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    Question

    Three individuals, R, S, and T, began a taxi service

    business with investments of Rs. (p-400), (p-1000), and (p+200), respectively. After four months, R reduced his investment by 50%, S increased his investment by 30%, and T decreased his investment by Rs. 'q'. At the end of one year, the profit share of S will be 50% more than the profit share of R. Additionally, at the end of one year, the profit share of S and T will be the same. The question seeks to establish the correct relationship between the values of 'q' and 'p'.
    A 5q = (p+800) Correct Answer Incorrect Answer
    B (p+200)/7 = q Correct Answer Incorrect Answer
    C 6q = (p+200) Correct Answer Incorrect Answer
    D (p+400)/6 = 1.5q Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    ATQ, Ratio of R, S and T investment with respect to the time ⇒ (p-400) x4+(100-50) % of (p-400) x8: (p-1000) x4+(100+30) % of (p-1000) x8: (p+200) x4+(p+200-q) x8    Eq.(i) The profit share of S will be 50% more than the profit share of R. profit share of S = (100+50) % of profit share of R profit share of S = 150% of profit share of R profit share of S = 1.5 x (profit share of R) profit share of S : profit share of R = 1.5/1 = 15/10 = 3:2    Eq.(ii) At the end of one year, the profit share of S and T will be the same. profit share of S: profit share of T ⇒ 3:3    Eq.(iii) From Eq.(ii) and Eq.(iii), profit share of R: profit share of S: profit share of T ⇒ 2:3:3    Eq.(iv) From Eq.(i) and Eq.(iv). [(p-400) x4+(100-50) % of (p-400) x8]/[(p-1000) x4+(100+30) % of (p-1000) x8] = 2/3 [(p-400) x4+50% of (p-400) x8]/[(p-1000) x4+130% of (p-1000) x8] = 2/3 [(p-400) x4+0.5(p-400)x8]/[(p-1000)x4+1.3(p-1000)x8] = ⅔ So p = 4000 Put the value of ‘p’ in Eq.(i) Ratio of R, S and T investment with respect to the time ⇒ (4000-400) x4+(100-50) % of (4000-400) x8: (4000-1000) x4+(100+30) % of (4000-1000) x8: (4000+200) x4+(4000+200-q) x8 ⇒ 3600x4+50% of 3600x8: 3000x4+130% of (3000) x8: (4200) x4+(4200-q) x8 ⇒ 3600+50% of 3600x2: 3000+130% of (3000) x2: (4200) +(4200-q) x2 ⇒ 3600+1800x2: 3000+3900x2: (4200) +(4200-q) x2 ⇒ 3600+3600: 3000+7800: (4200) +(8400-2q) ⇒ 7200: 10800: (12600-2q) Eq.(v) From Eq.(iii) and Eq.(v). 10800 / (12600-2q) = 3/3 10800 = (12600-2q) 10800 = (12600-2q) 2q = 12600-10800 = 1800 q = 900 a) 5q = (p+800) 5x900 = (4000+800) 4500 = 4800 The above equation is not satisfied. So, this is not the correct answer. b) (p+200)/7 = q (4000+200)/7 = 900 4200/7 = 900 600 = 900 The above equation is satisfied. So, this is the correct answer. c) 6q = (p+200) 6x900 = (4000+200) 5400 = 4200 The above equation is not satisfied. So, this is not the correct answer. d) (p+400)/6 = 1.5q (4000+400)/6 = 1.5x900 (4000+400)/6 = 1350 4400 = 8100 The above equation is not satisfied. So, this is not the correct answer.

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