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    Question

    The sum of (p/q)+(q/r)+(r/s)+(s/t) = (239/60). The value

    of ‘q’ is one less than the value of ‘s’. The average of ‘p’ and ‘s’ is 4. The value of ‘r’ is 3 less than the value of ‘s’. If the value of ‘r’ is the smallest prime number, then find out the value of (p+q+s+t)/4.
    A 5.5 Correct Answer Incorrect Answer
    B 4.5 Correct Answer Incorrect Answer
    C 6.5 Correct Answer Incorrect Answer
    D 3.5 Correct Answer Incorrect Answer
    E None of the above Correct Answer Incorrect Answer

    Solution

    The value of ‘q’ is one less than the value of ‘s’.

    q = (s-1)    Eq.(i)

    The average of ‘p’ and ‘s’ is 4.

    p+s = 4x2 = 8    Eq.(ii)

    The value of ‘r’ is 3 less than the value of ‘s’.

    r = (s-3)    Eq.(iii)

    If the value of ‘r’ is the smallest prime number.

    So r = 2 .

    Put the value of ‘r’ in Eq.(iii).

    2 = (s-3)

    s = 3+2

    s = 5

    Put the value of ‘s’ Eq.(ii).

    p+5 = 8

    p = 8-5

    p = 3

    Put the value of ‘s’ Eq.(i).

    q = (5-1)

    q = 4

    The sum of (p/q)+(q/r)+(r/s)+(s/t) = (239/60).

    Put the value of ‘p‘, ‘q‘, ‘r‘ and ‘s‘ in the above equation.

    (3/4)+(4/2)+(2/5)+(5/t) = (239/60)

    (63/20)+(5/t) = (239/60)

    (5/t) = (239/60)-(63/20)

    (5/t) = (239-189)/60

    (5/t) = (50/60)

    t = 6

    Value of (p+q+s+t)/4 = (3+4+5+6)/4

    = 18/4

    = 4.5

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