Quant Boosters - Hemant Yadav - Set 1


  • Being MBAtious!


    p(n) = (n + 5)!/(n - 3)!

    It is clear that p(n) is a product of 8 consecutive numbers and divisible by atleast 8!. That means minimum power of 2 in p(n) is 7, i.e, p(n) is divisible by 2^7.

    But last non-zero digit has to be an odd number, so p(n) should be divisible by 5^7.

    So 15625 (5^6) or its multiple has to be one of the consective numbers, and we have to make sure that there is no multiple of 16 among the 8 numbers.
    Multiple of 16 nearest to 15625 are 15616 and 15632.

    So, least value of n = 15620.

    p(n) = 15618 * 15619 * 15620 * 15621 * 15622 * 15623 * 15624 * 15625
    After removing (2^7)*(5^7), we will get
    7809 * 15619 * 781 * 15621 * 7811 * 15623 * 1953 * 1

    Unit digit = unit digit of 9 * 9 * 1 * 1 * 1 * 3 * 3 * 1 = 9


  • Being MBAtious!


    Q14) How many real number solutions are there for a, b, c and d such that a² + b² + c² + d² = a(b + c + d)

    [OA : Only one solution]


  • Being MBAtious!


    a² + b² + c² + d² = a(b + c + d)
    => a²/4 + (a²/4 + b² - ab) + (a²/4 + c² - ac) + (a²/4 + d² - ad) = 0
    => a²/4 + (a/2 - b)² + (a/2 - c)² + (a/2 - d)² = 0
    => a = b = c = d = 0 is the only solution


  • Being MBAtious!


    Q15) There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner. What is the probability that they don't collide?

    [OA : 1/4]


  • Being MBAtious!


    Each ants can go towards two direction. so total different possibilities are 2 * 2 * 2 = 8. Out of which 2 are favorable where they all goes clockwise or anticlockwise . Hence required probability = 1/4


  • Being MBAtious!


    Q16) A motor cyclist is travelling on a straight road at 70 kmph towards his house. His pet bird travels at 80 kmph from the motorcylist to the house and back in 1.6 hours. Find the distance between the house and the place from where the bird starts its journey towards the house?

    [OA : 120 km]


  • Being MBAtious!


    In 1.6 hours the bird flies 128 km and in the same time the motor cyclist goes 112 km and they travel a distance twice the the distance of the house. So distance is (128 + 112)/2 = 120 km


  • Being MBAtious!


    Let x = a, y + z = b

    we get (x + y + z)^2006 + (x - y - z)^2006 = k1 * a^2006 * b^0 + k2 * a^2004 * b^2 + .... + k1004 * b^2006

    => number of terms = sum of the number of terms in b^0, b^2, .... , b^2006
    = 1 + 3 + 5 + ... + 2007
    = 1004² = 1008016


  • Being MBAtious!


    a) boxes and balls are all distinct and only 1 ball in one box
    Its like permuting a 5 digit number, so 5! Ways

    b) boxes are same and balls are distict and only 1 ball per box
    Since all the boxes are same and one ball per box, so just one way

    c) balls are same and boxes are different and only 1 ball per box
    Here all the balls are same and every box has one ball, so again one way

    d) balls and boxes are same and only 1 ball per box
    Here also only only way

    e) repeat all above questions with any number of balls in any box

    i) 5^5 ways, as all the balls and all the boxes are different.

    ii) Here boxes are same but balls are different.
    (5, 0, 0, 0, 0) – 1 way
    (4, 1, 0, 0, 0) – C(5, 4) = 5 ways
    (3, 2, 0, 0, 0) – C(5, 3) = 10 ways
    (3, 1, 1, 0, 0) – C(5, 3) = 10 ways, we don’t have to choose one of other two as they will automatically choose any of the two boxes (all the boxes are similar)
    (2, 2, 1, 0, 0) – C(5, 2)*C(3, 2)/2 = 15, here we need to divide by 2 as as it doesn’t matter if I put the first pair in box 1 and second pair in box 2 or vice-versa
    (2, 1, 1, 1, 0) – C(5, 2) = 10 ways
    (1, 1, 1, 1, 1) – 1 way
    So, total = 1 + 5 + 10 + 10 + 15 + 10 + 1 = 52 ways

    iii) x1 + x2 + x3 + x4 + x5 = 5
    So, C(9, 5) = 126 ways

    iv) Since all balls and boxes are same, we just need to find what different combinations are possible:-
    (5, 0, 0, 0, 0)
    (4, 1, 0, 0, 0)
    (3, 2, 0, 0, 0)
    (3, 1, 1, 0, 0)
    (2, 2, 1, 0, 0)
    (2, 1, 1, 1, 0)
    (1, 1, 1, 1, 1)
    So, 7 ways


  • Being MBAtious!


    Q17) Find the largest possible value of k for which 3^11 is expressible as the sum of k consecutive positive integers. What will be the answer if 3^11 is expressed as sum of k consecutive integers.

    [OA : 486, 2 * 3^11]


  • Being MBAtious!


    Let the first of the consecutive integers be a + 1, then the consecutive integers will be a + 1, a + 2, ..., a + k
    So, (a + 1) + (a + 2) + .... + (a + k) = 3^11
    k(k + 2a + 1) = 2 * 3^11

    Now for the first part all the integers are positive, so both (k + 2a + 1) > k
    Hence maximum value of k will be 2 * 3^5 = 486

    For the second part there is no such constraint, so maximum possible value of k will be 2 * 3^11


  • Being MBAtious!


    Q18) Find the sum of 6 * 1! + 13 * 2! + 22 * 3! + 33 * 4! + 46 * 5! + ... n terms


  • Being MBAtious!


    It is (n +2)! + 2[(n + 1)!] - 4 = (n + 4)[(n + 1)!] - 4.
    Key lies in simply finding the nth term of the sequence which is (n + 1)[(n + 1)!] + 2n[n!].


  • Being MBAtious!


    Q19) How many natural numbers N < 40000 such that cube of sum of digits of N gives back the same number N.
    For eg, 8³ = 512, sum of digits is 8, same as cube root of 512.

    [OA : 6]


  • Being MBAtious!


    Since N < 40000, N < 35³

    It is clear from the given condition that the sum of digits and cube of sum of digits should leave same remainder when divided by 9.

    Hence sum of digits should be of form 9k + 1 or 9k - 1 or 9k

    1³ = 1 (sum of digits is 1, so possible)
    8³ = 512 (sum of digits is 8, so possible)
    9³ = 729 (sum of digits is 18, so not possible)
    10³ = 1000 (not possible)
    19³ = 6859 (sum of digit is not 19, so not possible)
    18³ = 5832 (sum of digits is 18, so possible)
    17³ = 4913 (sum of digit is 17, so possible)
    26³ = 17576 (possible)
    27³ = 19683 (possible)
    28³ = 21952 (not possible)

    So, 1, 512, 4913, 5932, 17576 and 19683 are the only such numbers


  • Being MBAtious!


    Q20) How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form i, j, k where i, j and k are positive integers not exceeding four?
    a) 60
    b) 72
    c) 64
    d) 76

    [OA : 76]


  • Being MBAtious!


    We can see that there are 12 planes (4 paralle to x-y plane, 4 parallel to y-z plane and 4 parallel to z-x plane) In every plane there would be 2 diagonals, so 24 lines. 4 diagonals of the cube and 16 lines parallel to every axis. So, total 4 + 24 + 48 = 76


  • Being MBAtious!


    Q21) Circles A, B and C are externally tangent to each other, and internally tangent to a bigger circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the center of D. What is the radius of circle B?

    [OA : 8/9]


  • Being MBAtious!


    O is the centre of the bigger circle. and A,B,C are the centres of the remaining circles. Circles with centres B and C have the same radius. Radius of circle with centre A = 1.
    Obviously, the radius of the bigger circle is 2.
    Let OD = x (D is the mid-point of BC)
    Then we have: r^2 + x^2 = (2-r)^2
    and (x+1)^2 + r^2 = (1+r)^2
    Taking the diff of the above equations gives x = 3r - 2
    using which we get r = 8/9


  • Being MBAtious!


    Q22) For any positive integer n , f(n) is the highest power of 2 that divides n!
    Find f(1) + f(2) + f(3) + ..... + f(1023)

    [OA : 518656]


 

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