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Percentage Increase/Decrease

Introduction

A very common type of percentage problem is when a value is increased or decreased by a certain percentage. These questions appear often in exams and in real life (prices, population, salaries, etc.).

The method is simple: multiply the original value by (1 ± percentage/100).

Pattern: Percentage Increase/Decrease

Pattern

The key idea: New value = Original value × (1 + x/100) for increase.

New value = Original value × (1 - x/100) for decrease.

Translate the sentence into math and apply directly.

Step-by-Step Example

Question

The price of a shirt is ₹200. It is increased by 10%. Find the new price.

Solution

  1. Step 1: Write original value and % change.

    Sentence: Original price = 200, Increase = 10%.

  2. Step 2: Apply the formula.

    Sentence: New value = Original × (1 + x/100).
    Math: New price = 200 × (1 + 10/100)

  3. Step 3: Simplify.

    New price = 200 × (1 + 0.10) = 200 × 1.10 = 220

  4. Step 4: Final Answer.

    The new price of the shirt = ₹220

  5. Step 5: Quick Check.

    10% of 200 = 20 → 200 + 20 = 220 ✅

Question

The population of a town is 50,000. It decreases by 5%. Find the new population.

Solution

  1. Step 1: Write original value and % change.

    Sentence: Original population = 50,000, Decrease = 5%.

  2. Step 2: Apply the formula.

    Sentence: New value = Original × (1 - x/100).
    Math: New population = 50,000 × (1 - 5/100)

  3. Step 3: Simplify.

    New population = 50,000 × (1 - 0.05) = 50,000 × 0.95 = 47,500

  4. Step 4: Final Answer.

    The new population = 47,500

  5. Step 5: Quick Check.

    5% of 50,000 = 2,500 → 50,000 - 2,500 = 47,500 ✅

Quick Variations

If price increases by x% and then decreases by y%: Apply separately step by step, or use the successive % change formula later.

If asked for % change in reverse: Eg: “Price increased from 200 to 220 → Increase = (20 ÷ 200) × 100 = 10%”.

Trick to Always Use

  • Step 1: Write original value and % change.
  • Step 2: Use multiplier: (1 + x/100) for increase, (1 - x/100) for decrease.
  • Step 3: Multiply and simplify.
  • Step 4: Quick check using direct % of original.

Summary

Summary

The Percentage Increase/Decrease pattern is solved by:

  • Increase: New = Original × (1 + x/100)
  • Decrease: New = Original × (1 - x/100)
  • Quick check: Add/subtract x% of the original.

This method makes increase/decrease questions very quick and accurate.

Practice

(1/5)
1. The price of a shirt is ₹200. It is increased by 10%. What is the new price?
easy
A. ₹220
B. ₹215
C. ₹210
D. ₹225

Solution

  1. Step 1: Identify original price and increase

    Original price = 200, Increase = 10%.

  2. Step 2: Apply increase multiplier

    New price = 200 × (1 + 10/100).

  3. Step 3: Compute the value

    200 × 1.10 = 220.

  4. Final Answer:

    ₹220 → Option A

  5. Quick Check:

    10% of 200 = 20 → 200 + 20 = 220 ✅
Hint: Increase → multiply by (1 + x/100).
Common Mistakes: Adding a fixed amount (e.g., ₹10) instead of the percentage.
2. The population of a town is 50,000. It decreases by 5%. What is the new population?
easy
A. 47,000
B. 47,500
C. 48,000
D. 47,250

Solution

  1. Step 1: Identify original value and decrease

    Original population = 50,000; Decrease = 5%.

  2. Step 2: Apply decrease multiplier

    New = 50000 × (1 - 5/100).

  3. Step 3: Compute the value

    50000 × 0.95 = 47,500.

  4. Final Answer:

    47,500 → Option B

  5. Quick Check:

    5% of 50000 = 2500 → 50000 - 2500 = 47500 ✅
Hint: Decrease → multiply by (1 - x/100).
Common Mistakes: Adding the percent instead of subtracting when decreasing.
3. The salary of a person is ₹40,000. It is increased by 15%. What is the new salary?
medium
A. ₹45,000
B. ₹46,500
C. ₹46,000
D. ₹44,000

Solution

  1. Step 1: Identify salary and increase

    Salary = 40,000; Increase = 15%.

  2. Step 2: Apply increase multiplier

    New salary = 40000 × (1 + 15/100).

  3. Step 3: Compute the value

    40000 × 1.15 = 46,000.

  4. Final Answer:

    ₹46,000 → Option C

  5. Quick Check:

    15% of 40000 = 6000 → 40000 + 6000 = 46000 ✅
Hint: Multiply by 1.15 for +15%.
Common Mistakes: Using 1.5 instead of 1.15 or miscalculating 15% of the base.
4. A company’s revenue was ₹800,000 last year. This year it decreased by 12%. What is the new revenue?
medium
A. ₹704,000
B. ₹720,000
C. ₹680,000
D. ₹700,000

Solution

  1. Step 1: Identify original revenue and decrease

    Original revenue = 800,000; Decrease = 12%.

  2. Step 2: Apply decrease multiplier

    New = 800000 × (1 - 12/100).

  3. Step 3: Compute the value

    800000 × 0.88 = 704,000.

  4. Final Answer:

    ₹704,000 → Option A

  5. Quick Check:

    12% of 800,000 = 96,000 → 800,000 - 96,000 = 704,000 ✅
Hint: Decrease by 12% → multiply by 0.88.
Common Mistakes: Forgetting to convert percent to decimal before multiplying.
5. The price of an article is ₹5,000. It is first increased by 20% and then decreased by 10%. What is the final price?
medium
A. ₹5,500
B. ₹5,600
C. ₹5,700
D. ₹5,400

Solution

  1. Step 1: Identify original price

    Original price = 5,000.

  2. Step 2: Apply the 20% increase

    After +20%: 5000 × 1.20 = 6,000.

  3. Step 3: Apply the 10% decrease

    Then -10%: 6000 × 0.90 = 5,400.

  4. Final Answer:

    ₹5,400 → Option D

  5. Quick Check:

    Net % = 20 - 10 - (20×10)/100 = 8% → 5000 × 1.08 = 5400 ✅
Hint: Apply multipliers stepwise: ×1.20 then ×0.90.
Common Mistakes: Assuming net change = 20% - 10% = 10% (incorrect).

Mock Test

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