Compound interest, reverse percentage and compound proportional change

Unit 7 Compound interest, reverse percentage and compound proportional change

Key unit competence: Solve real-world problems involving compound interest, reverse percentages and proportional change.

Key ideas

• Reverse percentage: if a value after a change of $p\%$ is known, divide by the multiplier to find the original: original $=\frac{\text{new}}{1\pm \frac{p}{100}}$.
• Compound interest: interest is added to the principal at the end of each period, and the next interest is calculated on the new amount.
• Formula: $A=P{\left(1+\frac{r}{100}\right)}^n$, where $A$ is the final amount, $P$ the principal, $r$ the rate per period, $n$ the number of periods. Compound interest $=A-P$.
• If interest is compounded $k$ times a year at $r\%$ per year for $t$ years: $A=P{\left(1+\frac{r}{100k}\right)}^{kt}$.
• Compound proportional change: growth $A=P(1+r{)}^n$, decay (depreciation) $A=P(1-r{)}^n$, with $r$ as a decimal.
• Continued proportions: several changes in a row multiply: $A=P\times m_1\times m_2\times \dots$ (never add the percentages).

Formulas to know

Common misconceptions

• "A price increased by 20% to 6 000 FRW, so the original was 6 000 − 20% of 6 000 = 4 800." No: original $=\frac{6000}{1.2}=5000$ FRW.
• "10% compound interest for 2 years is 20%." No: $1.1^2=1.21$, which is 21%.
• "A 10% rise then a 10% fall gives no change." No: $1.1\times 0.9=0.99$, a 1% fall.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. A bicycle costs 54 000 FRW after a 10% increase. Find the price before the increase.

   Show answer

   $54000÷1.1\approx 49091$ FRW.

   Common mistake: Subtracting 10% of 54 000.

2. The number of students in a school fell by 4% to 1 152. How many were there before?

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   $1152÷0.96=1200$.

   Common mistake: Adding 4% of 1 152.

3. Find the compound interest on 250 000 FRW at 6% per year for 2 years.

   Show answer

   $250000\times 1.06^2=280900$; interest $30900$ FRW.

   Common mistake: Giving the amount instead of the interest.

4. Find the amount when 300 000 FRW is invested at 12% per year compounded quarterly for 1 year.

   Show answer

   Quarterly rate 3%, 4 periods: $300000\times 1.03^4\approx 337653$ FRW.

   Common mistake: Using 12% for each quarter.

5. Compare simple and compound interest on 500 000 FRW at 10% for 3 years.

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   Simple: $150000$ FRW. Compound: $500000\times 1.331-500000=165500$ FRW. Compound earns 15 500 FRW more.

   Common mistake: Using $3\times 10\%$ for compound interest.

6. A shirt's price is reduced by 20% and later increased by 20%. Is it back to the original price? Explain with numbers.

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   No: $0.8\times 1.2=0.96$; a 10 000 FRW shirt ends at 9 600 FRW.

   Common mistake: Adding the percentages: $-20+20=0$.

7. A motorbike worth 1 800 000 FRW loses 15% of its value each year. Find its value after 2 years.

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   $1800000\times 0.85^2=1300500$ FRW.

   Common mistake: Subtracting 30% at once.

8. A population of bacteria doubles every hour. Starting with 500, how many are there after 6 hours?

   Show answer

   $500\times 2^6=32000$.

   Common mistake: Using $500\times 2\times 6$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. After two successive increases of 10%, a price is 6 050 FRW. Find the original price.	a. 102 010 FRW
   2. A population of 12 000 grows by 5% per year. Find it after 2 years.	b. 8 000 000 FRW
   3. A car worth 10 000 000 FRW depreciates by 20% per year. Its value after 1 year is:	c. 5 000 FRW
   4. 100 000 FRW earns 12% per year compounded monthly. Find the amount after 2 months.	d. 13 230

   Show answer

   1 → c (5 000 FRW) · 2 → d (13 230) · 3 → b (8 000 000 FRW) · 4 → a (102 010 FRW)

   • Step 1. Total multiplier: $1.1^2=1.21$.
     Step 2. $\frac{6050}{1.21}=5000$ FRW.
   • Step 1. $12000\times 1.05^2$.
     Step 2. $1.05^2=1.1025$, so $12000\times 1.1025=13230$.
   • $10000000\times 0.8$.
   • Step 1. Monthly rate $=\frac{12\%}{12}=1\%$.
     Step 2. $100000\times 1.01^2=102010$ FRW.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   100%  ·  75%  ·  1.32  ·  a 1% decrease
   1. A price rises by 10% then by 20%. The total multiplier is .
   2. A price rises 10% then falls 10%. The overall change is .
   3. A value decreases by 50% then increases by 50%. The final value compared with the start is .
   Show answer

   1: 1.32 · 2: a 1% decrease · 3: 75%

   • $1.1\times 1.2$.
   • $1.1\times 0.9=0.99$.
   • $0.5\times 1.5=0.75$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. 400 000 FRW is invested at 8% compound interest per year. Find the amount after 3 years (nearest FRW).	a. 150 000 FRW
   2. A population of 5 000 grows by 4% per year. Which expression gives it after $n$ years?	b. 5 000 FRW
   3. A price after a 20% increase is 6 000 FRW. What was the original price?	c. 503 885 FRW
   4. After a 12% pay rise, a salary is 168 000 FRW. What was it before?	d. $5000\times 1.04^n$

   Show answer

   1 → c (503 885 FRW) · 2 → d ($5000\times 1.04^n$) · 3 → b (5 000 FRW) · 4 → a (150 000 FRW)

   • Step 1. $A=400000\times 1.08^3$.
     Step 2. $1.08^3=1.259712$, so $A\approx 503885$ FRW.
   • Compound growth.
   • $6000÷1.2$.
   • $168000÷1.12$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   6 400 000 FRW  ·  33 100 FRW  ·  1%  ·  12%
   1. Simple interest on 100 000 FRW at 10% for 3 years is 30 000 FRW. Compound interest is .
   2. The same car after 2 years is worth .
   3. Interest at 12% per year compounded monthly has monthly rate .
   Show answer

   1: 33 100 FRW · 2: 6 400 000 FRW · 3: 1%

   • $100000\times (1.331-1)$.
   • $10000000\times 0.8^2$.
   • $12÷12$.

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