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    Question

    Quantity I  : A  Alone can complete a work in 21

    days. If be alone work for eight days and then leaves, then B alone can complete the remaining work in seven days.C  25% less efficient than A. If B&C work together for X days and then leave, then remaining work can be completed by A alone stays. Find the value of  X. Quantity II  . Train eight takes 24 seconds to cross a pole and 22nd to cross a man walking at the speed of 5.6 metre per second towards it. Train B travels 1 meter per second slower than train trainee and takes 25%. Less time than train need to cross a pole. If the speed of train B had been 7 meter per second less than  the time taken by it to cross 114 metre long platform is ‘5a’. Find the value of a
    A Quantity I > Quantity II Correct Answer Incorrect Answer
    B Quantity I < Quantity II Correct Answer Incorrect Answer
    C Quantity I ≤ Quantity II Correct Answer Incorrect Answer
    D Quantity I ≥ Quantity II Correct Answer Incorrect Answer
    E Quantity I = Quantity II or No relation Correct Answer Incorrect Answer

    Solution

    Let the efficiency of ‘A’ alone = y units/day Then, Total work = 21*y = 21y units Let the efficiency of B = z unit / day We have 8*z +7y = 21y z:y = 7:4 So the ratio of efficiency of B:A is 7:4. So the efficiency of B is = 7y/4 = 1.75y units / day Efficiency of C = 0.75y units/day Combined efficiency of B and C = 1.75y +0.75y = 2.5y units/day According to the question, 2.5y*x + 6*y = 21y 2.5x= 15 X=6   Quantity 2: Let the speed of the train A = y m/s Then the length of train ‘A’= 24*y = 24y mt. Relative speed of train A wrt man = (y+0.56) m/s Also length of the train A = (y+5.6)*20 = (20y+112) mt. So, 24y = 20y+112 Y=28 Speed of train A is 28 m/s. Speed of train B = 28-1 = 27m/s. Time taken by Train ‘B’ to cross pole = 24*0.75 = 18 sec So length of train ‘B’ = 18*27 =486mt. New speed of train ‘B’ = 27-7 = 20m/s Required time taken ‘B’ = (486+114)/20 = 30 seconds 5a=30sec a=6

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