Interest that earns interest — so the balance grows by a multiplier, not a fixed amount.
The problem
₱10,000 is invested at 5% compounded annually for 3 years. Find the final amount.
A = P(1 + r)ⁿ = 10,000(1.05)³ = 10,000 × 1.157625 = ₱11,576.25. D is simple interest (three lots of ₱500); the extra ₱76.25 is interest earned on earlier interest. A is the interest alone, not the amount.
Each year's interest is worked out on the new balance.
| Year | Balance at the start | 5% of it | New balance |
|---|---|---|---|
| 1 | 10,000 | 500.00 | 10,500.00 |
| 2 | 10,500 | 525.00 | 11,025.00 |
| 3 | 11,025 | 551.25 | 11,576.25 |
The yearly interest grows: 500, then 525, then 551.25.
Under simple interest it would have been ₱500 every year. The extra ₱76.25 is interest earned by interest.
Adding 5% is the same as multiplying by 1.05.
A = P(1 + r)t
A = 10000(1.05)³
three years, three multiplications
1.05³ = 1.157625
the growth factor
A = ₱11,576.25
matches the table
interest = 11576.25 − 10000
subtract the principal
= ₱1,576.25
the earnings
A already includes the original ₱10,000. If the question asks how much was earned, you still have to subtract P.
A = P(1 + rn)nt
n is how many times a year the interest is added
quarterly means n = 4
four times a year
rate per quarter = 0.05 ÷ 4 = 0.0125
divide the rate
periods = 4 × 3 = 12
multiply the time
A = 10000(1.0125)12 ≈ ₱11,607.55
slightly more
Both changes happen together, every time. Doing only one of them is the standard slip in this topic.
| ₱10,000 at 5% | Simple | Compounded yearly |
|---|---|---|
| after 3 years | 11,500 | 11,576.25 |
| after 10 years | 15,000 | 16,288.95 |
| after 20 years | 20,000 | 26,532.98 |
Over a short run the gap is small. Over twenty years compounding wins by more than half the original investment.
Compound interest is always the larger of the two after the first period, so if your compound answer comes out smaller than the simple one, something is wrong.
Depreciation is compounding with a minus sign.
a ₱500,000 car loses 15% a year
the setup
multiplier = 1 − 0.15 = 0.85
what is left each year
after 3 years = 500000(0.85)³
same structure
≈ ₱307,062
0.614125 of the price
The car keeps 61.4% of its value, not 55%. Each year's loss is taken from a smaller amount — the same reason successive discounts never add.
Time given in months. ₱8,000 is invested at 6% compounded semi-annually for 18 months. Find the amount.
rate per period = 0.06 ÷ 2 = 0.03
two periods a year
periods = 18 ÷ 6 = 3
one period is 6 months
1.03³ = 1.092727
1.0609 × 1.03
A = 8,000 × 1.092727 ≈ ₱8,741.82
interest ≈ ₱741.82
What is different: t is not a whole number of years. Count periods directly — 18 months is 3 half-years, not 18 periods and not 1.
1.0318 would mean 18 half-years, which is 9 years. The exponent counts compounding periods, so convert months into periods first.
Which offer is better? Bank X pays 12% compounded quarterly. Bank Y pays 12.5% compounded annually.
X: one year multiplies by 1.03⁴
3% a quarter, four quarters
1.03² = 1.0609, 1.0609² ≈ 1.1255
square twice
X ≈ 12.55% a year vs Y = 12.5%
compare the one-year growth
Bank X, by a little
on ₱10,000: ₱1,255 vs ₱1,250
What is different: the stated rates cannot be compared directly. Turn each into one year's multiplier first.
Divide the amount by the principal (or the factor) first — what is left is a power to recognise. All three below are compounded annually.
P(1.06)² = ₱16,854 → P = 16,854 ÷ 1.1236
find P: divide by the whole factor
P = ₱15,000
not 16,854 minus 12%
₱8,000 → ₱12,500 in 2 years: (1 + r)² = 1.5625
find r: divide, then take the root
1 + r = 1.25 → r = 25%
√1.5625 = 1.25
₱10,000 → ₱13,310 at 10%: 1.1ⁿ = 1.331
find n: 1.1, 1.21, 1.331 …
n = 3 years
count the multiplications
₱4,500 gained on ₱8,000 over 2 years “looks like” 4,500 ÷ 8,000 ÷ 2 ≈ 28%. That is the simple rate; compounding reaches the same amount at only 25%. Learn the squares 1.1² = 1.21, 1.2² = 1.44, 1.25² = 1.5625 — exam numbers are built on them.
Expect the amount or the interest (subtract P), semi-annual or quarterly periods (divide the rate, multiply the time), the compound-minus-simple gap, and the backwards versions above. Quick check: compound interest must beat simple interest at the same rate, but only slightly over 2–3 years.
The problem
₱8,000 is invested at 6% compounded annually for 2 years. Find the final amount.
8,000(1.06)² = 8,000 × 1.1236 = ₱8,988.80. A is simple interest (two lots of ₱480 = ₱960). The ₱28.80 difference is the second year's interest on the first year's interest — that gap is the whole idea of compounding.
