Quant Boosters - Hemant Yadav - Set 1


  • Being MBAtious!


    If h1, h2, h3 are the heights corresponding to the sides a, b, c, then

    a * h1 = b * h2 = c * h3 = 2A (where A is area of the triangle)

    => a = 2A/h1, b = 2A/h2 and c = 2A/h3

    Now, a + b > c
    => 2A/h1 + 2A/h2 > 2A/h3
    => 1/h1 + 1/h2 > 1/h3
    => 1/2 + 1/3 > 1/h3
    => h3 > 6/5

    Also, b + c > a
    => 2A/h2 + 2A/h3 > 2A/h1
    => 1/h2 + 1/h3 > 1/h1
    => 1/3 + 1/h3 > 1/2
    => h3 < 6

    So, 6/5 < h3 < 6
    => Integral values that h3 can take are 2, 3, 4, 5


  • Being MBAtious!


    minimum of |x|+|x-17|...+|x-440|
    median of 0,17,40,85,173,440 : (40+85)/2 = 62.5
    put x=62.5..so 641
    [swetabh kumar]


  • Being MBAtious!


    for 1st meet: ad/(a+b)= 400
    distance by A til 2nd meet: a/(a+b) * 3d = d + 300
    so 1200 = d + 300, d = 900
    [swetabh kumar]


  • Being MBAtious!


    ab - a - b = 35
    => (a - 1)(b - 1) = 36
    (a, b) = (2, 36), (3, 19), (4, 13), (5, 10), (7, 7)
    In last two cases a and b are not coprime


  • Being MBAtious!


    case 1: when there is no power of 2
    so 3 * 3 * 3=27 cases
    case 2: when there is only 1(2) in it
    one of the dice will have 2 or 6... and other two dices will have odd numbers.. so
    2 * 3 * 3 * 3c1=54
    case 3:
    when there are exactly two (2's)
    two dices will have (2,6) and one dice will have (1,3,5)
    so 2 * 2 * 3 * 3c1=36
    part2:
    exactly one dice will have 4 in it.. other two will have (1,3,5)
    so 3 * 3 * 3c1=27

    adding all 144 cases which will not have a multiple of 8.. remaining 72..
    so 72/216=1/3

    [jasneet dua]


  • Being MBAtious!


    17 * 20 * 23
    = (20-3) * 20(20+3)
    = 20 * (20² - 3²)
    = 8000 - 180
    (90 - 1)² = 8100 - 2 * 90 + 1 = 8000 - 180 + 101 is a perfect square
    => 101 is least such number


  • Being MBAtious!


    h(k1 + k2) = 1001
    1001 = 7 * 11 * 13
    Possible values of h : 2 * 2 * 2 = 8
    But if h = 1001 ; Sum = 1 => Not possible.
    Hence 8 - 1 = 7 possible values
    [Anubhav Sehgal]


  • Being MBAtious!


    d = x
    c = d + y = x + y
    b = c + z = x + y + z
    a = b + w = w + z + y + x, where x, y, z, w are natural numbers

    Add them to get 4x + 3y + 2z + w = 24
    x, y, z, w are natural numbers
    => 4x' + 3y' + 2z' + w' = 14
    (8 + 6 + 5 + 3 + 2) + (6 + 4 + 3 + 1) + (4 + 2 + 1) + (2) = 47
    So, 24 * 47 = 1128 solutions


  • Being MBAtious!


    Total number of 3 digit numbers = 900

    3-digit numbers having consecutive digits in first 2 places = (2*9 - 1)*10 = 170 ..as we can choose consecutive digits in 9 ways, but 0, is not allowed at first place)

    3-digit numbers having consecutive digits in last 2 places = (2*9)*9 = 162

    But above 2 cases have some number common. When 0 is middle digit, then we will have 1 common term(101), 1 as middle digit - 2 common terms (210, 212), 2 as middle digit - 4 common terms, when 3, 4, 5, 6, 7, 8 are middle digit then 4 common terms for each. For 9 - one common term

    SO total number of 3-digit number such that no adjacent digits are consecutive = 900 - 170 - 162 + 32 = 600


  • Being MBAtious!


    1/x + 1/y = 1/n
    => xy = nx + ny
    => (x - n)(y - n) = n²

    Now, n² must have 5 factors to get 5 ordered pairs
    => n must be of form p², where p is prime number

    Hence all primes less than 20


 

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