Aptitude · Profit, Loss and Discount · Reverse models

Running the discount backwards, and the models built on it

The harder half of the discount chapter is everything that runs the process in reverse: an unknown second discount, a marked price expressed in terms of the cost, a profit hidden behind a markup and a discount. All of it is one division you have to be willing to do.

Give two prices and one discount, and recover the other →

To find a hidden discount, divide the prices and subtract from 1. Working in rupees off the marked price is the error every one of these models is set to catch.

01 The idea

Every reverse question is a division

An article marked at ₹15,000 sells for ₹8,400 after two successive discounts, the first being 30%. What was the second? The instinct is to work in rupees: the total discount is ₹6,600, the first took ₹4,500, so the second took ₹2,100 — and then to call that 14% of ₹15,000. That is wrong.

The second discount is a percentage of the price the first one left, not of the marked price. After 30% off, the price is ₹10,500. The final price of ₹8,400 divided by ₹10,500 is 0.8, so 80% survived and the second discount was 20%. One division, and the base is correct by construction.

That single move — divide the price you ended at by the price you started that step from — unlocks the whole of this lesson. It finds an unknown discount, an unknown markup, or an unknown tax, because in each case the multiplier is exactly the ratio of the two prices.

The other family here is algebraic rather than arithmetic. “The marked price is ₹150 less than three times the cost price, and a 20% discount gives a selling price of ₹1,800.” Nothing to divide yet — you write the marked price in terms of the cost, apply the discount multiplier, set it equal to the selling price and solve. Same discipline about bases, one unknown instead of a ratio.

The multiplier of any step is the ratio of the two prices around it. Divide, then subtract from 1 to get the discount.

Reverse discountFinding a discount percentage from two known prices. Always the price after divided by the price before, subtracted from 1.

Intermediate priceThe price between two successive discounts. The second discount's base, and the figure that must be computed before anything else.

Marked price in terms of costQuestions where MP is given as an expression in CP, such as “₹150 less than three times the cost”. Solved by algebra, not by division.

02 Worked example

₹15,000 down to ₹8,400, first discount 30%

The source question, and the model for every reverse discount. The marked price of an article is ₹15,000. After two successive discounts it sold for ₹8,400. If the first discount was 30%, find the second.

Apply the discount you knowThirty per cent off ₹15,000 leaves 70% of it. Stop here — this is the second discount's base.15,000 × 70/100 = ₹10,500

Divide the final price by thatThe ratio of the two prices is exactly the second discount's multiplier.8,400 / 10,500 = 0.8

Read off the discountEighty per cent survived, so twenty per cent was taken off.1 − 0.8 = 0.2 ⇒ second discount = 20%

See the wrong routeWorking in rupees off the marked price gives a plausible and incorrect answer.(6,600 − 4,500)/15,000 = 14% ← wrong base

Check with the equivalent discountThe two discounts together should match the overall discount from ₹15,000 to ₹8,400.30 + 20 − 600/100 = 44%, and 1 − 8400/15000 = 44% ✓

The check in the last step is worth building in permanently. Compute the overall discount straight from the first and last prices, then confirm your two individual discounts combine to it via a + b − ab/100. It costs one division and it catches the wrong-base error immediately — here the incorrect 14% would combine with 30% to give 39.8%, not 44%.

03 The method

The five reverse models, and what each one needs

All of these are set regularly. The first three are divisions; the last two are one-line algebra.

Unknown second discount: divide the final price by the price after the first discount, then subtract from 1. Unknown markup: divide the marked price by the cost. Overall discount: 1 − SP/MP.

For the algebraic family, write everything in terms of CP and solve once. If MP = 3·CP − 150 and a 20% discount gives SP = 1800, then 0.8(3·CP − 150) = 1800, so 3·CP − 150 = 2250 and CP = ₹800. Resist computing intermediate rupee figures — one equation is less error-prone than three steps.

Model	What you are given	Route
Unknown second discount	MP, SP, first discount	`1 − SP/(MP×(1−d₁))`
Overall discount	MP and SP	`1 − SP/MP`
Markup from a discount and profit	discount%, profit%	`(100+p)/(100−d) − 1`
MP in terms of CP	an expression plus discount and SP	one equation in CP
Discount and profit amount	MP, discount%, profit in rupees	`CP = SP − profit`
Extra expense	purchase, expense, MP, discount	`CP = purchase + expense`
Rupees off the marked price	`wrong for a 2nd discount`	gives 14%, not 20%

05 Cheat sheet

Reverse models on one page

Three divisions and three algebraic set-ups. The last row is the error the whole lesson is about.

Want	Route	Worked
Second discount	`1 − SP/intermediate`	1 − 8400/10500 = 20%
Overall discount	`1 − SP/MP`	1 − 8400/15000 = 44%
Check the pair	`a + b − ab/100`	30+20−6 = 44% ✓
Markup for a target profit	`(100+p)/(100−d)`	120/80 = 1.5 → 50%
CP from MP, discount, profit₹	`CP = SP − profit`	1600×0.65 − 250 = 790
MP given in terms of CP	`one equation in CP`	0.8(3c−150)=1800 → 800
Rupees off the marked price	`wrong base`	gives 14%

Divide, do not subtract rupeesA second discount is a percentage of the intermediate price. The ratio of the two prices is its multiplier, so division gets the base right automatically.

Always cross-check with the overall discountCompute 1 − SP/MP independently and confirm your two discounts combine to it. One division, and it catches the wrong-base error every time.

Write one equation, not three stepsFor marked-price-in-terms-of-cost questions, set up a single equation in CP. Computing intermediate rupee figures multiplies the chances of a slip.

06 Where & why

Where these show up

This is the mains-paper end of the discount chapter, and the 16 models in the source material are almost all variations on the reverse move.

Bank PO Mains · SSC CGL

Unknown second discount

The single most set reverse question. The wrong answer from working in rupees off the marked price is always among the options.

TCS Digital · Infosys

Markup needed for a target profit

“What markup allows a 20% discount and 20% profit?” One division of the two multipliers gives 1.5, so 50%.

Marked price in terms of cost

“₹150 less than three times the cost”

Algebraic rather than arithmetic. One equation in CP, solved once, avoids the intermediate figures where errors creep in.

Profit amount rather than percentage

“makes a profit of ₹250”

Easier than it looks: apply the discount to get SP, then CP is simply SP minus the stated profit.

Every model in this lesson is the same instinct: identify which price is the base of the percentage you are missing, and get to it by dividing rather than by subtracting rupees from the wrong figure.

07 Interview questions

What gets asked

Ten, drawn from the harder models in the source material.

₹15,000 sells for ₹8,400 after two discounts, the first 30%. Find the second.

20%. After 30% the price is ₹10,500, and 8,400/10,500 = 0.8, so 80% survived and the second discount was 20%. Working in rupees off the marked price gives 14%, which uses the wrong base.

Why is 14% wrong there?

Because the second discount is a percentage of the ₹10,500 the first discount left, not of the original ₹15,000. The ₹2,100 removed by the second discount is 20% of ₹10,500 and 14% of ₹15,000 — and only the first of those is the discount.

How do you check a reverse-discount answer?

Compute the overall discount independently as 1 − SP/MP, then confirm your two discounts combine to it. Here 1 − 8400/15000 = 44%, and 30 + 20 − 600/100 = 44%. The wrong answer of 14% would give 39.8%, which fails immediately.

The marked price of an article is ₹1,600. After a 35% discount the dealer makes ₹250 profit. Find the cost price.

₹790. The selling price is 1,600 × 0.65 = ₹1,040, and the profit is a stated rupee amount, so the cost is simply 1,040 − 250 = ₹790. When the profit is given in rupees rather than as a percentage, no division is needed.

The marked price is ₹150 less than three times the cost, and a 20% discount gives a selling price of ₹1,800. Find the cost price.

₹800. Write MP = 3c − 150 and apply the discount: 0.8(3c − 150) = 1800, so 3c − 150 = 2250, giving 3c = 2400 and c = ₹800. One equation rather than three steps.

A stone costing ₹4,200 has ₹1,600 spent on it and is advertised at ₹9,000 with a 20% discount. Find the profit percentage.

About 24.14%. The cost price is 4,200 + 1,600 = ₹5,800, and the selling price is 9,000 × 0.8 = ₹7,200. So the profit is ₹1,400 and 1,400/5,800 × 100 = 24.14%. Forgetting the making charge is the intended error.

What markup allows a 20% discount and still gives a 20% profit?

Fifty per cent. You need the markup multiplier divided by the discount multiplier to give the profit multiplier: (100 + 20)/(100 − 20) = 120/80 = 1.5, so mark up 50% above cost.

An article marked ₹10,000 sells at successive discounts of 13% and 20%. Find the price.

₹6,960. Multiply: 10,000 × 0.87 × 0.80 = ₹6,960. The overall discount is 30.4%, which is less than the 33% you would get by adding, as it always must be.

Two successive discounts of 12% and 25% — what is the overall discount?

Hard

A shopkeeper wants to advertise two successive discounts of 20% and 25% while still making a 20% profit on cost. (a) Find the single equivalent discount of the two. (b) Find the percentage above cost at which he must mark his goods. (c) He then decides to advertise a third discount of 10% on top, keeping the same 20% profit. Find the new markup required, and state whether the extra discount forces the marked price up by more or less than 10%.

Follow-up

Part (c) is the real question. Adding a 10% discount does not require a 10% higher marked price, because the markup and the discounts compound rather than add — the same structural fact that runs through this entire module, now used to make a pricing decision.

Show the hint

Work in multipliers throughout: you need (markup multiplier) times (all the discount multipliers) to equal 1.2.