Profit, Loss, Discounts and Interest Previous Year Questions (PYQs) for CAT: 21+ Solved Questions with Step-by-Step Solutions

Solve 21+ Profit, Loss, Discounts and Interest previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Profit, Loss, Discounts and Interest

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Chapter Journey

Profit, Loss, Discounts and Interest

Step 1 • 9 CAT PYQs • Importance 0.57

🧾 Profit, Loss and Marked Price

Master CP, SP, MP, profit %, loss %, same-selling-price puzzles, overheads, and transaction chains.

Step 2 • 6 CAT PYQs • Importance 0.45

🏷️ Discounts, Pricing and Trade Schemes

Handle successive discounts, offers, bundles, and hidden pricing logic.

Step 3 • 10 CAT PYQs • Importance 0.62

📈 Simple and Compound Interest

Move from basic interest to compounding, installments, and rate-time traps.

Step 4 • 5 CAT PYQs • Importance 0.41

🤝 Investment, Returns and Business Shares

Use ratios, weighted returns, and partnership logic to split gains fairly.

By the end of this chapter: you should be able to convert every money question into a clean equation of cost, selling value, profit, loss, or return.

Topic Hero: Profit, Loss and Marked Price

CAT Quant • Arithmetic

Profit, Loss and Marked Price

The art of tracking money: what you paid, what you tagged, what you received, and what you gained or lost.

Cost Price

Marked Price

Selling Price

Core CAT skill: choose the correct base. Profit and loss are usually on CP. Discount is usually on MP. Rupee profit is SP minus CP.

Profit, Loss, Discounts and Interest: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to

A.
3221
B.
3180
C.
3150
D.
3096

Correct Answer:

Step-by-Step Solution

Key idea: this is a two-stage interest question, recognisable because it first gives two SI amounts to find the principal and rate, and then applies CI with half-yearly compounding.

Step 1: Use the SI data to find Principal ( $P$ ) and Rate ( $R$ ).

Amount after 3 years = 13920.
Amount after 6.5 years = 18960.

Step 2: The difference in time is $6.5 - 3 = 3.5$ years. The difference in amount is purely due to SI on $P$ for 3.5 years.

Step 3: SI for 3.5 years = $18960 - 13920 = 5040$ .

Step 4: SI for 1 year = $\frac{5040}{3.5} = 1440$ .

Step 5: SI for 3 years = $1440 \times 3 = 4320$ .

Step 6: Since Amount = $P$ + SI, $P + 4320 = 13920 ⟹ P = 9600$ .

Step 7: Find the annual rate $R$ : $1440 = \frac{9600 \times R}{100} ⟹ R = \frac{1440}{96} = 15%$ .

Step 8: Now, $P = 9600$ is invested for 2 years at 15% p.a., compounded half-yearly.

Step 9: For half-yearly compounding, the rate per period is $\frac{15}{2} = 7.5%$ , and the number of periods is $2 \times 2 = 4$ .

Step 10: Calculate the final Amount $A = 9600 \times (1 + 0.075)^{4} = 9600 \times (1.075)^{4}$ .

Step 11: $(1.075)^{2} = 1.155625$ . Squaring again: $(1.155625)^{2} \approx 1.335469$ .

Step 12: $A \approx 9600 \times 1.335469 \approx 12820.5$ .

Step 13: Total interest earned = $A - P = 12820.5 - 9600 = 3220.5$ .

Nearest to 3221.

Answer: Option A.

Question 2 · Quantitative Ability MCQ

Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 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

A.
35.42%
B.
52%
C.
31.25%
D.
26%

Correct Answer:

Step-by-Step Solution

Key idea: This is a transaction chain with effective cost question. The trigger words are "buys, sells, buys back, sells again" and "total profit from all transactions".

Why this method: We need to track cash flows through multiple transactions and distinguish between total profit and profit on a specific transaction.

Step-by-step working:

Track Transaction 1: Minu buys from original seller

Cash outflow: ₹1000
Minu now has the sunglasses
Running cash: $- 1000$

Track Transaction 2: Minu sells to Kanu at 20% profit

Selling price = $1000 \times 1.20 = 1200$
Cash inflow: ₹1200
Running cash: $- 1000 + 1200 = + 200$
Profit so far: ₹200

Track Transaction 3: Kanu sells back to Minu at 20% loss

Kanu's cost was ₹1200, sells at 20% loss = $1200 \times 0.80 = 960$
Minu pays ₹960 to buy back
Running cash: $200 - 960 = - 760$
Minu now has the sunglasses again

Determine Minu's effective cost for the final sale:

Minu's net cash position before final sale: $- 760$
This means Minu's effective cost basis for the sunglasses is ₹760
Alternatively: For the specific transaction of buying back at ₹960 and selling to Tanu, the cost is ₹960

Use total profit to find final selling price:

Total profit = Final cash position = ₹500
Final cash = $- 760 + S P_{3} = 500$
$S P_{3} = 1260$

Calculate profit percentage on the last transaction:

The question asks for profit % "when she sold to Tanu"
For this specific transaction, cost = ₹960 (buy-back price)
Selling price = ₹1260
Profit = $1260 - 960 = 300$
Profit % = $\frac{300}{960} \times 100% = 31.25%$

Answer: The percentage of profit made by Minu when she sold to Tanu is 31.25%.

Question 3 · Quantitative Ability MCQ

Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is

A.
20860
B.
20640
C.
20480
D.
20808

Correct Answer:

Step-by-Step Solution

This is a "same final amount, different route" pattern - you can tell because the question says "the amounts received by both at the end of 6 years are equal." Whenever two people end up with the same money through different investment paths, the method is: write an expression for each person's final amount, then set them equal and solve for the unknown.

Step 1 - Find Anil's final amount.

Anil invests Rs 22000 for 6 years at 4% per annum compounded half-yearly. "Compounded half-yearly" means the rate per period is half the annual rate, and the number of periods is doubled.

Rate per half-year = 4/2 = 2%. Number of half-years in 6 years = 12.

$A_{A ni l} = 22000 (1 + \frac{2}{100})^{12} = 22000 (1.02)^{12}$

Computing $(1.02)^{12} \approx 1.26824$ , so

$A_{A ni l} \approx 22000 \times 1.26824 \approx 27901.32$

Step 2 - Write Sunil's final amount in terms of his unknown investment P.

Sunil invests P in the SAME scheme (4% p.a., compounded half-yearly) for 5 years. That is 10 half-year periods at 2%.

$Value after 5 years = P (1.02)^{10}$

Using $(1.02)^{10} \approx 1.21899$ .

Step 3 - Apply the second stage.

This entire amount is then reinvested for 1 more year at 10% SIMPLE interest (not compound - this is the key trigger word to watch).

$A_{S u ni l} = P (1.02)^{10} \times (1 + \frac{10 \times 1}{100}) = P (1.02)^{10} (1.10)$

$A_{S u ni l} \approx P \times 1.21899 \times 1.10 \approx P \times 1.34089$

Step 4 - Equate the two final amounts (this is what the question tells us to do).

$P \times 1.34089 = 27901.32$

$P = \frac{27901.32}{1.34089} \approx 20808$

Trap to avoid: it's tempting to apply the same half-yearly compounding logic to Sunil's second year too, but the question clearly switches to simple interest for that final year - mixing this up gives a wrong answer close to, but not exactly, the correct one.

The initial investment made by Sunil is Rs 20808, option D.

Question 4 · Quantitative Ability MCQ

Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to

A.
929.48
B.
934.65
C.
931.72
D.
926.84

Correct Answer:

Step-by-Step Solution

Key idea: this is a compound-interest yearly-interest question, recognisable because the interests accrued during successive years are given.

Step 1: For compound interest compounded annually, the interest earned in successive years grows by the factor $(1 + r)$ , where $r$ is the annual rate in decimal form.

Step 2: Let the interest during the second year be $I_{2} = 806.25$ and during the third year be $I_{3} = 866.72$ .

Step 3: The growth factor from the second-year interest to the third-year interest is:

$1 + r = \frac{I _{3}}{I _{2}} = \frac{866.72}{806.25} .$

Step 4: Approximate the ratio. Since:

$806.25 \times 0.075 = 60.46875,$

the increase $60.47$ is very close to $7.5%$ of $806.25$ . Thus:

$1 + r \approx 1.075.$

Step 5: The fourth-year interest is the third-year interest multiplied by the same factor:

$I_{4} = I_{3} (1 + r) .$

Step 6: Substitute:

$I_{4} \approx 866.72 \times 1.075.$

Step 7: Calculate:

$866.72 \times 1.075 = 866.72 + 65.004 = 931.724.$

This is nearest to $931.72$ .

Answer: option C.

Question 5 · Quantitative Ability MCQ

An item with a cost price of Rs. 1650 is sold at a certain discount on a fixed marked price to earn a profit of 20% on the cost price. If the discount was doubled, the profit would have been Rs. 110. The rate of discount, in percentage, at which the profit percentage would be equal to the rate of discount, is nearest to

A.
16
B.
18
C.
14
D.
12

Correct Answer:

Step-by-Step Solution

Key idea: this is a linear pricing model question, recognisable because changing the discount percentage changes the profit by a fixed amount, allowing us to find the marked price.

Step 1: Let Marked Price be $M$ and original discount be $d %$ .

Step 2: Cost Price (CP) = 1650. Profit = 20% of 1650 = 330. Selling Price (SP1) = 1650 + 330 = 1980.

Step 3: We know $S P 1 = M (1 - \frac{d}{100}) = 1980$ .

Step 4: If discount is doubled to $2 d %$ , profit is 110. So $S P 2 = 1650 + 110 = 1760$ .

Step 5: We know $S P 2 = M (1 - \frac{2 d}{100}) = 1760$ .

Step 6: Subtract the SP2 equation from SP1: $M (\frac{d}{100}) = 1980 - 1760 = 220$ .

Step 7: Substitute $M (\frac{d}{100}) = 220$ into the first equation: $M - 220 = 1980 ⟹ M = 2200$ .

Step 8: Find original discount rate: $2200 \times \frac{d}{100} = 220 ⟹ 22 d = 220 ⟹ d = 10%$ .

Step 9: Now, we need a new discount rate $x %$ such that Profit% = $x$ .

Step 10: New SP = $2200 (1 - \frac{x}{100}) = 2200 - 22 x$ .

Step 11: Profit = SP - CP = $2200 - 22 x - 1650 = 550 - 22 x$ .

Step 12: Profit% on CP = $\frac{550 - 22 x}{1650} \times 100 = x$ .

Step 13: Solve for $x$ : $(550 - 22 x) \times 100 = 1650 x ⟹ 55000 - 2200 x = 1650 x ⟹ 3850 x = 55000$ .

Step 14: $x = \frac{55000}{3850} = \frac{100}{7} \approx 14.28%$ .

Nearest integer is 14.

Answer: Option C.

Profit, Loss, Discounts and Interest Previous Year Questions (PYQs) for CAT: 21+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Profit, Loss, Discounts and Interest

Profit, Loss, Discounts and Interest

🧾 Profit, Loss and Marked Price

🏷️ Discounts, Pricing and Trade Schemes

📈 Simple and Compound Interest

🤝 Investment, Returns and Business Shares

Topic Hero: Profit, Loss and Marked Price

Profit, Loss and Marked Price

Profit, Loss, Discounts and Interest: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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