6 votes 6 votes A person sold two different items at the same price. He made $10 \%$ profit in one item, and $10 \%$ loss in the other item. In selling these two items, the person made a total of$1 \%$ profit$2 \%$ profit$1 \%$ loss$2 \%$ loss Quantitative Aptitude gatecse2024-set2 quantitative-aptitude profit-loss + – Arjun asked Feb 16 • edited Feb 21 by makhdoom ghaya Arjun 4.0k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply vaishak2512 commented Feb 17 i edited by Arjun Mar 26 reply Follow Share Let's denote the cost price of each item as C. For the item in which the person made a 10% profit: Selling price = Cost price + 10% of cost price Selling price = C + 0.1*C = 1.1C For the item in which the person made a 10% loss: Selling price = Cost price -10% of cost price Selling price = C — O.1*C = 0.9C Given that the total profit is \$100, we can set up the equation: 1.1C + 0.9C = C + 100 2C = C + 100 C = 100 So, the cost price of each item is \$100. Now, let's calculate the total selling price: $I.IC = 1.1 \times 100 = \$110$ For the item with 10% profit: o.9C = 0.9 \times 100 = \$90 For the item with 10% loss: Total selling price = \$110 + \$90 = $200 The total profit is the difference between the total selling price and the total cost price: Total profit = \$200 - \$200 = \$0 So, the correct option is: C. 2% loss 0 votes 0 votes cseBiswajit commented Feb 18 reply Follow Share Question ache se poro... 2 bar na ho 3 times poro... den solve koro ... 1 votes 1 votes rhl commented Feb 25 reply Follow Share @vaishak2512Cost Price can not be same as we are already given that selling price is same and one item produces profit while other item results in loss. 0 votes 0 votes priyansh_1705 commented Mar 26 reply Follow Share Try to take numbers in such questions 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Solution using Product constancy priyansh_1705 answered Mar 26 • edited Mar 26 by priyansh_1705 priyansh_1705 comment Share Follow See all 0 reply Please log in or register to add a comment.