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Exam study plan

SSC CGL Preparation: Concepts and Practice

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Build your SSC CGL foundations with English and Hindi lessons, worked examples and explained practice across Quantitative Aptitude, General Intelligence and Reasoning, English Comprehension and General Awareness. Study arithmetic, algebra, geometry and trigonometry; practise analogy, classification, series, directions, ranking and clocks; strengthen grammar and constitutional basics. Use the linked topic tests to check understanding and review mistakes. Coverage is expanding subject by subject and does not yet represent the complete SSC CGL syllabus. See the module list and mock-test section for currently available material.

Lessons

Subject → Module → Lesson

English ComprehensionGeneral AwarenessGeneral Intelligence and ReasoningQuantitative Aptitude
Course outline 100
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Lessons 1–50 · Page 1 of 2
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8 lessons
  1. 1
    Natural Numbers, Integers, Rational and Irrational Numbers
  2. 2
    Place Value, Face Value and Number Comparison
  3. 3
    Prime, Composite and Co-prime Numbers
  4. 4
    Factors and Multiples
  5. 5
    Divisibility Tests and Missing Digits
  6. 6
    Remainders in Elementary Number Problems
  7. 7
    Unit Digits and Cyclic Patterns
  8. 8
    Counting Factors of a Number1 practice test

4 lessons
  1. 9
    Operations with Signed Integers
  2. 10
    BODMAS and Nested Brackets
  3. 11
    Simplifying Mixed Numerical Expressions
  4. 12
    Estimation and Checking Numerical Answers1 practice test

3 lessons
  1. 13
    Decimal notation and ordering
  2. 14
    Addition, subtraction, multiplication and division of decimals
  3. 15
    Terminating and recurring decimals

4 lessons
  1. 16
    Proper, improper, mixed and equivalent fractions
  2. 17
    Ordering and comparing fractions
  3. 18
    Arithmetic with fractions
  4. 19
    Converting fractions and decimals1 practice test

5 lessons
  1. 20
    Writing, simplifying and comparing ratios
  2. 21
    Proportion and continued proportion
  3. 22
    Direct and inverse proportion
  4. 23
    Dividing quantities in a ratio
  5. 24
    Compound ratios and changing ratios

7 lessons
  1. 25
    Meaning of percentage and fraction–decimal–percentage conversion
  2. 26
    Finding a percentage of a quantity
  3. 27
    Expressing one quantity as a percentage of another
  4. 28
    Percentage increase and decrease
  5. 29
    Successive percentage changes
  6. 30
    Finding the original quantity from a percentage change
  7. 31
    Population, income, expenditure and price applications

3 lessons
  1. 32
    Arithmetic Average of a Group
  2. 33
    Combined and Weighted Averages
  3. 34
    Average Changes After Addition, Removal or Replacement1 practice test

5 lessons
  1. 35
    Cost price, selling price, profit and loss
  2. 36
    Profit and loss percentages and reverse calculations
  3. 37
    Marked price, discount and markup
  4. 38
    Successive discounts
  5. 39
    Combined profit, loss and discount situations

3 lessons
  1. 40
    Principal, rate, time, simple interest and amount
  2. 41
    Finding an unknown principal, rate or time
  3. 42
    Comparing simple-interest arrangements

4 lessons
  1. 43
    Compound interest and successive accumulation
  2. 44
    Annual and subannual compounding
  3. 45
    Comparing simple and compound interest
  4. 46
    Growth and depreciation through repeated percentage changes

2 lessons
  1. 47
    Capital, time and profit-sharing ratios
  2. 48
    Partners joining, leaving or changing investment

2 lessons
  1. 49
    Concentration and weighted mixture averages
  2. 50
    Alligation for two-component mixtures

50 lessons across 12 modules

Lesson 46 of 100 | Quantitative Aptitude / Compound Interest / चक्रवृद्धि ब्याज

Growth and depreciation through repeated percentage changes

Learning outcome

Calculate repeated growth, depreciation and mixed changes, and recover an original value by reversing the full multiplier.

Concepts and assumptions

A percentage change uses the value immediately before that change. Growth by g% retains the old value and adds g/100 of it, giving multiplier 1 + g/100. Depreciation by d% removes d/100 of the current value, giving multiplier 1 - d/100.

For n complete periods at a constant rate:

Final value = original value × (1 + g/100)^n for growth.

Final value = original value × (1 - d/100)^n for depreciation.

These formulas repeat the same multiplication. Equal percentage depreciation produces decreasing monetary reductions because the base shrinks. It is not the same as subtracting a fixed amount every year.

For different rates or directions, multiply the appropriate factors in sequence. To find the net percentage change, divide final value minus original value by the original value and multiply by 100. A negative result indicates a decrease. To work backwards, divide the final value by the complete multiplier.

Assume positive initial values, no separate additions, removals or payments, and changes only at the stated period boundaries. Growth rates are nonnegative; depreciation rates range from 0% to below 100%, keeping values and reverse divisors positive. A 100% depreciation instead gives zero and cannot be reversed uniquely.

These are changes per named period, not nominal annual interest quotes. Do not divide an annual depreciation rate by 12 and assume an equivalent monthly model. Although a year has 12 months, fractional periods or days need an explicit model; no day-count rule is assumed here. Keep intermediate values exact, rounding final money to ₹0.01 only when needed, with half a paise rounded upward.

Worked examples

Example 1 — Repeated growth. A value of ₹18,000 grows by 5% annually for 2 years. First value = 18000 × 1.05 = ₹18,900. Second value = 18900 × 1.05 = ₹19,845. Total growth = 19845 - 18000 = ₹1,845; percentage growth = (1845 ÷ 18000) × 100 = 10.25%, not 10%.

Example 2 — Repeated depreciation. A machine valued at ₹62,500 depreciates 12% annually for 3 years. Successive values are 62500 × 0.88 = ₹55,000, then 55000 × 0.88 = ₹48,400, then 48400 × 0.88 = ₹42,592. Total depreciation = 62500 - 42592 = ₹19,908. Each reduction uses the current value.

Example 3 — Reverse mixed changes. After a 20% increase followed by a 15% decrease, a value is ₹24,480. Combined multiplier = 1.20 × 0.85 = 1.02. Original value = 24480 ÷ 1.02 = ₹24,000. Check: 24000 × 1.20 = ₹28,800, then 28800 × 0.85 = ₹24,480. Net growth is 2%, not 20% - 15% = 5%.

Common mistakes

Equal percentage increases and decreases do not cancel: (1 + x) × (1 - x) = 1 - x^2 for a decimal rate x between 0 and 1. This identity requires two equal opposite rates, each applied successively.

Practice questions

  1. ₹26,000 grows by 6% annually for 2 years. Find the final value.
  2. ₹45,000 depreciates by 10% annually for 2 years. Find the final value and total depreciation.
  3. ₹16,000 rises by 25%, then falls by 25%. Find the final value and net percentage change.
  4. After 2 annual depreciations of 8%, a value is ₹33,856. Find its original value.

Worked answers

  1. Values are 26000 × 1.06 = ₹27,560 and 27560 × 1.06 = ₹29,213.60.
  2. Values are 45000 × 0.90 = ₹40,500 and 40500 × 0.90 = ₹36,450. Total depreciation = 45000 - 36450 = ₹8,550.
  3. Values are 16000 × 1.25 = ₹20,000 and 20000 × 0.75 = ₹15,000. Decrease = ₹1,000, or (1000 ÷ 16000) × 100 = 6.25%.
  4. Combined multiplier = 0.92^2 = 0.8464. Original value = 33856 ÷ 0.8464 = ₹40,000. Checking gives ₹36,800 after one year and ₹33,856 after two.
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