Quant Boosters - Learn Quest - Set 1


  • Learn Quest


    Total number of ways in which 35 identical marbles can be distributed among 5 boys, such that each boy gets an odd number of marbles?

    Odd is of the form (2k+1)
    (2a+1) + (2b+1) ... + (2e+1) = 35
    2a + 2b +....+ 2e = 30
    a + b + ... + e = 15
    Whole number Solution = 19c4

    If N = 2^3 * 3^2 then how many sets of 2 distinct factors of N are co-prime to each other

    N => 2^3 * 3^2
    No. Of co primes => [(2 * 3 + 1) ( 2 * 2 + 1) - 1]/2
    (7 * 5 -1)/2
    => 17

    If N is of the form a^x * b^y, Number of co-primes = [ (2x +1) (2y+1) - 1]/2

    116 people participated in a singles tennis tournament of knock out format. The players are paired up in the first round, the winners of the first round are paired up in second round, and so on till the final is played between two players. If after any round, there is odd number of players, one player is given a bye, i.e. he skips that round and plays the next round with the winners. Find the total number of matches played in the tournament.

    116 ppl.
    1 winner 115 loser.
    In 1 match 1 loser is decided. So 115 matches to decide 115 losers and 1 winner.

    Find the one hundredth smallest positive integer that can be written using the digits 1,3 and 5 in base 7?

    1,3,5
    1 digit : 3
    2 digit: _ _ =>9
    3 digit : _ _ _ =>27
    Total 39 so far
    1 _ _ _ => 27
    3 _ _ _ => 27
    Total 93 so far
    5 1 1 _ => 3
    5 1 3 _ => 3
    So 99 done
    100th will be 5 1 5 1 Ans

    If ‘x’ and ‘y’ are real numbers such that 3x + 4y = 48. Then, what is the maximum possible value of xy?

    x + x + x + 4y/3 + 4y/3 + 4y/3 /6 >= (x^3y^3 * 64/27)^1/6
    8^6 * 27/64 >= (xy)^3
    2^12 * 3^3 >= (xy)^3
    xy = 2^4 * 3 = 48

    N! is ending with m zeroes.(N+2)! is ending with (m+2) zeroes. Also 90 < = N < = 190.Find possible values of N?

    Increment of 2 zeroes as result of multiple of 25
    Means if you check for 24! =>
    24/5 => 4
    25/5 => 5
    5/5 => 1
    So 5 +1 => 6
    Same for 26!

    So
    98! And 100!
    99! And 101!
    Similarly
    148! --> 150!
    149! --> 151!
    And
    173! --> 175!
    174! --> 176!

    Remember you don't have to consider 125! (As it results into increment of 3 zeroes )
    So 6 possibilities

    Which of the following statements is false?
    a) (23!)^2 > 23^23
    b) (20!∗19!∗18!) < 57!
    c) (33!)^4 < 33^60
    d) None of these

    B is clearly right since 20 + 19 + 18 = 57

    A is right coz
    Lhs=> 23!^2 = (23 * 1)(22 * 2)(21 * 3)...(2 * 22)(3 * 23)

    However in rhs each terms is =>23 * 23 * 23...
    Certainly lhs > rhs

    Now in 3rd
    33!^2 > 33^33
    So 33!^2 * 33!^2 > 33^33 * 33^33
    33!^4 > 33^66 hence 3rd is false

    OA:C

    In a plane there are 37 straight lines of which 13 passes through Point A and 11 passes through Point B. Besides, no line passes through 3 points and no line passes through both A and B and no two are parallel. Find the total number of point of intersections of the straight lines

    Total number of points of intersection of 37 lines is 37c2
    But 13 straight lines out of the given 37 straight lines pass through the same point A.
    Therefore instead of getting (13c2) points, we get only one point A.
    Similarly 11 straight lines out of the given 37 straight lines intersect at point B.
    Therefore instead of getting (11c2) points, we get only one point B.
    Hence the number of intersection points of the lines is = 37c2 - 13c2 -11c2 + 2 = 535

    Find the number of ways of selection 4 letters from the word EXAMINATION

    EXAMINATION has 11 letters in total
    And out of which
    E --> 1
    X --> 1
    M -->1
    T ---> 1
    O ---> 1
    (A) --> 2
    I --> 2
    N --> 2

    8 distinct letters.

    1. 4 letters selected, which are all distinct: 8C4 = 70
    2. 2 letters alike, and 2 distinct
      (eg: AAEM)
      = 3c1 x 7c2 = 63
    3. 2 letters alike, and 2 letters alike
      (eg: NNII)
      = 3c2 = 3

    So answer is,
    70 + 63 + 3 = 136

    Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

    Method:1 (Not recommended)
    _ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers
    342 numbers added till 999
    1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315
    3 0 _ _ : 49 added
    3 1 _ _ : 49 added
    3 3 _ _ : 49 added
    3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
    Total 294 more. Remaining 21
    3 9 0 _ : 7
    3 9 1 _ : 7
    3 9 3 _ : 7
    3939

    Method:2
    Base method approach
    0 -> 0
    1 -> 1
    2 -> 3
    3 -> 4
    4 -> 7
    5 -> 8
    6 -> 9
    1000th term : 1000 in base 7 = 2626 < - > 3939


Log in to reply
 

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