1
Four quantities, one formula.
I = Prt, and A = P + I.
P
principal
the amount borrowed or invested
r
rate per period, as a decimal
6% becomes 0.06
t
time in the SAME period
years with a yearly rate
Turn the percent into a decimal first
Leaving r as 6 rather than 0.06 multiplies every answer by a hundred. The quickest
check is size: interest should be a modest slice of the principal, not several
times it.
2
Substitute, then add if the question wants the total.
I = 8000 × 0.05 × 3
P, r, t in order
A = 8000 + 1200 = ₱9,200
the amount, if asked
Read the last line of the question again before answering. "How much interest" and
"how much will be in the account" are different numbers ₱8,000 apart.
3
Months and days must be turned into years.
9 months
t = 912 = 0.75
with a yearly rate
180 days
t = 180360 = 0.5
banker's year of 360 days
18 months
t = 1.5
more than a year is fine
Philippine banks usually quote an annual rate, so t is in years. If a problem states a
360-day year, use it — it is a convention, not an error.
4
Rearrange to find whichever quantity is missing.
2500 = 25000 × r × 2
substitute what is known
2500 = 50000r
tidy the right side
r = 0.05, so 5%
convert back to a percent
Finding r gives a decimal. The question asked for a rate, so the answer is 5%, not
0.05 — the last step is a translation, and it is worth a mark.
5
Find the time or the principal, even from the amount.
The same rearranging works for t and for P. When the question gives the final amount instead of the interest, use A = P(1 + rt).
How long does ₱15,000 take to earn ₱2,250 at 3% simple interest?
2250 = 15000 × 0.03 × t
I = Prt
2250 = 450t
one year's interest is ₱450
What principal grows to an amount of ₱13,800 in 3 years at 5%?
A = P + Prt = P(1 + rt)
the amount includes P
13800 = P(1 + 0.05 × 3) = 1.15P
the 15% is of P
P = 13800 ÷ 1.15 = ₱12,000
divide
12000 × 0.05 × 3 = 1800; 12000 + 1800 = 13800 ✓
check
Taking 15% off the amount
Working out 13,800 − 15% of 13,800 = ₱11,730 is the classic slip. The interest was 15% of P, not of A, so we must DIVIDE by 1.15 rather than subtract 15%.
6
Variations: days, a monthly rate, a fractional percent.
Time in days: ₱36,000 at 5% a year for 120 days
t = 120360 = 13
banker's year of 360 days
I = 36000 × 0.05 × 13
substitute
A third of a year earns a third of the yearly ₱1,800.
A monthly rate: ₱8,000 at 1.5% a month for 8 months
r = 0.015 per month, t = 8 months
the units already match
I = 8000 × 0.015 × 8 = ₱960
no conversion to years
When the rate is per month, t stays in months. Converting t to years here would give ₱80, far too little.
A fractional percent: ₱6,000 at 4½% for 2 years
4½% = 4.5% = 0.045
write the half as .5
I = 6000 × 0.045 × 2 = ₱540
substitute
Writing 4½% as 0.45 moves the decimal one place too far and gives ₱5,400.
7
Simple interest grows in a straight line.
year 1
₱900
6% of 15,000
year 2
₱900 again
still 6% of 15,000
year 5
₱900 again
the principal never changes
₱3,600 interest on the opening problem
That fixed ₱900 a year is the whole difference from the next lesson. Compound
interest charges on the balance, which keeps growing, so the yearly figure rises.
Simple interest charges on the principal, which never does.
On the exam
Periodical exams ask for the interest, the amount, or one missing value among P, r and t; entrance tests like time in months or days and the "find P from the amount" question. Write I = Prt, substitute everything you know, and solve for the one unknown. For a 10-second check, work out one year's interest in your head and multiply by the time — it must match.
Remember the flow
Percent to decimal · Time in the rate's period · I = Prt · Add to P only if the amount is asked
Interest and amount are different questions. Read the last line twice.
