A shopkeeper offers the following discount schemes for buyers. I. Two successive discounts of 15% and 20% II. Successive discount of 25 % and 10% III. A discount of 33% The selling price will be minimum under the scheme:

Correct — c. Turn each scheme into a single multiplier on the marked price. The smallest multiplier is the lowest selling price.

I: 15% then 20% → 0.85 × 0.80 = 0.68
II: 25% then 10% → 0.75 × 0.90 = 0.675
III: a flat 33% → 1 − 0.33 = 0.67

0.67 < 0.675 < 0.68, so scheme III leaves the least → option (c).

Why the others are wrong

• (a)
  II — Scheme II leaves 0.675 of the marked price, an effective discount of 32.5%. That beats scheme I but not the flat 33% of scheme III.
• (b)
  I or II — Scheme I leaves 0.68, the highest of the three, so it cannot share the lowest price with anything. Successive 15% and 20% are worth a single 32%, not 35%.
• (d)
  I — Scheme I is the weakest of the three: 0.85 × 0.80 = 0.68. It looks strongest only if 15 and 20 are added to 35%, which is not how successive discounts work.

Successive discounts multiply. They never add.

Discounts of a% and then b% leave (1 − a⁄100)(1 − b⁄100) of the marked price, so the equivalent single discount is a + b − ab⁄100 — always short of a + b by the ab⁄100 term.

That makes 15% and 20% worth 32%, and 25% with 10% worth 32.5%, both under the flat 33% of scheme III.

Once every scheme is one multiplier the comparison is a sort, and the order of the two discounts inside a scheme is irrelevant because multiplication commutes.

Key facts

• Successive discounts of a% and b% equal one discount of a + b − ab⁄100.
• 15% and 20% together are worth 32%, and 25% with 10% are worth 32.5%.
• A flat 33% leaves 0.67 of the marked price, less than either pair leaves.

Common traps

• Adding 15 and 20 to 35% and declaring scheme I the winner
• Believing the order of two discounts changes the result
• Reading 'minimum selling price' as 'minimum discount' and picking the weakest scheme

No directly related past PYQ was found.