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Part 1: Profit & Loss Practice Questions Q1. A shopkeeper bought an article for ₹800. He marked it up by 40\% above its cost price and offered a discount of 20\% on the marked price. Find his net profit percentage. Q2. The cost price of 15 items is equal to the selling price of 12 items. Find the profit percentage made by the seller. Q3. An article is sold at a profit of 15\%. If both the cost price and selling price are reduced by ₹200, the profit percentage becomes 20\%. Find the original cost price of the article. Q4. A shopkeeper allows two successive discounts of 10\% and 20\% on an article. If the final selling price of the article is ₹576, find the marked price of the article. Q5. A dishonest dealer sells goods at cost price but uses a false weight of 900\text{ g} instead of 1\text{ kg}. Find his actual profit percentage. Q6. The cost price of article A is 25\% more than the cost price of article B. Article A is sold at a profit of 20\% and article B is sold at a loss of 10\%. If the total profit earned on both articles together is ₹105, find the cost price of article B. Q7. A person sold two items for ₹4,800 each. On one item, he gained 20\% and on the other, he lost 20\%. Find his overall profit or loss amount in rupees. Q8. A trader marks his goods 50\% above the cost price and allows a discount of x\%. If he still makes a profit of 20\%, find the value of x. Q9. If an article is sold for ₹1,380, the loss percentage is equal to the profit percentage gained when the same article is sold for ₹1,620. Find the cost price of the article. Q10. The ratio of the marked price to the cost price of an item is 5 : 4. If a discount of ₹150 is given on the marked price, the shopkeeper earns a profit of 10\%. Find the cost price of the item. Part 1 Solutions: Profit & Loss S1. * * * * S2. * Given: 15 \times CP = 12 \times SP \implies \frac{SP}{CP} = \frac{15}{12} = \frac{5}{4} * S3. * Let initial CP = x. Then initial SP = 1.15x. * New CP = x - 200, New SP = 1.15x - 200. * New profit is 20\%: S4. * Let Marked Price be MP. * Single equivalent multiplier after 10\% and 20\% discount = 0.90 \times 0.80 = 0.72 * S5. * Formula: \text{Profit \%} = \frac{\text{Error}}{\text{True Value} - \text{Error}} \times 100 * Error = 1000\text{ g} - 900\text{ g} = 100\text{ g} * S6. * Let CP_B = 100x. Then CP_A = 125x. * Profit on A = 20\% \text{ of } 125x = 25x * Loss on B = 10\% \text{ of } 100x = 10x * Net Profit = 25x - 10x = 15x * Given 15x = 105 \implies x = 7 * S7. * * * Total CP = 4000 + 6000 = \text{₹}10,000 * Total SP = 4800 + 4800 = \text{₹}9,600 * Overall Loss = 10000 - 9600 = \mathbf{\text{₹}400 \text{ Loss}} S8. * Let CP = 100. Then MP = 150. * Since profit is 20\%, SP = 120. * Discount = MP - SP = 150 - 120 = 30 * Discount \% (x) = \frac{30}{150} \times 100 = \mathbf{20\%} S9. * Since Loss \% at ₹1,380 = Profit \% at ₹1,620: S10. * Let CP = 4x and MP = 5x. * * Discount = MP - SP = 5x - 4.4x = 0.6x * Given discount = 150 \implies 0.6x = 150 \implies x = 250 * Part 2: Mixture & Alligation Practice Questions Q11. In what ratio must a grocer mix two varieties of pulses costing ₹60 per kg and ₹85 per kg so that the resulting mixture is worth ₹70 per kg? Q12. A container contains 80\text{ liters} of pure milk. 16\text{ liters} of milk is taken out and replaced with water. This process is repeated one more time. Find the final quantity of pure milk remaining in the container. Q13. A mixture of 60\text{ liters} contains milk and water in the ratio 3 : 2. How much water should be added to this mixture so that the ratio of milk to water becomes 6 : 5? Q14. In Vessel A, milk and water are mixed in the ratio 4 : 1, and in Vessel B, they are in the ratio 3 : 2. In what ratio should liquid from Vessel A and Vessel B be mixed to form a new mixture containing milk and water in the ratio 7 : 3? 📌RRB PO 2016 pre topic wise practice📌✨

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