This MCQ module is based on: Compounding, Appreciation & Depreciation
TOPIC 3 OF 23
Compounding, Appreciation & Depreciation
🎓 Class 8
Mathematics
CBSE
Theory
Ch 1 — Percentages
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Compounding, Appreciation & Depreciation
Targeting Class 8 level in General Mathematics, with Basic difficulty.
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Profit and Loss — Worked Examples
Example 4: Snehal's Vase
Snehal buys a vase for ₹2650. Due to heavy rains, the vase gets damaged and she sells it at a loss of 18%. What was the sale amount of the damaged vase?
Solution:
Method 1: The loss amount is 18% of 2650. \(\frac{18}{100} \times 2650 = 477\). Selling price = \(2650 - 477 = ₹2173\).
Method 2: With respect to the buying price being 100%, the selling price would be 82% (100% − 18%). \(82\% \times 2650 = 0.82 \times 2650 = ₹2173\).
Method 2: With respect to the buying price being 100%, the selling price would be 82% (100% − 18%). \(82\% \times 2650 = 0.82 \times 2650 = ₹2173\).
Example: Manisha's Fertiliser
Manisha sells fertiliser. She buys 50 kg bags at ₹500 per bag and sells at ₹750 per bag.
Profit per bag = ₹750 − ₹500 = ₹250
Profit percentage = \(\frac{250}{500} \times 100 = 50\%\)
Profit percentage differs depending on whether calculated on CP (50%) or SP (33.33%). Standard practice: always use CP as base.
Important
Profit percentage is calculated based on the cost price (CP), not the selling price. "How much did I make compared to what I invested?" — that's why we use CP as the base.
🔵 A utensil store is offering a 35% discount on the cooker with an MRP of ₹1800. If the cost price was ₹900, what is the percentage profit made after the sale?
SP = ₹1800 − 35% × ₹1800 = ₹1800 − ₹630 = ₹1170. Profit = ₹1170 − ₹900 = ₹270. Profit % = \(\frac{270}{900} \times 100 = 30\%\).
SP = ₹1800 − 35% × ₹1800 = ₹1800 − ₹630 = ₹1170. Profit = ₹1170 − ₹900 = ₹270. Profit % = \(\frac{270}{900} \times 100 = 30\%\).
Growth and Compounding
You might have come across statements like, "1 year interest for FD (Fixed Deposit) in the bank @ 6% per annum" or "Savings account with interest @ 2.5% per annum." Interest? is the extra money paid by banks or post offices on money deposited (kept) with them. Interest is also paid by people or institutions when they borrow money.
Fixed Deposit (FD)
In a Fixed Deposit, you deposit a specific amount of money for a fixed period at a predetermined interest rate. The money remains locked for the chosen duration. At the end of the term, you receive both your original deposit and the interest earned.
Simple Interest vs Compound Interest
Consider an FD of ₹6000 at 10% per annum for 3 years. There are two options for how interest is calculated:
Option 1: Simple Interest
Interest is calculated on the original principal only, every year. The interest gained is NOT added back to the FD.
Year 1: \(10\% \times 6000 = ₹600\)
Year 2: \(10\% \times 6000 = ₹600\)
Year 3: \(10\% \times 6000 = ₹600\)
Total = ₹6000 + ₹1800 = ₹7800
Year 1: \(10\% \times 6000 = ₹600\)
Year 2: \(10\% \times 6000 = ₹600\)
Year 3: \(10\% \times 6000 = ₹600\)
Total = ₹6000 + ₹1800 = ₹7800
Option 2: Compound Interest
Interest is calculated on the principal + accumulated interest. The interest is added back to the FD each year.
Year 1: \(10\% \times 6000 = ₹600\) → Total ₹6600
Year 2: \(10\% \times 6600 = ₹660\) → Total ₹7260
Year 3: \(10\% \times 7260 = ₹726\) → Total ₹7986
Total = ₹7986
Year 1: \(10\% \times 6000 = ₹600\) → Total ₹6600
Year 2: \(10\% \times 6600 = ₹660\) → Total ₹7260
Year 3: \(10\% \times 7260 = ₹726\) → Total ₹7986
Total = ₹7986
| Interest added back (10% p.a.) | Amount in the FD | ||
|---|---|---|---|
| Beginning | ₹6,000 | ||
| Year 1 End | ₹600 | ₹6,600 | × 1.10 |
| Year 2 End | ₹660 | ₹7,260 | × 1.10 |
| Year 3 End | ₹726 | ₹7,986 | × 1.10 |
| Total amount received: | ₹7,986 | ||
We can see that with compounding?, the final amount is more (₹7986 vs ₹7800). The difference is ₹186.
Example 8: Comparing the Two Options
What percent is the total amount received with respect to the amount deposited in both options?
Solution:
Without compounding: \(\frac{7800}{6000} \times 100 = 130\%\). Gain = 30% over 3 years.
Total = \(6000 \times (1 + 0.1 + 0.1 + 0.1) = 6000 \times 1.3\).
With compounding: \(\frac{7986}{6000} \times 100 = 133.1\%\). Gain = 33.1% over 3 years.
Total = \(6000 \times 1.1 \times 1.1 \times 1.1 = 6000 \times 1.331\).
Total = \(6000 \times (1 + 0.1 + 0.1 + 0.1) = 6000 \times 1.3\).
With compounding: \(\frac{7986}{6000} \times 100 = 133.1\%\). Gain = 33.1% over 3 years.
Total = \(6000 \times 1.1 \times 1.1 \times 1.1 = 6000 \times 1.331\).
Formulas
Without compounding (Simple Interest):Amount = \(p \times (1 + rt)\) where \(p\) = principal, \(r\) = rate per period, \(t\) = number of periods.
With compounding:
Amount = \(p \times (1 + r)^t\)
The principal remains the same for all terms without compounding. With compounding, the principal grows each term.
🔵 Suppose we want to know the expression/formula to find the total interest amount gained at the end of the maturity period. What would be the formula for each of the two options?
Simple: Interest = \(p \times r \times t\). Compound: Interest = \(p \times (1+r)^t - p\).
Simple: Interest = \(p \times r \times t\). Compound: Interest = \(p \times (1+r)^t - p\).
Figure it Out (Compounding)
Q1. Bank of Yahapur offers an interest of 10% p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.
Without compounding: Interest = \(20000 \times 0.10 \times 2 = ₹4000\). Total = ₹24,000.
With compounding: Year 1: \(20000 \times 1.10 = ₹22000\). Year 2: \(22000 \times 1.10 = ₹24200\).
Difference: ₹200 more with compounding.
With compounding: Year 1: \(20000 \times 1.10 = ₹22000\). Year 2: \(22000 \times 1.10 = ₹24200\).
Difference: ₹200 more with compounding.
Q4. Jasmine invests amount 'p' for 4 years at an interest of 6% p.a. Which of the following expressions describe the total amount she will get after 4 years when compounding is not done?
(i) \(p \times 6 \times 4\) (ii) \(p \times 0.6 \times 4\) (iii) \(p \times \frac{0.6}{100} \times 4\) (iv) \(p \times \frac{6}{100} \times 4\) (v) \(p \times 1.6 \times 4\) (vi) \(p \times 1.06 \times 4\) (vii) \(p + (p \times 0.06 \times 4)\)
(i) \(p \times 6 \times 4\) (ii) \(p \times 0.6 \times 4\) (iii) \(p \times \frac{0.6}{100} \times 4\) (iv) \(p \times \frac{6}{100} \times 4\) (v) \(p \times 1.6 \times 4\) (vi) \(p \times 1.06 \times 4\) (vii) \(p + (p \times 0.06 \times 4)\)
Simple interest amount = \(p + p \times r \times t = p + p \times 0.06 \times 4 = p(1 + 0.24) = 1.24p\).
Checking each: (iv) gives interest = \(0.24p\) ✓ but not total. (vii) gives \(p + 0.24p = 1.24p\) ✓ — this is the total amount.
Correct answers: (iv) gives just the interest, (vii) gives the total amount.
Checking each: (iv) gives interest = \(0.24p\) ✓ but not total. (vii) gives \(p + 0.24p = 1.24p\) ✓ — this is the total amount.
Correct answers: (iv) gives just the interest, (vii) gives the total amount.
Q5. The post office offers 7% p.a. interest. How much interest would one get if investing ₹50,000 for 3 years without compounding? How much more would one get if it was compounded?
Simple: \(50000 \times 0.07 \times 3 = ₹10,500\). Total = ₹60,500.
Compound: \(50000 \times 1.07^3 = 50000 \times 1.225043 = ₹61,252.15\). Interest = ₹11,252.15.
Compounding gives ₹752.15 more.
Compound: \(50000 \times 1.07^3 = 50000 \times 1.225043 = ₹61,252.15\). Interest = ₹11,252.15.
Compounding gives ₹752.15 more.
Q7. Consider an amount ₹1000. If this grows at 10% p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth or linear growth?
Without compounding: Need interest = ₹1000. \(1000 \times 0.10 \times t = 1000\), so \(t = 10\) years.
With compounding: \(1000 \times 1.1^t = 2000\), so \(1.1^t = 2\). Testing: \(1.1^7 = 1.948\), \(1.1^8 = 2.144\). Doubles between 7 and 8 years.
Compounding is exponential growth (amount × 1.1 each year). Without compounding is linear growth (same ₹100 added each year).
With compounding: \(1000 \times 1.1^t = 2000\), so \(1.1^t = 2\). Testing: \(1.1^7 = 1.948\), \(1.1^8 = 2.144\). Doubles between 7 and 8 years.
Compounding is exponential growth (amount × 1.1 each year). Without compounding is linear growth (same ₹100 added each year).
Decline and Depreciation
Several items or materials lose financial value over time. A bike bought at ₹1,00,000 may be worth less after a few years. This reduction of value due to use and age of the item is called depreciation?.
Example 9: Car Depreciation
A car is purchased for ₹6,00,000. Its value depreciates by 10% every year. Find its value after 3 years.
Solution:
Each year the car retains 90% (= 100% − 10%) of its value.
After 1 year: \(6{,}00{,}000 \times 0.90 = ₹5{,}40{,}000\)
After 2 years: \(5{,}40{,}000 \times 0.90 = ₹4{,}86{,}000\)
After 3 years: \(4{,}86{,}000 \times 0.90 = ₹4{,}37{,}400\)
Or directly: \(6{,}00{,}000 \times 0.9^3 = 6{,}00{,}000 \times 0.729 = \mathbf{₹4{,}37{,}400}\).
After 1 year: \(6{,}00{,}000 \times 0.90 = ₹5{,}40{,}000\)
After 2 years: \(5{,}40{,}000 \times 0.90 = ₹4{,}86{,}000\)
After 3 years: \(4{,}86{,}000 \times 0.90 = ₹4{,}37{,}400\)
Or directly: \(6{,}00{,}000 \times 0.9^3 = 6{,}00{,}000 \times 0.729 = \mathbf{₹4{,}37{,}400}\).
Successive Percentage Changes
While comparing percentages, we have to be mindful that we are comparing fractions or proportions and not absolute values.
Example 12: Cakely vs Cakify
A bakery called Cakely is offering a 30% + 20% discount on all cakes. Another bakery called Cakify is offering a 50% discount on all cakes. Would you rather choose Cakely or Cakify if you want the cheaper cost?
Solution:
It seems like 30% + 20% = 50%, so both should be the same. But successive discounts don't add up!
Suppose a cake costs ₹200.
Cakely's 30% + 20%:
First discount: \(30\% \times 200 = ₹60\). Price after first discount = ₹140.
Second discount: \(20\% \times 140 = ₹28\). Final price = \(140 - 28 = ₹112\).
Cakify's 50%:
\(50\% \times 200 = ₹100\). Final price = \(200 - 100 = ₹100\).
Cakify is cheaper! ₹100 vs ₹112. The "30% + 20%" discount is actually \(1 - 0.7 \times 0.8 = 1 - 0.56 = 44\%\) overall, NOT 50%.
Suppose a cake costs ₹200.
Cakely's 30% + 20%:
First discount: \(30\% \times 200 = ₹60\). Price after first discount = ₹140.
Second discount: \(20\% \times 140 = ₹28\). Final price = \(140 - 28 = ₹112\).
Cakify's 50%:
\(50\% \times 200 = ₹100\). Final price = \(200 - 100 = ₹100\).
Cakify is cheaper! ₹100 vs ₹112. The "30% + 20%" discount is actually \(1 - 0.7 \times 0.8 = 1 - 0.56 = 44\%\) overall, NOT 50%.
Key Rule
Successive percentage changes are NOT additive.A 30% discount followed by a 20% discount is equivalent to a 44% total discount (not 50%).
Formula: Overall factor = \((1 - 0.30) \times (1 - 0.20) = 0.7 \times 0.8 = 0.56\), meaning the customer pays 56% of the original price (44% total discount).
Activity: Compound Growth Calculator
L3 Apply
Predict: If you invest ₹1000 at 10% compound interest, will it take exactly 10 years to double? More? Less?
- Start with ₹1000. Apply 10% growth each year: multiply by 1.1.
- Track the amount year by year in a table.
- Mark the year when the amount first exceeds ₹2000.
- Compare: how many years does simple interest (₹100/year) take to reach ₹2000?
Simple: 10 years (₹1000 + 10 × ₹100 = ₹2000).
Compound: After 7 years: ₹1948.72. After 8 years: ₹2143.59. Doubles between year 7 and 8 — about 7.27 years (the "Rule of 72": 72/10 = 7.2 years).
Competency-Based Questions
Scenario: Ramesh deposits ₹50,000 in a bank FD at 8% per annum compound interest. His friend Suresh deposits the same amount in another bank at 8% simple interest. Both for 3 years.
Q1. How much does Ramesh receive after 3 years?
L3 ApplyAnswer: (b) ₹62,985.60.
\(50000 \times 1.08^3 = 50000 \times 1.259712 = ₹62,985.60\).
\(50000 \times 1.08^3 = 50000 \times 1.259712 = ₹62,985.60\).
Q2. How much extra does Ramesh earn compared to Suresh over 3 years? Analyse why the gap increases with time.
L4 AnalyseSuresh (simple): \(50000 + 50000 \times 0.08 \times 3 = 50000 + 12000 = ₹62,000\).
Ramesh (compound): ₹62,985.60.
Difference: ₹985.60. The gap grows because each year, Ramesh earns interest on a larger base (previous interest is added), while Suresh's base stays at ₹50,000. This effect compounds — it's exponential vs linear growth.
Ramesh (compound): ₹62,985.60.
Difference: ₹985.60. The gap grows because each year, Ramesh earns interest on a larger base (previous interest is added), while Suresh's base stays at ₹50,000. This effect compounds — it's exponential vs linear growth.
Q3. Suresh claims that for just 1 year, simple and compound interest give the same result. Evaluate this claim.
L5 EvaluateSuresh is correct! For 1 year: Simple = \(p \times r \times 1 = 0.08p\). Compound = \(p \times 1.08^1 - p = 0.08p\). Both give exactly the same interest. The difference only appears from Year 2 onwards, when compound interest earns "interest on interest."
Q4. Design an investment plan: if you wanted to accumulate ₹1,00,000 in 5 years using compound interest at 8% p.a., how much should you deposit now? Show your working.
L6 CreateAnswer: We need \(p \times 1.08^5 = 1{,}00{,}000\).
\(1.08^5 = 1.469328\).
\(p = \frac{1{,}00{,}000}{1.469328} = ₹68{,}058.32\).
Deposit approximately ₹68,058 today to have ₹1,00,000 in 5 years at 8% compound interest.
\(1.08^5 = 1.469328\).
\(p = \frac{1{,}00{,}000}{1.469328} = ₹68{,}058.32\).
Deposit approximately ₹68,058 today to have ₹1,00,000 in 5 years at 8% compound interest.
Assertion–Reason Questions
Assertion (A): Two successive discounts of 20% and 30% are equivalent to a single discount of 44%.
Reason (R): Successive percentage changes multiply: \((1-0.20)(1-0.30) = 0.8 \times 0.7 = 0.56\), meaning the customer pays 56%.
Reason (R): Successive percentage changes multiply: \((1-0.20)(1-0.30) = 0.8 \times 0.7 = 0.56\), meaning the customer pays 56%.
Answer: (a) — Customer pays 56%, so discount = 100% − 56% = 44%. Both true, R explains A.
Assertion (A): A car depreciates by 10% each year. After 10 years, it will be worth 0%.
Reason (R): 10% depreciation per year for 10 years = 100% depreciation.
Reason (R): 10% depreciation per year for 10 years = 100% depreciation.
Answer: (d) — Both are false. After 10 years: value = \(P \times 0.9^{10} = P \times 0.3487 \approx 34.87\%\) of original — NOT zero. The reasoning that 10 × 10% = 100% is wrong because depreciation is compounded (applied to a shrinking base each year, not the original).
Assertion (A): Compound interest for 1 year is the same as simple interest for 1 year (at the same rate and principal).
Reason (R): In the first year, there is no accumulated interest to compound on.
Reason (R): In the first year, there is no accumulated interest to compound on.
Answer: (a) — Both true. In year 1, CI = \(p \times r\) and SI = \(p \times r\). They differ only from year 2 when compound interest earns "interest on interest." R correctly explains why.
Frequently Asked Questions — Chapter 1
What is Compounding, Appreciation & Depreciation in NCERT Class 8 Mathematics?
Compounding, Appreciation & Depreciation is a key concept covered in NCERT Class 8 Mathematics, Chapter 1: Chapter 1. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Compounding, Appreciation & Depreciation step by step?
To solve problems on Compounding, Appreciation & Depreciation, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Chapter 1?
The essential formulas of Chapter 1 (Chapter 1) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Compounding, Appreciation & Depreciation important for the Class 8 board exam?
Compounding, Appreciation & Depreciation is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Compounding, Appreciation & Depreciation?
Common mistakes in Compounding, Appreciation & Depreciation include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Compounding, Appreciation & Depreciation?
End-of-chapter NCERT exercises for Compounding, Appreciation & Depreciation cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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