Quant Boosters - Learn Quest - Set 2


  • Learn Quest


    One day Vikram was out bicycling. After entering a one-way tunnel and after having ridden one-fourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?

    Bus .......... A....... Vikram ...B
    AB is a tunnel
    And suppose person is at point C
    AC : CB = 1:3
    Now the time at which bus reaches A
    Vikram would reach from point C to point A
    And the time at which bus reaches to B
    Vikram reaches from C to B
    Let assume
    Length of tunnel is 4km
    So if Vikram moves 1 km forward towards B
    Bus would reach at A
    So remaining distance covered by Vikram is 2km
    And bus covered the whole distance 4km(length of tunnel) in the same time with a speed of 80km/hr
    So speed of Vikram = 80/2 => 40km/hr

    Find the ten's place digit of 625^246 - 441^128

    (a5)^even ends with 25
    Last two digits
    (41)^28
    => (81)^14
    => (61)^7
    => 21
    Or (41)^28
    => (4 * 8)1
    => 21

    So 25-21 = 04

    digit at tenth place = 0

    25 males and 12 females can complete a piece of work in 12 days. They worked together for 8 days and the women left the work.the remaining men completed the work in 6 days. Find out in how many days 15women can complete the entire work?

    They worked for 8 days
    I.e completed 2/3 work
    So Remaining (1/3) work is completed by men in 6 days
    I.e males complete this work in 18 days
    So 18 * X /(18+X) = 12
    => 3x /(18+X) = 2
    => 3x = 36 + 2x
    => X = 36
    I.e 12 women complete this work in 36 days
    So 15 women in 12 * 36/15
    I.e 144/5 days

    X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day one man whose capacity to do the work is twice that of X, joined. On the third day another man whose capacity is thrice that of X, joined and the process continues till the work is completed. In how many days will the work be completed, if everyone works for four hours a day?

    Total work = 20 * 7 = 140 units
    Efficiency of X => 1unit /hour

    First day
    1 * 4

    Second day
    Another man joined
    Whose efficiency is 2
    So total work on second day
    (1+2) * 4

    So basically
    4 [ 1 + 1+2 + 1+2+3 + 1+2+3+4 + 1+2+3+4+5 ] = 140
    So 5 days

    How many multiples of 18 are there which are less than 3500 and also 2 more than the square of a natural number

    18k = t^2 + 2 ---(1)
    Now t^2 mod 18
    Min Possibility 0
    and max possibility - 17
    so for t^2 +2 to be divisible by 18
    t^2 must give remainder as 16 .
    so t must be of type 18k+ 4 or 18k - 4 ..
    √3600 = 60
    √3500 = 59 (approx)
    So maximum value of t can be 59
    so t = 4 (18 * 0 +4)
    14 (18 * 1 - 4)
    22 ,
    32,
    40,
    50 ,
    58(18 * 3 + 4 )
    So 7

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    Area of triangle using medians
    M = (16+20+24)/2 => 30
    Area of triangle
    => 4/3 √ [30 * (30-16)(30-20)(30-24) ]
    => 4/3 √ [ 30 * 14 * 10 * 6]
    => 80 √7

    Now we know medians divide
    Triangle into 6 equal parts
    So area of one smaller triangle who height is BG
    Is 1/6 (80√7)
    And base = 1/3 of (16)
    So
    (1/2) * (1/3) * 16 * h = 1/6 (80√7)
    => H = 5√7

    If a > 0, b > 0 then the number of real roots of the equation 2x^7 - ax^5 - 3x^4 - bx^2 + 7 = 0 can be
    a) 2
    b) 3
    c) 4
    d) 0

    Use Descartes rule
    And see the sign changes in F(X)
    And F(-x)
    If Number of sign changes in F(X) =k
    number of Positive roots can be k , k-2 ,k-4 and so on
    Similarly
    Number of sign Changes in F(-x) = k
    number of negative roots = k , k-2 , k-4 and so on

    Find the largest 5 digit number which when divided by 8 leaves a remainder of 5 and when divided by 7 leaves a remainder of 2.

    7k +2 => 8p +5
    => 7k = 8p + 3
    For p = 4
    It satisfies

    I.e least number which satisfies the condition is
    8 * 4 +5 = 37
    And number is of the form
    56k + 37
    Now to find largest 5 digit number
    99999/56
    => 1785.xx
    So k = 1785
    56 * 1785 + 37
    => 99997

    From the first 25 natural numbers, how many arithmetic progressions of 6 terms can be formed such that common difference of the AP is a factor of the 6th term.

    Let the first term of the AP be a and the common difference be d.
    The sixth term of the series will be a+5d
    Given that d should be a factor of a+5d
    => a+5d is divisible by d
    => a should be divisible by d
    So the required cases are
    d = 1, a = 1,2,3.......20
    d= 2 , a = 2,4,6.......14
    d = 3, a = 3, 6,9
    d= 4, a = 4
    So the required number of AP’s are 20+7+3+1 = 31

    101010101010..... (94 digits) / 375. Find the remainder.

    375 => 125 * 3
    Use CRT
    With 3
    Sum of digits
    47 mod 3 = 2
    Now ,
    With 125
    You have to check only last three digits
    As 10^3 mod 125 = 0
    So last three digits would be 010
    So 125k+10
    3b +2 = 125k +10
    => For k = 2
    It satisfies
    So 260


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