Which of the following pair of symbols will define the

Solution

ATQ, Quantity I: ATQ; {(d - 80) /45} = 16 d - 80 = 720 'd' = 800 Again; {(800 + 40) /(s - 5) } = {(800 + 160) /(s + 5) } {840/(s - 5) } = {960/(s + 5) } 840(s + 5) = 960(s - 5) 840s + 4200 = 960s - 4800 9000 = 120s 's' = 75 Required time = {(d - 50) /s} = {(800 - 50) /75} = (750/75) = 10 hours So, Quantity I = 10 hours Quantity II: Let time taken by train 'P' to reach station 'Y' be 'h' hours Then time taken by train 'Q' to reach station 'Y' = (h - 2) hours ATQ; 64 X h = 72(h - 2) 64h = 72h - 144 8h = 144 'h' = 18 So, time taken by train 'P' to reach station 'Y' = 18 hours So, Quantity II = 18 hours Quantity III: Upstream speed of the Motarboat = 168 ÷ 14 = 12 kmph Downstream speed of the Motarboat = 528 ÷ 22 = 24 kmph Speed of Motarboat in still water = (12 + 24) /2 = 18 kmph Speed of the stream = (24 - 12) /2 = 6 kmph Increased speed of the Motarboat = 18 + 8 = 26 kmph Increased upstream speed = 26 - 6 = 20 kmph Increased downstream speed = 26 + 6 = 32 kmph Therefore, required time = (180/20) + (192/32) = 9 + 6 = 15 hours So, Quantity III = 15 hours