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Number and Algebra

Compound interest: watching ₱10,000 grow

Financial mathematics

Compound interest: paglago ng ₱10,000

A year-by-year table of a single deposit shows why compound growth starts slow and then accelerates, and how to read nominal versus effective rates.

Where it fits

Competencies

  • gm11.t3.w1.c1 Solve problems involving compound interest, including maturity value, future value, nominal rates, and number of terms.
  • gm11.t1.w5.c2 Determine the accumulated values of investment over time using sequences and series.

The data

₱10,000 at 6% per year, compounded annually
YearSimple interest valueCompound valueInterest this year (compound)
0₱10,000₱10,000—
1₱10,600₱10,600₱600
2₱11,200₱11,236₱636
3₱11,800₱11,910₱674
4₱12,400₱12,625₱715
5₱13,000₱13,382₱757
6₱13,600₱14,185₱803
7₱14,200₱15,036₱851
8₱14,800₱15,938₱902
9₱15,400₱16,895₱957
10₱16,000₱17,908₱1,013
Compound value by Year₱0₱8954₱18k0: 100001: 106002: 112363: 119104: 126255: 133826: 141857: 150368: 159389: 1689510: 17908012345678910
Compound value by Year

Source: Computed with A = P(1 + r)^t; rounding to the nearest peso.

Scenario and worked example

Lolo's time deposit

Elicit

When Nico was born his lolo placed ₱10,000 in a time deposit at 6% per year and never touched it. Nico is now in Grade 11. How much is in the account? Why is the yearly interest no longer ₱600?

Worked example

Explain
A=10,000(1.06)16≈10,000×2.5404=25,404A = 10{,}000(1.06)^{16} \approx 10{,}000 \times 2.5404 = 25{,}404Interest in year 16 alone: 25,404−23,966≈1,438\text{Interest in year 16 alone: } 25{,}404 - 23{,}966 \approx 1{,}438

Compound interest is a geometric sequence

Elaborate

Each year's value is the previous year's multiplied by the common ratio $1 + r$. The simple-interest column is an arithmetic sequence with common difference $Pr$.

Nominal and effective rates

Explain

A nominal 6% compounded monthly gives an effective rate of

(1+0.0612)12−1≈6.17%\left(1+\frac{0.06}{12}\right)^{12}-1 \approx 6.17\%

Discuss in class

  1. In which year does the compound value first exceed the simple-interest value by more than ₱1,000?
  2. Describe the two columns as sequences. Which is arithmetic? Which is geometric?
  3. Credit cards charge about 3% per month. What effective annual rate is that? Why is it dangerous?

Problems

5 problems · check your answer, then open the solution

  1. Problem 1 Basic

    Find the maturity value of ₱15,000 invested at 5% per year compounded annually for 3 years.

    Show step-by-step solution
    1. A=P(1+r)t=15,000(1.05)3A = P(1+r)^t = 15{,}000(1.05)^3.
    2. (1.05)3=1.157625(1.05)^3 = 1.157625.
    3. A=15,000×1.157625=17,364.38A = 15{,}000 \times 1.157625 = 17{,}364.38 pesos.

    Answer: 17364.38 ₱

  2. Problem 2 Standard

    ₱20,000 is invested at 8% per year compounded quarterly for 2 years. Find the maturity value to the nearest centavo.

    Show step-by-step solution
    1. Quarterly rate i=0.08/4=0.02i = 0.08/4 = 0.02; number of periods n=2×4=8n = 2 \times 4 = 8.
    2. A=20,000(1.02)8≈20,000×1.171659A = 20{,}000(1.02)^8 \approx 20{,}000 \times 1.171659.
    3. A≈23,433.19A \approx 23{,}433.19 pesos.

    Answer: 23433.19 ₱

  3. Problem 3 Basic

    Two accounts both advertise 6% per year. Account X compounds annually; Account Y compounds monthly. After one year, which is true?

    Choices
    Show step-by-step solution
    1. Monthly compounding adds interest on interest within the year: (1+0.06/12)12≈1.0617>1.06(1+0.06/12)^{12} \approx 1.0617 > 1.06.
    2. The effective rate of Y is about 6.17%, so Y earns more for any deposit.

    Answer: b

  4. Problem 4 Standard

    Parents want ₱50,000 for enrolment in 5 years. How much must they deposit now at 4% compounded annually? Round to the nearest peso.

    Show step-by-step solution
    1. Present value P=A(1+r)t=50,000(1.04)5P = \dfrac{A}{(1+r)^t} = \dfrac{50{,}000}{(1.04)^5}.
    2. (1.04)5≈1.216653(1.04)^5 \approx 1.216653.
    3. P≈41,096P \approx 41{,}096 pesos.

    Answer: 41096 ₱

  5. Problem 5 Challenge Non-routine

    An investment grew from ₱10,000 to ₱12,000 in 3 years with annual compounding. Find the annual rate, in percent, to two decimal places.

    Show step-by-step solution
    1. 12,000=10,000(1+r)3⇒(1+r)3=1.212{,}000 = 10{,}000(1+r)^3 \Rightarrow (1+r)^3 = 1.2.
    2. 1+r=1.21/3≈1.06271 + r = 1.2^{1/3} \approx 1.0627.
    3. r≈0.0627=6.27%r \approx 0.0627 = 6.27\%.

    Answer: 6.27 %

Compound interest calculator More practice: Compound InterestMore practice: Sequences in Finance

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