Profit & Loss for SSC GD Constable — Complete Concept to PYQ Guide

beginner 18 min read

Concept

Profit and Loss is a chapter about trade — what you paid vs. what you received. Everything in this chapter reduces to three prices and their relationships.

Cost Price (CP): What you paid to get the goods. This is your input.

Selling Price (SP): What you actually received when you sold them. This is your output.

Marked Price (MP): The price written on the tag before any discount is applied. The shopkeeper marks it artificially high so that even after giving a discount, they still make a profit.

The logic is simple:

Profit % and Loss % are always calculated on CP, not on SP. This is the single most exploited trap in SSC exams, and you need to tattoo it on your brain.

Profit %=SPCPCP×100\text{Profit \%} = \frac{SP - CP}{CP} \times 100

Loss %=CPSPCP×100\text{Loss \%} = \frac{CP - SP}{CP} \times 100

Think of it like this: you invest the CP. The return on that investment — profit or loss — is measured against what you invested, not what you recovered. A soldier who deployed 100 soldiers and recovered 120 made a 20% gain on deployment, not on the recovery.

The marked price introduces a two-step chain: first, the shopkeeper marks up from CP to MP; then gives a discount from MP to SP. The net effect on CP determines whether there is profit or loss.


Deep Dive

The Core Formulas (Use Multiplier Form — Always)

Don't write SP = CP + profit. Instead, think in multipliers. It eliminates separate steps.

| Situation | Relationship | |---|---| | 25% Profit | SP = CP × 1.25 | | 20% Loss | SP = CP × 0.80 | | 40% Markup | MP = CP × 1.40 | | 15% Discount | SP = MP × 0.85 |

This approach lets you chain two conditions in one multiplication, which is exactly what the MP-Discount problems require.

Finding CP from SP (Reverse Direction)

This trips candidates up the most. If SP = CP × 1.25, then:

CP=SP1.25=SP×100125=SP×45CP = \frac{SP}{1.25} = SP \times \frac{100}{125} = SP \times \frac{4}{5}

Look — when you get SP and profit%, convert the multiplier to a fraction and flip it. SP = 1200, profit = 25% means CP = 1200 × (100/125) = 960. Do not subtract 25% of SP from SP. That is wrong. It will give you 900, and 900 is wrong.

The MP-Discount Chain

This is a two-step problem that SSC GD repeats in every paper.

Setup: Shopkeeper marks goods m% above CP, then gives d% discount.

SP=CP×(100+m)100×(100d)100SP = CP \times \frac{(100 + m)}{100} \times \frac{(100 - d)}{100}

Net profit/loss % is determined by whether the final SP is above or below CP. The cleanest approach is to assume CP = 100 always. Then MP and SP follow automatically.

Example: 30% markup, 20% discount.

You never need a formula. Assume CP = 100, follow the chain.

Problems Where You Buy and Sell in Different Quantities

This is a classic SSC trap: "X fruits bought for Y paise, Z fruits sold for W paise."

The trick is to find per-unit CP and per-unit SP.

Alternatively, work with a common batch size (LCM of the buy-quantity and sell-quantity).

Mixed Lot Problems

"Sold 3/4 of goods at 20% profit and 1/4 at 10% loss."

Use weighted average. If CP of the whole lot is 100:

Finding Marked Price When CP and Profit% Are Given

You know CP and the final profit%. That means you know SP. You also know the discount%. Work backwards from SP to MP.

MP=SP×100100dMP = \frac{SP \times 100}{100 - d}

where d is the discount percentage. Remember: SP = MP × (100 − d)/100, so MP = SP reversed through that multiplier.


Memory Tricks & Shortcuts

patternMultiplier Chain for MP-Discount

Whenever you see markup % and discount % together, chain the two multipliers in one shot.

Net effect = (1 + m/100) × (1 − d/100) on CP.

Example: 40% markup, 25% discount = 1.40 × 0.75 = 1.05 → 5% profit.

Standard method: Set up two equations, 3 steps, ~50 seconds. Multiplier chain: One multiplication, ~12 seconds. The answer is the decimal beyond 1.00 — here 0.05 = 5%.

patternCP from SP — The Reverse Multiplier

When given SP and profit%, never subtract profit% of SP. Always use:

CP = SP × 100 / (100 + profit%)

For SP = 1200, profit = 25%: CP = 1200 × 100/125 = 1200 × 4/5 = 960.

Standard mistake path: 1200 − 25% of 1200 = 1200 − 300 = 900 (wrong, costs you the mark). Correct fraction flip: 2 steps, 10 seconds.

patternPer-Unit Method for Quantity Swap Problems

"11 items bought for 10 paise, 10 items sold for 11 paise" type questions.

Find CP per item and SP per item, then compute profit% directly.

  • CP per item = 10/11
  • SP per item = 11/10
  • Profit% = (11/10 − 10/11) ÷ (10/11) × 100
  • = (121 − 100) / 100 × 100 (after simplifying with LCM 110) = 21%

Alternatively: take LCM of 11 and 10 = 110 items. CP of 110 = 10 × 10 = 100 paise. SP of 110 = 11 × 11 = 121 paise. Profit = 21%. Standard method: 5 steps, ~60 seconds. LCM batch method: 3 steps, ~20 seconds.

patternCP of N Pens = SP of M Pens — Direct Formula

When "CP of A items = SP of B items":

Profit% = (A − B) / B × 100

CP of 12 pens = SP of 10 pens → Profit% = (12 − 10)/10 × 100 = 20%.

Standard method (assign unit values, compute SP per pen, calculate %): 4 steps, ~45 seconds. Direct formula: 1 step, ~8 seconds. Valid only when CP of more = SP of fewer (profit scenario) or CP of fewer = SP of more (loss scenario).

estimationMixed Lot — Weighted SP Directly

For mixed lot problems, convert each fraction of goods into its SP contribution and add.

Rule: Total SP = f₁ × CP₁ × (1 ± p₁/100) + f₂ × CP₂ × (1 ± p₂/100) where f is the fraction of total CP.

With CP = 6000: 3/4 at +20% → 4500 × 1.20 = 5400; 1/4 at −10% → 1500 × 0.90 = 1350. Total SP = 6750. Profit = 750/6000 = 12.5%.

This avoids setting up two separate profit/loss amounts and adding them — cuts 2 steps from the standard approach.


Fast-Solving Framework

When a Profit & Loss question appears, run this decision tree in 5 seconds:

Step 1 — Identify what's given and what's asked.

Step 2 — Assume CP = 100 for any question that doesn't give you actual rupee values. This converts every % problem into arithmetic on simple numbers.

Step 3 — Watch the base. Every percentage in this chapter is on CP unless the question explicitly says "on SP" (rare in SSC GD). Confirm before computing.

Step 4 — Quantity problems (N items bought for X, M items sold for Y): go straight to per-unit CP and SP, or use the LCM batch method.

Step 5 — Check answer against options. If your answer doesn't match any option, you likely took % on the wrong base (SP instead of CP or vice versa). Recompute.


Solved PYQs

Why this question: Tests the classic "buy-quantity ≠ sell-quantity" format where candidates typically try to set up algebraic equations and lose time.

Previous Year Questionपिछले वर्ष का प्रश्न2023
If 11 lichchus are bought for 10 paise and 10 lichchus are sold for 11 paise, the gain % is
  1. 10%
  2. 11%
  3. 20%
  4. 21%
Solutionसमाधान

Solving path: Use LCM batch method. LCM(11, 10) = 110. Buy 110 lichchus: cost = 10 × 10 = 100 paise. Sell 110 lichchus: revenue = 11 × 11 = 121 paise. Profit = 21 paise on CP of 100. Gain% = 21%. Total time under 20 seconds.


Why this question: Markup + Discount chain is the most repeated MP question type in SSC GD. Tests whether you can correctly chain two percentages instead of just adding/subtracting them.

Previous Year Questionपिछले वर्ष का प्रश्न
A dealer marks his goods 40% above cost price but gives a discount of 25%. What is his actual profit percentage?
एक दुकानदार अपने सामान पर लागत मूल्य से 40% ऊपर मार्क करता है, लेकिन 25% की छूट देता है। उसका वास्तविक लाभ प्रतिशत कितना है?
  1. 5%
  2. 10%
  3. 12%
  4. 15%
  1. 5%
  2. 10%
  3. 12%
  4. 15%
Solutionसमाधान
Let CP = 100. Marked Price = 140. After 25% discount, SP = 140 × 75/100 = 105. Profit = 105 - 100 = 5. Profit% = 5%.
माना क्रय मूल्य = 100। अंकित मूल्य = 140। 25% छूट के बाद, विक्रय मूल्य = 140 × 75/100 = 105। लाभ = 105 - 100 = 5। लाभ% = 5%।

Solving path: Assume CP = 100. MP = 140 (40% above CP). Discount 25% on MP: SP = 140 × 0.75 = 105. Profit = 5, so profit% = 5%. Option (a). Do not add markup and subtract discount directly — that gives 15%, which is wrong.


Why this question: Finding CP from SP is one of the most frequently answered incorrectly in mock data because candidates subtract profit% from SP instead of dividing properly.

Previous Year Questionपिछले वर्ष का प्रश्न
If selling price is ₹1200 and profit is 25%, what is the cost price?
यदि बिक्री मूल्य ₹1200 है और लाभ 25% है, तो लागत मूल्य क्या होगा?
  1. ₹900
  2. ₹960
  3. ₹1000
  4. ₹1050
  1. ₹900
  2. ₹960
  3. ₹1000
  4. ₹1050
Solutionसमाधान
SP = CP + 25% of CP = CP × 125/100. Therefore, 1200 = CP × 125/100. CP = 1200 × 100/125 = ₹960.
विक्रय मूल्य = क्रय मूल्य + क्रय मूल्य का 25% = क्रय मूल्य × 125/100। अतः, 1200 = क्रय मूल्य × 125/100। क्रय मूल्य = 1200 × 100/125 = ₹960।

Solving path: SP = CP × 125/100. So CP = 1200 × 100/125 = 1200 × 4/5 = ₹960. Option (b). The trap option ₹900 = 1200 × 0.75 — that's what you get if you wrongly subtract 25% of SP.


Why this question: Tests the mixed-lot problem where different portions are sold at different profit/loss rates. Requires weighted calculation, not simple averaging.

Previous Year Questionपिछले वर्ष का प्रश्न
A trader bought goods worth ₹6000 and sold 3/4 of it at 20% profit and remaining at 10% loss. What is his overall profit or loss percentage?
एक व्यापारी ने ₹6000 का सामान खरीदा और उसका 3/4 हिस्सा 20% लाभ पर तथा बाकी 10% हानि पर बेचा। उसका कुल लाभ या हानि प्रतिशत क्या है?
  1. 12.5% profit
  2. 13% profit
  3. 12% profit
  4. 15% profit
  1. 12.5% लाभ
  2. 13% लाभ
  3. 12% लाभ
  4. 15% लाभ
Solutionसमाधान
3/4 goods = ₹4500, sold at 20% profit = 4500 × 1.20 = ₹5400. 1/4 goods = ₹1500, sold at 10% loss = 1500 × 0.90 = ₹1350. Total SP = 6750, Total CP = 6000. Profit% = 750/6000 × 100 = 12.5%.
3/4 सामान = ₹4500, 20% लाभ पर बेचा = 4500 × 1.20 = ₹5400। 1/4 सामान = ₹1500, 10% हानि पर बेचा = 1500 × 0.90 = ₹1350। कुल विक्रय मूल्य = 6750, कुल क्रय मूल्य = 6000। लाभ% = 750/6000 × 100 = 12.5%।

Solving path: CP = 6000. Portion 1: ₹4500 sold at +20% → SP = 5400. Portion 2: ₹1500 sold at −10% → SP = 1350. Total SP = 6750. Profit = 750. Profit% = 750/6000 × 100 = 12.5%. Option (a).


Why this question: Fruit/item problems where buy-rate and sell-rate are given in different bundle sizes. Standard in SSC GD and easily solved with per-unit method.

Previous Year Questionपिछले वर्ष का प्रश्न
A man bought oranges at 5 for ₹10 and sold them at 4 for ₹10. What is his profit percentage?
एक आदमी ने संतरे 5 रुपये ₹10 में खरीदे और 4 रुपये ₹10 में बेचे। उसका लाभ प्रतिशत क्या है?
  1. 20%
  2. 25%
  3. 30%
  4. 35%
  1. 20%
  2. 25%
  3. 30%
  4. 35%
Solutionसमाधान
CP per orange = 10/5 = ₹2. SP per orange = 10/4 = ₹2.5. Profit per orange = 2.5 - 2 = ₹0.5. Profit% = (0.5/2) × 100 = 25%.
प्रति संतरे का क्रय मूल्य = 10/5 = ₹2। प्रति संतरे का विक्रय मूल्य = 10/4 = ₹2.5। प्रति संतरे लाभ = 2.5 - 2 = ₹0.5। लाभ% = (0.5/2) × 100 = 25%।

Solving path: CP per orange = 10/5 = ₹2. SP per orange = 10/4 = ₹2.50. Profit per orange = 0.50. Profit% = 0.50/2 × 100 = 25%. Option (b).


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