Quant Boosters - Vikas Saini - Set 5



  • Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source -



  • Q1) If 4n has 15 factors, 5n has 12 factors. Find number of factors in 3n



  • Important part here is multiplication of 4 & 5.
    4 = 2^2.
    Let n = 2^a x 5^b.
    4n = 2^(a+2) x 5^b.
    Total factors = (a+2+1) (b+1)
    15 = (a+3) (b+1) .... (i)
    5n = 2^a x 5^(b+1)
    Total factors = (a+1) (b+1+1)
    12 = (a+1) (b+2) ..... (ii)
    From equation (i) & (ii)
    a = 2 , b =2.
    n = 2^2 x 5^2
    3n = 2^2 x 3 x 5^2
    Total factors = 3 x 2 x 3 = 18.



  • Q2) For how many positive values of N, (N + 102) / (N +2) must be an integer



  • For such questions we need to work on factors only.
    (N+2+100) / (N+2)
    1 + 100 / N+2.
    100 = 2^2 x 5^2.
    Total factors = (2+1) (2+1) = 9.
    But here N+2 > 2.( N must be positive)
    We need to remove 1 & 2 from total factors.
    Required values = 9 – 2 = 7.
    Check :-
    Factors of 100 are 1,5,25,2,10,50,4,20,100.
    After removing 1,2 it can be checked by putting N+2 = 5,25,10,50,4,20,100.
    N=3,23,8,48,2,18,98.
    (N+102) / (N+2) = 21,5,11,3,26,6,2. (Total 7 values)



  • Q3) For how many total values of N, (N + 28) / (N + 8) must be an integer



  • We need to work on factors.
    (N+8+20) / (N+8 )
    1 + (20 / N+8 )
    60 = 2^2 x 5.
    Total factors = 3 x 2 = 6.
    But in question it is given that for total values, so factors can be negative too.
    Total = 2 x 6 = 12 .
    Check :-
    Factors of 20: – (1,2,4,5,10,20).
    Negative factors :- (-1,-2,-4,-5,-10,-20).
    N + 8 = -1,-2,-4,-5,-10,-20,1,2,4,5,10,20.
    N = -9,-10,-12,-13,-18,-28,-7,-6,-4,-3,2,12.
    (N+28 ) / (N+8 ) = -19,-9,-4,-3,-1,0,21,11,6,5,3,2.
    Total 12 integer values.



  • Q4) A composite number is given 150.
    (a) Find the sum of divisors.
    (b) Find the sum of even divisors.
    (c) Find the sum of odd divisors.



  • 150 = 2 x 3 x 5^2.
    (a) Sum of divisors
    (2^0 + 2) (3^0 + 3)(5^0 + 5 + 5^2)
    = (1+2)(1+3)(1+5+25)
    =3 x 4 x 31
    = 372.
    Check :-
    1+150+2+75+3+50+5+30+6+25+10+15 = 372.
    (b) sum of even divisors
    2 (1+3)(1+5+25)
    = 2 x 4 x 31
    = 248.
    Check :-
    150+2+50+30+6+10 =248.
    (c) sum of odd divisors
    (1+3)(1+5+25)
    = 4 x 31
    = 124.
    Check :-
    1+75+3+5+25+15=124.



  • Q5) How many divisors of 1296 are perfect square



  • 1296 = 36^2 = (2^2 x 3^2)^2
    1296 = 2^4 x 3^4
    Perfect square = [ (4/2) + 1] [(4/2) + 1] = 3 x 3 = 9.
    To find perfect square :-
    (1,2,4,8,16,32)(1,3,9,27,81)
    (1,4,16)(1,9,81)
    (1,9,81,4,36,324,16,144,1296)
    Total 9.



  • Q6) How many of the first 1200 natural numbers are not divisible by 2, 3 or 5 ?



  • Solve it by Euler’s theorem
    If N = p^a x q^b x r^c
    E(N) = N (1 – 1/p)(1 – 1/q) (1 – 1/r)
    1200 = 2^4 x 3 x 5^2.
    Numbers not divisible by 2,3,5
    = 1200 (1 – 1/2)(1 – 1/3) (1 – 1/5)
    = 1200 (1/2)(2/3)(4/5)
    = 320.



  • Q7) If (7x-3)/(2x+5) ≤ 3 then x lies between



  • (7x-3)/(2x+5) - 3 = < 0
    7x-3-6x-15 / 2x+5 = < 0
    x-18 / 2x+5 < = 0.
    x = 18, x = -2/5.
    on line -2/5 will lie on negative side while 18 will lie on positive side.
    -inf to -5/2 ~ positive value
    -5/2 to 18 ~ negative
    18 to inf ~ positive.
    So here on -5/2 value can be positive which will not satisfy to the question.
    While on 18 it will give perfect 0.
    Hence x € ( -5/2, 18]



  • Q8) If (x^3 - 10x^2 + 29x - 20)/(x^2 + 7x + 12) < 0 then x lies between



  • here for x = 1, numerator can give value 0.
    hence (x-1) is a factor of numerator.
    = > (x^3 - x^2 - 9x^2 + 9x + 20x - 20)/(x^2 + 3x + 4x + 12) < 0.
    = > [x^2(x-1) - 9x(x-1) + 20(x-1)] / [x(x+3) + 4(x+3)] < 0.
    = > (x-1) (x^2-9x+20) / (x+3)(x+4) < 0.
    = > (x-1)(x-4)(x-5)/(x+3)(x+4) < 0.
    = > x = 1,4,5,-3,-4.
    on line -3 & -4 will lie on left side while 1,4,5 will lie on right side.
    -inf to -4 ~ negative
    -4 to -3 ~ positive
    -3 to 1 ~ negative
    1 to 4 ~ positive
    4 to 5 ~ negative
    5 to inf ~ positive.
    x € (-inf,-4) U (-3,1) U (4,5).



  • Q9) If [(x+7)^2 (2x-3)^3] / [(x-5) (x+1)] < 0 then x lies between



  • x = -7.
    x = 3/2.
    x = 5.
    x = -1.
    -7 & -1 will lie left side of line while 3/2 & 5 will lie right side of line.
    -inf to -7 ~ positive
    -7 to -1 ~ positive
    -1 to 3/2 ~ positive
    3/2 to 5 ~ negative
    5 to inf ~ positive
    x € (-inf,-7)U(-7,-1)U(-1,3/2)U(5,inf).



  • Q10) How many positive integer values can x take that satisfy the inequality (x-8)(x-10)(x-12) ... (x-100) < 0


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