Profit and Loss for SBI PO — Cost Price, Selling Price, Discount & Markup

intermediate 18 min read

Concept

Profit and Loss is fundamentally about the relationship between three numbers: what you paid (Cost Price), what you charged (Selling Price), and what was originally advertised (Marked Price). The exam never asks you to memorise formulas — it asks you to see relationships fast.

Here is the core mental model that makes everything click: think of CP as the anchor at 100%. Every other value — SP, MP, discount, profit — is a percentage deviation from that anchor.

So when a question says "marked 30% above CP and sold at 10% discount", you don't need separate formulas. You convert directly:

That single multiplication gives you a 17% profit without invoking any named formula.

The analogy that works in the exam hall: imagine CP as the ground floor of a building. Markup takes you up some floors (MP). Discount brings you down some floors. Where you land is SP. If SP > CP, you're above ground = profit. If SP < CP, you're below ground = loss.

The key insight — discount is always on MP, profit/loss is always relative to CP. This distinction trips up more aspirants than any calculation error. Never take discount on CP, never measure profit on MP.

One more relationship worth internalising before you go deeper: if profit% = loss% in two transactions, the net result is always a loss. Specifically, if you sell two items each at ₹x, one at p% profit and one at p% loss, the combined loss% = p²/100. SBI PO has set questions on this pattern across multiple years.


Deep Dive

Core Formulas — as multipliers, not fractions

Stop writing Profit% = (SP - CP)/CP × 100 every time. Build the multiplier habit instead:

This means for a standard markup-discount problem:

SP = CP × (1 + m/100) × (1 - d/100)

where m = markup%, d = discount%. The profit% on CP is then just [(1 + m/100)(1 - d/100) - 1] × 100.

Successive Discounts

When two discounts d₁ and d₂ are applied one after the other, the equivalent single discount is:

Effective discount = d₁ + d₂ - (d₁ × d₂)/100

Example: 20% then 35% → 20 + 35 - (20 × 35)/100 = 55 - 7 = 48%. So the buyer pays 52% of MP.

This is directly tested in PYQs (see the 2016 question below). Don't do two separate multiplications in the exam — compute the effective multiplier once: 0.80 × 0.65 = 0.52, then divide the given SP by 0.52 to get MP.

Equal Profit Conditions

A question type that appeared in SBI PO 2025: two articles with different CPs, and the profit amount (not percent) is equal. This is an algebraic setup that resolves cleanly if you assign CP of one article a round number.

Standard move: let CP of the cheaper article = 100 (as a variable unit). Express everything else in terms of that. The "equal profit amount" condition then gives you the ratio of SPs, from which you can find the actual SP and hence profit%.

Allegation on Profit%

When two items are bought at equal price but sold at different profit percentages, and you know the combined profit%, you can use the allegation / weighted average approach to find the missing individual profit%. This is cleaner than setting up two equations. The 2019 PYQ (bags question) is a direct application — average of 10% and x% = 13%, so x = 16% by simple symmetric allegation.

The Symmetric SP Trick

If SP₁ gives a profit equal in rupees to the loss in SP₂:

SP₁ - CP = CP - SP₂ CP = (SP₁ + SP₂) / 2

This is the single fastest solve in profit-loss. The 2015 PYQ (article at ₹878 and ₹636) resolves in one line.

Working Backwards from SP

Many SBI PO questions give SP and ask for CP. The standard error is to apply the percentage on SP instead of CP. Look — if profit is 20%, then SP = 1.2 × CP, so CP = SP / 1.2, not SP × 0.8. The latter gives you 80% of SP, which is not the CP.

Build this as a reflex: always divide SP by the multiplier to get CP.

CP = SP / (1 ± profit% or loss% as decimal)


Memory Tricks & Shortcuts

patternMultiplier Chain — no formula needed

For any markup + discount problem, multiply the two multipliers directly on CP.

Markup 30%, discount 10%: SP = CP × 1.30 × 0.90 = CP × 1.17 → 17% profit.

Write the chain left to right in 5 seconds. Compare to standard method: write MP formula, substitute, write SP formula, substitute, compute profit — that's 4-5 steps (~40s). Multiplier chain: 2 steps (~10s).

patternSymmetric SP — average for CP

When profit rupees = loss rupees at two different SPs, CP = arithmetic mean of those two SPs.

CP = (SP₁ + SP₂) / 2. No algebra needed.

Example: profit at ₹878, loss at ₹636 → CP = (878 + 636)/2 = 1514/2 = ₹757. Standard algebraic route: 3 steps (~30s). This: 1 step (~8s).

patternSuccessive Discount — single multiplier

Two discounts d₁ and d₂ → net multiplier = (1 - d₁/100)(1 - d₂/100).

For 20% and 35%: 0.80 × 0.65 = 0.52. Buyer pays 52% of MP. To recover MP from SP: MP = SP / 0.52.

Avoids computing effective discount% separately and then applying it again — saves 2 arithmetic operations (~15s versus ~35s for two-stage approach).

eliminationAllegation for Equal-Cost, Different-Profit

Two items at same CP sold at different profit percentages p₁ and p₂, overall average profit = p_avg.

Draw allegation cross: (p_avg - p₁) and (p₂ - p_avg) give the ratio of quantities. When quantities are equal (one item each), the relationship simplifies to p_avg = (p₁ + p₂)/2.

So missing profit% = 2 × p_avg - p₁.

In the bags PYQ: 2 × 13 - 10 = 16%. Arrived in 3 seconds versus setting up and solving two equations (~50s).

substitution3% of CP = ₹90 — unit method for profit difference

When a question gives you the rupee difference between two profit scenarios and you have the profit percentages, divide rupee difference by the percentage-point difference to get 1% of CP, then scale.

In the bags PYQ: overall profit 13%, bag 1 at 10%, bag 2 at 16%. The difference between bag 2 and overall = 3%. Given that 3% of CP = ₹90, CP = ₹90/0.03 = ₹3000.

This is the unit method — find what 1% equals, then multiply. Works for any rupee-difference given in profit/loss questions. Saves you from writing full equations.


Fast-Solving Framework

In the exam hall, run this decision tree in under 10 seconds before writing anything:

Step 1 — Identify what's given and what's asked. Are you given CP and asked SP? Use multiplier. Given SP and asked CP? Divide by multiplier. Given MP and discount? Multiply by (1 - d/100) for SP.

Step 2 — Is profit/loss in rupees or percent? Rupees given → set up unit variable (let 1% of CP = ₹k). Percent given → use multiplier directly.

Step 3 — Is it a two-article or two-transaction problem? Same CP + different profit% → allegation. Same profit/loss amount at different SPs → symmetric average.

Step 4 — Successive discounts? Multiply the multipliers. Never add the percentages directly.

Step 5 — Sanity check. Markup > discount → profit. Markup < discount → loss. If your answer says 170% profit or 100% profit, don't second-guess — these are deliberately high values in SBI PO to test if you doubt yourself.


Solved PYQs

Why this question: Tests the equal-profit-amount condition with a ratio twist — a 2025-level difficulty pattern that requires you to handle CP ratio and SP ratio simultaneously.

Previous Year Questionपिछले वर्ष का प्रश्न2025
The cost price of article P is 30% more than that of article Q. The selling price of article Q is 90% of the selling price of article P. If the profit earned by selling article P is equal to the profit earned by selling article Q. Then find the percentage profit earned on selling article Q.
  1. 130
  2. 160
  3. 170
  4. 150
Solutionसमाधान
Setting CP of Q = 100 and CP of P = 130, and using equal profit condition gives SP ratio of 10:9. Solving yields SP of Q = 270 on CP of 100, giving 170% profit.

Solving path: Let CP of Q = 100 units, so CP of P = 130 units (30% more). Let SP of P = 10k. Then SP of Q = 9k (90% of SP of P). Equal profit condition: 10k - 130 = 9k - 100. Solving: k = 30. So SP of Q = 9 × 30 = 270, CP of Q = 100. Profit% on Q = (270 - 100)/100 × 100 = 170%.


Why this question: Classic markup-discount with a rupee-difference condition. Tests the unit-variable method under the markup framework.

Previous Year Questionपिछले वर्ष का प्रश्न2025
A toy seller marks his toy 50% above the cost price. His profit falls by ₹90 when he raises the discount from 15% to 25%. What is the cost price of the toy?
  1. ₹600
  2. ₹500
  3. ₹540
  4. ₹720
Solutionसमाधान
Let CP = 100x, MP = 150x. Profit at 15% discount = 127.5x - 100x = 27.5x. Profit at 25% discount = 112.5x - 100x = 12.5x. Difference = 15x = ₹90, so x = 6, CP = ₹600.

Solving path: Let CP = 100x, MP = 150x. SP at 15% discount = 150x × 0.85 = 127.5x. SP at 25% discount = 150x × 0.75 = 112.5x. Profit difference = 127.5x - 112.5x = 15x = ₹90. So x = 6. CP = 100 × 6 = ₹600.


Why this question: Multi-condition problem — original SP from markup/discount, then a hypothetical scenario that defines a new profit%. Two unknowns resolve from one equation.

Previous Year Questionपिछले वर्ष का प्रश्न2024
The cost price of a pen is ₹x and it is marked up 30% above its cost price. 10% discount was given on the marked price. If the cost price of the pen becomes ₹50 less and the selling price becomes ₹66 more, then profit received on selling the pen is 100%. Find the actual selling price of the pen (in ₹).
  1. 360
  2. 230
  3. 234
  4. 260
Solutionसमाधान
Selling price = 1.17x. With new cost (x−50) and new SP (1.17x+66), profit = 100% means 1.17x+66 = 2(x−50). Solving gives x = 200, so actual SP = 1.17×200 = ₹234.

Solving path: Actual SP = 1.30 × 0.90 × x = 1.17x. Under new conditions: new CP = x - 50, new SP = 1.17x + 66, profit = 100% means new SP = 2 × new CP. So 1.17x + 66 = 2(x - 50) = 2x - 100. Therefore 166 = 0.83x, giving x = 200. Actual SP = 1.17 × 200 = ₹234.


Why this question: Equal-cost, unequal-profit problem solved fastest via allegation. Tests whether you reach for equations or the smarter average-based approach.

Previous Year Questionपिछले वर्ष का प्रश्न2019
A person purchased two Bags at the same price and on selling the first Bag he makes a profit of 10%. Selling price of second Bag is ₹90 more than the selling price of the first Bag. Find the cost price of one Bag if his overall profit percent is 13%?
  1. ₹3000
  2. ₹2000
  3. ₹2400
  4. ₹1800
Solutionसमाधान
Using the allegation method with 10% and x% profit averaging to 13%, the ratio is 1:1, giving x=16%. Since 3% of cost price equals ₹90, cost price = ₹3000.

Solving path: Same CP for both bags. Bag 1: 10% profit. Overall: 13% profit. Since equal quantities, profit% on Bag 2 = 2 × 13 - 10 = 16%. Now, Bag 2's SP - Bag 1's SP = ₹90. SP of Bag 1 = 1.10 × CP. SP of Bag 2 = 1.16 × CP. Difference = 0.06 × CP = ₹90. CP = ₹3000 (per bag).


Why this question: Direct successive-discount reversal. Given final payment, find MP. Pure multiplier application.

Previous Year Questionपिछले वर्ष का प्रश्न2016
A trader gives an additional concession of 35% on an article which is already get discounted by 20% on the marked price. If the buyer pays an amount of 1300 for the article, then the marked price is
  1. 2700
  2. 2500
  3. 2600
  4. 2200
Solutionसमाधान
Marked Price = 1300 × 100 × 100 / (80 × 65) = 2500. The buyer pays after successive discounts of 20% and 35%, so working backwards gives a marked price of ₹2500.

Solving path: Net multiplier = 0.80 × 0.65 = 0.52. Buyer pays 52% of MP. MP = 1300 / 0.52 = ₹2500.


Why this question: The symmetric SP pattern in its purest form. One of the fastest solves in profit-loss.

Previous Year Questionपिछले वर्ष का प्रश्न2015
The profit earned after selling an article for ₹878 is the same as loss incurred after selling the article for ₹636. What is the cost price of the article?
  1. ₹787
  2. ₹757
  3. ₹797
  4. ₹767
Solutionसमाधान
Let CP = x. Then 878 − x = x − 636, so 2x = 1514, giving x = ₹757.

Solving path: Profit at ₹878 = Loss at ₹636 means CP is equidistant from both. CP = (878 + 636)/2 = 1514/2 = ₹757.


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