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Successive Discounts

Introduction

In many shopping problems, a product is sold at a discount, and then another discount is applied to the reduced price. These are called successive discounts.

Students often mistakenly add the discounts directly, but the correct method involves applying each discount one after the other or using the net discount formula.

Pattern: Successive Discounts

Pattern

When two discounts x% and y% are applied successively, the net discount is:

Net Discount = x + y - (x·y / 100)

Alternatively, you can apply them step by step: reduce the price by the first discount, then apply the second discount on the reduced price.

Step-by-Step Example

Question

A shirt is marked at ₹500. It is sold with successive discounts of 20% and 10%. Find the selling price.

Options:

  • A. ₹350
  • B. ₹360
  • C. ₹380
  • D. ₹400

Solution

  1. Step 1: Apply the first discount

    Apply the first discount of 20% → Price after 1st discount = 500 × (80/100) = ₹400.

  2. Step 2: Apply the second discount

    Apply the second discount of 10% → Price after 2nd discount = 400 × (90/100) = ₹360.

  3. Step 3: Use the formula method

    Net discount = 20 + 10 - (20×10)/100 = 28%. Final price = 500 × (72/100) = ₹360 (same result).

  4. Final Answer:

    ₹360 → Option B

  5. Quick Check:

    Net discount 28% → 28% of 500 = 140 → 500 - 140 = 360 ✅

Quick Variations

  • 1. 3 successive discounts x%, y%, z% → Net = x + y + z - (xy + yz + zx)/100 + (xyz)/10000
  • 2. Successive discounts work like successive percentage decrease problems.

Trick to Always Use

  • Use the formula x + y - (x·y / 100) to save time in exams.
  • For 3 or more discounts, extend the formula carefully.
  • Always confirm with a quick check using the net discount percentage.

Summary

Summary

  • Successive discounts must be applied sequentially, not added directly.
  • The second discount always applies on the reduced price.
  • Use the formula Net Discount = x + y - (x·y/100) to simplify calculations.
  • Cross-check using the net percentage to avoid mistakes.

Example to remember:
₹500 with discounts of 20% and 10% → Final Price = ₹360.

Practice

(1/5)
1. A shirt is marked at ₹600. Successive discounts of 10% and 20% are given. Find the selling price.
easy
A. ₹420
B. ₹432
C. ₹450
D. ₹480

Solution

  1. Step 1: Apply first discount

    Apply first discount 10% → Price = 600 × (90/100) = ₹540.

  2. Step 2: Apply second discount

    Apply second discount 20% on reduced price → Price = 540 × (80/100) = ₹432.

  3. Step 3: Verify with formula

    Net discount = 10 + 20 - (10×20)/100 = 28%. Selling price = 600 × (72/100) = ₹432.

  4. Final Answer:

    ₹432 → Option B

  5. Quick Check:

    28% of 600 = 168 → 600 - 168 = 432 ✅
Hint: Use net discount formula or multiply complements: 0.9 × 0.8 = 0.72.
Common Mistakes: Adding discounts directly (10% + 20% = 30%) instead of using successive method.
2. A laptop is marked at ₹50,000. It is sold with successive discounts of 15% and 5%. Find the selling price.
easy
A. ₹39,750
B. ₹40,375
C. ₹40,500
D. ₹42,000

Solution

  1. Step 1: Apply first discount

    First discount 15% → Price = 50,000 × (85/100) = ₹42,500.

  2. Step 2: Apply second discount

    Second discount 5% → Price = 42,500 × (95/100) = ₹40,375.

  3. Step 3: Verify with formula

    Net discount = 15 + 5 - (15×5)/100 = 19.25%. Selling price = 50,000 × (80.75/100) = ₹40,375.

  4. Final Answer:

    ₹40,375 → Option B

  5. Quick Check:

    19.25% of 50,000 = 9,625 → 50,000 - 9,625 = 40,375 ✅
Hint: Apply second discount on the reduced price (not on original).
Common Mistakes: Applying both discounts on original price.
3. A watch is marked at ₹2,000. It is sold with successive discounts of 25% and 10%. Find the selling price.
medium
A. ₹1,200
B. ₹1,300
C. ₹1,350
D. ₹1,400

Solution

  1. Step 1: Apply first discount

    First discount 25% → Price = 2,000 × (75/100) = ₹1,500.

  2. Step 2: Apply second discount

    Second discount 10% → Price = 1,500 × (90/100) = ₹1,350.

  3. Step 3: Verify with formula

    Net discount = 25 + 10 - (25×10)/100 = 32.5%. Selling price = 2,000 × (67.5/100) = ₹1,350.

  4. Final Answer:

    ₹1,350 → Option C

  5. Quick Check:

    32.5% of 2,000 = 650 → 2,000 - 650 = 1,350 ✅
Hint: Multiply complements: 0.75 × 0.90 = 0.675.
Common Mistakes: Adding percentages directly instead of successive calculation.
4. A shopkeeper gives successive discounts of 20% and 15% on a marked price of ₹800. Find the selling price.
medium
A. ₹540
B. ₹560
C. ₹544
D. ₹552

Solution

  1. Step 1: Apply first discount

    First discount 20% → Price = 800 × (80/100) = ₹640.

  2. Step 2: Apply second discount

    Second discount 15% → Price = 640 × (85/100) = ₹544.

  3. Step 3: Verify with formula

    Net discount = 20 + 15 - (20×15)/100 = 32%. Selling price = 800 × (68/100) = ₹544.

  4. Final Answer:

    ₹544 → Option C

  5. Quick Check:

    32% of 800 = 256 → 800 - 256 = 544 ✅
Hint: Use net discount formula for quick verification.
Common Mistakes: Forgetting second discount applies to reduced price.
5. A camera is sold with successive discounts of 30% and 20% on a marked price of ₹10,000. Find the selling price.
medium
A. ₹5,400
B. ₹5,600
C. ₹5,800
D. ₹6,000

Solution

  1. Step 1: Apply first discount

    First discount 30% → Price = 10,000 × (70/100) = ₹7,000.

  2. Step 2: Apply second discount

    Second discount 20% → Price = 7,000 × (80/100) = ₹5,600.

  3. Step 3: Verify with formula

    Net discount = 30 + 20 - (30×20)/100 = 44%. Selling price = 10,000 × (56/100) = ₹5,600.

  4. Final Answer:

    ₹5,600 → Option B

  5. Quick Check:

    44% of 10,000 = 4,400 → 10,000 - 4,400 = 5,600 ✅
Hint: Compute complements product: 0.70 × 0.80 = 0.56.
Common Mistakes: Treating discounts as additive without correction term.

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