Quant Boosters - Rajesh Balasubramanian - Set 5


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Given, x^2 + 5x – 7 = 0 has roots a and b. We know that, Sum of roots in a quadratic equation = a + b = (−5)/1 = -5.

    Product of the roots = ab = (−7)/1 = -7.

    Now, The second equation 2x^2 + px + q = 0 has roots a + 1 and b + 1.

    Sum of the roots = a + 1 + b + 1 = a + b + 2 = (−p)/2 = -5 + 2 = -3 = (−p)/2 => -p = -6 => p = 6.

    Product of the roots = (a + 1)(b + 1) = ab + a + b + 1 = q/2. We know the values of ab and a+b. Substituting this, we get, -7 + (-5) + 1 = q/2 => -11 = q/2 => q = - 22.

    Hence, p = 6 and q = -22. => p + q = 6 + (-22) = -16.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q11) Sum of the roots of a quadratic equation is 5 less than the product of the roots. If one root is 1 more than the other root, find the product of the roots?
    a) 6 or 3
    b) 12 or 2
    c) 8 or 4
    d) 12 or 4


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Sum of the roots = (product of the roots) – 5
    Let r1 and r2 be the two roots
    r1 = r2 + 1

    Therefore,
    r1 + 1 + r2 = (r1r2) ∗ r2 − 5
    r2 + 1 + r2 = (r2 + 1) ∗ r2 − 5
    2r2 + 6 = (r2)^2 + r2
    (r2)^2 − r2 − 6 = 0
    Solving, we get r2 = 3 or -2
    Therefore, r1 = 4 or -1
    Hence product of the two roots is either 12 or 2.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q12) How many real solutions are there for the equation x^2 – 7|x| - 30 = 0?
    a) 3
    b) 1
    c) 2
    d) none


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Case 1: x > 0
    x^2 – 7x - 30 = 0
    (x - 10 ) (x - 3 ) = 0
    x = 10, x = -3
    But x = -3 is not possible as we have considered x > 0, thus 1 solution for this case.

    Case 2: x < 0
    x^2 + 7x - 30 = 0
    (x + 10 )(x - 3)=0
    x = -10 and x = 3
    Only x = -10 is permissible.
    Thus this equation has 2 real solutions

    Alternatively, we can think of the above as a quadratic in |x|
    x^2 – 7|x| - 30 = 0 can be factorized as
    (|x| -10) ( |x| + 3) = 0
    |x| cannot be -3, |x| can only be 10. x can take 2 real values.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q13) If (3x + 2y - 22)^2 + (4x - 5y + 9)^2 = 0 and 5x - 4y = 0. Find the value of x + y.
    a) 7
    b) 9
    c) 11
    d) 13


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Do not jump and start expanding the squares.
    In equation (1), both the squares will yield either positive number or 0. Since sum of two positive number cannot be 0, each of the term must be 0
    ∴ 3x + 2y - 22 = 0
    Multiplying by 2, and adding both equation
    6x + 4y - 44 = 0
    5x - 4y = 0
    =) x = 4
    Putting the value of x in the equation
    5 * 4 – 4y = 0
    y = 5
    ∴ x + y = 9


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q14) Let x^3 - x^2 + bx + c = 0 has 3 real roots which are in A.P. which of the following could be true
    a) b = 2, c = 2
    b) b = 1, c = 1
    c) b = -1, c = 1
    d) b = -1, c = -1


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Given the roots are in A.P. so let a-d, a, a+d be the roots
    From equation, sum of roots = 1
    Sum of two roots taken at a time = +b
    Product of two roots = -c

    ∴ (a-d) + (a) + (a+d) = 1
    =) 3a = 1
    =) a = 1/3

    Also, (a-d)a+ a(a+d)+ (a-d)(a+d) = b
    =) a^2 – ad + a^2 + ad + a^2 – d^2 = b
    =) 3a^2 – d^2 = b
    =) 3 * 1/9 – d^2 = b
    =) d^2 = 1/3 - b

    Now, since d is a real number,
    1/3 − b > 0 =) b < 1/3
    Also, (a-d)(a)(a+d) = -c
    =) a(a^2 – d^2) = -c
    =) 1/3 ∗ (1/9 − d^2) = −c
    =) c = d^2/3 − 1/27
    =) c = (9d^2 − 1)/27
    =) d^2 = 3c + 1/9

    Again, 3c + 1/9 > 0
    c > −1/27 > 0
    ∴ Only option (c) b = -1 , c = 1 could be true .


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q15) If f(y) = x^2 + (2p + 1)x + p^2 - 1 and x is a real number, for what values of ‘p' the function becomes 0?
    a) p > 0
    b) p > -1
    c) p ≥ -5/4
    d) p ≤ 3/4


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    The function f(y) is a quadratic equation.
    It is given that x is real.
    So the discriminant of f(y) ≥ 0
    i.e. D = b^2 - 4ac ≥ 0 or
    (2p + 1)^2 – 4(p^2 - 1) ≥ 0
    4p^2 + 4p + 1 – 4(p^2 - 1) ≥ 0
    4p + 5 ≥ 0
    Or p ≥ - 5/4


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q16) A merchant decides to sell off 100 articles a week at a selling price of Rs. 150 each. For each 4% rise in the selling price he sells 3 less articles a week. If the selling price of each article is Rs x, then which of the below expression represents the rupees a week the merchant will receive from the sales of the articles?
    a) 175 - x/2
    b) x/3 + 156
    c) 350 - x^2/2
    d) x^2 - 2x + 75


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    The selling price per article is Rs. x
    The price of the article is Rs (x - 150) more than Rs 150

    4% rise from Rs 150 ==> Rs 6
    For each Rs 6 increase in price over Rs.150, he sells 3 articles lesser.

    So, the reduction in number of articles = (x – 150) / 6 * 3
    So, in the new price of Rs.x he sells 100 - (x – 150) / 6 * 3
    Or he sells 100 – (x – 150)/2
    = (200 – x + 150) / 2
    = (350 – x)/ 2
    = 175 – x/2


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q17) How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?
    a) 6
    b) 12
    c) 24
    d) 48


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    288 = 2^5 × 3^2. So it has 6 × 3 = 18 factors. Or, there are 9 ways of writing this number as a product of two positive integers.

    Let us list these down.
    1 × 288, 2 × 144, 3 × 96, 4 × 72, 6 × 48, 8 × 36, 9 × 32, 12 × 24 and 16 × 18

    Now, this is where the question gets interesting. If a, b are integers either a + b and a – b have to be both odd or a+ b and a –b have to be both even. So, within this set of possibilities
    1 × 288, 3 × 96 and 9 × 32 will not result in integer values of a, b. So, there are 6 sets of numbers that work for us.
    Moving on to these six sets; let us start with one example and see how many possibilities we can generate from this.

    Let us consider the set 2 × 144.
    Let us solve for this for a, b being natural numbers first, then we will extend this to integers.
    When a, b are natural numbers
    a + b > a – b
    So, a + b = 144, a – b = 2; a = 73 and b = 71

    Now, if a = 73, b = 71 holds good. We can see that a = 73, b = –71 also holds good. a = –73, b = 71 works and so does a = –73, b = –71.

    There are 4 possibilities.
    a = 73, b = 71
    a = 73, b = –71
    a = –73, b = 71
    a = –73, b = –71
    So, for each of the 6 products remaining, we will have 4 possibilities each. Total number of (a, b) that will satisfy this equation = 6 × 4 = 24.

    Answer choice (c)


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q18) If x^4 – 8x^3 + ax^2 – bx + 16 = 0 has positive real roots, find a – b.
    a) -8
    b) 6
    c) -12
    d) -14


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Let p, q, r, s be the roots of the equation
    p + q + r + s = 8
    pqrs = 16
    This happens only when p = q = r = s = 2

    (p + q + r + s)/4 = 2.
    (pqrs)^1/4 = 2

    Arithemtic mean is equal to Geometric Mean. This is possible only when all the numbers are equal. (Note that the question says all roots are positive and real)

    p = q = r = s = 2
    pq + pr + ps + qr + qs + rs = a
    24 = a
    pqr + pqs + prs+ qrs = b
    32 = b
    So, a – b = 24 – 32 = – 8
    Answer choice (a)


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q19) x^3 – 18x^2 + bx – c = 0 has positive real roots, p, q and z. If geometric mean of the roots is 6, find b.
    a) 36
    b) -216
    c) 108
    d) -72


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    p + q + z = 18
    pq + qz + zp = b
    pqz = c
    According to the question, (pqz)^1/3 = 6. Hence, pqz = c = 216.
    This is possible only when p = q= z = 6. {Arithmetic mean of p, q and z = 6, GM is also 6. This is possible only when all three numbers are equal.}
    So, pq + qz + zp = 36 × 3 = 108 = b.
    Hence, b = 108.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    Q20) x^3 – 4x^2 + mx – 2 = 0 has 3 positive roots, two of which are p and 1/p. Find m.
    a) 5
    b) -11
    c) 8
    d) -2


 

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