A cooker is marked at ₹12,500. The dealer allows successive discounts of 6%, 6% and 4% on it. What is the net selling price of the cooker?

Correct — C. Successive discounts multiply; they are never added. Each one leaves a fraction of the price that reached it: 94%, then 94%, then 96%.

12,500 × 0.94 = 11,750
11,750 × 0.94 = 11,045
11,045 × 0.96 = ₹10,603.20 → option (c)

One line instead of three: 12,500 × 0.94 × 0.94 × 0.96 = 12,500 × 0.848256.

Why the others are wrong

• (a)
  ₹12,320.60 — ₹12,320.60 is only 1.4% below the ₹12,500 marked price, while the three discounts together remove about 15% of it. No step in the chain leaves the price that close to the marking.
• (b)
  ₹10,306 — ₹10,306 is ₹10,603 with the 3 and the 6 swapped, and it also drops the paise. The third step gives 11,045 × 0.96 = 10,603.20 exactly.
• (d)
  ₹12,230 — ₹12,230 is 2.16% below the marked price. Like option (a) it treats the reductions as trivial instead of compounding to roughly 15%.

A discount of d% leaves a multiplier of (1 − d⁄100). Chain three of them and the multipliers multiply:

0.94 × 0.94 × 0.96 = 0.848256

So the single equivalent discount is 1 − 0.848256 = 15.1744%, not the 16% you get by adding 6 + 6 + 4. Successive discounts always come to less than their sum, because each later cut is taken on an already reduced price.

Multiplication commutes, so the order of the three discounts makes no difference to the final price.

Key facts

• Successive discounts of 6%, 6% and 4% leave 0.94 × 0.94 × 0.96 = 0.848256 of the marked price.
• That is a single equivalent discount of 15.1744%, less than 6 + 6 + 4 = 16%.
• 12,500 × 0.848256 = ₹10,603.20.
• Selling price = marked price − discount, with the cost price playing no part in this question.

Common traps

• Adding 6 + 6 + 4 to 16% and taking 84% of ₹12,500, which gives ₹10,500.
• Applying the 4% to the marked price instead of to the twice-reduced price.
• Rounding the paise away and settling for a whole-rupee option.

No directly related past PYQ was found.