Amount of Re 1 and amount of Re 1 per annum: compound…

Core facts

Symbols used in all valuation tables:
- i = rate of interest per year, written as a decimal (6% → 0.06)
- n = number of years
Amount of Re 1 (compound interest factor): A = (1 + i)^n
- Re 1 at 10% for 3 years → (1.10)^3 = 1.331
- ₹100 at 10% for 5 years → 100 × 1.6105 = ₹161 (ATP 2026)
Amount of Re 1 per annum (future value of an annuity): [(1 + i)^n − 1] / i
- Re 1 p.a. at 5% for 10 years = 12.578
- Instalments are paid at the end of each year, so the last instalment earns no interest (ATP 2013).
Reciprocals (exam favourite):
- Amount of Re 1 ↔ Present value of Re 1: 1/(1 + i)^n
- Amount of Re 1 p.a. ↔ Annual sinking fund: i/[(1 + i)^n − 1]
Both tables are always more than 1 (for n ≥ 1 and i > 0); the Amount of Re 1 p.a. is always at least n.

Explained

Why compound interest? A valuer compares money at different dates. A rupee today is worth more than a rupee later, because it can be invested and earn interest on interest. All six standard valuation tables are built from one expression, (1 + i)^n.

Amount of Re 1

After year 1: 1 × (1 + i). After year 2: (1 + i)^2. After n years: (1 + i)^n.
Use it to find the future value of a lump sum: future value = present sum × (1 + i)^n.
Example (ATP 2018 pattern): ₹100 invested now at 10% for 2 years = 100 × 1.21 = ₹121.

Amount of Re 1 per annum

Re 1 is put aside at the end of each year for n years and left to accumulate.
The first rupee earns interest for (n − 1) years; the last rupee earns nothing.
Sum of the series 1 + (1 + i) + (1 + i)^2 + … + (1 + i)^(n−1) = [(1 + i)^n − 1] / i.
Short cut: Amount of Re 1 p.a. = (Amount of Re 1 − 1) / i.

Worked example 1: two-year check

Re 1 p.a. at 5% for 2 years.
Formula: (1.1025 − 1)/0.05 = 2.05.
Long hand: first rupee grows to 1.05; second rupee (paid at the end of year 2) is 1.00. Total 2.05.

Worked example 2: regular savings

An owner puts aside ₹10,000 at the end of each year for 10 years at 6%.
(1.06)^10 = 1.7908. Amount of Re 1 p.a. = 0.7908/0.06 = 13.181.
Fund after 10 years = 10,000 × 13.181 = ₹1,31,810 (about).

Worked example 3: lump sum

₹1,00,000 invested for 5 years at 8%.
(1.08)^5 = 1.4693. Future value = ₹1,46,933.

Specimen values (recomputed; round to four decimals)

i	n	Amount of Re 1	Amount of Re 1 p.a.
4%	10	1.4802	12.0061
5%	10	1.6289	12.5779
6%	20	3.2071	36.7856
10%	5	1.6105	6.1051

Where these tables are used in valuation

Growing a sinking fund (see VAL-VT-03): accumulated fund after m years = annual instalment × Amount of Re 1 p.a. for m years.
Depreciation by the sinking fund method (see VAL-VM-08) uses exactly this product.
Cost of holding land: the cost of land bought now, carried to a future date, is cost × (1 + i)^n.
Deriving the present value of an annuity (YP): Amount of Re 1 p.a. ÷ Amount of Re 1 (ATP 2026: 18.286 ÷ 2.00 = 9.143).

Common traps

✗ "Amount of Re 1 = 1/(1 + i)^n." ✓ That is the present value of Re 1. Amount of Re 1 = (1 + i)^n.
✗ "Amount of Re 1 p.a. = i/[(1 + i)^n − 1]." ✓ That is the annual sinking fund. Amount of Re 1 p.a. = [(1 + i)^n − 1]/i.
✗ "In the Amount of Re 1 p.a. table the last instalment earns a full year's interest." ✓ Payments are at the end of each year, so the last instalment earns no interest.
✗ "₹100 at 10% for 3 years = ₹130." ✓ That is simple interest. Compound: 100 × 1.331 = ₹133.10.
✗ "Amount of Re 1 and Amount of Re 1 p.a. are reciprocals." ✓ Amount of Re 1 ↔ PV of Re 1; Amount of Re 1 p.a. ↔ annual sinking fund.

Quick revision

Amount of Re 1 = (1 + i)^n.
Amount of Re 1 p.a. = [(1 + i)^n − 1]/i.
Amount of Re 1 p.a. = (Amount of Re 1 − 1)/i.
(1.10)^3 = 1.331; (1.10)^5 = 1.6105; (1.05)^10 = 1.6289.
Re 1 p.a. at 5% for 10 years = 12.578.
Last instalment earns no interest (end-of-year payments).
Reciprocals: Amount ↔ PV of Re 1; Amount p.a. ↔ sinking fund.
Sinking fund growth uses the accumulative (safe) rate.
Rule of 72: 7.2% doubles money in about 10 years.

Go deeper: Deeper, Mains writing (beyond what MPSC ATP usually asks)

Deeper Advanced

Ordinary annuity vs annuity due. Valuation tables assume payments at the end of each period (ordinary annuity). If payments are at the start (annuity due), each payment earns one more year: amount = [(1 + i)^n − 1]/i × (1 + i).
Rule of 72. Money doubles in about 72 ÷ (rate in %) years. At 7.2%, (1.072)^10 = 2.004, so Re 1 doubles in about 10 years. It is an estimate, not a table value.
More frequent compounding. If interest is added m times a year, the factor is (1 + i/m)^(mn). Valuation tables assume annual rests unless stated.
Statement-type traps.
- The Amount of Re 1 p.a. grows faster than n because interest is earned on earlier instalments.
- The Amount of Re 1 for n years equals 1 + i × (Amount of Re 1 p.a. for n years). Check: 1 + 0.05 × 12.5779 = 1.6289.
- Raising i raises both tables; raising n raises both tables.
Remunerative vs accumulative rate. When the Amount of Re 1 p.a. is used to grow a sinking fund, the rate is the low, safe accumulative rate (say 3–4%), not the investor's remunerative rate (see VAL-VT-04).

Mains writing Mains writing

Frame: time value of money is the base of income valuation; every table is a rearrangement of (1 + i)^n.
Points:
- two families: lump sum (Amount of Re 1, PV of Re 1) and annual series (Amount of Re 1 p.a., sinking fund, YP, annuity)
- pairs of reciprocals reduce memory load and help check answers
- choice of rate (remunerative or accumulative) matters more than arithmetic
Example: show that Re 1 p.a. at 5% for 2 years = 2.05 by both formula and long hand; examiners reward the check.

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17.1.1 Amount of Re 1 and amount of Re 1 per annum: compound interest tables

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