Quant Boosters - Sibanand Pattnaik - Set 1


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    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


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    Q17) How many pairs of factors of N = 360 will be coprime to each other?


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    360 = 36 * 10 = 2^3 * 3^2 * 5
    ORDERED PAIRS = (2 * 3 + 1)(2 * 2 + 1) (2 * 1 + 1) = 105
    UNORDERED PAIR = [(2 * 3 + 1)(2 * 2 + 1) (2 * 1 + 1) -1 ]/2 = 52
    As in this case variables are implicit so Its UNORDERED PAIRS .. so 52


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    Q18) How many ways can 360 be expressed as Product of 2 coprimes?


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    Let us suppose you have to express the number 360 as product of 2 co primes i,e a and b
    So 360 = 2^3 * 3^2 *5
    SO TOTAL 3 PRIMES Are there in 360 ..
    We have to express as product of 2 co-primes "a" and "b"
    So each of the primes i.e 2 , 3 and 5 has just 2 choices i.e either they can go "a" or to "b" .. so 2 * 2 * 2 = 8
    there these 8 include cases like (1. 360 ) and (360,1) i.e "repeat cases
    so 8/2 = 4
    the formula is (2^n)/2 = 2^(n-1)


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    Q19) Real madrid and Manchester united play a football match in which REAL beats MANCHESTER 4 - 2 ..In how many Ways can the goals be scored provided MANCHESTER never had a lead over REAL during the match?


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    if Manchester never had a lead means 1st goal was scored by REAL .. So score card at this juncture would ve been 1 - 0 ..The only way MANCHESTER could take lead is by scoring next 2 goals , i.e a pattern like RMMRRR.. Except this one case , MANCHESTER Can no way take lead as such ..
    So (5!)/3! *2! = 10
    NOW excluding this one case of RMMRRR ... We have 10 - 1 = 9


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    Q20) How many integers P are there between 0 and 10^100 such that units digit of P^3 is 1 ?


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    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


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    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


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    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


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    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%


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    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 %


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


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    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


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    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 ?


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    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 ?


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    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) ..


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    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?


 

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