Every shopkeeper writes a price tag on an article — that number is the Marked Price (MP), also called the List Price. When the shopkeeper reduces that price to attract buyers, the amount cut off is the Discount. What the buyer actually pays is the Selling Price (SP).
Think of it like a cinema ticket. The booking window shows ₹250 (marked price). The app gives you ₹50 off (discount). You pay ₹200 (selling price). The discount is always calculated on the marked price — not on the cost price, not on the selling price. This is the single most important thing to anchor in your mind before touching any discount question.
The three core relationships:
where d is the discount percentage.
A successive discount means the shopkeeper applies one discount, then applies another discount on the already-reduced price — not on the original marked price. This is where most candidates lose marks: they add the two discount percentages directly. You cannot. If you get 30% off and then 70% off, you do NOT get 100% off. You will see exactly why in the Deep Dive.
The analogy that sticks: successive discounts are like two tax slabs applied one after another — each slab bites into a smaller base, so the total effect is always less than the simple sum.
This is straightforward. Given MP and discount %:
Given SP and discount %, find MP:
Example: SP = ₹255, discount = 15%. Then MP = 255 × 100 / 85 = ₹300.
When two discounts d₁ and d₂ are given successively on a marked price MP:
Step 1: After first discount: SP₁ = MP × (100 - d₁)/100
Step 2: After second discount on SP₁: SP₂ = SP₁ × (100 - d₂)/100
Combined: SP = MP × (100 - d₁)/100 × (100 - d₂)/100
For three discounts d₁, d₂, d₃:
Why you can't just add: With discounts 30% and 70% on ₹100:
For two discounts d₁ and d₂, the single equivalent discount is:
This is the shortcut formula you must memorise. It comes directly from algebra:
Example: Discounts of 10% and 15%:
Net = 10 + 15 - (10×15)/100 = 25 - 1.5 = 23.5%
For three or more discounts, use the multiplier method directly — don't try to extend the formula in your head. Calculate on 100 as the base.
Example: Discounts 20%, 15%, 10% on MP = 100:
This is a two-step problem type that appears in SSC MTS. The structure is always:
x% above cost price → find MP in terms of CPd% on MP → find SPStandard setup: Let CP = ₹100, markup = 80% → MP = ₹180. Discount = 30% on MP → SP = 180 × 0.70 = ₹126. Since CP = 100 and SP = 126, profit = 26%.
Here's the trick that lets you skip two calculation steps in combined problems: always let CP = 100. The arithmetic stays clean and the profit/loss % reads off directly as SP − 100.
| Term | Meaning | |------|---------| | Marked Price (MP) | Price written on the tag | | List Price | Same as MP | | Discount | MP − SP (always on MP) | | Trade Discount | Discount given to bulk buyers / retailers | | Selling Price (SP) | Amount actually paid |
When you see exactly two successive discounts, don't multiply two fractions — use this direct formula. For d₁ = 20% and d₂ = 10%: Net = 20 + 10 − (20×10)/100 = 30 − 2 = 28%. Standard method (multiply 0.80 × 0.90 = 0.72, then 1−0.72 = 28%) takes ~40 seconds. This formula: ~10 seconds. The subtracted term (d₁×d₂/100) is always the "correction" that stops you from double-counting. Remember: the formula gives you less than d₁+d₂, always.
For three or more discounts, write the survival multipliers as fractions and multiply straight across on a base of 100. Discounts 20%, 15%, 10% → survival fractions 80/100, 85/100, 90/100. Multiply: 80 × 85 = 6800; 6800 × 90 = 612,000; divide by 10,000 = 61.2. Net discount = 38.8%. This chain method takes 4 steps. Trying to apply the two-discount formula repeatedly (first to 20% and 15%, then apply result to 10%) takes 6 steps and risks rounding error. Use the chain.
When SP and discount % are given and you need MP, never set up an equation. Instead, recognise that SP = (100 − d)% of MP. So MP = SP × 100 / (100 − d). SP = ₹255, discount = 15%: MP = 255 × 100 / 85 = 255 × 20/17 = 300. Key: reduce 100/(100−d) to its simplest fraction first — it avoids large multiplication. Standard long-division approach: ~50 seconds. Fraction simplification: ~15 seconds.
Any question mixing markup % on CP and discount % on MP: set CP = ₹100 immediately. This collapses the problem — you never need variable algebra. Markup 80% → MP = 180. Discount 30% → SP = 180 × 0.7 = 126. Profit % = 26% (read directly as SP − 100). Without the anchor you'd write equations for two unknowns. With it: 3 arithmetic steps, under 25 seconds.
In an MCQ with two discounts that add to 100% (like 30% + 70%), option "100%" will always be present as a trap. Eliminate it immediately — successive discounts can never give 100% net discount unless one of them is literally 100%. The real answer is always less than the sum. This single insight eliminates one wrong option in under 3 seconds, improving your odds from 25% to 33% before any calculation.
Read the question and identify which variable is unknown:
Unknown = SP?
→ Use SP = MP × (100−d)/100. For multiple discounts, multiply all survival multipliers on MP.
Unknown = MP?
→ Use MP = SP × 100 / (100−d). Simplify the fraction first.
Unknown = Net Discount % (two discounts given)?
→ Use the formula d₁ + d₂ − (d₁×d₂)/100 directly. Done in one line.
Unknown = Net Discount % (three+ discounts given)? → Assume MP = 100, apply each discount sequentially, subtract final value from 100.
Combined markup + discount problem? → Set CP = 100, find MP, apply discount, read off profit/loss as SP − 100.
Decision check before solving: Is discount calculated on MP or CP? If the question says "discount on marked price" — standard. If it says "discount on cost price" — that's actually equivalent to reduced profit, recalculate accordingly (rare in SSC MTS but possible).
Why this question: The classic trap — two discounts summing to 100%. Tests whether you apply successive discount logic or fall for the additive shortcut.
Solving path: Assume MP = ₹100. After 30% discount: SP₁ = ₹70. After 70% discount on ₹70: SP₂ = 70 × 30/100 = ₹21. Net discount = (100 − 21)/100 × 100 = 79%. Option "100%" is the trap — eliminate first.
Why this question: Three successive discounts with a non-round marked price. Tests your ability to chain multipliers without error.
Solving path: SP = 9000 × (70/100) × (80/100) × (85/100). Step by step: 9000 × 0.7 = 6300; 6300 × 0.8 = 5040; 5040 × 0.85 = 4284. Answer: ₹4,284.
Why this question: Asks for single equivalent discount for two successive discounts — direct formula application.
Solving path: Use d₁ + d₂ − (d₁×d₂)/100 = 10 + 15 − (150/100) = 25 − 1.5 = 23.5%. This is a one-line question if you have the formula memorised.
Why this question: Find SP from MP with a single discount. Foundation-level — get this in under 20 seconds.
Solving path: SP = 3040 × (80/100) = 3040 × 0.8. Multiply: 3040 × 8 = 24320, shift decimal → ₹2,432.
Why this question: Reverse calculation — SP and discount % are given, find MP. Tests whether you can work backwards.
Solving path: MP × (85/100) = 255 → MP = 255 × 100/85 = 255 × 20/17 = 15 × 20 = ₹300. The key step: simplify 255/85 = 3 first, then × 100 = 300.
Why this question: Combined markup + discount → find profit %. The most exam-relevant variant of discount problems.
Solving path: Set CP = 100. Markup 80% → MP = 180. Discount 30% on 180 → SP = 180 × 0.7 = 126. Profit = 126 − 100 = 26%. Anchor trick works perfectly here.
Why this question: Three successive discounts — equivalent discount %. Tests the multiplier chain method.
Solving path: Base 100. After 20%: 80. After 15% on 80: 80 × 0.85 = 68. After 10% on 68: 68 × 0.9 = 61.2. Net discount = 100 − 61.2 = 38.8%.
Why this question: Three successive discounts — find actual SP. Same chain method, just don't convert to equivalent discount first.
Solving path: SP = 5000 × 0.8 × 0.85 × 0.9. Step by step: 5000 × 0.8 = 4000; 4000 × 0.85 = 3400; 3400 × 0.9 = ₹3,060.
Adding successive discounts directly. Discounts of 30% and 70% do not give 100% net discount. Every successive discount applies to the already-reduced price, not the original. The net is always less than the arithmetic sum.
Calculating discount on cost price instead of marked price. Discount is always on MP. Profit/markup is on CP. Mixing these two bases is the single most common conceptual error in combined problems.
Using the two-discount formula for three discounts. The formula d₁ + d₂ − (d₁×d₂)/100 works only for exactly two discounts. For three, use the multiplier chain on a base of 100.
Forgetting to simplify fractions in reverse-MP problems. When finding MP from SP, the calculation SP × 100/(100−d) often simplifies cleanly. Missing the simplification leads to messy arithmetic and arithmetic errors. Always reduce first.
Confusing "discount on MP" with "discount on SP". Some poorly worded problems say "10% discount" without specifying the base. In all standard SSC problems, discount is on MP unless explicitly stated otherwise.
In markup + discount problems, computing markup on SP instead of CP. "Marked 80% above cost" means MP = CP + 80% of CP = 1.8 × CP. It does not mean MP = 0.8 × SP or any other interpretation. Fix CP = 100 and the confusion disappears.