📑 Topics on this page (Jump to Topic):
📌 1. Basic Formulas
📘 Formulas:
- Profit = SP − CP; Loss = CP − SP
- Profit% = Profit/CP × 100 (always on CP)
- SP = CP × (100 + profit%)/100; SP = CP × (100 − loss%)/100
- Discount = MP − SP; Discount% = Discount/MP × 100 (always on MP)
- MP/CP = (100 + profit%)/(100 − discount%)
⚠️ Exam Traps:
- Profit% is on CP and discount% is on MP — mixing the base is a common mistake.
📌 2. Special Cases
| Case | Result |
|---|---|
| Two items at the same SP, x% profit on one, x% loss on the other | Always a loss = x²/100 % |
| Successive discounts a% and b% | Net discount = a + b − ab/100 |
| Dishonest dealer weighs w g for 1000 g (sells at CP) | Profit% = (1000 − w)/w × 100 |
| CP of x items = SP of y items | Profit% = (x − y)/y × 100 |
| Buy n get m free | Discount% = m/(n + m) × 100 |
💡 Shortcut Trick: Fraction method: 20% profit = CP : SP = 5 : 6; 25% loss = 4 : 3. Think in ratios to solve faster.
📌 3. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. An article bought for ₹1200 is sold at 15% profit. The selling price is: Practice
Correct answer: (B)
Solution: SP = 1200 × 115/100 = ₹1380.
Solution: SP = 1200 × 115/100 = ₹1380.
Q2. A shopkeeper sells two watches at the same price, gaining 30% on one and losing 30% on the other. The overall result is: Practice
Correct answer: (C)
Solution: Same SP with ±x% → always a loss of x²/100 = 30²/100 = 9%.
Solution: Same SP with ±x% → always a loss of x²/100 = 30²/100 = 9%.
Q3. Two successive discounts of 30% and 40% are equivalent to a single discount of: Practice
Correct answer: (D)
Solution: a + b − ab/100 = 30 + 40 − 1200/100 = 58%.
Solution: a + b − ab/100 = 30 + 40 − 1200/100 = 58%.
Q4. A dishonest dealer claims to sell at cost price but uses a 960 g weight instead of 1 kg. His profit percent is: Practice
Correct answer: (A)
Solution: Profit% = (1000 − 960)/960 × 100 = 4.17%.
Solution: Profit% = (1000 − 960)/960 × 100 = 4.17%.
Q5. The cost price of 18 articles equals the selling price of 15 articles. Find the profit percent. Practice
Correct answer: (B)
Solution: Profit% = (x − y)/y × 100 = (18 − 15)/15 × 100 = 20%.
Solution: Profit% = (x − y)/y × 100 = (18 − 15)/15 × 100 = 20%.
Q6. A shop offers 'Buy 7 Get 3 Free'. The effective discount percent is: Practice
Correct answer: (C)
Solution: 10 items for the price of 7 → discount = 3/10 × 100 = 30%.
Solution: 10 items for the price of 7 → discount = 3/10 × 100 = 30%.
Q7. By selling an article for ₹550, a profit of 10% is made. Its cost price is: Practice
Correct answer: (D)
Solution: CP = SP × 100/(100 + 10) = 550 × 100/110 = ₹500.
Solution: CP = SP × 100/(100 + 10) = 550 × 100/110 = ₹500.
Q8. An article bought for ₹640 is sold at 25% profit. The selling price is: Practice
Correct answer: (A)
Solution: SP = 640 × 125/100 = ₹800.
Solution: SP = 640 × 125/100 = ₹800.
Q9. A shopkeeper sells two watches at the same price, gaining 10% on one and losing 10% on the other. The overall result is: Practice
Correct answer: (B)
Solution: Same SP with ±x% → always a loss of x²/100 = 10²/100 = 1%.
Solution: Same SP with ±x% → always a loss of x²/100 = 10²/100 = 1%.
Q10. A dishonest dealer claims to sell at cost price but uses a 950 g weight instead of 1 kg. His profit percent is: Practice
Correct answer: (C)
Solution: Profit% = (1000 − 950)/950 × 100 = 5.26%.
Solution: Profit% = (1000 − 950)/950 × 100 = 5.26%.
Q11. The cost price of 12 articles equals the selling price of 10 articles. Find the profit percent. Practice
Correct answer: (D)
Solution: Profit% = (x − y)/y × 100 = (12 − 10)/10 × 100 = 20%.
Solution: Profit% = (x − y)/y × 100 = (12 − 10)/10 × 100 = 20%.
Q12. By selling an article for ₹1650, a profit of 10% is made. Its cost price is: Practice
Correct answer: (A)
Solution: CP = SP × 100/(100 + 10) = 1650 × 100/110 = ₹1500.
Solution: CP = SP × 100/(100 + 10) = 1650 × 100/110 = ₹1500.
Q13. An article marked at ₹4000 is sold after successive discounts of 20% and 15%. The selling price is: Practice
Correct answer: (B)
Solution: SP = 4000 × 80/100 × 85/100 = ₹2720.
Solution: SP = 4000 × 80/100 × 85/100 = ₹2720.
Q14. An article bought for ₹800 is sold at 12.5% profit. The selling price is: Practice
Correct answer: (C)
Solution: SP = 800 × 112.5/100 = ₹900.
Solution: SP = 800 × 112.5/100 = ₹900.
Q15. An article is bought for ₹500 and sold for ₹450. Find the loss percent. Practice
Correct answer: (D)
Solution: Loss = ₹50; loss% = 50/500 × 100 = 10% (on CP).
Solution: Loss = ₹50; loss% = 50/500 × 100 = 10% (on CP).
Q16. A shopkeeper sells two watches at the same price, gaining 15% on one and losing 15% on the other. The overall result is: Practice
Correct answer: (A)
Solution: Same SP with ±x% → always a loss of x²/100 = 15²/100 = 2.25%.
Solution: Same SP with ±x% → always a loss of x²/100 = 15²/100 = 2.25%.
Q17. An article costs ₹600. To earn 20% profit even after a 10% discount, what should the marked price be? Practice
Correct answer: (B)
Solution: MP = CP × (100 + profit%)/(100 − discount%) = 600 × 120/90 = ₹800.
Solution: MP = CP × (100 + profit%)/(100 − discount%) = 600 × 120/90 = ₹800.
Q18. A dishonest dealer claims to sell at cost price but uses a 940 g weight instead of 1 kg. His profit percent is: Practice
Correct answer: (C)
Solution: Profit% = (1000 − 940)/940 × 100 = 6.38%.
Solution: Profit% = (1000 − 940)/940 × 100 = 6.38%.
Q19. A shop offers 'Buy 3 Get 1 Free'. The effective discount percent is: Practice
Correct answer: (D)
Solution: 4 items for the price of 3 → discount = 1/4 × 100 = 25%.
Solution: 4 items for the price of 3 → discount = 1/4 × 100 = 25%.
Q20. An article marked at ₹2400 is sold after successive discounts of 25% and 20%. The selling price is: Practice
Correct answer: (A)
Solution: SP = 2400 × 75/100 × 80/100 = ₹1440.
Solution: SP = 2400 × 75/100 × 80/100 = ₹1440.