Profit, Loss & Discount
🔒 Log in to trackThe highest-yield arithmetic block in Tier 1. CP–SP basics, successive discounts, the marked-price chain and dishonest-dealer cases appear in nearly every shift; Tier 2 layers on trade chains and price-shift comparisons. Everything reduces to multiplier chips.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
69 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (69 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Direct profit/loss per cent
very commonCP and SP are given (or the CP follows in one step, e.g. with a repair cost) and the profit or loss per cent is asked.
How to solve: Profit or loss = SP − CP; divide by CP (never by SP) and multiply by 100. Overheads join the CP first. Fraction chips (12.5% = 1/8) skip the arithmetic.
Example: A cycle bought for ₹850 is sold for ₹1,020. Find the profit per cent.
Profit = 170 → 170/850 × 100 = 20%.
CP or SP reconstruction
very commonOne of CP/SP is given with the profit or loss per cent — find the other price.
How to solve: Multiply or divide by the chip (100 ± x)/100. 'Had it been sold for ₹k more, gain b% instead of loss a%' ⇒ k = (a + b)% of CP.
Example: A trader sells a scooter at a loss of 5%. Had he sold it for ₹960 more, he would have gained 15%. Find the CP.
20% of CP = 960 → CP = ₹4,800.
Successive discounts / price changes
very common'Two successive discounts of 20% and 10%', or a price that rises and then falls by given per cents.
How to solve: Chips multiply; equivalent single discount = d1 + d2 − d1d2/100. Same % up and down ends at −x²/100%. Never add discounts.
Example: The marked price of a sofa is ₹1,600. It is sold after two successive discounts of 20% and 12.5%. Find the selling price.
1600 × 4/5 × 7/8 = ₹1,120.
MP–discount–profit chain
very commonMarked x% above cost, a discount of y% allowed — find the gain per cent, or work back to the MP for a target gain.
How to solve: MP/CP = (100 + g)/(100 − d) fixes any one of markup, discount, gain from the other two. Trace CP = 100 through the chain to check.
Example: A trader allows a discount of 25% on the marked price and still gains 20%. The marked price is what per cent above the cost price?
(120/75 − 1) × 100 = 60%.
Dishonest dealer / false weight
commonFalse weights, short measures, or a claimed loss with cheating on the quantity.
How to solve: gain% = (W − w)/w × 100 for short weight at CP; for claims compare money per true unit: (100 − L)/(100 − c). Cheating at both ends multiplies chips (100 + b)/(100 − s).
Example: A milkman sells at the cost price but gives only 800 mL per litre. Find his gain per cent.
200/800 × 100 = 25%.
Same SP, one profit one loss
commonTwo items sold at the same price with equal ±x% — the net result in per cent or rupees.
How to solve: Net loss% = x²/100 of the total CP. For rupees, reconstruct both CPs by dividing the common SP by the chips.
Example: Two articles are sold at ₹1,200 each, one at 25% profit and the other at 25% loss. Find the overall loss per cent.
(6.25%). CPs 960 and 1600, total 2560; loss ₹160 = 6.25% of CP.
Multi-level markup (trade chain)
occasionalManufacturer → wholesaler → retailer → customer, each adding a profit per cent.
How to solve: Multiply the markup chips end to end; divide to walk backwards to the base cost. Each trader's profit acts on his own purchase price.
Example: A manufacturer sells at 20% profit to a wholesaler, who sells at 25% profit to a retailer, who sells at 10% profit to a customer. Overall rise over the manufacturing cost?
1.2 × 1.25 × 1.1 = 1.65 → 65%.
Cost of goods sold / goods-based cases
occasional'The CP of 15 articles equals the SP of 12', 'buys 4 for ₹15, sells 5 for ₹24', free-item offers.
How to solve: Fix a common count of articles (LCM of the counts) and compare CP and SP; for counts, profit% = extra ÷ articles given × 100. Free-item offers: discount = free ÷ received.
Example: If the cost price of 15 articles is equal to the selling price of 12 articles, find the profit per cent.
(15 − 12)/12 × 100 = 25%.