Profit & Loss in CAT 2026: Only 3 Formulas Needed for All Variations

Introduction to Profit and Loss for CAT

To understand the topic, you must understand the basic definitions first, like:

Also, you should understand the question types that are being asked in the CAT:

Why Profit and Loss Is Important in CAT Quant

The number of Profit and Loss questions varies across CAT years and slots, ranging in difficulty from medium to high level. So, this topic should be practised more.

From the analysis of CAT previous years' question papers, we noticed that, along with direct questions from Profit and Loss, Caselet Data Interpretation based on this is also asked.

CAT Profit and Loss Questions: Previous Year Trends (2020-2025)

We have analysed the previous 6 years' papers of CAT to find out the weightage and CAT exam pattern of the questions from profit and loss that are asked in the exam. Here is a summary of the weightage of questions of profit and loss year-wise and slot-wise.

CAT Profit and Loss: Can You Find the Original Price When Only the Final Transaction Is Given?

Some CAT questions provide the final selling price, profit percentage or discount but hide the original cost price or marked price. Instead of calculating forward from the cost price, candidates must work backwards to identify the starting value.

Work Backwards from the Final Selling Price

Suppose an article is sold for ₹1,080 after a discount of 10%. Find its marked price.

$MP = \frac{1080 \times 100}{90} = ₹1,200$

If the same article was marked 20% above its cost price, its original cost price would be:

$CP = \frac{1200 \times 100}{120} = ₹1,000$

Reverse Multiple Price Changes

Consider an article whose price is increased by 25% and then reduced by 20%. The final selling price is ₹1,000. What was its original price?

Let the original price be $x$.

$x \times 1.25 \times 0.80 = 1000$

$x = \frac{1000}{1} = ₹1,000$

The increase and decrease cancel each other in this particular case. However, this does not happen with every pair of percentage changes.

CAT Question-Solving Tip

When the final value is given, reverse the operations in the opposite order:

This approach is useful when a question contains multiple price changes and asks for the original value.

The Only 3 Profit and Loss Formulas You Need for CAT

There are 3 basic formulas in profit and loss that need to be understood to solve any question in Profit and Loss. The rest are the variations of the questions.

Formula 1: Relating SP, CP, and Profit/Loss

$\text{Profit %} = \frac{\text{Profit}}{\text{CP}} \times 100$, where CP is the cost price.

$\text{Loss %} = \frac{\text{Loss}}{\text{CP}} \times 100$, where CP is the cost price.

Profit $= SP - CP$, Loss $= CP - SP$

Other variations of this formula:

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

Key Usage: Basic P&L questions, cost-price/selling-price direct problems.

Example: A shopkeeper buys a book for ₹400 and sells it at ₹460. Find profit %.

Profit $= 60$, CP $= 400$

$\text{Profit %} = \frac{60}{400} \times 100 = 15%$

Formula 2: Relating SP, MP, and Discount

$\text{Discount %} = \frac{\text{Discount}}{\text{MP}} \times 100$, where MP is the marked price.

Discount $= MP - SP$

Other variation:

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

Key Usage: Marked price, successive discounts, successive profit/loss.

Example: A shopkeeper gives two successive discounts of $40%$ and $30%$ respectively. Find the effective discount %.

Let MP $= 100$. After the 1st discount, price $= 60$. After the second discount of $30%$, final SP $= 60 \times 0.7 = 42$.

$\text{Net Discount %} = \frac{100 - 42}{100} \times 100 = 58%$

Alternate solution (Trick):

$\text{Net discount %} = 40 + 30 - \frac{40 \times 30}{100} = 70 - 12 = 58%$

Formula 3: Relating MP, CP, and Markup Value

$\text{Markup %} = \frac{\text{Markup value}}{\text{CP}} \times 100$

Other variation:

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

Key Usage: Marked price, hike in the price of value, successive increase in the price.

Example: A man marked up the price on one article by $20%$ and lost $10%$ on the marked price while selling. Find net % gain/loss.

Let CP $= 100$. After an increase of $20%$, price $= 120$. After the second transaction, final SP $= 120 \times 0.9 = 108$.

$\text{Net gain %} = \frac{108 - 100}{100} \times 100 = 8%$

How to Solve Profit and Loss in CAT Using These Formulas?

A few points must be remembered before solving questions using these 3 formulas:

Also, there are some prerequisite conditions:

• Hands-on percentage to fraction conversion and vice versa.

Step-by-Step Approach to Profit & Loss CAT Questions

Consider the following example:

Question: A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by $p$ percent so as to make an overall profit of 15%. Then $p$ is nearest to: [CAT 2024]

Solution:

Step 1: First, understand the problem statement and the information given in the question.

In this question, the markup percent is to be calculated.

Given:

• Targeted Profit % $= 20%$
• Sold 5 kg at the targeted price.
• Sold 15 kg at a discount of 10%.
• 3 kg of sugar is wasted.

Now, we need to find the percentage increase in the price of the remaining sugar so that the total profit is 15%.

Step 2: Identify the concept required to solve the question.

Formula 1 should be used in between:

$\text{Profit/Loss %} = \frac{\text{Profit or Loss}}{\text{CP}} \times 100%$, where CP is the cost price.

Formula 2 used in between:

$\text{Discount %} = \frac{\text{Discount}}{\text{MP}} \times 100%$, where MP is the marked price.

Formula 3 used in the end:

$\text{Markup %} = \frac{\text{Markup Value}}{\text{CP}} \times 100%$, where CP is the cost price.

Step 3: Apply the concept/formula to find the unknown.

Final Solution:

Let the cost price of 1 kg of sugar $= ₹100$.

The total cost price of 35 kg $= ₹3500$.

Marked price per kg $= ₹120$.

Given, the final profit is 15%. So, final SP of 35 kg $= 3500 \times 1.15 = ₹4025$.

First 5 kg are sold at 20% marked-up price:

$SP_1 = 5 \times 100 \times 1.2 = ₹600$

Next 15 kg are sold after giving 10% discount:

$SP_2 = 15 \times 100 \times 1.2 \times 0.9 = ₹1620$

3 kg of sugar got wasted.

23 kg of sugar was sold at:

$₹(600 + 1620) = ₹2220$

The remaining 12 kg should be sold at:

$₹(4025 - 2220) = ₹1805$

$\Rightarrow \text{SP of 1 kg} = \frac{1805}{12} \approx ₹150$

Hence, the seller should further mark up by:

$\frac{150 - 120}{120} \times 100 = 25%$

Identifying Shortcut Opportunities in CAT Quant

Here are a few shortcuts that can be helpful to solve the questions effectively and save time during the exam.

Solved Past 5 Years’ Questions with Solutions of Profit and Loss for CAT

Q.1) Bina incurs 19% loss when she sells a product at Rs. 4860 to Shyam, who in turn sells this product to Hari. If Bina would have sold this product to Shyam at the purchase price of Hari, she would have obtained 17% profit. Then, the profit, in rupees, made by Shyam is:

Solution: [Expected time to be taken: 45 to 60 seconds, Level of Difficulty: Moderate]

Let the cost price for Bina be $x$.

Bina sells to Shyam at a 19% loss:

$4860 = x - \frac{19}{100}x = \frac{81}{100}x$

So, $x = \frac{4860 \times 100}{81} = 6000$

Now, let the selling price to Hari be $y$.

If Bina had sold to Shyam at Hari's price and made 17% profit:

$y = x + \frac{17}{100}x = \frac{117}{100} \times 6000 = 7020$

So, Shyam bought at Rs. 4860 and sold at Rs. 7020.

Therefore, his profit is:

$7020 - 4860 = 2160$

Hence, the correct answer is 2160.

Q.2) Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is:

Solution: [Expected time to be taken: 45 seconds, Level of Difficulty: Easy]

Let the cost price of the product be $100x$.

Gopi wants to make a $20%$ profit, so the marked price (MP) is:

$\text{MP} = 100x \times \left(1 + \frac{20}{100}\right) = 120x$

Ravi gets 10% discount on the marked price, so the selling price (SP) is:

$\text{SP} = 120x \times \left(1 - \frac{10}{100}\right) = 120 \times \frac{90}{100} = 108x$

Now, the discount Ravi gets:

$120x - 108x = 12x$

But the question says Ravi saves Rs 15, so:

$12x = 15$

$\Rightarrow x = \frac{15}{12} = \frac{5}{4}$

The profit made by Gopi:

$108x - 100x = 8x = 10$ rupees.

Hence, the correct answer is option 3.

Q.3) Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to:

Solution: [Expected time to be taken: 75 to 90 seconds, Level of Difficulty: Moderate to Hard]

Let the original selling price of both A and B be Rs 100.

So, the cost price of A, with a profit of 20%, is:

$\frac{5}{6} \times 100 = \frac{250}{3}$

And, the cost price of B, with a loss of 10%, is:

$\frac{100}{0.9} = \frac{1000}{9}$

After an increase in the selling price by some value, the profit on object B is 10%.

So, the new selling price is 110% of the cost price of B:

$= \frac{1000}{9} \times \frac{110}{100} = \frac{1100}{9}$

Profit of A:

$\frac{1100}{9} - \frac{250}{3} = \frac{350}{9}$

Profit percentage of A:

$\frac{\frac{350}{9}}{\frac{250}{3}} \times 100 = 46.67%$

Nearest to $47%$.

Hence, the correct answer is option (2).

Q.4) Minu purchased a pair of sunglasses for Rs. 1000 and sold them to Kanu for 20% profit. Then, Kanu sells it back to Minu at a 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs. 500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is:

Solution: [Expected time to be taken: 60 to 75 seconds, Level of Difficulty: Moderate]

Initial Cost Price for Minu $= 1000$

Cost Price to Kanu $= 120%$ of $1000 = 1200$

Profit made by Minu $= 1200 - 1000 = 200$

Again, Minu purchased at 80% of 1200:

$= 960$

The total profit made by Minu is 500. So, profit made when she sold it to Tanu:

$= 500 - 200 = 300$

Final Selling Price for Minu when sold to Tanu:

$= 960 + 300 = 1260$

Profit percent:

$\frac{300}{960} \times 100 = 31.25%$

Hence, the correct answer is option (2).

Q.5) Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of Rs. 1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of Rs. 744. Then the amount, in rupees, that she had spent in buying almonds is:

Solution: [Expected time to be taken: 90 to 120 seconds, Level of Difficulty: Hard]

It is given, $7C = 30P = 9A$ and Ankita bought $4C$, $14P$ and $6A$.

Let:

$7C = 30P = 9A = 630k$

$\Rightarrow C = 90k$, $P = 21k$, and $A = 70k$

Cost price of $4C$, $14P$ and $6A$:

$= 4(90k) + 14(21k) + 6(70k) = 1074k$

Marked-up price:

$= 1074k + 1752$

So,

$\text{S.P.} = \frac{1}{6}(1074k + 1752) + \left(\frac{4}{5}\right)\left(\frac{5}{6}\right)(1074k + 1752) = \frac{5}{6}(1074k + 1752)$

$\Rightarrow \text{S.P.} - \text{C.P.} = \text{profit}$

$\Rightarrow 1460 - \frac{1074k}{6} = 744$

$\Rightarrow \frac{1074k}{6} = 716$

$\Rightarrow k = 4$

Money spent on buying almonds:

$= 420k = 420 \times 4 = ₹1680$

The correct answer is Rs. 1680.

Q.6) Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at Rs 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in rupees per kg, is:

Solution: [Expected time to be taken: 90 to 120 seconds, Level of Difficulty: Hard]

Let the price of juice be Rs. $x$ per kg.

Since the cost price of syrup is 20% less than the cost price of juice, the cost price of syrup is $0.8x$ per kg.

Total cost price of syrup:

$= 110 \times 0.8x = 88x$

Total cost price of juice:

$= 120 \times x = 120x$

Total cost price:

$= 88x + 120x = 208x$

Since the overall profit percentage is 64%:

Total profit $= 0.64(208x) = 133.12x$

Profit generated by selling 10 kg of syrup, which costs Rs $0.8x$ per kg, at 10% profit:

$= 0.1 \times 10 \times 0.8x = 0.8x$

Profit generated on selling 20 kg of juice, which costs Rs $x$ per kg, at 20% profit:

$= 0.2 \times 20x = 4x$

The remaining profit:

$133.12x - (0.8x + 4x) = 128.32x$

This is generated by selling 100 kg of syrup and 100 kg of juice at Rs 308.32 per kg.

The total selling price of 100 kg of syrup and 100 kg of juice:

$= 200 \times 308.32 = 2 \times 30832$

The cost price of 100 kg of syrup:

$= 0.8x \times 100 = 80x$

Cost price of 100 kg of juice:

$= 100x$

Total cost price:

$= 80x + 100x = 180x$

Profit = Selling price – Cost price

$\Rightarrow 128.32x = 2 \times 30832 - 180x$

$\Rightarrow 308.32x = 2 \times 30832$

$\Rightarrow x = 200$

Cost price of syrup per kg:

$= 0.8 \times 200 = ₹160$

Hence, the correct answer is option (1).

Common Mistakes to Avoid in Profit & Loss CAT Questions

It is important to understand the common mistakes in the CAT exam that a student makes while solving the Profit and Loss questions.

CAT Topicwise Study Materials

The most important CAT quantitative aptitude topics, along with the CAT quantitative aptitude practice questions, are provided in the table below. The candidates are encouraged to practise CAT quantitative aptitude questions from these topics as they have a high probability of being asked in the CAT examination.

CAT Profit and Loss: What Happens When a Seller Changes the Price After Selling Part of the Stock?

A seller may sell one portion of the stock at a particular price and change the price for the remaining quantity. Such questions require candidates to calculate the revenue from each transaction separately before finding the overall profit.

Calculate the Overall Profit After Two Selling Prices

A shopkeeper purchases 50 notebooks at ₹40 each. He sells 20 notebooks at ₹50 each and the remaining 30 notebooks at ₹45 each. Find his overall profit percentage.

$\text{Profit %} = \frac{350}{2000} \times 100 = 17.5%$

Find the Selling Price Needed to Reach a Target Profit

Suppose the shopkeeper wants to earn an overall profit of 20% instead. How much should he charge for each of the remaining 30 notebooks if the first 20 are already sold at ₹50 each?

Required total revenue:

$2000 \times 1.20 = ₹2,400$

Revenue already earned:

$20 \times 50 = ₹1,000$

Revenue required from the remaining notebooks:

$2400 - 1000 = ₹1,400$

Required selling price per notebook:

$\frac{1400}{30} = ₹46.67$

What Makes This CAT Question Pattern Tricky?

The profit percentage is calculated on the total cost of all 50 notebooks, not just the cost of the remaining stock. Candidates should keep track of:

CAT Profit and Loss: How Do Multiple Buyers and Sellers Change the Final Profit?

In some questions, an article passes through two or more people before reaching the final buyer. Each person purchases the article at one price and sells it at another. The profit of one seller becomes the cost price of the next seller.

Track the Price at Every Transaction

Suppose A purchases an article for ₹800 and sells it to B at a 25% profit. B then sells it to C at a 10% loss. Find the final selling price.

The final selling price is ₹900.

Although A makes a profit of 25%, B incurs a loss of 10%. The overall change from A's original purchase price is:

$\frac{900 - 800}{800} \times 100 = 12.5%$

Find an Unknown Profit Percentage in the Chain

Suppose A purchases an article for ₹1,000 and sells it to B at a 20% profit. B then sells it to C. If C pays ₹1,320, find B's profit percentage.

A's selling price:

$1000 \times 1.20 = ₹1,200$

B's profit:

$1320 - 1200 = ₹120$

B's profit percentage:

$\frac{120}{1200} \times 100 = 10%$

Use Multiplication to Handle Long Transaction Chains

For multiple transactions, represent each profit as a multiplier and each loss as a reduction factor.

For example:

The final price can be calculated as:

$\text{Final Price} = \text{Initial Price} \times 1.20 \times 0.90 \times 1.15$

This method avoids calculating every intermediate profit percentage separately and is useful when an article passes through several buyers and sellers.

CAT 2026 Preparation Resources by Careers360

Download free CAT 2026 preparation resources from Careers360, including five-year PYQ question banks for all three sections, sample papers, formula sheets and revision plans. These eBooks and practice materials help you practise section-wise questions and revise key concepts before the exam.