Profit & Loss — Advanced Exam Problems

🎯 Learning Objective

Handle multi-layer profit questions — two selling scenarios, double-cheating dealers and part-by-part sales — without losing the base.

💡 Concept

• 'CP of x articles = SP of y articles' → profit% = (x − y)/y × 100; it is a loss when y > x
• Two-scenario pattern: the ₹ gap between two selling prices = (gap between the two profit %s) of CP
• Dealer cheating while buying AND selling → net factor = (100 + a)/(100 − b), never a + b
• Goods sold in parts at different profits → fix the target total SP first, then work backwards
• Marked-price chains: CP × (markup MF) × (discount MFs) = SP — one long multiplication

🧮 Key Formulas

CP of x = SP of y → gain% = (x − y) × 100/y

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SP gap = (g₂ − g₁)% of CP

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Double cheat: gain% = [(100 + a)/(100 − b) − 1] × 100

✏️ Easy Example

Q. The cost price of 12 pens equals the selling price of 10 pens. Find the profit per cent.

1. Let CP of one pen = ₹1, so CP of 12 pens = ₹12 — assuming ₹1 per unit turns the statement into plain numbers
2. SP of 10 pens = CP of 12 pens = ₹12, so SP of one pen = 12/10 = ₹1.20
3. Profit per pen = 1.20 − 1 = ₹0.20 on a CP of ₹1
4. Profit% = (0.20/1) × 100 = 20

Answer: 20%

🇮🇳 Real-Life Example

Wholesale traders juggle markup, discount and short-weighing together — their real margin is exactly this chained-factor maths.

📝 Exam-Level Example

Q. A man sells a table at a loss of 10%. Had he sold it for ₹450 more, he would have gained 5%. Find the cost price of the table.

1. First SP = 90% of CP (10% loss → MF 0.90); second SP = 105% of CP (5% gain → MF 1.05)
2. The ₹450 gap between the two selling prices is the ONLY difference between the scenarios
3. So 1.05 CP − 0.90 CP = 450 — both SPs sit on the same CP base, which is why the difference is clean
4. 0.15 CP = 450
5. CP = 450/0.15 = ₹3,000; check: SPs are 2700 and 3150, gap 450 ✓

Answer: ₹3,000

📝 Exam-Level Example

Q. A dishonest dealer buys goods using a weight 10% more than stated and sells using a weight 10% less than stated, at cost price. Find his overall gain per cent.

1. While buying he takes 1100 g but pays for 1000 g; while selling he gives only 900 g but charges for 1000 g — write both cheats separately
2. Assume goods cost ₹1 per gram, so he spends ₹1,000 and holds 1100 g
3. Selling 1100 g in 900-gram 'kilos' brings in (1100/900) × 1000 = ₹11000/9 = ₹1222.22
4. Gain = 1222.22 − 1000 = ₹222.22 on a cost of ₹1,000
5. Gain% = (222.22/1000) × 100 = 22.22% — same as the shortcut (110/90 − 1) × 100 = 200/9 %

Answer: 22.22% (200/9 %)

📝 Exam-Level Example

Q. A trader buys 100 kg of rice for ₹4,000. He sells 40 kg at a 10% profit. At what profit per cent must he sell the remaining 60 kg to gain 19% on the whole?

1. Fix the finish line first: target total SP = 4000 × 1.19 = ₹4,760 — then work backwards
2. CP per kg = 4000/100 = ₹40; the first 40 kg cost 40 × 40 = ₹1,600 and sold at 10% profit → SP = 1600 × 1.10 = ₹1,760
3. SP still needed from the last 60 kg = 4760 − 1760 = ₹3,000
4. CP of those 60 kg = 60 × 40 = ₹2,400
5. Profit needed = 3000 − 2400 = ₹600 → (600/2400) × 100 = 25%

Answer: 25%

🪄 Memory Trick

Two-scenario questions in one line: CP = (money gap) × 100/(% gap). Here 450 × 100/15 = ₹3,000 — no equations needed.

⚠️ Common Mistakes

• ❌ Adding the two cheating percentages (10 + 10 = 20%) instead of multiplying 110/90
• ❌ Taking the ₹450 gap as a percentage of SP — both scenarios share the same CP base
• ❌ Averaging the profit percents of two parts without weighting them by cost

🏆 Exam Tips

• ✅ In 'CP of x = SP of y', profit% = (x − y)/y × 100 — gap on top, y below
• ✅ Assume ₹1 per gram or ₹1 per unit whenever quantities are being cheated
• ✅ Part-sales: target SP minus earned SP = SP still needed — three subtractions, done

📌 Summary

• Two scenarios share one CP — equate through it
• CP of x = SP of y → (x − y)/y × 100
• Buy-and-sell cheating multiplies: 110/90 → 22.22%
• Part sales: fix total target SP, then walk backwards