SSC CGL Maths Arithmetic Percentage: Concepts, Tricks, Formulas and Questions

Percentage is one of the most important topics in SSC CGL Quantitative Aptitude. Questions from percentage are frequently asked directly and also form the foundation for Profit and Loss, SI-CI, Ratio, Alligation, and Data Interpretation. Understanding percentage concepts such as pass-fail calculations, Venn diagrams, fresh and dry fruit problems, and population growth can help candidates solve arithmetic questions quickly and accurately. 

Problem 1: Calculating Passing Percentage

Scenario: A student scores 20% marks and fails by 10 marks. Another student scores 35% marks and got 20 marks more than the passing score.

Objective: Find the minimum passing percentage.

Solution:

1. Equate the expressions for passing marks:
   (20% of Total Marks) + 10 = (35% of Total Marks) - 20

2. Rearrange the equation:
   35% - 20% = 10 + 20
   15% = 30 marks

3. Calculate the value of 1%:
   1% = 2 marks

4. Determine Total Marks:
   Total Marks (100%) = 200

5. Calculate Passing Marks:
   Using the first student's score: (20% of 200) + 10 = 40 + 10 = 50 marks.

6. Calculate Passing Percentage:
   Passing Percentage = (50 / 200) * 100% = 25%.
   The minimum passing percentage is 25%.

Problem 2: Calculating Maximum Marks and Minimum Passing Marks

Scenario: A candidate scores 36% marks and fails by 55 marks. Another candidate scores 71% marks and got 15 marks more than the passing score.

Objective: Find the Maximum Marks and the Minimum Passing Marks.

Solution:

1. Equate passing marks:
   (36% of Total Marks) + 55 = (71% of Total Marks) - 15

2. Rearrange:
   71% - 36% = 55 + 15
   35% = 70 marks

3. Calculate the value of 1%:
   1% = 2 marks

4. Determine Maximum Marks:
   Maximum Marks (100%) = 200.

5. Calculate Minimum Passing Marks:
   Using the first candidate: (36% of 200) + 55 = 72 + 55 = 127 marks.

Venn Diagram Concept

The Venn Diagram is a crucial concept for analyzing overlapping sets, particularly useful in percentage problems involving multiple conditions.

Methodology for 3-Set Venn Diagrams:

1. Draw three overlapping circles representing the subjects.

2. Fill the common portion of all three subjects first (innermost intersection).

3. Fill the intersection of two subjects by subtracting the triple intersection value.

4. Fill the "only" portion of each individual subject by subtracting all relevant intersection values.

5. Sum all these unique region percentages to find the total percentage (e.g., total failures).

6. Subtract this total from 100% to find the complement (e.g., total passes).

Problem 1: Calculating Total Pass Percentage

Scenario: In an exam, 2% of students fail in all three subjects (Maths, English, Hindi).

• Individual Subject Failures: M: 20%, E: 15%, H: 25%.

• Intersection Failures: M&H: 15%, E&H: 10%, M&E: 5%.
  Objective: Calculate the total percentage of students who passed.

Solution:

1. Fail in M, E, & H (All three): 2%

2. Fail in M & H only: 15% - 2% = 13%

3. Fail in E & H only: 10% - 2% = 8%

4. Fail in M & E only: 5% - 2% = 3%

5. Fail in Only H: 25% - (13% + 2% + 8%) = 25% - 23% = 2%

6. Fail in Only E: 15% - (3% + 2% + 8%) = 15% - 13% = 2%

7. Fail in Only M: 20% - (13% + 2% + 3%) = 20% - 18% = 2%

8. Total Percentage of Students Who Failed:
   2% (M) + 2% (E) + 2% (H) + 13% (M&H) + 8% (E&H) + 3% (M&E) + 2% (All three) = 32%.

9. Total Percentage of Students Who Passed:
   100% - 32% = 68%.

Problem 2: Calculating Total Pass Percentage (Reinforcement)

Scenario: 10% of students fail in all three subjects.

• Individual Subject Failures: M: 30%, E: 25%, H: 40%.

• Intersection Failures: M&H: 25%, E&H: 20%, M&E: 15%.
  Objective: Calculate the total percentage of students who passed.

Solution:

1. Fail in M, E, & H (All three): 10%

2. Fail in M & H only: 25% - 10% = 15%

3. Fail in E & H only: 20% - 10% = 10%

4. Fail in M & E only: 15% - 10% = 5%

5. Fail in Only H: 40% - (15% + 10% + 10%) = 40% - 35% = 5%

6. Fail in Only E: 25% - (5% + 10% + 10%) = 25% - 25% = 0%

7. Fail in Only M: 30% - (15% + 10% + 5%) = 30% - 30% = 0%

8. Total Percentage of Students Who Failed:
   0% (M) + 0% (E) + 5% (H) + 15% (M&H) + 10% (E&H) + 5% (M&E) + 10% (All three) = 45%.

9. Total Percentage of Students Who Passed:
   100% - 45% = 55%.

(Memory Tip: Always be careful to provide the answer requested by the question (e.g., passing percentage) rather than stopping at the intermediate calculation (e.g., failure percentage), as examiners often include both options to trick students.)

Miscellaneous Percentage Problems

Problem 1: Algebraic Manipulation with Percentages

Scenario: (x - y) * 60% = (x + y) * 45%. It is stated that y is K% of x.

Objective: Find 21% of K.

Solution:

1. Simplify the equation:
   (x - y) * 60 = (x + y) * 45
   Divide by 15: (x - y) * 4 = (x + y) * 3

2. Expand and solve for x and y:
   4x - 4y = 3x + 3y
   x = 7y

3. Express y as a fraction of x:
   y = (1/7)x

4. Relate to K%:
   Given y = K% of x, so y = (K/100) * x.
   Comparing, K/100 = 1/7, which means K = 100/7.

5. Calculate 21% of K:
   21% of (100/7) = (21/100) * (100/7) = 21/7 = 3.

Problem 2: Complex Percentage Calculation

Scenario: Evaluate (25% of 50% of 30% of 150) / (40% of 2250).

Solution:

1. Rewrite the expression, converting 50% to its fractional equivalent (1/2):
   (25% * (1/2) * 30% * 150) / (40% * 2250)
   To simplify, retain one '%' in the numerator and cancel another.
   (25% * (1/2) * (30/100) * 150) / ((40/100) * 2250)
   Cancel one 100 from the numerator and one from the denominator:
   (25% * 1/2 * 30 * 150) / (40 * 2250)

2. Simplify terms:
   (25% * 15 * 150) / (40 * 2250) (after cancelling a zero from 30 and 40)
   (25% * 2250) / (40 * 2250) (since 15 * 150 = 2250)

3. Cancel 2250:
   25% / 40

4. Final Calculation:
   25 / 40 = 0.625.
   Therefore, the result is 0.625%.

Fresh Fruit / Dry Fruit Problems

Concept: Constant Pulp

In fresh fruit and dry fruit problems, the pulp (the solid, non-water content) remains constant regardless of the water content.

(Memory Tip: Always solve these problems by equating the quantity of pulp, not the water content.)

Problem 1: Calculating Dry Fruit Yield

Scenario: 100 kg of fresh fruit contains 60% water. How much dry fruit can be obtained if dry fruit contains 20% water?

Solution:

1. Pulp in Fresh Fruit:
   60% water means 40% pulp.
   Quantity of pulp = 100 kg * 40% = 40 kg.

2. Pulp in Dry Fruit:
   20% water means 80% pulp.

3. Equate Pulp Quantities:
   Let 'x' be the amount of dry fruit.
   40 kg = x kg * 80%
   40 = x * (80/100)
   x = 40 * (100/80) = 50 kg.
   Thus, 50 kg of dry fruit can be obtained.

Problem 2: Dry Fruit Yield (Practice)

Scenario: 180 kg of fresh fruit contains 50% water. How much dry fruit can be obtained if dry fruit contains 10% water?

Solution:

1. Pulp in Fresh Fruit:
   50% water means 50% pulp.
   Quantity of pulp = 180 kg * 50% = 90 kg.

2. Pulp in Dry Fruit:
   10% water means 90% pulp.

3. Equate Pulp Quantities:
   Let 'x' be the amount of dry fruit.
   90 kg = x kg * 90%
   90 = x * (90/100)
   x = 90 * (100/90) = 100 kg.
   Thus, 100 kg of dry fruit can be obtained.

Fraction Problems (Numerator and Denominator Changes)

Concept: Percentage Change in Fractions

When the numerator or denominator of a fraction changes by a percentage, the initial value is considered 100%. An increase of N% means the new value is (100 + N)% of the original, while a decrease means (100 - N)%. These questions frequently appear in competitive examinations.

Problem 1: Finding the Original Fraction

Scenario: The numerator of a fraction is increased by 60%, and the denominator is increased by 40%. The resulting fraction is 16/63.

Objective: Find the original fraction.

Solution:

1. Let the original fraction be x/y.

2. New Numerator = x * (100% + 60%) = x * 160%.

3. New Denominator = y * (100% + 40%) = y * 140%.

4. Set up the equation:
   (x * 160%) / (y * 140%) = 16/63

5. Simplify:
   (x/y) * (160/140) = 16/63
   (x/y) * (16/14) = 16/63
   (x/y) * (8/7) = 16/63

6. Solve for x/y:
   x/y = (16/63) * (7/8) = (2/9) * 1 = 2/9.
   The original fraction is 2/9.

Problem 2: Finding the Original Fraction (Practice)

Scenario: The numerator of a fraction is increased by 50%, and the denominator is decreased by 28%. The resulting fraction is 25/36.

Objective: Find the original fraction.

Solution:

1. Let the original fraction be x/y.

2. New Numerator = x * (100% + 50%) = x * 150%.

3. New Denominator = y * (100% - 28%) = y * 72%.

4. Set up the equation:
   (x * 150%) / (y * 72%) = 25/36

5. Simplify:
   (x/y) * (150/72) = 25/36

6. Solve for x/y:
   x/y = (25/36) * (72/150) = 25 * (2/150) = 50/150 = 1/3.
   The original fraction is 1/3.

Problem 3: Finding the Original Fraction (Advanced Percentage Increase)

Scenario: The numerator of a fraction is increased by 200%, and the denominator is increased by 350%. The resulting fraction is 5/12.

Objective: Find the original fraction.

Solution:

1. Let the original fraction be x/y.

2. New Numerator: An increase of 200% means (100% + 200%) = 300% of the original. So, x * 300%.
   (Memory Tip: An increase of 200% means the new value is (100% + 200%) = 300% of the original. Similarly, a 350% increase means (100% + 350%) = 450%.)

3. New Denominator: An increase of 350% means (100% + 350%) = 450% of the original. So, y * 450%.

4. Set up the equation:
   (x * 300%) / (y * 450%) = 5/12

5. Simplify:
   (x/y) * (300/450) = 5/12
   (x/y) * (2/3) = 5/12

6. Solve for x/y:
   x/y = (5/12) * (3/2) = 15/24 = 5/8.
   The original fraction is 5/8.

Time, Savings, and Purchasing Power

Concept: Inverse Proportionality of Time and Rate

For a fixed purchase price, if the time available decreases, the rate of saving or effort must increase proportionally to meet the deadline.

(Memory Tip: To solve such problems efficiently, assume a total value (e.g., car price, total work) that is the Least Common Multiple (LCM) of the given time periods.)

Problem: Increasing Savings to Reduce Purchase Time

Scenario: An individual plans to save for a car within 1 year (12 months). If they wish to purchase the same car in 9 months, by what percentage must they increase their monthly savings?

Solution:

1. Time Periods: 12 months and 9 months.

2. Assume Car Price: LCM of 12 and 9 is 36. Let the car's price be 36 units.

3. Original Monthly Savings (for 12 months):
   36 units / 12 months = 3 units/month.

4. New Monthly Savings (for 9 months):
   36 units / 9 months = 4 units/month.

5. Increase in Savings:
   4 units - 3 units = 1 unit.

6. Percentage Increase in Savings:
   (1 / 3) * 100% = 33 1/3%.
   The individual must increase their monthly savings by 33 1/3%.

Population Growth (Alligation Method)

Concept: Alligation Method for Population Changes

The Alligation Method is effective when two groups (e.g., males and females) change by different percentages, resulting in an overall percentage change for the total group. It helps determine the ratio of the initial quantities of the two groups.

Problem: Calculating the Male Population after Growth

Scenario: A town has a total population of 8000. The male population increased by 10%, and the female population increased by 8%. The total population increased by 9%.

Objective: Find the number of males in the town.

Solution using Alligation Method:

1. Identify Components:

• Percentage increase in Males = 10%

• Percentage increase in Females = 8%

• Overall percentage increase in Total Population = 9%

2. Apply Alligation Rule:
   Males (10%) Females (8%) \ / Overall (9%) / \ (9 - 8)% (10 - 9)% 1% 1%

3. Determine Ratio of Males to Females:
   The ratio of (Males : Females) is 1 : 1.

4. Calculate Number of Males:
   Since the ratio is 1:1, the male and female populations are equal.
   Number of Males = 8000 / 2 = 4000.