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How Many Apples Did Naman Originally Buy? Solution Let the original price per apple be x and the number of apples bought initially be n . Step 1: Initially, the number of apples is: n = 720 / x Step 2: After the price reduction, the new price is (x - 2) , and the number of apples becomes: n + 4 = 720 / (x - 2) Step 3: Substitute n = 720 / x into the second equation: (720 / x) + 4 = 720 / (x - 2) Step 4: Simplify the equation: 720 / (x - 2) - 720 / x = 4 (720 * 2) / [x(x - 2)] = 4 1440 / [x(x - 2)] = 4 Step 5: Multiply through by x(x - 2) : 1440 = 4x(x - 2) Step 6: Simplify the quadratic equation: 1440 = 4x² - 8x x² - 2x - 360 = 0 Step 7: Solve for x by factorization: (x - 20)(x + 18) = 0 x = 20 (as price m...

Solving the Rice Price Problem Alok is looking to buy rice, and with a 24% reduction in price, he can buy 10 kg more rice for ₹2,500. Let's solve the problem and find out the reduced rate of rice per kg. Step 1: Defining Variables Let the original price of rice per kg be ₹x. After a 24% reduction, the new price per kg is 76% of the original price, which is 0.76x . Step 2: Setting Up the Equation At the original price, Alok can buy: Quantity at original price = 2500 / x At the reduced price, Alok can buy: Quantity at reduced price = 2500 / (0.76x) The difference in the quantity of rice he can buy is 10 kg, so we have the equation: (2500 / (0.76x)) - (2500 / x) = 10 Step 3: Simplifying the Equation Factor out 2500 from the equation: 2500 * (1 / (0.76x) - 1 / x) = 10 Now, simplify the expression inside the parentheses: (1 ...

Problem Statement: If a + b + c = 6 and a² + b² + c² = 14, find the value of (a-b)² + (b-c)² + (c-a)². Solution: The formula for (a-b)² + (b-c)² + (c-a)² is: (a-b)² + (b-c)² + (c-a)² = 2(a² + b² + c²) - 2(ab + bc + ca). From the problem: a + b + c = 6 → (a + b + c)² = 36. Expand (a + b + c)²: a² + b² + c² + 2(ab + bc + ca) = 36. Substitute a² + b² + c² = 14: 14 + 2(ab + bc + ca) = 36. 2(ab + bc + ca) = 22 → ab + bc + ca = 11. Now substitute values into the formula: (a-b)² + (b-c)² + (c-a)² = 2(a² + b² + c²) - 2(ab + bc + ca). = 2(14) - 2(11) = 28 - 22 = 6. Final Answer: The value is 6 .

Election Problem Solution By TecScholar Let's solve a problem step by step where an election result is analyzed based on given conditions. Problem In an election, 2% of registered voters did not participate, and 500 votes were invalid. Candidates A and B contested the election, and A defeated B by 200 votes. If 43% of registered voters voted for A, find the total number of votes cast. Solution Let the total number of registered voters = x . Voters who participated = 98% of x , i.e., 0.98x . Invalid votes = 500 . Therefore, valid votes = 0.98x - 500 . Candidate A defeated B by 200 votes. Thus, A's votes = B's votes + 200 . 43% of registered voters voted for A. Hence, A's votes = 0.43x . Equations: Total valid votes: A's votes + B's votes = 0.98x - 500 ...

Solving the Equation: 8^x + 2^x = 130 Solving the Equation: 8 x + 2 x = 130 We are tasked with solving the equation: 8 x + 2 x = 130 We begin by expressing 8 x as a power of 2. Since 8 = 2 3 , we can rewrite the equation as: (2 3 ) x + 2 x = 130 This simplifies to: 2 3x + 2 x = 130 Let y = 2 x . Then, the equation becomes: y 3 + y = 130 Now, we need to solve the cubic equation: y 3 + y - 130 = 0 We can try some values for y to find a root. If we substitute y = 5 , we get: 5 3 + 5 = 125 + 5 = 130 So, y = 5 is a solution. Since y = 2 x , we now have: ...

Integration of x^2024 - 1 / x^506 - 1 Solving the Integral: ∫ (x^2024 - 1) / (x^506 - 1) dx The integral we are solving is: ∫ (x^2024 - 1) / (x^506 - 1) dx We start by simplifying the expression using polynomial division. Dividing x 2024 - 1 by x 506 - 1 gives: x 2024 - 1 = (x 506 - 1) * (x 1518 + x 1012 + x 506 + 1) This allows us to rewrite the integral as: ∫ (x 1518 + x 1012 + x 506 + 1) dx Now, we can integrate each term: ∫ x 1518 dx = x 1519 / 1519 ∫ x 1012 dx = x 1013 / 1013 ∫ x 506 dx = x 507 / 507 ∫ 1 dx = x Thus, the final result is: x 1519 / 1519 + x 1013 / 1013 + x 507 / 507 + x + C Where C is the constant of inte...