One formula, and one thing to watch: time is always in years.
The problem
₱20,000 is invested at 6% simple interest for 3 years. Find the interest.
I = Prt = 20,000 × 0.06 × 3 = ₱3,600. D is the total amount rather than the interest — check which one the question asked for. B is what compounding would produce, and simple interest never compounds.
I = P · R · T
P — principal
₱20,000
R — rate per year
0.06
T — years
3
6% means 6 per hundred, so R = 0.06. Leaving it as 6 makes the interest a hundred times too big — and that value is usually one of the choices.
I = 20000 × 0.06 × 3
the formula
= 1200 × 3
interest for one year
= ₱3,600
over three years
A = 20000 + 3600
the total amount
= ₱23,600
what you withdraw
“How much interest” wants 3,600. “How much will he have” wants 23,600. Both appear in the choices, so read the last line of the question twice.
The rate is per year, so the time has to match it.
| Given | T | As a decimal |
|---|---|---|
| 6 months | 612 | 0.5 |
| 8 months | 812 | 0.667 |
| 3 quarters | 912 | 0.75 |
| 90 days | 90360 | 0.25 |
| 18 months | 1812 | 1.5 |
Philippine problems usually take a banker's year of 360 days. If the question says 365, use 365.
₱20,000 at 6% for 8 months gives 20000 × 0.06 × ⅔ = ₱800 — not ₱9,600.
| Looking for | Formula |
|---|---|
| interest | I = PRT |
| principal | P = I ÷ (RT) |
| rate | R = I ÷ (PT) |
| time | T = I ÷ (PR) |
₱5,000 earns ₱600 in 2 years
find the rate
R = 600 ÷ (5000 × 2)
divide by both
= 0.06 = 6%
turn it back into a percent
If the question gives the final amount, subtract first: ₱4,000 growing to ₱4,960 at 8% has I = ₱960, and each year earns ₱320, so T = 960 ÷ 320 = 3 years.
No amounts at all? A sum that triples in 20 years has earned interest of 2P, so R = 2P ÷ (P × 20) = 0.10 = 10%. The P cancels. And a rate found from one deal carries to another: once R is known, run I = PRT again with the new P and T.
₱50,000 is invested, part at 3% and the rest at 6% simple interest. The total interest after one year is ₱2,400. How much is at each rate?
x at 6%, so 50,000 − x at 3%
one letter for both parts
0.06x + 0.03(50,000 − x) = 2,400
interest + interest = total
0.03x + 1,500 = 2,400
expand and collect
x = 900 ÷ 0.03 = ₱30,000 at 6%
and ₱20,000 at 3%
Check: 0.06(30,000) + 0.03(20,000) = 1,800 + 600 = 2,400 ✓. Both ₱30,000 and ₱20,000 will be choices — let x stand for the part asked about, so the answer is x itself.
If instead the two parts earn the SAME interest, set the two interests equal. ₱18,000 split at 5% and 4%, equal interest: 0.05x = 0.04(18,000 − x), so 0.09x = 720 and x = ₱8,000 at 5%, leaving ₱10,000 at 4%. Check: 400 = 400 ✓.
Only the final amount is known. What principal grows to ₱11,200 in 2 years at 6% simple interest?
A = P + PRT = P(1 + RT)
factor out P
11,200 = P(1 + 0.06 × 2) = 1.12P
12% interest in total
P = 11,200 ÷ 1.12 = ₱10,000
check: 10,000 × 0.12 = 1,200 ✓
What is different: neither P nor I is given, so I = PRT cannot be used alone. Divide by 1.12.
Taking 12% off ₱11,200 charges interest on the interest. The 12% was earned on P, not on A, so it has to be undone by dividing.
Time in days. Find the simple interest on ₱18,000 at 8% for 75 days (banker's year).
T = 75 ÷ 360
days over 360
I = 18,000 × 0.08 × 75 ÷ 360
one year earns 1,440
= 1,440 × 75 ÷ 360 = ₱300
multiply before dividing: no rounding
What is different: T is an awkward fraction (524). Keep it as 75360 to the end instead of rounding to 0.21.
A rate per month. A loan of ₱4,000 charges 2% simple interest per month. How much interest is due after 1 year 3 months?
T = 12 + 3 = 15 months
the rate is per month, so T is in months
I = 4,000 × 0.02 × 15
80 a month
= ₱1,200
same as 24% a year for 1.25 years ✓
What is different: “years” is not the rule; matching is. Using 2% with T = 1.25 years gives ₱100, far too small.
The interest is worked out on the original amount every single year.
| Year | Interest earned | Running total |
|---|---|---|
| 1 | 1,200 | 21,200 |
| 2 | 1,200 | 22,400 |
| 3 | 1,200 | 23,600 |
The same ₱1,200 every year. The interest already earned never earns anything itself.
That is the whole difference from compound interest, where year two would earn 6% of ₱21,200 instead.
Expect I = PRT run backwards (find the rate or the time from an amount) and the two-rate split. Work out one year's interest first — it makes time and rate questions a single division. For a split, test the choices: the right one makes the two interests add to the total.
The problem
Find the simple interest on ₱15,000 at 8% for 9 months.
Nine months is 0.75 of a year, so I = 15,000 × 0.08 × 0.75 = ₱900. A charges a full year by ignoring the conversion. The rate is per annum, so the time has to be in years as well.
