Profit & Loss – AFCAT Study Notes
(Ideal for IIT‑JEE & competitive exam preparation)
AFCAT – Armed Forces Common Admission Test
Topic – Quantitative Aptitude – Profit & Loss
These notes combine concise theory, derivations, key formulae, and a challenging practice problem that mirrors the style of AFCAT questions.
1. Core Definitions
| Term | Symbol | Formula | Interpretation |
|---|---|---|---|
| Cost Price (CP) | (C) | – | Price paid to buy the item. |
| Selling Price (SP) | (S) | – | Price at which the item is sold. |
| Profit | (P) | (P = S - C) | Amount gained over cost. |
| Loss | (L) | (L = C - S) | Amount lost over cost. |
| Profit/Loss Percentage | (%) | (\dfrac{P}{C}\times 100) or (\dfrac{L}{C}\times 100) | Relative change with respect to cost. |
| Markup | – | (\dfrac{S-C}{C}\times 100) | Same as profit % (source‑based). |
| Discount | – | (\dfrac{C-S}{C}\times 100) | Same as loss % (source‑based). |
| Revenue | – | (S \times Q) | Total income from selling (Q) units. |
| Profit on Total Sale | – | ((S-C)\times Q) | Total profit from (Q) units. |
2. Fundamental Relationships
2.1 Profit / Loss in Percent
[ \boxed{ \text{Profit %} = \frac{S-C}{C}\times 100 } ]
[ \boxed{ \text{Loss %} = \frac{C-S}{C}\times 100 } ]
Derivation
Starting with (P = S-C) and normalizing by cost:
[ \text{Profit %} = \frac{P}{C}\times 100 = \frac{S-C}{C}\times 100 ]
Similarly for loss.
2.2 Reverse Calculation – Finding CP or SP from Percentage
| Given | Formula |
|---|---|
| Profit % = (x%) and CP = (C) | (S = C(1 + \frac{x}{100})) |
| Loss % = (x%) and CP = (C) | (S = C(1 - \frac{x}{100})) |
| SP = (S) and Profit % = (x%) | (C = \dfrac{S}{1 + \frac{x}{100}}) |
| SP = (S) and Loss % = (x%) | (C = \dfrac{S}{1 - \frac{x}{100}}) |
2.3 Multiple Items – Weighted Average
If two items with CPs (C_1, C_2) and quantities (q_1, q_2) are sold at the same SP (S), then
[ \text{Overall Profit %} = \frac{(S-C_{\text{avg}})}{C_{\text{avg}}}\times 100 ] where [ C_{\text{avg}} = \frac{C_1q_1 + C_2q_2}{q_1+q_2} ]
3. Advanced Topics & Tricks
| Scenario | Key Insight | Formula |
|---|---|---|
| Price increases by (x%) and then decreases by (y%) | Net change ≠ (x-y) (non‑linear) | (\text{Net factor} = (1+\frac{x}{100})(1-\frac{y}{100})) |
| Buying at a discount | Discount is taken from CP | (C_{\text{net}} = C(1-\frac{d}{100})) |
| Selling at a discount on marked price | Discount on SP | (S = M(1-\frac{d}{100})) |
| Profit on multiple transactions | Use total CP and total SP | (\text{Total Profit} = \sum S_i - \sum C_i) |
| Compound profit/loss | Two successive operations | (S_{\text{final}} = C(1+\frac{p_1}{100})(1-\frac{l_2}{100})) |
| Profit on selling at different rates | Weighted average of rates | (S_{\text{avg}} = \frac{S_1q_1 + S_2q_2}{q_1+q_2}) |
| Break‑even point | SP = CP | (S_{\text{avg}} = C_{\text{avg}}) |
| Mark‑up vs. Mark‑down | Different bases | Mark‑up uses CP, Mark‑down uses SP |
3.1 Common Tricky Points
- Profit % vs. Mark‑up % – They are numerically equal but conceptually different.
- Loss % vs. Discount % – Same numeric value but different reference points.
- Percentage change in price – For successive changes, multiply the factors.
- Multiple items with different CPs – Always compute the average CP before applying the profit %.
4. Worked Example – Quick Derivation
Problem
A shopkeeper sells an item at a profit of 20 % on his cost price. If he had sold it at a 10 % loss instead, he would have earned ₹200 less. Find the cost price.
Solution
Let CP = (C).
Profit 20 % → SP₁ = (C(1+0.20) = 1.2C).
Loss 10 % → SP₂ = (C(