Quant Boosters by Shubham Ranka, IIM Calcutta


  • Being MBAtious!


    6! - 2 * x
    x = D _ _ _ _ _ = 5! = 120
    = _ D _ _ _ _ = 3c1 (Sice A/B can not come in the first place now) * 4! = 72
    = _ _ D _ _ _ = 3c2 * 3! * 2! = 36
    = _ _ _ D _ _ = 3! * 2! = 12
    == > x = 120 + 72 + 36 + 12 = 240
    == > Total no. of ways = 720 = 480 = 240


  • Being MBAtious!


    Q 12) 2 years ago, fathers age was 6 times of his son's age. 6 years hence the ratio between the ages of father and son is 10:3. What is fathers present age?
    a) 40
    b) 42
    c) 44
    d) 48


  • Being MBAtious!


    2 years ago, let ages be: F S
    F = 6s
    F+8/S+8 = 10/3
    3(6S+8 ) = 10(S+8 )
    S = 7
    F = 42
    == > Current age is 44


  • Being MBAtious!


    Q 13) A square is divided into 108 identical rectangles. The ratio of length to breadth of the rectangle is 3:4. If the diagonal of the rectangle measures 1 cm, then the length of the diagonal of the square is closest to
    a) 9 cm
    b) 10 cm
    c) 11 cm
    d) 12 cm


  • Being MBAtious!


    Let the side of square be a, and rectangle be 3y, 4y
    == > a^2 = 108 * 12y^2 --- (1)
    Length of diagonal, 5y=1 == > y=1/5
    == > length of diagonal of square = rt(2)a = rt(2 * 108 * 12) * 1/5 = 10.18 --- > (B )


  • Being MBAtious!


    Q 14) If 3, 2 and 2 are three altitudes of a triangle then find the area & perimeter of the triangle.


  • Being MBAtious!


    Corresponding sides will be: 2x,3x,3x
    s = 4x
    == > A = rt(4x * 2x * x * x) = 2rt(2)x^2 = 1/2 * 3 * 2x
    == > x = 3/2rt(2)
    A = 3x = 9rt(2)/4
    p = 2s = 8x = 6rt(2)


  • Being MBAtious!


    Q 15) The earnings of a and b are in ratio 3:7 and that of b and c is 4:9 and that of d and c is ,7:6 if the sum of the earnings of a b c and d is 52950 then what are the earnings of d ?


  • Being MBAtious!


    a:b = 3:7
    b:c = 4:9
    c:d = 6:7
    a:b:c:d=72:168:378:441
    TRICK:
    [a:b:c:d = 3 * 4 * 6: 7 * 4 * 6 : 7 * 9 * 6 : 7 * 9 * 7]
    a+b+c+d=52950
    d = 52950 * 441/1059 = 22050


  • Being MBAtious!


    Q 16) f(x)=ax^2+bx+c (b > a) & f(x) ≥ 0 for all x. Find the minimum value of (a+b+c)/(b-a)


  • Being MBAtious!


    a > 0, b^2 < = 4ac
    By symmetry:- a=c== > b=2a (D=0)
    4/1== > 4

    Explanation:

    We know, a > 0, b^2 < =4ac. So, c > 0. Now sice b > a == > a,b,c > 0
    Now, Considering the boundary case:- b^2 = 4ac. And, in this eqn: interchanging a and c does not change my final outcome. So, I safely assumed a=c to check for the boundary case. Now, it could either be minima or maxima. Check for other values to figure out that.


  • Being MBAtious!


    Q 17) a1 = p, b1=q.
    a(n)=pb(n-1) and b(n) = qb(n-1) for all even n > 1
    a(n)=pa(n-1) and b(n) = qa(n-1) for all odd n > 1
    Find sigma[a(n)+b(n)] for even n.


  • Being MBAtious!


    a1=p, b1=q
    a2=pq, b2 = q^2
    a3 = p^2q, b3 = pq^2
    a4 = p^2q^2, b4 = pq^3
    a5 = p^3q^2, b5 = p^2q^3
    a6 = p^3q^3, b6 = p^2q^4
    ...
    an = p^n/2.q^n/2,
    bn = sigma[a(n)+b(n)] = pq(1+pq+p^2q^2+....(pq)^n/2 -1) + q^2 (1+pq+p^2q^2 + ...(pq)^n/2 -1)
    = (pq+q^2)* pq^(n/2-1)/(pq-1)
    = q(p+q)[pq^(n/2)-1]/ (pq-1)


  • Being MBAtious!


    Q18 ) Find the minimum value of |x^2 - 2x + 5| + |2x^2 - 4x + 9| + |3x^2 - 6x + 7| .


  • Being MBAtious!


    |(x-1)^2 + 4 | + |2(x-1)^2 + 7| + |3(x-1)^2 + 4|
    min. @ x=1 --- > 15
    PS: Assuming x to be a real no.


  • Being MBAtious!


    Q 19) If a + b + c = 6 where a,b,c are positive then maximum value of (a+1)(b+2)(c+3) is


  • Being MBAtious!


    AM > = GM
    (a+1 + b+2 + c+3)/3 > = (a+1)(b+2)(c+3)^1/3
    4 > = (a+1)(b+2)(c+3)^1/3
    (a+1)(b+2)(c+3) < = 4^3
    (a+1)(b+2)(c+3) < = 64
    Maximum value is 64

    Alternate approach:
    a + 1 =x, b + 2 = y and c + 3 = z
    x + y + z = 12
    For maximum value equate the terms x = y = z = 4
    So product max is 4 x 4 x 4 = 64


  • Being MBAtious!


    Q 20) If 2x^2 + 3y^2 = 1. If M is the maximum value and N is the minimum value of x^2 + 3xy + 2y^2. What is M+N - MN


  • Being MBAtious!


    Let x^2 + 3xy + 2y^2 = k
    Given that 1 = 2x^2 + 3y^2
    So, x^2 + 3xy + 2y^2 = k = k*1 = k(2x^2 + 3y^2)
    == > x^2 (1-2k) + 3xy + y^2 (2-3k) = 0
    Dividing by y^2:-
    (x/y)^2 (1-2k) + (x/y)*3 + (2-3k) = 0
    For the above quadratic eqn in (x/y) to have a real solution:-
    D > = 0 thus
    9 - 4(1-2k)(2-3k) > = 0'
    so 24k^2 - 28k - 1 < = 0. Thus the interval is [N,M] so N and M are roots of the equation, thus M + N = 7/6 and MN = -1/24
    so M + N - MN= 7/6 - 1/24 = 27/24 = 9/8


  • Being MBAtious!


    Q 21) Find the value of the series 3/1! + 7/2! + 15/3! + 27/4! + 43/5! + ...


 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.