Profit and Loss for SSC CHSL — Complete Guide with Shortcuts and PYQs

intermediate 18 min read

Concept

Profit and Loss is fundamentally a chapter about the gap between what something costs and what it sells for. Every transaction in this chapter has three values that matter: the Cost Price (CP) — what the seller paid to acquire the item, the Selling Price (SP) — what the buyer actually pays, and the Marked Price (MP) — the label price before any discount is applied.

Here is the core logic: if SP > CP, you made a profit. If SP < CP, you took a loss. The percentage is always calculated on the CP (unless a question explicitly tells you otherwise).

Think of it like this — you are a vegetable vendor at Sarojini Nagar market. You bought tomatoes at ₹30/kg (that is your CP). You write ₹50/kg on your sign (that is your MP). A customer bargains, so you agree to ₹40/kg (that is your SP after the discount). Your profit is ₹10 on a CP of ₹30, which is 33.3%.

The marked price and discount layer is where SSC CHSL loves to hide difficulty. Questions will give you the MP and a discount percentage, ask you to find the profit percentage, or flip it entirely — give you the profit percentage and ask what the MP:CP ratio must be. That chain — CP → MP → Discount → SP → Profit% — is the backbone of 60% of the questions in this chapter.

One more concept worth locking in early: dishonest shopkeeper problems. Here the "loss" or "gain" does not come from price manipulation but from weight manipulation. The shopkeeper sells less than the stated quantity. The gain formula for this is specific and frequently tested, so it gets its own trick block below.

Keep the vocabulary straight:

Deep Dive

The Core Formulas

Profit=SPCP(when SP > CP)\text{Profit} = SP - CP \quad \text{(when SP > CP)}

Loss=CPSP(when CP > SP)\text{Loss} = CP - SP \quad \text{(when CP > SP)}

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

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

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

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

You must be able to rearrange these in your head. If SP and Profit% are given, find CP:

CP=SP×100100+Profit%CP = \frac{SP \times 100}{100 + \text{Profit\%}}

The MP-Discount Chain

SP=MP×100Discount%100SP = MP \times \frac{100 - \text{Discount\%}}{100}

The chain question always has this structure: CP is the base, MP is set above CP by some markup, then a discount brings SP below MP. You need to figure out where SP lands relative to CP.

Suppose CP = 100, markup = 80%, so MP = 180. Discount = 48%, so:

SP=180×52100=93.6SP = 180 \times \frac{52}{100} = 93.6

Since SP < CP, there is a loss of 6.4 on CP of 100, giving a loss% of 6.4% = 6\frac{2}{5}\%.

Look — the key insight here is that markup and discount do not cancel neatly. An 80% markup followed by a 48% discount does not give you 32% profit. The discount is calculated on the marked price, not on CP. This is the most common trap in this entire chapter.

For the combined effect of markup (m%) and discount (d%):

Net%=mdm×d100\text{Net\%} = m - d - \frac{m \times d}{100}

If this is positive, it is profit. If negative, it is loss. This formula saves 30-40 seconds on combined markup-discount questions.

Dishonest Shopkeeper — Weight Manipulation

When a shopkeeper uses a false weight w grams instead of the true 1000 grams (1 kg), and sells at cost price:

Gain%=1000ww×100\text{Gain\%} = \frac{1000 - w}{w} \times 100

This is derived from the fact that he gives only w grams worth of goods but charges for 1000 grams. The error is (1000 - w) grams. His gain is on an "investment" of only w grams of actual goods.

For a 920 gm weight used instead of 1000 gm:

Gain%=80920×100=8.695%8.70%\text{Gain\%} = \frac{80}{920} \times 100 = 8.695\% \approx 8.70\%

The MP:CP Ratio Question

SSC CHSL loves asking: "After giving x% discount, profit is y%. What is MP:CP?"

The approach: let CP = 100. Then SP = 100 + y. Since SP = MP × (100-x)/100:

MP=SP×100100x=(100+y)×100100xMP = \frac{SP \times 100}{100 - x} = \frac{(100+y) \times 100}{100 - x}

Then MP:CP = that value : 100. Simplify. This is a one-step ratio extraction once you have the SP.

Successive Discounts

If two discounts a% and b% are offered one after another on the same item:

Effective Discount%=a+bab100\text{Effective Discount\%} = a + b - \frac{ab}{100}

This is the same structure as the markup-discount net formula. Pattern recognition across these two will save you re-deriving from scratch.

Loss on Total = SP of Some Items

A specific question pattern (tested in 2025 CHSL): "Sells X items for ₹P and suffers a loss equal to the SP of Y items." Here:

This structure looks unfamiliar but solves cleanly in under 60 seconds once you recognize it.

Memory Tricks & Shortcuts

patternNet Effect Formula for Markup + Discount

When a shopkeeper marks up by m% and gives a discount of d%, the net profit or loss percentage is:

Net% = m − d − (m × d)/100

Positive = profit, negative = loss.

Example: markup 25%, discount 20%. Net% = 25 − 20 − (25×20)/100 = 5 − 5 = 0%. Break-even.

Example: markup 80%, discount 48%. Net% = 80 − 48 − (80×48)/100 = 32 − 38.4 = −6.4%. Loss of 6.4%.

Standard method (set CP=100, compute MP, apply discount, subtract): 5 steps, ~50 seconds. This formula: 3 operations, ~15 seconds.

patternDishonest Weight Gain% — Error Over True Given

For a false weight problem: Gain% = (Error ÷ Weight actually given) × 100.

Error = (True weight − False weight). Weight given = False weight.

920 gm used instead of 1 kg: Error = 80, Given = 920. Gain% = 80/920 × 100 = 8.70%.

Students often divide by 1000 (true weight) by instinct — that gives 8%, which is wrong. The denominator is always what the shopkeeper actually gives out, not what he claims.

Standard error (dividing by 1000): wrong answer in same time. This anchor: correct answer, ~10 seconds.

substitutionSP-Loss in Terms of Pen/Item Count

When loss = SP of Y items and total sold = X items at total SP = P:

Step 1: SP per item = P/X. Step 2: Loss = Y × (P/X). Step 3: CP total = P + Loss. Step 4: CP per item = CP total / X.

For "12 pens sold for ₹480, loss = SP of 3 pens": SP/pen = ₹40. Loss = 3×40 = ₹120. CP total = 600. CP/pen = ₹50.

Standard approach (setting up equation with unknown CP, solving): ~70 seconds. This step-by-step substitution: ~25 seconds — no algebra needed.

substitutionMP:CP Ratio via SP as Bridge

Given: discount = d%, profit = p%. Find MP:CP.

Set CP = 100. SP = (100+p). MP = SP × 100/(100−d).

Ratio MP:CP = [100×(100+p)/(100−d)] : 100 = (100+p):(100−d).

Example: 10% discount, 20% profit → MP:CP = 120:90 = 4:3.

You do not need to write any equations. Just plug into the ratio directly. Standard method (equation with variables): 3 variable steps, ~45 seconds. This pattern: 2 arithmetic operations, ~12 seconds.

patternUsual SP from Reduced SP + Known Profit%

If an item is sold at (n/m) of its usual SP and still makes p% profit:

Let CP = 100. Then (n/m) × usual SP = (100+p). Usual SP = (100+p) × (m/n). Profit at usual SP = usual SP − 100.

Example: sold at 9/10 of usual SP → 20% profit. (9/10) × usual SP = 120 → usual SP = 120 × 10/9 = 133.33. Profit% = 33.33%.

Standard method (two variables): ~60 seconds. This one-line derivation: ~18 seconds.

Fast-Solving Framework

When you see a Profit and Loss question in the exam hall, run this decision tree in 10 seconds before writing anything:

Step 1 — Identify what is given and what is asked. Is CP given? Is MP given? Is Discount% given? Is Profit% given?

Step 2 — Spot the question type:

Step 3 — Set CP = 100 as default unless the problem gives you an actual CP value. Working with 100 as base eliminates one multiplication step and lets you read off percentages directly.

Step 4 — Avoid quadratic equations. Every question in this chapter at CHSL level solves linearly. If you are setting up , you have gone off track — restart with CP = 100.

Solved PYQs

Why this question: This is the classic "loss = SP of some items" pattern that appeared in CHSL 2025. Recognizing the structure immediately tells you the solve path.

Previous Year Questionपिछले वर्ष का प्रश्न2025
A person sells ₹12 pens for ₹480 and suffers a loss equal to the selling price of 3 pens. What is the cost price of one pen?
  1. ₹55
  2. ₹50
  3. ₹30
  4. ₹60
Solutionसमाधान
SP of 12 pens = ₹480, SP per pen = ₹40. Loss = SP of 3 pens = 3 × 40 = ₹120. CP of 12 pens = 480 + 120 = ₹600. CP per pen = 600/12 = ₹50.

Solving path: SP per pen = 480/12 = ₹40. Loss = SP of 3 pens = 3 × 40 = ₹120. Since loss means CP = SP + Loss: CP of 12 pens = 480 + 120 = ₹600. CP per pen = 600/12 = ₹50. The trick is recognizing that "loss equals SP of 3 pens" gives you the loss amount directly — no algebraic equation needed.


Why this question: Tests the MP:CP ratio derivation under a combined discount and profit condition, which is one of the three most tested structures in CHSL Profit and Loss.

Previous Year Questionपिछले वर्ष का प्रश्न2023
In a show room, the marked prices of various items are displayed in such a manner that after giving a 10% discount to the customers, the owner of the show room earns a profit of 20%. What is the ratio of the marked price and cost price of each item?
  1. 4:3
  2. 3:4
  3. 5:2
  4. 2:5
Solutionसमाधान
Let CP = 100, then SP = 120 (20% profit). SP = MP × 0.9 = 120, so MP = 120/0.9 = 400/3. Ratio MP:CP = (400/3):100 = 4:3.

Solving path: Set CP = 100. Profit = 20%, so SP = 120. Discount = 10%, so SP = MP × 0.9 = 120, giving MP = 120/0.9 = 400/3. Ratio MP:CP = (400/3):100 = 4:3. Using the shortcut: MP:CP = (100+20):(100−10) = 120:90 = 4:3.


Why this question: This 2024 question uses the "drop in SP = percentage increase in loss" structure to back-calculate the CP — a reverse-engineering approach that trips students who try to forward-solve.

Previous Year Questionपिछले वर्ष का प्रश्न2024
A seller decreased the selling price of each item from ₹5,000 to ₹4,680, by which his loss percentage increased by 4%. If he has to get 4% profit, then the selling price of the item should be:
  1. ₹8,320
  2. ₹7,280
  3. ₹8,840
  4. ₹7,800
Solutionसमाधान
The drop from ₹5000 to ₹4680 (₹320 difference) represents a 4% increase in loss, so cost price = 320/0.04 = ₹8,000. For 4% profit: SP = 8000 × 1.04 = ₹8,320.

Solving path: The SP dropped from ₹5,000 to ₹4,680, a drop of ₹320. This ₹320 drop caused the loss percentage to increase by 4%. So 4% of CP = ₹320, meaning CP = 320/0.04 = ₹8,000. For 4% profit: SP = 8000 × 1.04 = ₹8,320. The key step is translating "4% increase in loss%" into a rupee value linked to CP.


Why this question: Split-discount problems appeared in 2024 and test whether you can track total discount in rupees versus percentages — a bookkeeping question more than a formula question.

Previous Year Questionपिछले वर्ष का प्रश्न2024
What is the discount that Rohan should offer on the remaining ₹8,000 of a laptop priced at ₹48,000, given that, he has already given a 12% discount on the first ₹28,000 and 8% discount on the next ₹12,000 to match the discount amount of 9.5% given on the total price?
  1. ₹402
  2. ₹240
  3. ₹204
  4. ₹420
Solutionसमाधान
Total discount needed = 9.5% of 48000 = ₹4560. Discount on first ₹28000 = 12% × 28000 = ₹3360. Discount on next ₹12000 = 8% × 12000 = ₹960. Remaining discount = 4560 − 3360 − 960 = ₹240.

Solving path: Total discount required = 9.5% of 48,000 = ₹4,560. Discount on first ₹28,000 at 12% = ₹3,360. Discount on next ₹12,000 at 8% = ₹960. Remaining discount = 4,560 − 3,360 − 960 = ₹240. Work entirely in rupees — do not convert back to percentages mid-way.


Why this question: The dishonest shopkeeper with false weights appears in almost every CHSL cycle. This 2023 question is a clean test of the gain% formula, and the wrong-denominator trap is live here.

Previous Year Questionपिछले वर्ष का प्रश्न2023
A dishonest shopkeeper professes to sell grains at the cost price, but he uses a weight of 920 gm for 1 kg weight. Find his gain percentage (round up to two decimal places).
  1. 8.50%
  2. 7.80%
  3. 7.90%
  4. 8.70%
Solutionसमाधान
Gain% = (Error/(True value − Error))×100 = (80/920)×100 = 8.695% ≈ 8.70%.

Solving path: True weight = 1000 gm, false weight used = 920 gm. Error = 80 gm. Gain% = (80/920) × 100 = 8.695% ≈ 8.70%. The denominator is 920 (what the shopkeeper actually gives), not 1000 (what he claims to give). Using 1000 gives 8% — tempting but wrong.


Why this question: The "reduced SP, known profit → find profit at usual SP" pattern requires a clean chain from CP to usual SP via the given fraction. It appeared in CHSL 2023 and is a pure pattern-recognition question.

Previous Year Questionपिछले वर्ष का प्रश्न2023
If an article is sold at 9/10th of its usual selling price, 20% profit would be made on it. Find the profit percentage which would have been made on it on selling it at its usual selling price.
  1. 20%
  2. 25%
  3. 33.33%
  4. 30%
Solutionसमाधान
Let CP=100. At 9/10 of SP, profit=20%, so 9/10×SP=120, SP=133.33. Profit at usual SP = 133.33–100=33.33, so profit% = 33.33%.

Solving path: Let CP = 100. Selling at 9/10 of usual SP gives 20% profit, so (9/10) × usual SP = 120. Usual SP = 120 × (10/9) = 133.33. Profit at usual SP = 133.33 − 100 = 33.33. Profit% = 33.33%. The fraction 9/10 is your entry point — multiply both sides by 10/9 immediately.

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