Quant Boosters - Vikas Saini - Set 5



  • 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



  • x = 8,10,12 ... 100.
    total terms = (100-8 )/2 + 1 = 47.
    value of x for which total product should be negative.
    so for 8-10,if we put x = 9 then one value will be positive,rest 46 values will be negative. Product of negative nos(even no of negative terms) give a positive value.
    For x = 11, 2 positive & 45 negative
    hence product negative.
    so negative value of x = {(100-8 )/2}/2 = 23.
    Hence 23 values of x for which product < 0.



  • Q11) Find the number of integer values of x for which (x+1)(x-3)(x+5)(x-7) ... (x+97)(x-99) < 0



  • Here x = -1,-5,-9 ... -97.
    x = 3,7 ... 99
    for x € (-1,-99) product will come negative.
    total terms = (-1+97)/4 + 1
    = 25.
    but in between two terms there will be few values for which product will become negative.
    Let's take any two term -1 & -5.
    in between -1 & -5, 3 values x=-2,-3,-4 product will become negative. It means only 3 values in between any two terms.
    Hence total values = 3 x 25 = 75.



  • Q12) Find the number of integer values of x for which (x-1)(x+5)(x-9) ... (x+61) < 0



  • x = 1,9....57.
    x = -5,-13.....-61.
    here terms of x which will give product in negative.
    total term = (-5+61)/8 + 1
    = 7 + 1
    = 8.
    total values in between -5 & -13 which will give negative values are :-
    -6,-7,-8,-9,-10,-11,-12
    total 7 values.
    Total = 8 x 7 = 56.



  • Q13) Find the number of integer values of x for which (x+1)(x-2)(x+3) ... (x-98) < 0



  • x = -1,-3 ... -97
    x = 2,4 .... 98
    so total terms which will give negative value = (-1+97)/2 + 1
    = 49.
    The value in between two -1 & -3 which will give negative value is only one -2.
    so total values = 49 x 1 = 49.



  • Q14) Find the sum of all values of x which satisfy the equation 2|x-3| + 3|x+2| + 5|x-2| = 40.



  • Here values of x = 2,3,-2.
    Firstly we take extreme value of x = 3.
    At x = 3
    2|x-3|+3|x+2|+5|x-2| = 20.
    At x = 4,
    it will increase by 10,because 2+3+5 = 10.
    2|4-3|+3|4+2|+5|4-2| = 30.
    At x = 5,
    it will give value 40.
    2|5-3|+3|5+2|+5|5-2| = 40.
    So max value of x = 5.
    For min value of x = -2.
    at x = -2,
    2|x-3|+3|x+2|+5|x-2| = 30.
    at x = -3
    It will give 40.
    So sum of values of x = 5 - 3 = 2.



  • Q15) Find the range of x if |x+2| + |x-3| + |x-5| < 15.



  • First we take extreme value of x = 5.
    At x = 5,
    |x+2|+|x-3|+|x-5| = 9.
    at x = 6,
    |x+2|+|x-3|+|x-5| = 12.
    at x = 7,
    |x+2|+|x-3|+|x-5| = 15.
    For minimum take x = -2(min)
    at x = -2,
    |x+2|+|x-3|+|x-5| = 12.
    at x = -3,
    |x+2|+|x-3|+|x-5| = 15.
    x € (-3,7).


 

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