Quant Boosters by Hemant Malhotra - Set 19


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


    Find number of integral solutions of 2/x + 3/y = 1/10

    Concept:
    a/x + b/y=1/k
    where a, b and k are positive integers and we want number of value of (x,y) satisfying this equation
    Approach - first find number of factors of a * b * k^2 let number of factors=F
    a) total number of positive integral solutions=F
    b) total integral solutions = 2 * F-1
    c)total number of negative solution= zero ( bcz if both x and y will be negative than lhs will be negative but rhs is positive so not possible )

    solution:

    First find total number of factors of 2 * 3 * 10^2
    2^3 * 3 * 5^2 so (4 * 2 * 3) = 24 total number of factors
    so positive integral solutions = 24
    total number of integral solution = 2 * 24 - 1 = 47

    Find the smallest positive integer such that (n-13)/(5n+6) is a non-zero reducible fraction.

    hcf(n-13,5n+6)
    hcf((n-13,5n+6-5n+65))
    hcf(n-13,71))
    so reducible form for min n=
    n - 13 = 71
    so n = 84

    Find the minimum possible value of a + b + c if abc + bc + c = 2014, where a, b, c are positive integers.

    c * (ab+b+1)=2014
    c * (ab+b+1) = 2 * 19 * 53
    so c=19
    ab+b+1=106
    ab+b=105
    b * (a+1)=7*15
    so b=7 and a=14
    so sum =40

    Six jars of cookies contain 18 , 19 , 21 , 23 , 25 and 34 cookies respectively. One jar contains almond cookies only. The other 5 jars contain no almond cookies. A takes three of the jars and B takes two of the others. Only the jars of almond cookies remains. If A gets twice as many cookies as B, how many almond cookies are there ?

    A gets twice So he has even number of cookies
    even = even + even + even
    even = odd + odd + even
    three even is not here so he will choose two odd and one even
    so 34 + 19 + 21 or 34 + 19 + 23 or 34 + 19 + 25 or 34 + 21 + 23 or 34 + 23 + 25
    so 74 or 76 or 78 or 78 or 82
    now B has 37 or 38 or 39 or 41
    37 = 18+19 (19 is already used so not possible)
    38 is not possible
    39 = 18+21 (this is possible combination)
    41 = 23 + 18
    so 23 almond cookies are there

    If x = 1! + 2! +3! +4! + ... + n!, how many value of n, x is a perfect square?

    3,5,6 any perfect square will not leave this remainder when div by 7
    so 1!+2!+3!+4!=33 mod 7=5
    or 1!+2!+3!+4!+5!=153 mod 7=6
    or 1!+2!+3!+4!+5!+6! mod 7=153+620 mod7=6+4=10 mod 7=3
    so no value of n>=4 will give perfect square
    so check for n=1,2,3
    when n=1 then 1!=1 so perfect square
    when n=2 then 1!+2!=3 not a perfect square
    when n=3 then 1!+2!+3!=9=perfect square
    so n=1 and 3 only two values

    Find the number of integral solutions to |x| + |y| + |z| = 15
    a. 902
    b. 728
    c. 734
    d. 904

    x + y + z = 15
    so positive solutions = 15-1c3-1 = 14c2 = 7 * 13 = 91
    now
    when all negative =91
    when 2 positive one negative 3c2 * 91
    same for 2 negative one positive
    91 + 91 + 3 * 91 + 3 * 91
    8 * 91= 728
    now case when one is zero
    |y|+|z|=15
    then
    14 cases
    again 14 * 4=56
    so 3c2 * 56 = 3 * 56= 168
    now when 2 are zero
    |z|=15
    so 2 solutions
    so 3c2 * 2=6 solutions
    so total 728+174 =902

    If x² + xy +y² = 41, y² + yz + z² = 73 and z² + xz + x² = 9, find (x+y)/z.

    (x^3 - y^3) = 41 * (x - y)
    y^3 - z^3 = 73 * (y - z)
    z^3 - x^3 = 9 * (z - x)
    add all three
    41x - 41y + 73y - 73z + 9z - 9x = 0
    32x + 32y = 64z
    so (x+y)/z = 2

    Find all primes p, q, so that p^2 − 2q^2 = 1.

    p^2-2q^2=1
    (p^2-1) = 2 * q^2
    (p-1)(p+1) = 2 * q^2
    when p-1=1
    then p+1 = 2q^2
    so 2q^2=3 not possible
    when p-1=2
    then p+1=2q^2
    4=2q^2 so q^2=2 no possible
    when p+1=2 and p-1=q^2
    p=1 and 0=q^2 not possible
    now when p-1=2q
    and p+1=q
    so 2p=3q
    so p=3q/2
    so q=2 and p=3 will satisfy

    Let F(x) =x^5+x^4+x^3+x^2+x+1 . Find remainder when F(x^12) is divided by F(x)

    x^n-a^n=(x-a)(x^(n-1)+x^(n-2)+ .... a^(n-1)
    f(x^12)=1+(x^12)+(x^12)^2+(x^12)^3+(x^12)^4+(x^12)^5
    f(x^12)=1+(((x^12 -1)+1+(x^12)^2 -1 +1 +((x^12)^3 -1 +1 +(x^12)^4 -1 +1 +(x^12)^5 -1+1
    f(x^12)=6+((x^12-1)+(x^12)^2-1+(x^12)^3 -1+(x^12)^4 -1 +(x^12)^5-1
    now x^12-1=(x^6-1)(x^6+1)
    x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)
    so every term will contain x^6-1 and will be divisible by 1+x+x^2+x^3+x^4+x^5
    so OA=6

    The sum of first 2 terms of an infinite GP is 18. Also, each term in the series is seven times the sum of the terms that follow. Find the first term and common ratio of the GP

    a = 7 * ( ar + ar^2 + ar^3 ... )
    a=7(ar/1-r)
    so a-ar=7ar
    so 1-r=7r so 8r=1 so r=1/8
    now a+ar=18
    a(1+1/8)=18
    so a * 9/8=18
    so a=16


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.