Quant Boosters - Sagar Gupta, CAT Quant 99.2 Percentile - Set 2


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    Author : Sagar Gupta - MBA student at Symbiosis Institute of Operations Management. 99.2 Percentile in CAT 2015 (Quant)

    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.

    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

    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

    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

    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?

    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 !

    17!=355687ab8096000. Find the value of ab

    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)

    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.

    Each number is of the form 3^0, 3^1, 3^1 + 3^0, 3^2, 3^2+3^0, 3^2 + 3^2 + 3^2+3^1+3^0
    convert in base 3 :
    1,10,11,100,101 and so on
    100 in base 2 is 1100100
    100th term will be 3^6 + 3^5 + 3^2 = 729 + 243 + 9 = 981

    There exist three positive integers P, Q and R such that P is not greater than Q, Q is not greater than R and the sum of P, Q and R is not more than 10. How many distinct sets of the values of P, Q and R are possible?

    q = p + x, r = q + y, where p is positive integer and x, y are non-negative integers
    => y + 2x + 3p < = 10
    => y + 2x + 3p' < = 7 (p' = p + 1)
    So we have (4 + 3 + 1) + (4 + 2 + 1) + (3 + 2) + (3 + 1) + (2 + 1) + (2) + (1) + (1)
    = 31 solutions

    India and Brazil play a football match in which India defeats Brazil 5-2. In how many different ways could the goals have been scored if Brazil never had a lead over India during the match ?

    First goal always India will score : 1-0
    Now :
    I / B B I I I I
    Only 1 way brazil can take a lead :
    ( 6! / 2! 4! ) -1
    14

    For any positive integer n, P(n) is the product of digits of n, then find the value of P(1) + P(2) + ...... + P(999).
    Note:- P(1) = 1, P(6) = 6, P(23) = 6, P(900) = 9 and so on

    0-9 : 1 + ( 1+2+3+.....9)
    10-19 : 1 + ( 1+2+3+.....9)
    20-29 : 2 + 2 ( 1+2+3......9)
    .
    .
    90-99 : 9 + 9 ( 1+2+3...... 9 )
    total =46 + 46 *45 = 46^2

    100-109 : 1 + ( 1+2+3...9)
    110 - 119 : 1 + ( 1+2+3+.....9)
    120 - 129 : 2 + 2( 1+2+3......9)
    .
    .
    190-199 : 9+ 9 ( 1+2 + 3 +.........9)
    total = 46 + 46 *45 = 46^2

    200-209 : 2 + 2 [ 1 + 2 +.......9 ]
    210 -219 : 2 + 2 [ 1 + 2 +........9 ]
    220 - 229 : 4 + 4 [ 1 + 2 +......... 9 ]
    .
    .
    290 -299 : 18 + 18 [ 1 +2 +3 ...... 9
    total = 92 + 92 [ 45 ] = 92 [ 46 ] = 2*46^2

    If you see the pattern :
    Series will go like :
    46^2 + ( 46^2 + 2 * 46^2 +.........9 * 46^2 )
    46^2 + 46^2 [ 45 ] = 46^3
    Now subtract 1 from it as we considered 0 in the beginning
    Answer = 46^3 - 1 = 97735

    The numbers a1, a2,...,a108 are written on a circle such that the sum of any 20 consecutive numbers equals 1000.
    If a1 = 1, a19 = 19, and a50 = 50, find a100

    HCF of 108 and 20 is 4, so terms will start repeating after every 4th term
    So, a100 will be 1000/5 - 50 - 19 - 1 = 130

    Find the number of integral solutions of x^2 - 3y^2 = -2 , 0 < x < 20

    x^2 - 3y^2 = -2
    x^2 = 3y^2 - 2
    x^2 mod 3 = -2
    x^2 = 3k - 2 = 3k + 1
    3y^2 = 3k + 1 + 2 = 3(k+1)
    y^2 = k+1
    k = -1,0,3,8,15,24,35,48,63,80,99,120
    x^2 = -2,1,10,25,46,73,106,145,190,241,298,361
    x = 1,5,19 at k = 0,8,120 = y^2
    Points : 1,1 ; 1,-1 ; 5,3 ; 5,-3 ; 19,11 ; 19,-11
    6 solutions


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