Profit & Loss – Maharashtra TET (Quantitative Aptitude)
Why it matters?
In the TET syllabus, Profit & Loss questions form a large chunk of the Quantitative Aptitude section. Mastering the concepts, formulas, and quick tricks will give you a solid edge.
1. Simple Explanation
| Term | Meaning |
|---|---|
| Profit | The excess of selling price over cost price. |
| Loss | The excess of cost price over selling price. |
| Rate of Profit (R.P.) | (Profit / Cost Price) × 100 |
| Rate of Loss (R.L.) | (Loss / Cost Price) × 100 |
| Profit % | P / C × 100 |
| Loss % | L / C × 100 |
| Selling Price (S.P.) | Cost Price + Profit (if profit) or Cost Price – Loss (if loss) |
Quick formula cheat‑sheet
| Scenario | Formula |
|---|---|
| Profit % | P = (R.P./100) × C |
| Loss % | L = (R.L./100) × C |
| Cost Price | C = S.P. / (1 ± R.P./100) |
| Selling Price | S.P. = C × (1 ± R.P./100) |
| Profit/Loss amount | P = S.P. – C or L = C – S.P. |
Tip: Always keep the sign of the rate (+ for profit, – for loss).
2. Key Points (What you must remember)
| # | Key Point | Why it’s important |
|---|---|---|
| 1 | Use percentages as decimals – Convert 20 % → 0.20. | Simplifies multiplication/division. |
| 2 | Avoid “trick” numbers – A 50 % profit halves the price; 200 % triples it. | Saves time in mental calculations. |
| 3 | When multiple rates apply – Use successive application: | |
NewPrice = OldPrice × (1 ± r1/100) × (1 ± r2/100)… |
Handles multiple profit/loss stages (e.g., wholesaler → retailer). | |
| 4 | Reverse calculations – For “what was the original price?” use division by (1 ± rate). |
Quick back‑tracking. |
| 5 | Profit & Loss are not additive – BE careful with “profit of 20 % on cost, then loss of 10 % on selling price” → compute sequentially. | Prevents common pitfalls. |
| 6 | Check units – All rates are per 100 units of cost price. | Keeps formulas consistent. |
| 7 | Use ratios – Profit:Loss = R.P. : R.L. | Helps in “mixing” and “change of price” problems. |
3. Worked Example (Step‑by‑Step)
Problem
A shopkeeper bought a batch of goods for ₹12,000. He sold them at a 20 % profit. Later, the goods were sold again at a 25 % loss on the selling price. What was the final selling price?
Step 1: Compute first selling price
- Cost Price (C) = ₹12,000
- Profit % = 20 % →
1 + 0.20 = 1.20
[ S.P._1 = C \times 1.20 = 12{,}000 \times 1.20 = 14{,}400 ]
Step 2: Apply the second operation (loss)
- Loss % = 25 % →
1 – 0.25 = 0.75
[ S.P._2 = S.P._1 \times 0.75 = 14{,}400 \times 0.75 = 10{,}800 ]
Answer: The final selling price is ₹10,800.
Verification
- Profit on initial cost: ₹14,400 – ₹12,000 = ₹2,400 (20 %) ✔️
- Loss on first selling price: ₹14,400 – ₹10,800 = ₹3,600 (25 %) ✔️
4. Quick Revision Summary
| Concept | Formula | Quick Tip |
|---|---|---|
| Profit/Loss Percentage | P% = (P/C) × 100 |
Think “extra amount ÷ base”. |
| Selling Price after Profit | S.P. = C × (1 + R.P./100) |
Simple multiplier. |
| Selling Price after Loss | S.P. = C × (1 – R.L./100) |
Same as profit but subtract. |
| Reverse (Cost from Selling Price) | C = S.P. / (1 ± R/100) |
Use division. |
| Sequential Rates | Final = Original × (1 ± r1) × (1 ± r2)… |
Chain multipliers. |
| Profit : Loss Ratio | Profit : Loss = R.P. : R.L. |
Helpful for “mixing” problems. |
- Always write down the given data first.
- Convert percentages to decimals early.
- Check the sign of each rate (profit +, loss –).
- Use the chain‑multiplier trick for multiple stages.
Final Thought
- Master the formulas → you’ll solve any profit & loss question in under a minute.
- Practice sequential problems (multiple profit/loss stages) – they appear frequently in TET.
- Keep a small “cheat‑sheet” of the formulas handy for quick reference during the exam.
Good luck!