Money has a peculiar power: when you save it or lend it, it can grow by itself, and when you borrow it, it costs you extra to pay back. That extra is called interest, and understanding how it works is one of the most valuable life skills there is — it's the difference between a savings account quietly making you richer and a loan quietly making you poorer. The Civil Service Exam tests this practical money-math directly, and the key is one distinction: simple interest versus compound interest. Let's master both.
Interest problems describe money earning or owing extra over time, and ask for the final amount, the original amount (principal), the rate, or the time. The heart of the topic is the difference between simple interest (which stays flat, always figured on the original amount) and compound interest (which snowballs, because interest starts earning its own interest).
On the Civil Service Exam, interest questions are a dependable part of Numerical Reasoning, usually 2 to 4 items, and they build directly on the Percentage lesson. Because interest also underlies loans, savings, and investments in real life, it's a topic that pays off far beyond the exam.
The difficulty is intermediate. The formulas are short; the exam tests whether you can tell simple from compound, convert the rate to a decimal, adjust for compounding more than once a year, and rearrange the formula to find a missing quantity. Get those and this is steady, high-value marks.
After completing this lesson you will be able to:
You should be comfortable with:
A quick refresher on the piece that trips people most: the rate must be a decimal in these formulas. "5%" means 5 per hundred, which is 5 ÷ 100 = 0.05. So 8% is 0.08, 12% is 0.12, and 2.5% is 0.025. Using the whole number 5 instead of 0.05 makes an answer a hundred times too big — always convert first.
Interest is the math of money over time, and it shapes almost every financial decision Filipinos make:
Knowing interest lets you judge whether a loan is fair or a savings plan is worthwhile — a genuinely empowering, exam-tested skill.
We build from simple interest to compound, then to the real-world twists.
Simple interest is always calculated on the original principal, so each period earns the same peso amount.
Simple Interest = Principal × Rate × Time. (I = P × R × T.) Total Amount = Principal + Interest = P × (1 + R × T).
₱10,000 at 5% simple interest for 3 years: Interest = 10,000 × 0.05 × 3 = ₱1,500. Amount = 10,000 + 1,500 = ₱11,500.
Each of the three years earns exactly ₱500, because the interest is always figured on the same ₱10,000.
Analogy: Simple interest is like a plant that grows the same number of leaves every year, no matter how big it already is. The growth never accelerates.
Compound interest adds each period's interest to the principal before the next period's interest is figured, so the money grows on an ever-larger base.
Amount = Principal × (1 + Rate)^Time. (A = P × (1 + R)^T.) The interest is Amount − Principal.
₱10,000 at 5% compound interest for 3 years: Amount = 10,000 × (1.05)³ ≈ 10,000 × 1.1576 = ₱11,576.
Notice this beats the simple-interest total of ₱11,500 — by ₱76 here — because in years two and three, the interest earned interest.
Analogy: Compound interest is a snowball rolling downhill: the bigger it gets, the more snow it picks up, so it grows faster and faster. This "interest on interest" is the most powerful idea in personal finance.
In the first period, simple and compound interest are identical (both figure on the original principal). From the second period on, compound interest earns on the accumulated interest too, so it pulls ahead — and the gap widens the longer the money stays invested. Over decades, the difference is enormous. This is the mathematical backbone of "the earlier you start saving, the better."
Interest sometimes compounds semi-annually (twice), quarterly (four times), or monthly (twelve times) a year. Two adjustments go together: divide the annual rate by the number of periods per year (n), and multiply the number of years by n.
Amount = P × (1 + R/n)^(n × T).
₱10,000 at an annual rate of 8%, compounded quarterly, for 2 years: rate per quarter = 0.08 ÷ 4 = 0.02; number of periods = 2 × 4 = 8. Amount = 10,000 × (1.02)⁸ ≈ ₱11,717.
More frequent compounding produces a slightly higher final amount at the same stated rate, because interest starts earning interest sooner. Both adjustments must be made together — changing only one is a classic error.
The same formulas solve for whichever variable is unknown. For simple interest, A = P(1 + RT), so:
Principal = A ÷ (1 + R × T); Rate = I ÷ (P × T); Time = I ÷ (P × R).
How much must be invested now, at 6% simple interest, to have ₱6,360 in 4 years? 6,360 = P × (1 + 0.06 × 4) = P × 1.24 → P = 6,360 ÷ 1.24 ≈ ₱5,129.
For two years, the extra that compound interest earns over simple interest is exactly P × R² (principal times rate squared).
₱10,000 at 10% for 2 years: the compound-minus-simple difference = 10,000 × (0.10)² = 10,000 × 0.01 = ₱100. (Check: simple = ₱2,000, compound = 10,000 × 1.21 − 10,000 = ₱2,100; difference ₱100. ✓)
This is a favorite exam question, and the shortcut skips computing both totals.
A tidy fact: under simple interest, money doubles when the total interest equals the principal, i.e. when R × T = 1. So at 10% simple interest, doubling takes 1 ÷ 0.10 = 10 years.
With the toolkit built, let's picture, tabulate, and drill.
| Concept | Formula |
|---|---|
| Simple interest | I = P × R × T |
| Amount (simple) | A = P × (1 + R × T) |
| Amount (compound, annual) | A = P × (1 + R)^T |
| Amount (compound, n/year) | A = P × (1 + R/n)^(n × T) |
| Interest (compound) | A − P |
| Principal (from simple amount) | P = A ÷ (1 + R × T) |
| Rate (simple) | R = I ÷ (P × T) |
| Time (simple) | T = I ÷ (P × R) |
| Compound − simple (2 years) | P × R² |
| Doubling (simple) | R × T = 1 |
Rate reminder: every R above is a decimal (5% = 0.05). And in the "n per year" formula, remember to change both the rate (÷ n) and the exponent (× n) together.
| When the problem says… | Do this… |
|---|---|
| "simple interest" | I = P × R × T |
| "compound interest" / "compounded annually" | A = P × (1 + R)^T |
| "compounded quarterly / monthly / semi-annually" | divide rate by n, multiply time by n |
| "how much to invest now" | solve for principal |
| "at what rate / how long" | rearrange for R or T |
| "difference between simple and compound for 2 years" | P × R² |
| "how long to double (simple)" | T = 1 ÷ R |
| a percentage rate like 6% | convert to 0.06 first |
Step 1 — Identify the type: simple or compound? (And if compound, how often?) ↓ Step 2 — Convert the rate to a decimal, and match the time unit to the rate's period. ↓ Step 3 — Adjust for compounding frequency if it's not annual (rate ÷ n, exponent × n). ↓ Step 4 — Plug into the right formula, or rearrange it for the missing quantity. ↓ Step 5 — Sanity-check: the amount must exceed the principal (for an investment); compound should beat simple over multiple years.
Why Step 1 matters most: using the simple formula on a compound problem (or vice versa) is the defining error here. Read for "simple," "compound," and the compounding frequency before computing.
Example 1. Find the simple interest on ₱8,000 at 5% for 2 years. Solution: I = 8,000 × 0.05 × 2 = ₱800. Difficulty: ★☆☆☆☆
Example 2. Find the total amount on ₱5,000 at 4% simple interest for 3 years. Solution: I = 5,000 × 0.04 × 3 = ₱600; Amount = 5,000 + 600 = ₱5,600. Difficulty: ★☆☆☆☆
Example 3 (compound, annual). Find the amount on ₱20,000 at 10% compound interest for 2 years. Solution: A = 20,000 × (1.10)² = 20,000 × 1.21 = ₱24,200 (interest ₱4,200). Difficulty: ★★☆☆☆
Example 4 (find rate, simple). ₱5,000 earns ₱900 in simple interest over 3 years. Find the annual rate. Solution: R = I ÷ (P × T) = 900 ÷ (5,000 × 3) = 900 ÷ 15,000 = 0.06 = 6%. Difficulty: ★★★☆☆
Example 5 (find time, simple). At 5% simple interest, how long does ₱8,000 take to earn ₱1,200? Solution: T = I ÷ (P × R) = 1,200 ÷ (8,000 × 0.05) = 1,200 ÷ 400 = 3 years. Difficulty: ★★★☆☆
Example 6 (compounded quarterly). Find the amount on ₱10,000 at 8% per year, compounded quarterly, for 2 years. Solution: rate per quarter = 0.02; periods = 8. A = 10,000 × (1.02)⁸ ≈ ₱11,717. Difficulty: ★★★☆☆
Example 7 (compounded semi-annually). Find the amount on ₱10,000 at 10% per year, compounded semi-annually, for 1 year. Solution: rate per half-year = 0.05; periods = 2. A = 10,000 × (1.05)² = ₱11,025 (versus ₱11,000 with annual compounding — the extra ₱25 comes from mid-year compounding). Difficulty: ★★★★☆
Example 8 (find principal). What principal, at 8% simple interest, grows to ₱7,440 in 5 years? Solution: 7,440 = P × (1 + 0.08 × 5) = P × 1.40 → P = 7,440 ÷ 1.40 = ₱5,314.29 (≈ ₱5,314). Difficulty: ★★★★☆
Example 9 (simple vs. compound difference). Find the difference between the compound and simple interest on ₱15,000 at 10% for 2 years. Solution: Shortcut: difference = P × R² = 15,000 × (0.10)² = 15,000 × 0.01 = ₱150. (Check: simple = ₱3,000; compound = 15,000 × 1.21 − 15,000 = ₱3,150; difference ₱150. ✓) Difficulty: ★★★★☆
Example 10 (loan). Aling Nena borrows ₱50,000 at 12% simple interest for 3 years. How much must she repay in total? Solution: Interest = 50,000 × 0.12 × 3 = ₱18,000; total repayment = 50,000 + 18,000 = ₱68,000. Difficulty: ★★★☆☆
Example 11 (doubling). At what number of years will ₱20,000 double under 8% simple interest? Thinking: Money doubles when R × T = 1. Solution: T = 1 ÷ 0.08 = 12.5 years. Difficulty: ★★★★☆
Example 12 (three-year compound). Find the amount on ₱8,000 at 5% compound interest for 3 years. Solution: A = 8,000 × (1.05)³ = 8,000 × 1.157625 = ₱9,261 (to the nearest peso). Difficulty: ★★★★☆
Interest is extra money over time, and the master distinction is simple versus compound. Simple interest (I = P × R × T) stays flat, always figured on the original principal. Compound interest (A = P × (1 + R)^T) snowballs, because interest earns interest — so it always beats simple interest beyond the first period, and the gap widens over time. For compounding more often than yearly, divide the rate by n and multiply the exponent by n (both together). Always convert the rate to a decimal, rearrange the formulas to find a missing principal, rate, or time, and remember the P × R² shortcut for the two-year difference. Classify the type first, convert the rate, and this becomes reliable, real-world-useful marks.
| Item | Key point |
|---|---|
| Simple interest | I = P × R × T |
| Amount (simple) | A = P(1 + RT) |
| Amount (compound) | A = P(1 + R)^T |
| n times a year | A = P(1 + R/n)^(nT) |
| Rate | always a decimal (5% = 0.05) |
| 2-year difference | P × R² |
| Double (simple) | T = 1 ÷ R |
| Rule of 72 (compound) | ≈ 72 ÷ rate% years to double |
| Check | amount > principal; compound > simple over years |
What's the real difference between simple and compound interest? Simple interest is always calculated on the original principal, so it's the same each period. Compound interest is calculated on the principal plus previously earned interest, so it grows faster over time.
Why must I convert the rate to a decimal? The formulas multiply by the rate directly. "5%" means 5 per 100 = 0.05; using 5 makes the answer 100 times too large.
How do I handle "compounded quarterly/monthly"? Divide the annual rate by the number of periods per year and multiply the number of years by that same number. Both adjustments happen together.
Does more frequent compounding really earn more? Yes, but only slightly at the same stated rate — interest starts earning interest sooner. Quarterly beats annual, monthly beats quarterly, by small amounts.
What is the rule of 72? A quick estimate: money roughly doubles in 72 ÷ (rate as a percent) years under compound interest. At 6%, that's about 12 years. It's an approximation, not an exact formula.
Tick them all and interest becomes dependable exam points — and a life skill that protects your savings and your loans.
Put it to the test with 1,373 practice questions on this topic.