounded once per payment period, so where compounding is more frequent the nominal rate must first be converted into the effective rate over the payment period. Using the nominal rate directly understates the interest, and the error grows with the frequency of compounding.

Answer: B

Where the convention is that no interest is credited between compounding dates, the payments within a period are simply summed. The alternative is to find an equivalent rate for the shorter payment period, which the problem must specify.

Answer: C

Correct answer: decreases as the loan is repaid

Interest is charged on the balance outstanding, and that balance falls with every payment, so the interest portion falls and the principal portion rises even though the instalment itself is constant. This is why the early payments of a long loan repay so little principal, and why an amortisation schedule is worth drawing out.

Answer: D

Correct answer: the capital recovery factor

The instalment is the annual equivalent of the present sum borrowed, which is precisely what \( (A/P, i, n) \) gives. Multiplying the principal by it produces the payment that repays the loan with interest over the term, and the same factor gives the annual equivalent of any capital cost.