Quant Boosters - Sibanand Pattnaik - Set 1


  • QA/DILR Mentor | Be Legend


    So , each number from 1 - 10 has a distinct unit digit for its cube ..
    occurrence of each digit in cube is 1/10 times
    So 0 - 10^100 it will be 10^100 * 1/10 = 10^99


  • QA/DILR Mentor | Be Legend


    Q21) 20 groups of five shooters each compete in an shooting championship. A shooter finishing in Nth position contributes N points to his team, and there are no ties. The team that wins will be the one that has the least score . Given that , the 1st position team's score is not the same as any other team, the number of winning scores that are possible is


  • QA/DILR Mentor | Be Legend


    The group’ scores must sum to 1 + 2 + . . . + 99 + 100 = 5050.
    The winning group can be ATMOST 1/20 * 5050 = 252.5 and is at least 1 + 2 + 3 + 4 + 5 = 15.
    However, not all scores between 15 and 252 inclusively are possible because all teams must have integer scores and no team can tie the winning team.
    number of scores possible = 251 - 15 + 1 = 237


  • QA/DILR Mentor | Be Legend


    Q22) X varies inversely as y and x varies directly as the square of z. If y decreases by 43.75% and z decreases by 75%, by what percent does x change?
    a. 88.88%
    b. 50%
    c. 45.45%
    d. 54.54%
    e. 44.44%


  • QA/DILR Mentor | Be Legend


    Here my approach would be
    X varies inversely as Y so XY is Constant
    X varies directly with z^2 so x/z^2 is a constant
    so (XY)/Z^2 = constant
    X = (constant * Z^2 )/Y
    So multiplying factor of X = (1 * multiplying factor of Z^2 )multiplying factor of Y
    => multiplying factor of X = [1 * (1/4)^2] / (9/16)
    => multiplying factor of X = 1/9
    As mulpliying factor of x is 1/9 so change is "x" is 8/9 .. i.e 11.11 * 8 = 88.88 %


  • QA/DILR Mentor | Be Legend


    Q23) Total positive integral solutions of 4x + 5y + 2z = 111?


  • QA/DILR Mentor | Be Legend


    Answer by Bruce Wayyne
    y needs to be odd, else LHS would become even.
    say y = 2k-1 k is positive
    => 4x + 5 (2k-1) + 2z = 111
    2x + 5k +z = 58
    Let k = 1,
    2x+z= 53 so (1, 51)...(26, 1) so 26
    k=2
    2x+z=48 so (23, 2).....(1, 46) so 23
    k=3
    2x+z=43 so (1,41)....(21, 1) so 21
    k=4
    2x+z= 38 so (1, 36)....(18,2) so 18
    k=5
    2x+z= 33 so (1,31)....(16,1) so 16
    Hence the pattern is +3, +2, +3, +2....so on.
    so (26+21+16+11+6+1) + (23+18+ 13 + 8 + 3) = 146


  • QA/DILR Mentor | Be Legend


    Q24) The volume of a mixture of milk and water was increased by 68.75% by adding pure milk to it..if the ratio of milk n water in the resultant solution is 3:1.. find the ratio of milk n water in the original solution ?


  • QA/DILR Mentor | Be Legend


    68.75 = 11/ 16
    let Vol 1= 16 then vol 2 = 27
    We know water proportion initially * Volume 1 = Water proportion finally *Volume 2
    => water proportion initially * 16 = (1/4) * 27
    =>water proportion initially = 27/64
    so water : milk = 27: 37
    or
    milk : water = 37: 27


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    Q25) HCF of 3 numbers is 6 and their sum is 120. How many such triplets exist ?


  • QA/DILR Mentor | Be Legend


    6x + 6y + 6z = 120
    => x + y + z = 20
    19c2 cases

    HCF > 1
    only possible case for HCF = Either 2 and 5 ..
    case -1 (when HCF =2)
    2x + 2y + 2z = 20
    => x + y + z = 10
    9c2

    Case -2 , (When HCF = 5)
    5x + 5y + 5z = 20
    => x + y + z = 4
    so 3c2

    Therefore , required cases = 19c2 - 9c2 - 3c2 = 132 (ORDERED )

    Now remove the 2 same 1 cases where HCF is 1
    cases , 1 , 1 ,18 ; 3, 3 , 14 ; 7,7,6; 9,9,2 ... each of these arranged in 3!/2! ways so , 12 ways ..
    so (132 - 12)/6 + 4 = 24
    (for un ordered) ..


  • QA/DILR Mentor | Be Legend


    Q26) Ram bought a few mangoes and apples spending an amount of at most 2000.If each mango cost 4 and Apple 6 and Ram bought at least 1. Find the different possible amounts could have spent in purchasing fruits?


  • QA/DILR Mentor | Be Legend


    Sum of 3 number is even so either all three are even or 2 odd and 1 even ..
    7^4 > 2002 .. so it must be less than this ... 5^4 is 3 digit number so we MUST TAKE 6^4 else how can you form a 4 digit number .. ??
    that way 6^4 is included , now only 2 cases are possible
    either the other 2 number ll be even or they have to be odd ... By common sense , we ignore 4 and 2 so it must be 3^4 and 5^4
    so 6 + 5 + 3 = 14


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    Q27) If 1/a + 1/b + 1/c + 1/d = 2, where a , b , c , d are distinct natural numbers, what is the value of a + b + c + d ?


  • QA/DILR Mentor | Be Legend


    If sum of Factors of N excluding N is equal to N then N is called a perfect number ..
    for Eg 28 = 1 + 2 + 4 + 7 + 14
    496 = 1 +2 +4 +8 + 16 + 31 + 62 + 124 + 248
    PERFECT NUMBERS show another property
    The sum of the reciprocal of the factors of a perfect number INCLUDING THE NUMBER ITSELF = 2
    Again i repeat "INCLUDING THE NUMBER ITSELF"
    for eg : 28 is a perfect number
    1+ 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 2


  • QA/DILR Mentor | Be Legend


    Q28) Find the largest value of a for which 100 + a^3 becomes perfectly divisible by 10 + a


  • QA/DILR Mentor | Be Legend


    by Factor theorem, if 100 + a^3 is divisible by 10 + a then substitute a = -10 in 100 + a^3 = 100 - 1000 = 900
    Now 10 + a must be a factor of 900
    so let K * (10 + a) = 900
    Now to obtain the greatest value of "a", we have to put K as 1
    so 10 + a = 900
    => a = 890


  • QA/DILR Mentor | Be Legend


    Q29) How many ways can 22 identical balls be distributed in 3 identical boxes ?


  • QA/DILR Mentor | Be Legend


    It is unordered distribution of a+b+c = 22
    so total cases = 24c2
    2 same n 1 different cases .. will be .. 0 0 22 ; 1 1 20 ....till 11 11 0 .
    so total 0 - 11 that is 12 cases and each of them ll be arranged in 3!/2! i.e 3 ways
    why ?? because they are of type A A B ..
    so( 24c2 - 36 ) / 6 + 12 = 52


  • QA/DILR Mentor | Be Legend


    Q30) X and Y have some chocolates with them which they wish to sell . The cost of each chocolate is equal to the number of total chocolates with both of them together initially. Together they sell all the chocolates and after that they start distributing the money collected in this particular fashion. First X takes a 10 rupee note, then Y takes a 10 rupee note and so on. In end it's turn of Y who didn't get any more 10 rupees. How much rupees Y get in his last turn?


 

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