Exercise 5.3 — Simple and Compound Interest

Simple Interest and Compound Interest problems.

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Exercise 5.3 – Simple Interest, Compound Interest & Depreciation

Exercise 5.3 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) introduces two of the most practical concepts in everyday finance: Simple Interest (S.I.) and Compound Interest (C.I.). You'll also learn about depreciation — how the value of items like TVs, cars, and machinery reduces over time — and how to apply compound interest formulas to real situations like loans, population growth, and bacteria growth.

The most important skill in this exercise is recognizing which formula to use. Simple interest is calculated on the same principal every time, while compound interest is calculated on a principal that keeps changing (growing or shrinking) after each time period. Most questions in this exercise are direct applications of one of the four core formulas below.

Key Formulas You Must Know

ConceptFormulaUsed In
Simple InterestI = (P × T × R) / 100Q1, Q5, Q9, Q11, Q12
Amount (Simple Interest)A = P × (100 + TR) / 100Q1, Q5, Q9
Amount (Compound Interest)A = P × (100 + R)n / 100nQ3, Q4, Q6, Q8–Q15
Compound InterestI = A − PQ3, Q4, Q7, Q9, Q12, Q15
Value After DepreciationA = P × (100 − R)n / 100nQ2, Q13
Compounding Period (n)Annually: n = T; Half-yearly: n = 2T, R = R/2; Quarterly: n = 4T, R = R/4Q8, Q15
💡 The golden rule: In compound interest, each new time period's interest is calculated on the amount from the end of the previous period — not on the original principal. This is exactly why compound interest grows faster than simple interest over time.
Question 1
Finding Monthly Repayments on a Simple Interest Loan
Simple Interest — Monthly Instalments

Sudhakar borrows ₹15,000 at 9% per annum simple interest for 8 years. We first find the total amount he must repay, then divide it equally over all the monthly instalments for 8 years.

SymbolMeaningValue
PPrincipal₹15,000
RRate of interest9% p.a.
TTime8 years

Step 1: Total Amount to Repay

A = P × (100 + TR)/100 = 15000 × (100 + 8×9)/100
= 15000 × (100 + 72)/100 = 150 × 172
= ₹25,800

Step 2: Monthly Repayment

Number of monthly instalments = 8 × 12 = 96
Monthly payment = Total Amount / Number of instalments = 25800 / 96
∴ Monthly repayment = ₹268.75
Question 2
Finding the Depreciated Value of a TV After 1 Year
Depreciation

A TV bought for ₹21,000 depreciates (loses value) by 5% after 1 year. Depreciation works exactly like compound interest, except the value decreases instead of increasing — so we use (100 − R) instead of (100 + R).

Value after depreciation = Original Price × (100 − R%) / 100
Value after 1 year = 21000 × (100 − 5)/100 = 210 × 95
∴ Value of TV after 1 year = ₹19,950
Question 3
Finding Amount and Compound Interest Over 2 Years (Same Rate)
Compound Interest — Basic

A sum of ₹8000 is invested at 5% per annum for 2 years, compounded annually. This is the most direct application of the compound interest formula — apply (100 + R)/100 twice (raised to the power n = 2).

A = P × [(100 + R)/100]n
A = 8000 × (105/100)2 = 8000 × (105/100) × (105/100)
= 4 × 21 × 105
Amount = ₹8820
Compound Interest = A − P = 8820 − 8000
∴ Compound Interest = ₹820
Question 4
Compound Interest with Different Rates Each Year
Compound Interest — Variable Rate

A sum of ₹6500 earns 5% in the first year and 6% in the second year, compounded annually. Since the rates differ each year, we cannot use the formula with a single power n — instead, we calculate the amount for the first year, then use that result as the new principal for the second year.

Step 1: First Year (Rate = 5%)

A = P × (100 + TR)/100 = 6500 × (100 + 1×5)/100 = 65 × 105
Amount after 1st year = ₹6825

Step 2: Second Year (Rate = 6%, P = ₹6825)

A = 6825 × (100 + 1×6)/100 = 6825 × 106/100 = 273 × 106/4
Amount after 2nd year = ₹7234.50
Compound Interest = A − P = 7234.50 − 6500
∴ Compound Interest = ₹734.50
💡 Key idea: Whenever the rate of interest changes between years, treat each year as a separate simple-interest calculation, where the amount from one year becomes the principal for the next. This "principal carries forward" idea is central to all compound interest problems.
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Question 5
Total Repayment and Monthly Instalments on a Car Loan
Simple Interest — Monthly Instalments

Prathibha borrows ₹47,000 at 17% simple interest for 5 years. This question follows the exact same two-step pattern as Question 1: first find the total repayable amount, then divide it across the monthly instalments.

(a) Total Amount After 5 Years

A = P × (100 + TR)/100 = 47000 × (100 + 5×17)/100
= 47000 × (100 + 85)/100 = 470 × 185
∴ Amount to repay = ₹86,950

(b) Monthly Repayment

Number of instalments = 5 × 12 = 60
Monthly payment = 86950 / 60
∴ Monthly repayment ≈ ₹1449.17
Question 6
Population Growth Using Compound Interest Formula
Compound Growth — Real-Life Application

The population of Hyderabad was 68,09,000 in 2011, growing at 4.7% per year. We need the population at the end of 2015 — that is, after 4 years. Population growth, like compound interest, multiplies by the same growth factor every year, so we apply the compound interest formula with n = 4.

A = P × (100 + R)n / 100n = 6809000 × (104.7/100)4
= 6809000 × (1047/1000)4
= 6809000 × (1047/1000) × (1047/1000) × (1047/1000) × (1047/1000)
≈ 81,82,199 (approximately)
💡 Compound growth formulas aren't just for money — they apply to anything that grows by a fixed percentage of its current value at regular intervals: population, bacteria, plant growth, and investments all follow the same mathematical pattern.
Question 7
Compound Interest for Part of a Year (1 Year 3 Months)
Compound Interest — Mixed Time Period

A sum of ₹10,000 is invested at 8½% per annum compounded annually for 1 year and 3 months. When the time period includes a fraction of a year, the compounding formula is applied for the whole year(s) first, and then simple interest is applied for the remaining months on the new principal.

Step 1: Compound Interest for 1 Full Year

R = 8½% = 17/2%
A = 10000 × (100 + 17/2)/100 = 10000 × (217/200) = 50 × 217
Amount after 1 year = ₹10,850

Step 2: Simple Interest for Remaining 3 Months (= ¼ year)

A = 10850 × (100 + (1/4)×(17/2)) / 100 = 10850 × (100 + 17/8)/100
= 10850 × (817/800)
≈ ₹11,080.56
Compound Interest = A − P = 11080.56 − 10000
∴ Compound Interest ≈ ₹1080.56
📐 For the "extra" period that's less than a full compounding cycle (here, 3 months out of a year), the additional interest is calculated as simple interest on the amount accumulated so far — this is the standard convention in CBSE/Telangana/AP textbooks.
Question 8
Comparing Annual vs Half-Yearly Compounding
Compounding Frequency Comparison

Arif takes a loan of ₹80,000 at 10% per annum for 1½ years. We find the final amount under two different compounding methods — annual and half-yearly — and compare the difference. The key change for half-yearly compounding: the rate is halved and the number of periods is doubled.

Method 1: Compounded Annually

For 1st year: A = 80000 × (100 + 10)/100 = 80000 × 110/100
= ₹88,000
For next 6 months (½ year, simple interest on ₹88,000 at 10%):
A = 88000 × (100 + (1/2)×10)/100 = 88000 × 105/100
∴ Amount for 1½ years (annual compounding) = ₹92,400

Method 2: Compounded Half-Yearly

R = 10/2 = 5% per half-year; T = 1½ years → n = 3 half-year periods
A = 80000 × (100 + 5)3 / 1003 = 80000 × (105/100)3
∴ Amount for 1½ years (half-yearly compounding) = ₹92,610

Difference Between the Two Methods

Difference = 92610 − 92400
∴ Difference = ₹210
💡 More frequent compounding (half-yearly, quarterly, monthly) always results in a slightly higher final amount than annual compounding at the same nominal annual rate — because interest starts earning interest sooner.
Question 9
Extra Amount Payable If Compound Interest Is Charged Instead of Simple Interest
Simple vs Compound Interest Comparison

A sum of ₹12,000 is borrowed at 6% per annum for 2 years. We compare the total amount payable under simple interest versus compound interest (compounded annually), and find the extra amount due to compounding.

Simple Interest Amount

A = P × (100 + TR)/100 = 12000 × (100 + 2×6)/100 = 120 × 112
A (simple interest) = ₹13,440

Compound Interest Amount

A = 12000 × (100 + 6)2/1002 = 12000 × (106/100) × (106/100)
A (compound interest) = ₹13,483.20

Extra Amount Due to Compounding

Extra amount = 13483.20 − 13440
∴ Extra amount = ₹43.20
Question 10
Bacteria Count Growth Using Compound Interest Formula
Compound Growth — Science Application

A bacteria culture starts at 5,06,000 and grows at 2.5% per hour. We find the count after 2 hours. Since the growth compounds every hour, n = 2 — the exact same structure as the population growth in Question 6, just with a smaller initial value and a shorter time unit (hours instead of years).

A = P × (100 + R)n/100n = 506000 × (102.5/100)2
= 506000 × (1025/1000)2
= 506000 × (1025/1000) × (1025/1000)
≈ 5,31,216 (approximately)
Question 11
Compound Interest for Years Plus Extra Months
Compound Interest — Mixed Time Period

Kamala borrows ₹26,400 at 15% per annum compounded yearly, and needs to repay it after 2 years and 4 months. Just like Question 7, we apply compound interest for the whole years first, then simple interest for the extra months on the new amount.

Step 1: Compound Interest for First 2 Years

A = 26400 × (100 + 15)2/1002 = 26400 × (115/100)2
= 26400 × (115/100) × (115/100)
Amount after 2 years = ₹34,914

Step 2: Simple Interest for Next 4 Months (= 1/3 year)

I = PTR/100 = 34914 × (1/3) × 15 / 100 = 34914 × 5/100
Interest for 4 months = ₹1745.70
Total amount = Amount + Interest = 34914 + 1745.70
∴ Amount payable after 2 years 4 months = ₹36,659.70
Question 12
Comparing Simple Interest vs Compound Interest for Two Borrowers
Simple vs Compound Interest Comparison

Bharathi borrows ₹12,500 at 12% simple interest for 3 years, while Madhuri borrows the same amount at 10% compound interest for the same period. We need to find who pays more interest, even though their rates and methods are completely different.

Bharathi's Interest (Simple Interest at 12%)

I = PTR/100 = (12500 × 3 × 12)/100
Bharathi's interest = ₹4500

Madhuri's Interest (Compound Interest at 10%)

A = 12500 × (100 + 10)3/1003 = 12500 × (110/100)3
= 12500 × (110/100) × (110/100) × (110/100)
A = ₹16,637.50
Compound Interest = A − P = 16637.50 − 12500
Madhuri's interest = ₹4137.50

Comparison

Excess interest = 4500 − 4137.50
∴ Bharathi pays ₹362.50 more interest than Madhuri
💡 Even though Bharathi's rate (12%) is higher than Madhuri's (10%), the way compound interest accumulates means the comparison isn't obvious without calculation. Always compute both fully before comparing — never guess based on the rate alone.
Question 13
Machinery Value After Depreciation (1 Year)
Depreciation

Machinery worth ₹10,000 depreciates by 5%. We find its value after 1 year — this is the same depreciation formula used in Question 2, applied to a different item and value.

A = P × (100 − R)n/100n = 10000 × (100 − 5)/100, where n = 1
= 10000 × 95/100
∴ Value of machinery after 1 year = ₹9500
Question 14
City Population Growth Over 2 Years
Compound Growth

A city's current population is 12 lakh (12,00,000), growing at 4% per year. We find the population after 2 years using the same compound growth formula seen in Questions 6 and 10, here with whole numbers that simplify nicely.

A = P × (100 + R)n/100n = 1200000 × (104/100)2, where n = 2
= 1200000 × (104/100) × (104/100)
= 120 × 104 × 104
∴ Population after 2 years = 12,97,920
Question 15
Compound Interest with Quarterly Compounding
Compounding Frequency — Quarterly

A sum of ₹1000 is invested for 1 year at 10% per annum, but the interest is compounded quarterly (every 3 months). When compounding quarterly, the annual rate is divided by 4 and the number of periods n becomes 4 (for 1 year).

Quarterly compounding: Rquarter = R ÷ 4, n = T × 4
R per quarter = 10/4 = 2.5%; n = 1 × 4 = 4
A = 1000 × (100 + 2.5)4/1004 = 1000 × (1025/1000)4
= 1000 × (1025/1000) × (1025/1000) × (1025/1000) × (1025/1000)
≈ ₹1103.81
Compound Interest = A − P = 1103.81 − 1000
∴ Compound Interest ≈ ₹103.81

Common Mistakes to Avoid in Exercise 5.3

  • Confusing simple interest and compound interest formulas: Simple interest uses (100 + TR)/100 with the original principal throughout. Compound interest uses [(100 + R)/100]n, where the principal effectively grows every period.
  • Getting the compounding frequency wrong: For half-yearly compounding, halve the rate and double the number of periods. For quarterly compounding, divide the rate by 4 and multiply the number of periods by 4.
  • Mishandling fractional time periods: When time includes extra months (like 1 year 3 months or 2 years 4 months), apply compound interest for the full years first, then simple interest for the remaining months on the new amount — not the original principal.
  • Using (100 + R) for depreciation: Depreciation reduces value, so always use (100 − R), not (100 + R). This is the most common sign error in this exercise.
  • Forgetting to subtract P at the end: Many questions ask for the interest, not the amount. Always check whether the answer required is A or (A − P).
📐 Board exam tip: In CBSE, Telangana, and Andhra Pradesh exams, write out the formula first, clearly substitute the values of P, R, T, and n, and show each step of simplification — partial marks are awarded even if the final numeric answer has a small error.

What Exercise 5.3 Prepares You For

Simple and compound interest are foundational for understanding exponential growth and decay, a theme that reappears throughout higher mathematics. The percentage techniques practised in Exercise 5.2 on profit, loss, discount, and GST combine directly with the compounding ideas here to solve more advanced financial mathematics problems.

The exponent-based formulas — (100 + R)n — also connect to Exponents and Powers, since compound interest is really just repeated multiplication expressed using powers. In Class 9 and Class 10, similar growth and decay models appear in Direct and Inverse Proportions and various real-world application problems.

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