A shopkeeper marked an article at ₹5,000. The shopkeeper allows successive discounts of 20%,15% and 10%. The selling price of the article is:

Correct — A. Successive discounts apply to what is left after the previous one, so multiply the survivors instead of adding the rates.

After 20%: 5,000 × 0.80 = ₹4,000
After 15%: 4,000 × 0.85 = ₹3,400
After 10%: 3,400 × 0.90 = ₹3,060

Same thing in one line: 5,000 × 0.8 × 0.85 × 0.9 = 5,000 × 0.612 = ₹3,060 — option (a).

Why the others are wrong

• (b)
  ₹2,750 — ₹2,750 is ₹5,000 less 45%, the three rates added. Chained discounts multiply: 0.8 × 0.85 × 0.9 = 0.612, so the real cut is 38.8%.
• (c)
  ₹3,000 — ₹3,000 is ₹4,000 less 25% — the last two rates folded into one and applied together. Taken one at a time they give 4,000 → 3,400 → 3,060.
• (d)
  ₹2,800 — ₹2,800 is 56% of the marked price, i.e. a 44% cut. The chain leaves 61.2% of ₹5,000, so no arrangement of these three rates reaches ₹2,800.

Successive discounts are charged on the running amount, not on the marked price, so they compose multiplicatively.

Rates a% and b% together leave (1 − a/100)(1 − b/100) of the price. For 20%, 15% and 10% that is 0.8 × 0.85 × 0.9 = 0.612, so the buyer pays 61.2% of ₹5,000.

The single discount equivalent to the chain is therefore 38.8% — always less than the arithmetic sum of 45%, because each later rate bites on a smaller base.

Key facts

• Two successive discounts of a% and b% equal a single discount of (a + b − ab/100)%.
• Discounts of 20%, 15% and 10% leave 0.612 of the marked price, a single equivalent discount of 38.8%.
• The equivalent single discount is always smaller than the sum of the individual rates.
• ₹5,000 × 0.612 = ₹3,060.

Common traps

• Adding 20 + 15 + 10 to 45% and answering ₹2,750
• Applying every rate to ₹5,000 rather than to the running amount
• Folding two of the three rates together, which under-counts the price by ₹60 here

No directly related past PYQ was found.