Pariksha SaathiSuccessive Discount Solved Questions
When two or more discounts are applied one after another, they do NOT simply add up — each discount is calculated on the price that’s already been reduced by the one before it, so successive discounts are always worth less to the shopper than their sum suggests. This page covers that core idea, combining three discounts two at a time, and reverse problems (given the final price, find one of the discounts), with several fully worked examples.
Written by Mahesh Srivastava, Founder of Pariksha Saathi. Published: 2026-10-05. Editorial standards & corrections
The concept, quickly
- Apply each discount one at a time, in order, to whatever price is left after the previous discount — never add the percentages together and apply them once.
- The single discount equivalent to two successive discounts d1% and d2% is d1 + d2 − (d1 × d2)/100 — always a little LESS than the simple sum d1 + d2, because the second discount is applied to an already-reduced price.
- Successive discounts almost always leave the buyer paying MORE than a single discount equal to their sum would — "25% and 10% off" is a smaller saving than a flat "35% off," even though 25 + 10 = 35.
- Two successive 50% discounts do NOT make an item free — the second 50% is off the already-halved price, so the combined effective discount is 75%, not 100%.
- For THREE successive discounts, combine them two at a time using the same formula repeatedly: first combine d1 and d2 into one equivalent discount, then combine that result with d3 — this is easier than tracking three percentages in your head at once.
- Reverse problems (given the final price, or the overall equivalent discount, find one individual discount) just run the multiplication backwards: divide the overall multiplier by the known discount's multiplier to isolate the unknown one.
Worked examples
1. An item with MRP ₹2000 is sold after two successive discounts of 20% and 10%. Find the final selling price.
Solution
- After the first discount: ₹2000 × (100 − 20)/100 = ₹1600.
- After the second discount: ₹1600 × (100 − 10)/100 = ₹1440.
Answer: ₹1440
2. Using the same item (MRP ₹2000, successive discounts of 20% and 10%), find the single discount percentage that would produce the same final price.
Solution
- Single equivalent discount = d1 + d2 − (d1 × d2)/100 = 20 + 10 − (20 × 10)/100 = 30 − 2 = 28%.
- Check: ₹2000 × (100 − 28)/100 = ₹1440, which matches the final price from successive discounts.
Answer: 28%
3. An item with MRP ₹4000 is sold after three successive discounts of 10%, 20% and 25%. Find the final selling price.
Solution
- After 10%: ₹4000 × 0.90 = ₹3600.
- After 20%: ₹3600 × 0.80 = ₹2880.
- After 25%: ₹2880 × 0.75 = ₹2160.
Answer: ₹2160
4. A shopkeeper offers two successive discounts of 25% and 10% on an item with MRP ₹4000, instead of a single flat discount of 30%. Which option is cheaper for the customer, and by how much?
Solution
- Successive discounts: ₹4000 × 0.75 × 0.90 = ₹2700.
- Single 30% discount: ₹4000 × 0.70 = ₹2800.
- The successive discounts give a lower price (₹2700 vs ₹2800) — cheaper for the customer by ₹100.
Answer: Successive discounts, by ₹100
5. Find the single discount equivalent to three successive discounts of 10%, 20% and 25%, by combining two at a time.
Solution
- Combine 10% and 20% first: 10 + 20 − (10×20)/100 = 30 − 2 = 28%.
- Now combine that 28% with the remaining 25%: 28 + 25 − (28×25)/100 = 53 − 7 = 46%.
- Check using multipliers: 0.90 × 0.80 × 0.75 = 0.54, which corresponds to a 46% discount (1 − 0.54 = 0.46). Matches.
Answer: 46%
6. An item with MRP ₹5,000 is sold after two successive discounts, the first being 20%, for a final price of ₹3,400. Find the second discount percentage.
Solution
- After the first discount: 5,000 × 0.80 = ₹4,000.
- Let the second discount be d2: 4,000 × (1 − d2/100) = 3,400, so (1 − d2/100) = 3,400/4,000 = 0.85.
- d2 = 15%.
Answer: 15%
7. Two successive discounts are applied to an item, the first being 10%. If the overall equivalent single discount is 28%, find the second discount.
Solution
- Overall multiplier = 1 − 0.28 = 0.72. First discount's multiplier = 1 − 0.10 = 0.90.
- Second discount's multiplier = 0.72 / 0.90 = 0.80, so the second discount = 20%.
Answer: 20%
Try these yourself
Apply each discount one at a time, then check against the answer.
1. An item with MRP ₹1500 is sold after two successive discounts of 10% and 20%. Find the final selling price.
Answer: ₹1080
2. Find the single discount percentage equivalent to two successive discounts of 15% each.
Answer: 27.75%
3. An item with MRP ₹1000 is sold after two successive discounts of 50% each. Find the final selling price (it is NOT free).
Answer: ₹250
4. Find the single discount equivalent to three successive discounts of 10%, 10% and 10%, by combining two at a time.
Answer: 27.1%. Combine first two: 19%; combine 19% with 10%: 19+10−1.9 = 27.1%. Check: 0.9³ = 0.729.
5. An item with MRP ₹3,000 is sold after two successive discounts, the first being 25%, for a final price of ₹1,800. Find the second discount.
Answer: 20%. After first discount: ₹2,250; 2,250×(1−d2/100)=1,800 gives d2=20.
6. Two successive discounts are applied, the first being 20%. If the overall equivalent discount is 28%, find the second discount.
Answer: 10%. Overall multiplier 0.72 ÷ first multiplier 0.8 = 0.9.
7. An item with MRP ₹6,000 is sold after successive discounts of 15% and 20%. Find the final selling price.
Answer: ₹4,080. 6,000 × 0.85 × 0.8.
All 14 practice questions on Successive Discount, with answers →
Where this comes up
Successive Discount is tested directly in the Quantitative/Numerical Aptitude section of exams Pariksha Saathi covers — SSC CGL, SSC MTS, SSC CHSL, IBPS PO, IBPS Clerk, SBI PO and SBI Clerk — as a standard extension of Profit and Loss questions.
Keep practising
Solve the questions above before opening the answers. Explain each step in your own words, then revisit any mistakes.
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