Quant Boosters by Hemant Malhotra - Set 18


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


    Let f be a one-to one function from the set of natural numbers to itself such that f(mn) = f(m) x f(n) for all natural numbers m and n. What is the least possible value of f(111)

    one to one so single image for every element
    now f(1 * 1)=f(1) * f(1)
    so f(1)(f(1)-1)=0
    so f(1)=1 bcz f(1)=0 not possible we are taking about natural number
    now f(111)=f(3 * 37) = f(3) * f(37)
    we want f(3) and f(37) as min as possible and different from each other
    so put f(3)=2 and f(37)=3
    so 2 * 3=6

    If a three digit number ‘abc’ has 3 factors, how many factors does the 6-digit number ‘abcabc’ have?

    square of prime
    so abc = k^2
    now abcabc = abc * 1000 + abc
    abc * 1001
    so k^2 * 7 * 11 * 13
    now when k=any prime other that 11,13 till 31 then 24 factors
    when k is prime 11 or 13 let k=11 then 11^3 * 7 * 13 = 4 * 2 * 2 = 16

    Find number of real values of x satisfying 3[x]+2{x}=x where [x] greatest integral & {x}= fractional part
    3I + 2F = I + F
    so F = -2I
    so 0 < = F < 1
    so 0 < = -2I < 1
    so -1/2 < I < = 0
    so I = 0 is possibility
    so F = 0
    so x = 0 is only value

    A fare dice is thrown 5 times. What is the probability that a prime number appear even number of times

    2,3,5 three prime numbers
    so out of 6 we have 3 prime number
    so prob of prime=1/2 and prob of non prime=1/2
    so 5c0 * (1/2)^0 * (1/2)^55c2 * (1/2)^2 * (1/2)^3 + 5c4 * (1/2)^4 * (1/2)
    132 + 10 * 1/4 * 1/8 + 5 * 1/16 * 1/2
    16/32=1/2

    Find the largest integer N , if N + 496 = a² , N + 224 = b².

    a^2-496=b^2-224
    so (a-b)(a+b)=272
    now N will be max when a and b are as max possible
    a=69 and b=67
    so N=69^2-496=4265

    2x + y + z = 20 Find number of positive integral solution.

    y and z both even then y=2k and z=2m
    then x+k+m=10 so 9c2=36 cases
    when y and z both odd then y=2m-1 and z=2k-1
    then 2x+2m+2k=22
    so x+m+k=11 so 10c2=45
    so 36+45=81

    In a right angled triangle with integral sides, smallest side is four times of difference of other two sides and sum of all sides is 120 then find area of this triangle
    a. 60
    b. 120
    c. 180
    d. 240
    e. None of the above

    Method 1
    S = 4 * (H-M)
    Smallest side is multiple of 4 and S+M+H=120 so triplet will be
    S+M+(S+4M)/4=120
    so 5S+8M=480
    so M=(60)-((5S/8))
    so S=24 and M=45 so H=51 so area=540

    Method 2
    S = 4 * (H-M)
    so 8 = 4 * (17-15)
    so multiple of this will give our ans
    so 24, 45, 51 will be triplet

    gcd(1,55) + gcd(2,55) + ... + gcd(55,55) = ?

    E(55)=40
    now factors of 5 and 11
    till 50 , multiple of 5 are 10 so 5 * 10 = 50
    and multiple of 11 will be 11,22,33,44 so 11 * 4 = 44
    now 55 also
    so 40 + 50 + 44 + 55
    = 90 + 99 = 189

    For how many integral x , (3x+8)^2/(3x-2) is also an integer

    3x-2=y
    so 3x+8=y+10
    so (y+10)^2/y
    y+100/y +20
    100/y
    now factors of 100=2^2 * 5^2
    so 9

    How many integral values of (x, y) will satisfy the following inequality :-
    √(x² - 6x + 25) + √(y² - 8y + 25) ≤ 7
    a) 0
    b) 1
    c) 2
    d) 3
    e) more than 3

    x^2 - 6x + 25 = (x-3)^2 + 16
    so min of sqrt (x^2-6x+25) = 4
    y^2 - 8y + 25 = (y-4)^2 + 9
    so min of sqrt (y^2-8y+25) will be 3
    so min sum will be 7
    for x=3 and y=4


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