Question
Quantity I: X: Aman, Bhuvan, and
Chintu started a business with initial investments of Rs. (4a + 544), Rs. (2a + 896), and Rs. (6a – 192), respectively. After 10 months, Aman withdrew Rs. 200, Bhuvan added Rs. 400, and Chintu withdrew Rs. 400. At the end of 20 months, their total profit was Rs. 94960. If Bhuvan's profit share from the total is Rs. 33920, then Arjun’s profit share is denoted as 'X'. Quantity II: Y: A rectangular field has a surrounding path of 4 meters in width on the inside. The perimeter of the field is (9a + 20) meters, and its length is (4a – 10) meters. Given that the total area of the path is 376 m², the cost of fencing the path on both sides at Rs. 175 per meter is represented as 'Y'. In the question, two Quantity I and Quantity II are given. You have to solve both the Quantities to establish the correct relation between Quantity I and Quantity II.Solution
ATQ, Investment of Aman = (4a + 544) × 20 – 200 × 10 = 80a + 10880 – 2000 = 80a + 8880 Investment of Bhanu = (2a + 896) × 20 + 400 × 10 = 40a + 17920 + 4000 = 40a + 21920 Investment of Chintu = (6a – 192) × 20 – 400 × 10 = 120a – 3840 – 4000 = 120a – 7840 Total Investment = (80a + 8880) + (40a + 21920) + (120a – 7840) = 240a + 22960 (40a + 21920)/(240a + 22960) = 33920/94960 = 424/1187 1187(a + 548) = 424(6a + 574) 1187a + 650476 = 2544a + 243376 1357a = 407100 a = 300 Profit of Aman = X = 94960 × (80 × 300 + 8880)/(240 × 300 + 22960) = 94960 × 32880/94960 = 32880 - Quantity I Length = (4a – 10) m, Breadth = (9a + 20)/2 – (4a – 10) = 4.5a + 10 – 4a + 10 = 0.5a + 20 m Area of path = 374 = (4a – 10)(0.5a + 20) – (4a – 18)(0.5a + 12) 374 = 2a2 – 5a + 80a – 200 – 2a2 + 9a – 48a + 216 376 = 36a + 16 36a = 360 a = 10 Length = 4 × 10 – 10 = 30 m, Breadth = 0.5 × 10 + 20 = 25 m Cost of Fencing = Y = 175 × 2 × [(30 + 25) + (22 + 17)] = 350 × (55 + 39) Y = 350 × 94 = 32900 - Quantity II Hence, Quantity I < Quantity II
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