Depositing 20 000 annually for 10 years at 8 per cent accumulates to about
Answer: C
Correct answer: 289 700
\[ F = 20\,000 \times \frac{(1.08)^{10} - 1}{0.08} = 20\,000 \times 14.487 \] \[ = 289\,700. \] The deposits total 200 000, so nearly 90 000 is interest - a much larger share than in the five-year case, because the earlier deposits have twice as long to compound.
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A perpetual annual cost of 5000 at 8 per cent has a capitalised cost of
Answer: D
Correct answer: 62 500
Capitalised cost of a perpetual annual amount is that amount divided by the interest rate: \[ \frac{5000}{0.08} = 62\,500. \] Invested at eight per cent, 62 500 yields exactly 5000 a year for ever without touching the capital, which is what makes it the equivalent present sum.
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To accumulate 500 000 in 15 years at 9 per cent requires an annual deposit of about
Answer: A
Correct answer: 17 020
\[ A = 500\,000 \times \frac{0.09}{(1.09)^{15} - 1} = 500\,000 \times 0.03406 \] \[ = 17\,030 \approx 17\,020. \] The deposits total about 255 000 against 500 000 accumulated, so more than half the fund is interest - the effect of fifteen years of compounding.
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A series of 5 annual receipts of 10 000 beginning at the end of year 4, at 10 per cent, has a present worth at time zero of about
Answer: B
Correct answer: 28 480
The series present worth factor places its value one period before the first payment, so with the first receipt at the end of year four the value sits at the end of year three: \[ 10\,000 \times 3.791 = 37\,910 \ \text{at } t = 3. \] Discounting that back three years, \[ \frac{37\,910}{(1.10)^3} = \frac{37\,910}{1.331} = 28\,480. \] Forgetting the second step and reporting 37 910 is the classic error with a deferred series.
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A cost of 1000 in year 1 rising by 200 each year for 5 years is treated as a uniform 1000 plus a gradient of
Answer: C
The gradient part is zero in the first period by definition and takes the value G in the second, so the decomposition matches how the factor is derived. Applying the gradient from year one would double-count the increase.
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Match the factor with what it gives.
1. \( (A/P, i, n) \) 2. \( (A/F, i, n) \) 3. \( (F/A, i, n) \)
P. Annual deposit to reach a future sum Q. Instalment that repays a present sum R. Accumulation of a uniform series
Answer: D
Correct answer: 1-Q, 2-P, 3-R
Read each notation as a fraction and the answer follows. \( (A/P) \) finds the annual amount given a present sum, which is the loan instalment. \( (A/F) \) finds the annual amount given a future sum, which is the sinking fund deposit. And \( (F/A) \) finds the future accumulation of a uniform series.
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Match each factor with its reciprocal.
1. \( (P/A, i, n) \) 2. \( (F/A, i, n) \) 3. \( (F/P, i, n) \)
P. \( (A/F, i, n) \) Q. \( (A/P, i, n) \) R. \( (P/F, i, n) \)
Answer: A
Correct answer: 1-Q, 2-P, 3-R
Each factor is the reciprocal of the one with its symbols reversed, since finding P from A and finding A from P are inverse operations. So \( (P/A) \) pairs with \( (A/P) \), \( (F/A) \) with \( (A/F) \), and \( (F/P) \) with \( (P/F) \). Recognising the pairing halves what has to be looked up.
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Put the steps of valuing a deferred uniform series in order.
1. Discount that value to time zero 2. Identify the period one before the first payment 3. Apply the series present worth factor to place the value there 4. Count the number of payments in the series
Answer: B
Correct answer: 4, 2, 3, 1
Count the payments first, since the factor needs n. Identify the period one before the first payment, because that is where the series present worth factor places its result. Apply the factor to put the value there. Then discount that single sum back to time zero with the single-payment present worth factor. Omitting the last step is the usual mistake.
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Consider the following statements about interest factors.
1. \( (P/A) \) places its value one period before the first payment 2. \( (A/P) \) exceeds \( (A/F) \) by the interest rate 3. \( (F/A) \) places its value one period after the last payment
Answer: C
Correct answer: 1 and 2 only are correct
Statements 1 and 2 are correct: \( (P/A) \) places its value one period before the first payment, and the capital recovery factor exceeds the sinking fund factor by exactly the interest rate. Statement 3 is wrong - \( (F/A) \) places its value at the date of the last payment, not a period beyond it.
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Which of the following factors would NOT be used to find an annual equivalent?
Answer: D
Correct answer: \( (F/P, i, n) \)
An annual equivalent is a value of A, so the factor must have A before the slash. \( (A/P) \), \( (A/F) \) and \( (A/G) \) all produce one, from a present sum, a future sum and a gradient respectively. \( (F/P) \) produces a future single sum and has no A in it at all.
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