Question 4 (b)

Replacing three payments by a single sum: finding its due date at 10% effective

6 marks

Time to write: About 12 minutes
Topic: Depreciation and average due date

Question

Question 4 (b) · 6 marks

The sums of ₹2400, ₹3600 and ₹4800 fall due at the end of 3, 5 and 9 years respectively. It is proposed to replace this series of payments with a lump sum of ₹10800 payable at the end of n years. With the rate of interest at 10% per annum effective, find the value of n.

Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

Before you write

Read off the question and its marks

12 minfor 6 marks

Read from the question

findFormula first, then the working, then the answer with its unit (₹, %, units).

What the paper leaves open

Part of "Solve any three"; 6 marks.

Write it in this order

Read the data; note exactly what is asked 1 min
1 Equation of value 2 marks 3 min
2 Present value of the three payments 2 marks 3 min
3 Logarithmic solution for n, with check 2 marks 3 min
Check: arithmetic, units (₹, %), final answer underlined 2 min

123

Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.

Solution

Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.

Principle (equation of value). The single payment replaces the series fairly only if both have the same value at any one date. At 10% a year effective, a sum S due at the end of t years is worth S/(1.10)^t now, the present value of a single sum, P = A/(1 + i)ⁿ (from A = P(1 + i)ⁿ). Taking the comparison date as today:

10,800/(1.10)ⁿ = 2,400/(1.10)³ + 3,600/(1.10)⁵ + 4,800/(1.10)⁹

Present value of the given payments

Payment (₹)	Due at end of year	(1.10)^t	Present value (₹)
2,400	3	1.331000	1,803.1555
3,600	5	1.610510	2,235.3168
4,800	9	2.357948	2,035.6686
Total			6,074.1409

Solving for n

(1.10)ⁿ = 10,800/6,074.14 = 1.778029

n log 1.10 = log 1.778029 ⟹ n × 0.041393 = 0.249939 ⟹ n = 6.04 years.

The lump sum of ₹10,800 should be paid at the end of about 6.04 years (a little over 6 years, roughly 6 years and 2 weeks).

Check: 10,800/(1.10)^6.04 = ₹6,074.14, equal to the present value of the three payments.

(The lump sum equals the total of the three payments, so its due date falls between the first and the last payment dates. It comes earlier than the simple weighted average of the dates, (2,400 × 3 + 3,600 × 5 + 4,800 × 9)/10,800 = 6.33 years, because compound discounting weighs the early payments more heavily.)

Where the 6 marks sit

#	Component	Marks
1	Equation of value	2
2	Present value of the three payments	2
3	Logarithmic solution for n, with check	2
	Total	6

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