Question
Rajiv, Sanju and Tanu started a business with the
investment of Rs. (z-400), (z-1000) and (z+200) respectively. After four months, Rajiv decreased his investment by 50% and Sanju increased his investment by 30% and Tanu decreased his investment by Rs. тАШyтАЩ. At the end of one year, the profit share of Sanju will be 50% more than the profit share of Rajiv. At the end of one year, the profit share of Sanju and Tanu will be the same. Which of the following options represents the correct relation between the values of тАШyтАЩ and тАШzтАЩ?Solution
Ratio of Rajiv, Sanju and Tanu investment with respect to the time =(z-400) x4+(100-50) % of (z-400) x8 : (z-1000)x4+(100+30)% of (z-1000)x8 : (z+200)x4+(z+200-y)x8┬а ┬а Eq.(i) The profit share of Sanju will be 50% more than the profit share of Rajiv. profit share of Sanju = (100+50) % of profit share of Rajiv the profit share of Sanju = 150% of the profit share of Rajiv profit share of Sanju = 1.5 x (profit share of Rajiv) the profit share of Sanju: profit share of Rajiv = 1.5/1 = 15/10 = 3:2┬а ┬а Eq.(ii) At the end of one year, the profit share of Sanju and Tanu will be the same. profit share of Sanju: profit share of Tanu тЗТ 3:3┬а ┬а Eq.(iii) From Eq.(ii) and Eq.(iii), the profit share of Rajiv: profit share of Sanju: profit share of Tanu тЗТ 2:3:3┬а ┬а Eq.(iv) From Eq.(i) and Eq.(iv). [(z-400)x4+(100-50)% of (z-400)x8]/[(z-1000)x4+(100+30)% of (z-1000)x8] = 2/3 [(z-400)x4+50% of (z-400)x8]/[(z-1000)x4+130% of (z-1000)x8] = 2/3 [(z-400) x4+0.5(z-400) x8]/[(z-1000)x4+1.3(z-1000)x8] = тЕФ So,┬а z = 4000 Put the value of тАШzтАЩ in Eq.(i) The ratio of Rajiv, Sanju, and Tanu investment with respect to the time тЗТ (4000-400) x4+(100-50) % of (4000-400) x8: (4000-1000)x4+(100+30)% of (4000-1000)x8 : (4000+200)x4+(4000+200-y)x8 тЗТ 3600x4+50% of 3600x8: 3000x4+130% of (3000) x8: (4200) x4+(4200-y) x8 тЗТ 3600+50% of 3600x2: 3000+130% of (3000) x2: (4200) +(4200-y) x2 тЗТ 3600+1800x2: 3000+3900x2: (4200) +(4200-y) x2 тЗТ 3600+3600: 3000+7800: (4200) +(8400-2y) тЗТ 7200: 10800: (12600-2y) тАж. Eq.(v) From Eq.(iii) and Eq.(v). 10800 / (12600-2y) = 3/3 10800 = (12600-2y) 10800 = (12600-2y) 2y = 12600-10800 = 1800 y = 9 00 a) 5y = (z+800) 5x900 = (4000+800) 4500 = 4800 The above equation is not satisfied. So, this is not the correct answer. b) (z+200)/7 = y (4000+200)/7 = 900 4200/7 = 600 600 = 900 The above equation is satisfied. So, this is not the correct answer. c) 6y = (z+200) 6x900 = (4000+200) 5400 = 4200 The above equation is not satisfied. So this is not the correct answer. d) (z+400)/6 = 1.5y (4000+400)/6 = 1.5x900 (4000+400)/6 = 900 4400 = 8100 The above equation is not satisfied. So, this is not the correct answer.
рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рд╡рд╛рдХреНрдпрд╛рдВрд╢ рд╣реЗрддреБ рд╕рд╛рд░реНрдердХ рд╢рдмреНрдж рдХреА рдкрд╣рдЪрд╛рди рдХрд░реЗрдВред
рдЬя┐╜...
рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рдореЗрдВ рд╕реЗ 'рдХрд░реНрддреГрд╡рд╛рдЪреНрдп' рдХрд╛ рдЙрджрд╛рд╣рд░рдг рдХреМрдирд╕рд╛ рд╣реИ?
тАШрдкреНрд░рд╛рдкреНрддреНрдпрд╛рд╢рд╛тАЩ рдХрд╛ рд╕рдиреНрдзрд┐ рд╡рд┐рдЪреНрдЫреЗрдж рдХреНрдпрд╛ рд╣реЛрдЧрд╛ ?
тАШрдЕрдВрдзреЗ рдХреЗ рд╣рд╛рде рдмрдЯреЗрд░ рд▓рдЧрдирд╛тАЩ рд▓реЛрдХреЛрдХреНрддрд┐ рдХрд╛ рд╕рд╣реА рдЕрд░реНрде рд╣реИ -
рджрд┐рдпреЗ рдЧрдпреЗ рд╢рдмреНрдж рдХрд╛ рдкрд░реНрдпрд╛рдпрд╡рд╛рдЪреА -
рдЕрд╢реНрд╡-
рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рдореЗрдВ рд╕реЗ рдХреМрди рд╕рд╛ рд╕рд╣реА рд╕реБрдореЗрд▓рд┐рдд рдпреБрдЧреНрдо рдирд╣реАрдВ рд╣реИ┬а
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рдХрд┐рд╕реА (1) / рднреВрдЦреЗ рдирдВрдЧреЗ рдХреА (2)/ рд╣реИ (3) / рд╕рд╣рд╛рдпрддрд╛ (4) / рдкрд╛рдк (5) / рди рдХрд░рдирд╛ (6) рдХреНрд░рдо рд╕рдВя┐╜...
'рд╕реБрд▓рдн' рдХрд╛ рдЙрдкрдпреБрдХреНрдд рд╡рд┐рдкрд░реАрддрд╛рд░реНрдердХ рд╢рдмреНрдж рд╣реИ
рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рдореЗрдВ рд╕реЗ рдХреМрди тАУ рд╕рд╛ рд╕рд╣реА рд╕реБрдореЗрд▓рд┐рдд рдпреБрдЧреНрдо рдирд╣реАрдВ рд╣реИрдВ?