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    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.
    A Quantity I > Quantity II Correct Answer Incorrect Answer
    B Quantity I < Quantity II Correct Answer Incorrect Answer
    C Quantity I = Quantity II or no relation can be established Correct Answer Incorrect Answer
    D Quantity I ≥ Quantity II Correct Answer Incorrect Answer
    E Quantity I ≤ Quantity II Correct Answer Incorrect Answer

    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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