Quant Boosters by Hemant Malhotra - Set 11


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    How many positive integral solutions does the following equation has : a + b + c + d = a × b × c × d

    All these types of questions can be solved by taking upper bound of the variables.
    Suppose all a, b, c and d are more than 1, then minimum product is 16, which is more than 2+2+2+2, taking bigger numbers will increase the product even further. Thus one of them is 1
    then the equation reduces to 1 + a + b+ c = abc.
    Again all a, b, c cannot be more than 1 as abc becomes a larger number, thus one of them is also 1,
    so 1 + 1 + a+b =ab or ab -a -b= 2 or (a -1)b-1) = 3,
    so (a, b) = (2, 4)
    so (1,1,2,4)

    Find number of real roots of x^4-4x^3+6x^2-4x+1=0

    Pascal triangle (1,4,6,4,1 ) for 4 degree so (x-1)^4 so OA=4

    In the expansion of (a + b + 2)^6, what will be the coefficient of a^2b^2?

    General term of (x+y+z .)^n is n!/(p!) * (q!) * (r!) * (x)^p * (y)^q*(z)^r where p+q+r=n
    same for 4,5,6 and so on variables
    now 6!/(p!) * (q!) * (r!) * (a)^p * (b)^q * (2)^r
    so p=2 and q=2 so r=2
    so coefficient
    6!/(2! * 2! * 2!) * 2^2

    A real valued function f is such that f(x + y) = x + f(f(y)) find f(9)

    let f(y) is quadratic equation then f(f(y) will be biquadratic which is not possible bcz RHS will not be biquadratic
    so F(y) will be linear
    so f(y)=ay+b
    so f(x+y)=a(x+y)+b
    and x+f((ay+b)=x+a*(ay+b)+b=x+a^2y+ab+b
    so ax+ay+b=x+a^2y+ab+b
    so ax+ay=x+a^2y+ab
    so ab=0 so a=0 or b=0 but a=0 not possible so b=0
    so ax+ay=x+a^2y
    so a=1
    so f(x)=ax+b=x
    so f(9)=9

    If |x – 1| + |x – 5| + |x – 9| = 8, what is the average of all the possible values of x?

    at x=9 , 8+4+0=12 so x > = 9 will not satisfy
    at x=1 ,0+4+8=12 again x < =1 will not satisfy
    now check for 1 < = x < = 5 and 5 < = x < = 9
    when 1 < = x < = 5
    then x-1-x+5-x+9=8
    so -x+13=8 so x=5 will satisfy
    now for 5 < x < 9
    x-1+x-5-x+9=8
    so 2x+4=8
    so x=2 which is not satisfying so only x=5 will satisfy

    Consider sets containing 3 elements, Sn = { n, 2n + 1, 3n + 2}. For how many values of n, where n is a natural number and n < 1000, both sets Sn and S(n+1) contain exactly one element that is divisible by 7?

    S={n,2n+1,3n+2} Sn+1={n+1,2n+3,3n+5}
    exact one element that is divisible by 7
    when n=7k form
    then {7k,14k+1,21k+2,7k+1,14k+3,21k+5} so exact one element is div by 7 so 7k < 1000
    so k=143
    now n=7k+1
    then Sn={7k+1,14k+3,21k+5,7k+2,14k+5,12k+12 so no element is exact div by 7
    now check till 7k+6 and find answer.

    If all the roots of y^5 – 10y^4 + py^3 + qy^2 + ry – 32 = 0 are positive, find the value of (p+q)/r ?

    Sum of roots=10
    Average=2
    Product of roots=32
    Geometric mean = 2
    so AM = GM
    so all roots equal. Roots are 2, 2,2,2,2

    A is the absolute difference between a three digit natural number and the sum of the digit of the number. If ‘A’ is a multiple of 12 and greater than 504, then how many values can A assume?

    let three digit number=100x+10y+z
    |100x+10y+z-x-y-z|=A
    so |99x+9y|=A
    so 9|11x+y|=A
    if A is multiple of 12 so A=12k
    9|11x+y|=12k
    so 12 k > 504
    so k > 42
    3|11x+y|=4k
    so k will be multiple of 3 and greater than 42
    so k=45
    then 11x+y=60
    when k=48 then 11x+y=64
    till 11x+y=108

    Find the remainder if 8533^859^361 divided by 19?

    8533 mod 19=2
    so 2^(859^361) mod 19
    E(19)=18
    so 859^361 mod 18=13
    so 2^13 mod 19=(16)^3 * 2 mod 19=9 * 32 mod 19=9 * 13 mod 19=3

    From the numbers 1, 2, 3, .., 9, we can choose six numbers in exactly two ways such that their product is a perfect square, let’s say sum are m^2 and n^2, where m and n are distinct positive integers.
    Find (m + n).
    a) 108
    b) 11
    c) 61
    d) 56
    e) 144

    The number can never have 5 and 7 so 1 * 2 * 3 * 4 * 6 * 8 * 9
    Now we can remove numbers which are x and x*n^2 once each.
    Only such pair is 2 & 8
    So one number is 36^2 and another one is 72^2
    so 108


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