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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 6 of 100 | Quantitative Aptitude / Number System / संख्या पद्धति

Remainders in Elementary Number Problems

Learning outcome

Express division with a valid remainder and simplify elementary remainder problems involving sums, differences and products.

Concepts and reasons

Here the dividend N is a nonnegative (zero or positive) integer and divisor d is a positive integer. Write N = d × q + r: q is the whole-number quotient and r is the remainder, with 0 ≤ r < d.

The remainder is what remains after removing the maximum number of complete groups of size d. It cannot equal or exceed d: that would allow another complete group. A remainder of 0 means exact divisibility. When N < d, the quotient is 0 and remainder is N.

For the same divisor, replace numbers by their remainders when finding the remainder of a sum, difference or product. Then reduce the result again into the allowed range.

Why does replacement work? Write A = d × a + r and B = d × b + s. Adding or subtracting leaves r + s or r - s apart from multiples of d. Multiplication leaves r × s apart from multiples of d. Those complete multiples contribute no remainder.

A negative intermediate difference is not the final remainder under our convention. Add d to put it into the allowed range.

Worked examples

Example 1. Divide 157 by 12. Since 12 × 13 = 156, write 157 = 12 × 13 + 1. The remainder is 1, which lies between 0 and 11.

Example 2. A and B leave remainders 5 and 4 when divided by 7. Their sum leaves the remainder of 5 + 4 = 9, namely 2. Their product leaves the remainder of 5 × 4 = 20, namely 6.

Example 3. Suppose A ≥ B and their remainders upon division by 7 are 2 and 5. For A - B, the provisional difference is -3. Add 7: the required remainder is 4, not -3.

Common traps

A remainder is not a quotient or a decimal part. Divide oversized results again to find the remainder. Do not combine remainders obtained using different divisors, and never divide by zero.

Practice questions

  1. Find the remainder when 246 is divided by 11.
  2. N leaves remainder 5 upon division by 8. Find the remainder of 3N + 7 upon division by 8.
  3. N leaves remainder 3 upon division by 5. Find the remainder of (N + 1)(N + 2) upon division by 5.

Answers and explanations

  1. Since 11 × 22 = 242, write 246 = 11 × 22 + 4. The remainder is 4, smaller than 11.
  2. Replace N by 5: 3 × 5 + 7 = 22. Since 22 = 8 × 2 + 6, the remainder is 6.
  3. N + 1 leaves 4, while N + 2 leaves 0. Their product therefore leaves 4 × 0 = 0.
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