1
Simple interest is flat. Compound interest is not.
| Year | Simple (5% of ₱20,000) | Compound |
|---|
| 1 | ₱21,000 | ₱21,000 |
| 2 | ₱22,000 | ₱22,050 |
| 3 | ₱23,000 | ₱23,152.50 |
| 10 | ₱30,000 | ₱32,577.89 |
The two are identical for one year, then separate. Under compounding the second year's 5% is charged on ₱21,000, not on the original ₱20,000 — and the gap widens every year.
Simple interest has its own formula: I = P × r × t. For ₱20,000 at 5% for 3 years, I = 20,000 × 0.05 × 3 = ₱3,000, which is the ₱23,000 in the table. The gap between the two columns is the extra that compounding earns.
2
The formula, and what each letter is.
A = P × (1 + r)ⁿ
A = final amount
P = principal
r = rate as a decimal
n = number of periods
P = 20,000, r = 0.05, n = 3
5% becomes 0.05
A = 20,000(1.05)³
1 + r = 1.05
= ₱23,152.50
interest = ₱3,152.50
A is the total, not the interest
The formula returns everything you have, principal included. If the question asks
how much interest was earned, subtract P at the end.
3
Compounding more often than yearly.
A = P × (1 + rk)kt — divide the rate, multiply the periods.
| Compounded | k | ₱20,000 at 5% for 3 years |
|---|
| annually | 1 | ₱23,152.50 |
| semi-annually | 2 | ₱23,193.87 |
| quarterly | 4 | ₱23,215.15 |
| monthly | 12 | ₱23,229.67 |
More frequent compounding always pays more, but with rapidly diminishing returns —
the jump from yearly to monthly here is about ₱77 over three years.
4
Depreciation is the same formula with a minus.
A motorcycle costs ₱90,000
losing 15% of its value a year
A = 90,000(1 − 0.15)⁴
= 90,000(0.85)⁴
= 90,000 × 0.52200625
≈ ₱46,980.56
Losing 15% a year for four years is not losing 60%. Each year's 15% is taken from a smaller value, so about 48% is lost in total — which is why the vehicle keeps more than you might expect.
5
Working backwards: the principal, or the time.
A = ₱60,000 after 2 years at 8%
find P
60,000 = P(1.08)²
substitute
P = 60,000 ÷ 1.1664
divide, do not subtract
≈ ₱51,440.33
check: × 1.1664 ≈ 60,000 ✓
How many years does ₱10,000 at 10% take to reach ₱13,310?
13,310 ÷ 10,000 = 1.331
the growth factor (1.1)ⁿ
1.1, 1.21, 1.331
multiply by 1.1 until it matches
n = 3 years
check: 10,000 × 1.331 = 13,310 ✓
Divide first to get the factor, then count how many times 1 + r was multiplied in.
Growth
multiply by (1 + r)
each period
Undo growth
divide by (1 + r)
never subtract the percent
Remember the flow
Rate as a decimal · 1 + r for growth, 1 − r for loss · Divide the rate and multiply the periods if compounding more often · Subtract P for the interest alone
Compound growth is repeated multiplication, so undoing it means dividing.
6
Variations: monthly compounding, an unknown rate, months instead of years.
The same formula; each case changes what we feed into it.
Monthly: ₱12,000 at 6% compounded monthly for 1 year
rk = 0.06 ÷ 12 = 0.005, kt = 12
a monthly rate of 0.5%
A = 12,000(1.005)¹² ≈ 12,000 × 1.0616778
use a calculator for the power
≈ ₱12,740.13
simple interest gives ₱12,720
Different: the yearly rate is split twelve ways, and compounding adds ₱20.13 over simple interest.
The rate is unknown: ₱25,000 grows to ₱36,000 in 2 years
(1 + r)² = 36,000 ÷ 25,000 = 1.44
divide first
1 + r = √1.44 = 1.2
undo the square
r = 0.2 = 20%
check: 25,000 × 1.2 × 1.2 = 36,000 ✓
Different: the total gain is 44%, but the yearly rate is 20%, not 22%.
Time in months: ₱30,000 at 8% compounded semi-annually for 18 months
18 months = 1.5 years, so kt = 2 × 1.5 = 3
three half-year periods
A = 30,000(1.04)³ = 30,000 × 1.124864
8% ÷ 2 = 4% a period
= ₱33,745.92
exact, no rounding needed
Different: count periods, not years — 18 months is 3 periods here, not 18 and not 1.5.
On the exam
Expect the amount after n years, the interest alone, a depreciation item, or an expression to choose for quarterly or monthly compounding. The fastest route is to write the growth factor first — 1 + r, 1 − r, or 1 + rk — and count the periods. The 10-second check: compound must beat simple interest for growth (and a depreciated value must stay above the "subtract it all at once" figure).
