Quant Boosters - Sagar Gupta - Set 1


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    C(58, 29) can be written as
    C(29, 0)^2 + C(29, 1)^2 + .... + C(29, 29)^2
    Except first and last term rest all terms are divisible by 29
    So, remainder will be 2.


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    Q6) If abc not equal to 0 and (2a^2) + (17b^2) + (8c^2) - 6ab - 20bc = 0, then what is the value of (a+b-c)/(a+b+c) ?


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    (2a^2-6ab+18b^2/4)+(50b^2/4-20bc+8c^2)=0
    2(a-3b/2)^2+2(5b/2-2c)=0
    so a=3b/2 and 5b/2 = 2c
    a = 3b/2 = 6c/5
    Solve and answer is 1/3


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    Q7) A diamond expert cuts a huge cubical diamond into 960 identical diamond pieces in minimum no. of 'n' cuts. If he wants to maximize the no. of identical pieces making same no. of cuts to it. Then maximum no. of such pieces are ?


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    960 = 8 * 10 * 12.
    total cuts (8-1) + (10-1) + (12-1) = 27 = 9 + 9 + 9.
    max pieces. 10 x 10 x 10 = 1000.


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    Q8) What is the remainder when { 2^11 + 5^11 + 8^11 + 11^11 + .......... + 59^11} is divided by 610 ?


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    Direct property: a^n + b^n + ... mod (a + b + c ..) = 0 if a, b, c... are in AP and n is odd


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    Q9) If the expression ax^2 + bx + c is equals to 4 when x = 0 leaves a reminder 4 when divided by x+1 and a reminder 6 when divided by x+2, then the value of a,b and c are respectively
    a) 1,1,4
    b) 2,2,4
    c) 3,3,4
    d) 4,4,4
    e) 2,3,4


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    x = 0 => c = 4
    x = -1 => a - b + 4 = 4 => a = b
    x = -2 => 4a - 2b + 4 = 6 => a = b = 1
    a,b,c = 1,1,4


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    Q10) Find the number of negative integral solution in (x +1)/x + |x + 1| = (x + 1)^2/x
    a) 0
    b) 1
    c) 2
    d) 3


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    Case 1 : x + 1 > 0 => x > -1
    Would not give any negative integral solution.
    Case 2 : x + 1 < 0 => x < -1
    Expression becomes :
    -(x + 1) = (x + 1)^2 / x
    => -x = x + 1 => x = -1/2 => No negative integral
    Case 3 : x = -1 satisfies. So 1 solution only.


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    Q11) If one root of the equation (I-m)x^2 + Ix +1 = 0 is double of the other and is real, find the greatest value of m.


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    Lets root be x and 2x
    Sum of roots => I/(m-I) = 3x ------ (1)
    Prod of roots => 1/(I-m) =2x^2 ------ (2)
    (2) /(1) => x = -3/2I
    Substitute in (1) => 2I^2-9I+9m=0
    D >= 0 for m to be max, D=0
    81=72m => m =9/8


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    Q12) The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is
    (1) b >= a^4
    (2) a >= 4b^4
    (3) a >= b^4
    (4) b >= 4a^4
    (5) none of these


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    x^2+y^2+2xy = a^2
    Now x^2+y^2 >=2xy
    => 2(x^2 + y^2) >=a^2
    Similarily working on 2(x^2+y^2) >=a^2
    we get a^4/8 < = (x^4 + y^4)
    => a^4/8 None of these


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    Q13) If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set S2 are divisible by at least four distinct prime numbers that are elements of the set S1?


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    Case 1 : 30p (2.3.5)
    30 * 7, 30 * 11, 30 * 13, 30 * 17, 30 * 19, 30 * 23 => 6 elements
    Also,30 * 7 * 2, 30 * 7 * 3 => 2 elements
    Case 2 : 42p (2.3.7)
    42 * 11, 42 * 13 => 2 elements
    Case 3 : 66p (2.3.11)
    66 * 13 => 1 element
    Total : 11 elements !


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    Q14) 17!=355687ab8096000. Find the value of ab


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    Divisibility of 11
    33 + a - (24 + b) = 0 or 11
    9 + a - b = 0 or 11
    9 + a - b = 11
    a - b = 2 ......(1)
    Divisibility of 9
    57 + a + b = 6 or 15 .....(2)
    a + b = 15 => a = 17/2 Not possible
    a + b = 6 => a = 8/2 , b=4/2 => a,b=(4,2)


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    Q15) Consider the increasing sequence 1, 3, 4, 9, 10, 12, 13… and so on. The sequence consists of all those positive integers which are powers of 3 or sum of distinct powers of 3. Find the 100th term of the sequence.


 

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