The word problem is familiar. "A shirt is sold for ₹560 after a 30\% discount. What was the original price?" You need a reflex that triggers instantly on the phrase "original price" paired with "after X% discount" — because the wrong reflex, "add 30\% to ₹560," is the most common trap in the whole of school arithmetic.

This article is not about re-deriving why the reflex works (that is done in How Do You Find the Original Price When Only the Discounted Price and Discount % Are Given?). It is about building the one-second recognition pattern, so that when you see this shape, the correct move is automatic.

The recognition pattern

Scan the problem for this exact shape:

"... final / discounted / sale price is F, after a discount of d\%. Find the original price."

The disguises are many — "selling price," "marked down," "reduced by," "off," "on sale for" — but they all mean the same thing: you have the after number and you want the before number.

The moment you recognise this shape, reach for the formula

P = \frac{F}{1 - \tfrac{d}{100}}

where P is the original price, F is the final (discounted) price, and d is the discount percentage. That is it. Don't compute F + d\% of F. Don't set up \tfrac{d}{100} = \tfrac{F}{P}. Don't reach for cross-multiplication. Just divide F by the remaining fraction (1 - \tfrac{d}{100}).

Recognition flowchart for reversing a discountA three-step flowchart. The first box reads the problem statement saying discounted price is F after d percent discount find original. An arrow labelled recognise leads to the middle box showing the formula P equals F divided by open paren one minus d over one hundred close paren. A second arrow labelled apply leads to the last box showing the answer P. final price = F discount = d% find original recognise P = F ÷ (1 − d/100) apply original P for a 30% discount on ₹560: P = 560 ÷ 0.70 = ₹800 not 560 × 1.30 = ₹728 — that's the trap
The recognition-and-apply loop for reversing a discount. The only move is to divide by the remaining fraction $(1 - d/100)$. Adding the discount percentage back always undershoots the true original, because the two percentages act on different bases.

Why the "add it back" reflex fails

The trap is worth naming so you can recognise it and shut it down. Your brain may want to say:

30\% was taken off. I'll put 30\% back. Multiply by 1.30.

This gives 560 \times 1.30 = 728, which is wrong by ₹72. The reason is that the original 30\% was taken off the original price (30\% of 800 = 240), while your "30\% back" would add 30\% of the discounted price (30\% of 560 = 168). Two different bases, two different rupee amounts, and they don't cancel.

Why: percentages are multiplications, not additions. Going forward, P \times 0.70 = F. Going backward, F \div 0.70 = P — not F \times 1.30. Division by a multiplier undoes the multiplication exactly; adding the same percentage undoes it approximately (and only for small d).

The sanity check

After you compute P, check that the answer is larger than F. The original price must be larger than the discounted price — always. If your answer is smaller, you divided the wrong way. If your answer is bigger than F but looks suspiciously like F \times (1 + d/100), you have fallen into the trap.

For the ₹560 at 30\% off problem:

The combined check is: answer larger than F, and obtained by dividing, not multiplying.

Three more problem shapes to recognise

The same pattern recognises and the same reflex applies.

Shape 1 — reverse a price hike. "After a 15\% price hike, a packet costs ₹230. What was the old price?"

P = 230 / 1.15 = 200. The remaining fraction is now (1 + \tfrac{d}{100}) because the price went up, not down — but the logic is identical: divide by the multiplier, don't subtract the percentage.

Shape 2 — reverse a GST add-on. "A bill including 18\% GST comes to ₹1180. What was the pre-GST amount?"

P = 1180 / 1.18 = 1000. Same structure.

Shape 3 — chained reverses. "After a 20\% discount and then an 8\% cashback, you paid ₹736. What was the original?"

P = 736 / (0.80 \times 0.92) = 736 / 0.736 = 1000. Each percentage step contributes its own multiplier; reversing multiplies by the reciprocals, which is the same as dividing by the product.

The four-word reflex

When you see "original price after discount," say these four words in your head:

Divide by remaining fraction.

That's the whole skill. The remaining fraction after a d\% discount is (1 - \tfrac{d}{100}); after a d\% hike it is (1 + \tfrac{d}{100}). Divide. Don't multiply.

Worked example: triple-check the trap

A laptop is selling for ₹42{,}500 after a 15\% discount. Naive student writes 42{,}500 \times 1.15 = ₹48{,}875. Correct computation is 42{,}500 / 0.85 = ₹50{,}000. Difference: ₹1{,}125. The naive answer is too low, because the "15\% back" was computed on the discounted ₹42{,}500 instead of on the original ₹50{,}000.

A second check: 15\% of ₹50{,}000 = ₹7{,}500, and 50{,}000 - 7{,}500 = 42{,}500. ✓ The original-and-discount pair is consistent; the naive answer is not.

The generalisation

Every "undo a percentage change" problem in school arithmetic has this exact shape. If the forward operation was multiplication by m (where m = 1 - d/100 for discounts, 1 + d/100 for hikes, and a product of these for chains), the backward operation is division by m. Never multiplication, never addition of a percentage.

Once this becomes reflex, reversing a discount is no harder than applying one. The five-second recognition is what separates students who solve these problems quickly from students who fall for the trap every time.

Related: Percentages and Ratios · How Do You Find the Original Price When Only the Discounted Price and Discount % Are Given? · Price Up 20% Then Down 20% — Why You End Up Below Where You Started · See Successive Percentage Changes → Multiply Factors, Never Add Them