Percentages, Profit and Loss, Discounts and Interest Previous Year Questions (PYQs) for XAT: 8+ Solved Questions with Step-by-Step Solutions

Solve 8+ Percentages, Profit and Loss, Discounts and Interest previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Percentages, Profit & Loss, Discounts and Interest

Chapter Journey

XAT · QA & DI

1. Profit, Loss & Revenue OptimizationCurrent

Core concepts · Maxima and minima · High weightage

You are here: foundations of commercial arithmetic and optimization.

2. Discount Coupons & Pricing Strategies

Successive discounts · Coupon comparison logic

3. Interest, Loans & Percentage Applications

Simple vs compound interest · Repayment schedules

Topic Hero: Beyond Basic Profit

The Exam Perspective

Every problem in this topic reduces to one identity:

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

. Everything else is built on top of it.

Accounting identity

Profit = Revenue − Cost. Any transaction, however complex, collapses to this line.

Percentage framework

The base of every percentage matters. Profit on CP and profit on SP are different numbers for the same deal.

Optimization

Choose the price or allocation that maximizes profit or revenue, subject to demand and capacity rules. This is the layer advanced exams test most.

Mindset shift: do not memorize shortcuts such as "profit on SP". Derive them from the definition every time — that single habit removes the most common trap options.

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

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

The market value of beams, made of a rare metal, has a unique property: the market value of any such beam is proportional to the square of its length. Due to an accident, one such beam got broken into two pieces having lengths in the ratio 4:9. Considering each broken piece as a separate beam, how much gain or loss, with respect to the market value of the original beam before the accident, is incurred?

A.
74.23% gain
B.
42.60% loss
C.
31.77% loss
D.
57.40% loss
E.
No gain or loss

Correct Answer:

Step-by-Step Solution

Key idea: This is a profit/loss problem involving a non-linear relationship (value proportional to the square of length). The key is to use the ratio of lengths to find the ratio of values.

Step 1: Understand the relationship.

Value $V \propto L^{2}$ .

This means $V = k \cdot L^{2}$ for some constant $k$ .

Step 2: Define the original beam.

The beam is broken into two pieces in the ratio 4:9.

Let the lengths of the pieces be $4 x$ and $9 x$ .

The original length of the beam was the sum of the pieces: $L_{or i g ina l} = 4 x + 9 x = 13 x$ .

Step 3: Calculate the original value.

$V_{or i g ina l} = k \cdot (13 x)^{2} = 169 k x^{2}$ .

Step 4: Calculate the new total value.

The beam is now considered as two separate beams.

Value of piece 1: $V_{1} = k \cdot (4 x)^{2} = 16 k x^{2}$ .

Value of piece 2: $V_{2} = k \cdot (9 x)^{2} = 81 k x^{2}$ .

Total new value: $V_{n e w} = V_{1} + V_{2} = 16 k x^{2} + 81 k x^{2} = 97 k x^{2}$ .

Step 5: Calculate the loss percentage.

Since $V_{n e w} < V_{or i g ina l}$ , there is a loss.

Loss = $V_{or i g ina l} - V_{n e w} = 169 k x^{2} - 97 k x^{2} = 72 k x^{2}$ .

Loss % = $\frac{Loss}{V _{or i g ina l}} \times 100 = \frac{72 k x ^{2}}{169 k x ^{2}} \times 100$ .

Loss % = $\frac{72}{169} \times 100 \approx 42.6035%$ .

Rounding to two decimal places, the loss is 42.60%.

Answer: 42.60% loss

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Common Description: A store offers a choice of ve different discount coupons to its customers, described as follows: Coupon A: A at discount of Rs. 250 on a minimum spend of Rs. 1200 in one transaction.
Coupon B: A 15% discount on a minimum spend of Rs. 500 in one transaction, up to a maximum discount of Rs. 300.
Coupon C: A at discount of Rs. 100 on a minimum spend of Rs. 600 in one transaction.
Coupon D: A 10% discount on a minimum spend of Rs. 250 in one transaction, up to a maximum discount of Rs. 100.
Coupon E: A at discount of Rs. 50 on a minimum spend of Rs. 200 in one transaction.
The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction. A family wanted to purchase four products worth Rs. 1000 each, and another product worth Rs. 300. They were told that they could: I) pay for the ve products through one or more transactions in any way they wanted, as long as the purchase amount of any one product would not get split into different transactions, and
II) use the same discount coupon repeatedly for separate transactions, if they opt for more than one transaction.
What was the maximum discount that they could obtain for their purchase?

A.
Rs. 600
B.
Rs. 645
C.
Rs. 650
D.
Rs. 700
E.
None of the remaining options is correct.

Correct Answer:

Step-by-Step Solution

Key idea: This is a combinatorial optimization problem involving discount coupons. The goal is to maximize the total discount by splitting the purchase into transactions that best utilize the coupon conditions (thresholds, caps, and flat vs percentage benefits).

Step 1: Analyze the items and total value.

Items: Four products of Rs. 1000 each, one product of Rs. 300.

Total Value = $4 \times 1000 + 300 = 4300$ .

Constraint: A single product cannot be split across transactions. We can group these 5 items into any number of transactions.

Step 2: Analyze the coupons to find the most efficient ones.

Coupon A: Flat 250 off min spend 1200. Effective rate $\approx 20.8%$ at threshold.

Coupon B: 15% off min spend 500, max discount 300. Max benefit achieved at spend $S$ where $0.15 S = 300 \Rightarrow S = 2000$ . For $S < 2000$ , discount is $0.15 S$ .

Coupon C: Flat 100 off min spend 600. Low value.

Coupon D: 10% off min spend 250, max discount 100. Max benefit at $S = 1000$ .

Coupon E: Flat 50 off min spend 200. Low value.

Comparison:

Coupon B is very strong for bills between 500 and 2000. At bill 1000, discount is 150. At bill 2000, discount is 300.
Coupon A gives 250 only if bill $\geq 1200$ .
Coupon D gives 100 only if bill $\geq 1000$ (since $10%$ of $1000 = 100$ , which is the cap).

Step 3: Evaluate transaction strategies.

We have items: ${1000, 1000, 1000, 1000, 300}$ .

Strategy 1: One single transaction.

Total Bill = 4300.

Best coupon?

A: 250 (min 1200 met).
B: $15%$ of $4300 = 645$ , but capped at 300. So discount = 300.
Others are lower.

Max Discount = 300. (Too low).

Strategy 2: Split into two transactions.

Possible splits of items:

Case 2a: ${1000, 1000}$ and ${1000, 1000, 300}$ .

Transaction 1 (Bill 2000): Use Coupon B. Discount = $min (15% \times 2000, 300) = 300$ .

Transaction 2 (Bill 2300): Use Coupon B. Discount = $min (15% \times 2300, 300) = 300$ .

Total Discount = $300 + 300 = 600$ .

Case 2b: ${1000, 1000, 1000}$ and ${1000, 300}$ .

Transaction 1 (Bill 3000): Use Coupon B. Discount = 300 (capped).

Transaction 2 (Bill 1300): Use Coupon B. Discount = $15% \times 1300 = 195$ . Or Coupon A? Min 1200 met, flat 250. $250 > 195$ . So use A. Discount = 250.

Total Discount = $300 + 250 = 550$ .

Case 2c: ${1000, 1000, 1000, 300}$ and ${1000}$ .

Transaction 1 (Bill 3300): Coupon B. Discount = 300.

Transaction 2 (Bill 1000): Coupon D ( $10%$ of $1000 = 100$ , cap 100). Discount = 100. Coupon B ( $15%$ of $1000 = 150$ ). Discount = 150.

Total Discount = $300 + 150 = 450$ .

Strategy 3: Split into three transactions.

We need to form groups that maximize individual discounts.

High value coupons require higher spends or specific thresholds.

Let's try to get the max 300 from Coupon B twice? That requires two bills $\geq 2000$ . Total needed 4000. We have 4300.

Remaining 300.

Can we make two bills $\geq 2000$ ?

Items: 1000, 1000, 1000, 1000, 300.

Group 1: ${1000, 1000} = 2000$ . Discount (Coupon B) = 300.

Group 2: ${1000, 1000, 300} = 2300$ . Discount (Coupon B) = 300.

Wait, this uses four 1000s and one 300. This is exactly all items.

This is actually Strategy 2 Case 2a. Total = 600.

Can we do better with 3 transactions?

Maybe use Coupon A (250) more effectively?

Coupon A needs min 1200.

If we make a bill of exactly 1200-1999, Coupon A gives 250. Coupon B gives $15%$ .

Break-even for A vs B: $250 = 0.15 S \Rightarrow S = 1666.6$ .

If $S < 1666$ , Coupon A (250) is better than B ( $< 250$ ).

If $S > 1666$ , Coupon B is better (up to cap 300 at 2000).

Let's try to create transactions in the "Sweet Spot" for Coupon A (1200 to 1666) or high efficiency for B.

Try grouping:

T1: ${1000, 300} = 1300$ .

Best Coupon for 1300:

A: 250 (Min 1200 met).
B: $15% \times 1300 = 195$ .

Choose A. Discount = 250.

Remaining items: ${1000, 1000, 1000}$ .

T2: ${1000, 1000} = 2000$ .

Best Coupon for 2000:

B: $15% \times 2000 = 300$ (Cap 300).
A: 250.

Choose B. Discount = 300.

T3: ${1000} = 1000$ .

Best Coupon for 1000:

B: $15% \times 1000 = 150$ .
D: $10% \times 1000 = 100$ (Cap 100).
A: Min 1200 not met.

Choose B. Discount = 150.

Total Discount = $250 + 300 + 150 = 700$ .

Let's check if we can improve T3.

Is there a better way to split the remaining 3000 (three 1000s)?

Current split: 2000 (Disc 300) + 1000 (Disc 150) = 450.

Alternative split for 3000:

Option X: One transaction 3000. Disc 300 (Cap B). Lower.

Option Y: Three transactions of 1000. Disc $150 \times 3 = 450$ . Same.

So with T1 fixed as 1300 (Disc 250), the rest yield 450. Total 700.

Can we improve T1?

What if we don't pair 300 with a 1000?

If 300 is alone: Bill 300.

Coupons:

E: Min 200. Flat 50.
D: Min 250. $10%$ of $300 = 30$ .
B: Min 500. No.

Max disc for 300 is 50 (Coupon E).

Remaining: Four 1000s.

Best way to split four 1000s?

Two transactions of 2000: Disc $300 + 300 = 600$ .

Total = $600 + 50 = 650$ . (Lower than 700).

Four transactions of 1000: Disc $150 \times 4 = 600$ .

Total = $600 + 50 = 650$ .

One 3000, One 1000: Disc $300 + 150 = 450$ .

Total = $450 + 50 = 500$ .

What if we pair 300 with two 1000s?

T1: ${1000, 1000, 300} = 2300$ .

Best Coupon: B. Disc 300.

Remaining: Two 1000s.

T2: ${1000, 1000} = 2000$ . Disc 300.

Total = 600.

What if we pair 300 with three 1000s?

T1: 3300. Disc 300.

T2: 1000. Disc 150.

Total = 450.

Let's re-evaluate the split that gave 700.

T1: 1300 (1000+300). Coupon A gives 250.

T2: 2000 (1000+1000). Coupon B gives 300.

T3: 1000. Coupon B gives 150.

Sum = 700.

Is there any constraint violation?

"At most one coupon in one transaction". Checked.

"Same coupon repeatedly". Checked.

"Purchase amount of any one product would not get split". Checked.

Can we get more than 700?

Max possible discount per item roughly:

1000 item: Max disc 150 (via B).

300 item: Max disc 50 (via E) or part of a larger bill.

If we treat them independently: $4 \times 150 + 50 = 650$ .

But combining allows accessing Coupon A (250 off 1200).

Efficiency of A on 1300 bill: $250/1300 \approx 19.2%$ .

Efficiency of B on 1000 bill: $150/1000 = 15%$ .

Efficiency of B on 2000 bill: $300/2000 = 15%$ .

In the 700 scenario:

Bill 1300 gets 250. (Efficiency 19.2%)

Bill 2000 gets 300. (Efficiency 15%)

Bill 1000 gets 150. (Efficiency 15%)

Weighted average is higher because the 300 item "piggybacked" on a 1000 item to cross the 1200 threshold for Coupon A, gaining a flat 250 instead of just small discounts.

Let's check if we can use Coupon A twice.

Need two bills $\geq 1200$ .

Items: 1000, 1000, 1000, 1000, 300.

T1: 1000 + 300 = 1300. Disc 250 (A).

Remaining: 1000, 1000, 1000.

Can we make another bill $\geq 1200$ ?

T2: 1000 + 1000 = 2000.

For 2000, Coupon A gives 250. Coupon B gives 300. So we use B. Disc 300.

T3: 1000. Disc 150.

Total 700.

What if we split differently to get two A coupons?

T1: 1000 + ? No other small items.

We only have one 300.

To get a second bill $\geq 1200$ using only 1000s, we must combine at least two 1000s.

T2: 1000 + 1000 = 2000.

As seen, B is better than A for 2000.

What if we make a bill between 1200 and 1666 using two 1000s? Impossible, sum is 2000.

So we can only trigger Coupon A efficiently on the 1300 bill.

Is there a way to get more than 150 on the last 1000?

No, max discount on 1000 is 150 (Coupon B).

Is there a way to get more than 300 on the 2000 bill?

No, cap is 300.

Is there a way to get more than 250 on the 1300 bill?

Coupon B gives 195. Coupon A gives 250.

So 700 seems to be the maximum.

Answer: Rs. 700.

Wait, let me double check Option B: 645.

And Option C: 650.

And Option D: 700.

Let's re-read carefully.

Coupon B: 15% discount ... up to a maximum discount of Rs. 300.

Coupon A: Flat 250 ... min spend 1200.

My calculation:

T1: 1300. Coupon A -> 250.

T2: 2000. Coupon B -> 300.

T3: 1000. Coupon B -> 150.

Total: 700.

Is there any hidden trap?

"Customers are allowed to use at most one coupon in one transaction".

"Use the same discount coupon repeatedly".

Let's check if Coupon D can help.

Coupon D: 10% max 100.

On 1000, D gives 100. B gives 150. B is better.

Let's check if Coupon C can help.

Coupon C: Flat 100 min 600.

On 1000, C gives 100. B gives 150. B is better.

Let's check if mixing 300 differently helps.

If we put 300 with two 1000s: 2300.

B gives 300 (cap).

Remaining two 1000s.

Split 1000, 1000.

Each gives 150.

Total $300 + 150 + 150 = 600$ .

If we put 300 with three 1000s: 3300.

B gives 300.

Remaining 1000.

Gives 150.

Total 450.

If we keep 300 separate: 300.

E gives 50.

Four 1000s.

Best split for four 1000s?

Two pairs of 2000: $300 + 300 = 600$ .

Total $600 + 50 = 650$ .

So 700 is indeed the maximum.

Answer: Rs. 700.

Looking at options:

A: 600

B: 645

C: 650

D: 700

E: None

The calculated answer is 700, which corresponds to Option D.

Wait, I selected "B" in the initial thought trace header? No, I must output the correct option letter.

Option D is "Rs. 700".

Let me re-read the options provided in the JSON.

"options": [

"Rs. 600",

"Rs. 645",

"Rs. 650",

"Rs. 700",

"None of the remaining options is correct."

]

Index 0: 600

Index 1: 645

Index 2: 650

Index 3: 700

Index 4: None

So the correct option is the 4th one, which is usually 'D' in A,B,C,D,E format.

Final Answer: Option D.

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Common Description: A store offers a choice of ve different discount coupons to its customers, described as follows: Coupon A: A at discount of Rs. 250 on a minimum spend of Rs. 1200 in one transaction.
Coupon B: A 15% discount on a minimum spend of Rs. 500 in one transaction, up to a maximum discount of Rs. 300.
Coupon C: A at discount of Rs. 100 on a minimum spend of Rs. 600 in one transaction.
Coupon D: A 10% discount on a minimum spend of Rs. 250 in one transaction, up to a maximum discount of Rs. 100.
Coupon E: A at discount of Rs. 50 on a minimum spend of Rs. 200 in one transaction.
The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction. Four customers used four different discount coupons for their respective transactions in such a way that nobody used any discount coupon sub-optimally.(A discount coupon is used sub-optimally if using another discount coupon could have resulted in a higher discount for the same transaction.) What was the minimum combined spend(before application of any discount)?

A.
Rs. 2250
B.
Rs. 2500
C.
Rs. 2350
D.
Rs. 2300
E.
Rs. 1550

Correct Answer:

Step-by-Step Solution

Key idea: This is a coupon optimization problem where we must find the minimum bill amount for which each coupon is the strictly optimal (or joint-optimal) choice. The condition "nobody used any coupon sub-optimally" means the chosen coupon must yield a discount greater than or equal to all other valid coupons for that specific bill amount.

Step 1: Analyze the discount functions for each coupon.

Coupon A: Flat 250 (Valid for $S \geq 1200$ )
Coupon B: $15%$ of $S$ , capped at 300 (Valid for $S \geq 500$ )
Coupon C: Flat 100 (Valid for $S \geq 600$ )
Coupon D: $10%$ of $S$ , capped at 100 (Valid for $S \geq 250$ )
Coupon E: Flat 50 (Valid for $S \geq 200$ )

Step 2: Determine the optimal range for each coupon.

To minimize the total spend, we need the lowest possible $S$ where each coupon is the best choice.

Coupon E: For $S \in [200, 250)$ , E gives 50, others are invalid. E is optimal. Min $S = 200$ .
Coupon D: For $S \geq 250$ , D gives $0.1 S$ . But E gives 50 for $S < 500$ . Since $0.1 S < 50$ for $S < 500$ , D is never better than E. For $S \geq 500$ , B gives $0.15 S$ , which is always $> 0.1 S$ . Thus, Coupon D is NEVER the optimal choice.
Coupon B: B becomes valid at $S = 500$ . At $S = 500$ , B gives $0.15 \times 500 = 75$ . D gives 50, E gives 50. C is invalid. So B is the best. Min $S = 500$ .
Coupon C: C becomes valid at $S = 600$ . At $S = 600$ , C gives 100. B gives $0.15 \times 600 = 90$ . So C is better than B. Min $S = 600$ .
Coupon A: A becomes valid at $S = 1200$ . At $S = 1200$ , A gives 250. B gives $0.15 \times 1200 = 180$ . C gives 100. So A is the best. Min $S = 1200$ .

Step 3: Calculate the minimum combined spend.

The 4 customers must have used A, B, C, and E (since D is never optimal).

Minimum spends:

E: 200

B: 500

C: 600

A: 1200

Total minimum spend = $200 + 500 + 600 + 1200 = 2500$ .

Answer: Rs. 2500

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

The cost of running a movie theatre is Rs. 10,000 per day, plus additional Rs. 5000 per show. The theatre has 200 seats. A new movie released on Friday. There were three shows, where the ticket price was Rs. 250 each for the rst two shows and Rs. 200 for the late-night show.
For all shows together, total occupancy was 80%. What was the maximum amount of pro t possible?

A.
Rs. 1,20,000
B.
Rs. 87,000
C.
Rs. 95,000
D.
Rs. 91,000
E.
Rs. 1,16,000

Correct Answer:

Step-by-Step Solution

Key idea: This is a revenue maximization problem with fixed and variable costs, and a global occupancy constraint. The key is to allocate the total sold tickets to the highest-priced shows first to maximize revenue, then subtract the total costs.

Step 1: Calculate total capacity and total tickets sold.

Total seats per show = 200.

Number of shows = 3.

Total available seats = $200 \times 3 = 600$ .

Total occupancy = 80%.

Total tickets sold = $80%$ of $600 = 480$ .

Step 2: Maximize the revenue.

Ticket prices: Show 1 = 250, Show 2 = 250, Show 3 = 200.

To maximize revenue, we must fill the higher-priced shows first.

Max capacity for Show 1 and Show 2 = $200 + 200 = 400$ seats.

Since we need to sell 480 tickets in total, we fill Show 1 and Show 2 completely (400 seats).

The remaining $480 - 400 = 80$ seats must be sold in Show 3.

Step 3: Calculate maximum revenue.

Revenue from Show 1 & 2 = $400 \times 250 = 100, 000$ .

Revenue from Show 3 = $80 \times 200 = 16, 000$ .

Total Maximum Revenue = $100, 000 + 16, 000 = 116, 000$ .

Step 4: Calculate total costs.

Fixed daily cost = 10,000.

Variable cost per show = 5,000.

Total variable cost for 3 shows = $3 \times 5, 000 = 15, 000$ .

Total Cost = $10, 000 + 15, 000 = 25, 000$ .

Step 5: Calculate maximum profit.

Profit = Total Revenue - Total Cost

Profit = $116, 000 - 25, 000 = 91, 000$ .

Answer: Rs. 91,000

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Jose borrowed some money from his friend at simple interest rate of 10% and invested the entire amount in stocks. At the end of the rst year, he repaid 1/5th of the principal amount. At the end of the second year, he repaid half of the remaining principal amount. At the end of third year, he repaid the entire remaining principal amount. At the end of the fourth year, he paid the last three years’ interest amount. As there was no principal amount left, his friend did not charge any interest in the fourth year. At the end of fourth year, he sold out all his stocks. Later, he calculated that he gained Rs. 97500 after paying principal and interest amounts to his friend. If his invested amount in the stocks became double at the end of the fourth year, how much money did he borrow from his friend?

A.
250000
B.
200000
C.
150000
D.
125000
E.
None of the above

Correct Answer:

Step-by-Step Solution

Key idea: This is a year-by-year reducing principal simple interest problem. The key is to track the outstanding principal at the end of each year to calculate the interest for the next year, and then sum the total interest paid.

Step 1: Define the initial principal.

Let the borrowed amount be $P$ .

Interest rate $R = 10%$ per annum.

Step 2: Track Year 1.

Interest for Year 1 = $10%$ of $P = 0.1 P$ .

Principal repaid at the end of Year 1 = $\frac{1}{5} P = 0.2 P$ .

Remaining principal = $P - 0.2 P = 0.8 P$ .

Step 3: Track Year 2.

Interest for Year 2 = $10%$ of $0.8 P = 0.08 P$ .

Principal repaid at the end of Year 2 = half of remaining = $0.5 \times 0.8 P = 0.4 P$ .

Remaining principal = $0.8 P - 0.4 P = 0.4 P$ .

Step 4: Track Year 3.

Interest for Year 3 = $10%$ of $0.4 P = 0.04 P$ .

Principal repaid at the end of Year 3 = entire remaining = $0.4 P$ .

Remaining principal = 0.

Step 5: Calculate total interest and total amount paid.

Total interest paid over the 3 years = $0.1 P + 0.08 P + 0.04 P = 0.22 P$ .

Total principal repaid = $P$ .

Total amount paid to the friend = $P + 0.22 P = 1.22 P$ .

Step 6: Relate to the stock investment gain.

The invested amount $P$ became double at the end of Year 4, so the selling price = $2 P$ .

Net gain = Selling Price - Total Amount Paid

$97500 = 2 P - 1.22 P$

$97500 = 0.78 P$

$P = \frac{97500}{0.78} = 125000$ .

Answer: 125000

Percentages, Profit and Loss, Discounts and Interest Previous Year Questions (PYQs) for XAT: 8+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Percentages, Profit & Loss, Discounts and Interest

Chapter Journey

Topic Hero: Beyond Basic Profit

The Exam Perspective

Percentages, Profit and 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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