Percentages, Profit and Loss, Discounts and Interest Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Percentages, Profit and Loss, Discounts and Interest notes for XAT: 32 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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.

The Base Trap: Profit on CP vs SP

Distinguishing the Base

Statement	Meaning	Formula
"20% Profit"	On Cost Price (default)	$SP = 1.2 \times CP$
"20% Profit on SP"	On Selling Price	$Profit = 0.2 \times SP$
"20% Margin"	Usually on SP	$CP = 0.8 \times SP$

Conversion identity — if profit is

x %

on Selling Price:

Profit % on CP = \frac{x}{100 - x} \times 100

Quick check: 20% profit on SP

\to \frac{20}{80} \times 100 = 25%

profit on CP. Same deal, two different-looking answers — the base decides which one is right.

Method: Weighted Average for Mixed Profits

Portfolio Profit Calculation

Scenario: two or more items with different quantities, unit costs and profit rates.

Compute total cost value of each component

V_{A} = Q_{A} \times P_{A}, V_{B} = Q_{B} \times P_{B}

Compute individual profits on each value

Π_{A} = V_{A} \times r_{A}, Π_{B} = V_{B} \times r_{B}

Combine for overall profit percentage

Net % = \frac{Π _{A} + Π _{B}}{V _{A} + V _{B}} \times 100

Watch out: ratios given in questions are often weight ratios, not value ratios. Always convert to value using the unit price before mixing percentages.

Percentages, Profit and Loss, Discounts and Interest: Solved Questions with Step-by-Step Explanations (2 Problems)

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

In a revenue optimization model where Profit $Π (x)$ is a quadratic function of price change $x$ , given by $Π (x) = k (x - a) (b - x)$ with $a < b$ , at what value of $x$ does the maximum profit occur?

A.
$\frac{a + b}{2}$
B.
$\frac{b - a}{2}$
C.
$ab$
D.
$a + b$

Correct Answer:

Step-by-Step Solution

Key idea: This is a conceptual question about the properties of quadratic functions in optimization.

Step 1: Recognize the function form.

The profit function is given in intercept form: $Π (x) = k (x - a) (b - x)$ .

The roots of this quadratic equation (where Profit = 0) are $x = a$ and $x = b$ .

Step 2: Apply the vertex property.

For any parabola, the vertex (maximum or minimum point) lies exactly on the axis of symmetry.

The axis of symmetry is always at the midpoint of the roots.

Step 3: Calculate the midpoint.

$x_{v er t e x} = \frac{Root _{1} + Root _{2}}{2} = \frac{a + b}{2}$

Answer: $\frac{a + b}{2}$

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

A food stall has a capacity constraint stating that "no more than 400 plates can be produced". If the stall currently plans to produce $P$ plates, which of the following expressions represents the maximum number of additional plates they can produce?

A.
$400 - P$
B.
$P - 400$
C.
$400 + P$
D.
$400 im es P$

Correct Answer:

Step-by-Step Solution

Key idea: This is a constraint translation question, converting a verbal limit into an algebraic expression for remaining capacity.

Step 1: The total capacity limit is 400 plates.

Step 2: The number of plates already planned is $P$ .

Step 3: The remaining capacity is the difference between the total limit and the planned amount.

Step 4: Remaining Capacity = $Total Limit - Planned Amount = 400 - P$ .

Step 5: This expression represents the maximum additional plates that can be produced without exceeding the constraint.

Answer: $400 - P$

More notes in this unit

chapter

Percentages, Profit and Loss, Discounts and Interest Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Percentages, Profit and Loss, Discounts and Interest notes for XAT: 32 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

A question from this chapter

Question 1

In a revenue optimization model where Profit Π (x) is a quadratic function of price change x, given by Π (x) = k (x - a) (b - x) with a < b, at what value of x does the maximum profit occur?

Question 2

A food stall has a capacity constraint stating that "no more than 400 plates can be produced". If the stall currently plans to produce P plates, which of the following expressions represents the maximum number of additional plates they can produce?

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