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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 38 of 100 | Quantitative Aptitude / Commercial Arithmetic / वाणिज्यिक अंकगणित

Successive discounts

Learning outcome

Calculate successive discounts, compare equivalent offers and reverse the calculation without confusing the changing percentage bases.

Why successive discounts multiply

With successive discounts, each reduction applies to the price immediately before it. The first uses marked price (MP); the second uses the already reduced price, not MP again. Unless otherwise stated, this is what “successive” means.

A discount of a% retains the fraction 1 - a/100. Two successive discounts therefore give:

SP = MP × (1 - a/100) × (1 - b/100).

The equivalent single discount is the total reduction expressed as a percentage of the original MP. For two discount rates a% and b%, its percentage is a + b - (a × b)/100. The subtracted term corrects the overcount that occurs when the rates are merely added: the second discount acts on a smaller price.

For three or more stages, multiply all retained fractions, then subtract the resulting fraction from 1. Multiply by 100 to express the overall discount as a percentage.

All prices here are positive; discount rates are nonnegative and below 100%. Calculations are exact, with no added fees or intermediate rounding. Under these assumptions, exchanging the order of percentage discounts leaves the final price unchanged because multiplication is commutative, although the separate discount amounts can change.

To recover MP, divide SP by the product of the retained fractions. To find an unknown later discount, compare the final price with the price immediately before that discount.

Worked examples

Example 1 — Two reductions. MP is ₹2,800, with discounts of 15% and then 10%. First price = 2800 × 0.85 = ₹2,380. Final SP = 2380 × 0.90 = ₹2,142. Total discount = ₹658, so equivalent discount = (658 ÷ 2800) × 100 = 23.5%, not 25%.

Example 2 — Three reductions. MP is ₹3,600, with successive discounts of 20%, 10% and 5%. Prices become 3600 × 0.80 = ₹2,880, then 2880 × 0.90 = ₹2,592, then 2592 × 0.95 = ₹2,462.40. Retained fraction = 0.80 × 0.90 × 0.95 = 0.684. Equivalent discount = (1 - 0.684) × 100 = 31.6%.

Example 3 — Reverse the discounts. SP is ₹2,772 after discounts of 12% and 25%. Retained fraction = 0.88 × 0.75 = 0.66. Therefore MP = 2772 ÷ 0.66 = ₹4,200. Multiplying ₹4,200 by both retained fractions reproduces ₹2,772.

Common mistakes

Do not add successive discount rates. Do not apply the second rate to MP. When comparing offers, use the same MP; a larger equivalent discount means a lower SP only for that common base.

Practice questions

  1. MP is ₹1,750. Find SP after successive discounts of 8% and 5%.
  2. For the same MP, compare discounts of 25% then 16% with one discount of 35%. Which gives the lower SP?
  3. SP is ₹2,376 after discounts of 10% and 12%. Find MP.
  4. MP is ₹2,400. After a 25% discount and a second discount, SP is ₹1,476. Find the second discount rate.

Worked answers

  1. First price = 1750 × 0.92 = ₹1,610. Final SP = 1610 × 0.95 = ₹1,529.50.
  2. Retained fraction = 0.75 × 0.84 = 0.63, giving 37% discount. The successive offer is cheaper: its discount is 2 percentage points larger, saving an additional 2% of MP.
  3. MP = 2376 ÷ (0.90 × 0.88) = 2376 ÷ 0.792 = ₹3,000.
  4. After the first discount, price = 2400 × 0.75 = ₹1,800. The next reduction is 1800 - 1476 = ₹324, so its rate = (324 ÷ 1800) × 100 = 18%.
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