Quant Boosters - Rajesh Balasubramanian - Set 5


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


    Number of Questions - 30
    Topic - Permutation & Combination
    Solved ? - Yes
    Source - 2IIM


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


    Q1) 3x + 4|y| = 33. How many integer values of (x, y) are possible?
    a) 6
    b) 3
    c) 4
    d) More than 6


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


    Let us rearrange the equation:
    3x = 33 – 4|y|

    Since x and y are integers, and since |y| is always positive regardless of the sign of y, this means that when you subtract a multiple of 4 from 33, you need to get a multiple of 3.

    Since 33 is already a multiple of 3, in order to obtain another multiple of 3, you will have to subtract a multiple of 3 from it. So, y has to be a positive or a negative multiple of 3.

    y = 3, -3, 6, -6, 9, -9, 12, -12...etc.
    For every value of y, x will have a corresponding integer value.
    So there are infinite integer values possible for x and y.


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


    Q2) (|x| - 3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?
    a) 4
    b) 6
    c) 10
    d) 8


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


    If x and y are integers, so are |x| –3 and |y| + 4. So, we start by finding out in how many ways 12 can be written as the product of two integers.

    12 can be written as 12 × 1, or 6 × 2, or 3 × 4. To start with, we can eliminate the possibilities where the two terms are negative as |y| + 4 cannot be negative.

    Further, we can see that |y| + 4 cannot be less than 4. So, among the values, we can have |y| +4 take values 4, 6 or 12 only, or |y| can take values 0, 2 and 8 only.

    When |y| = 0, |x| - 3 = 3, |x| = 6, x can be +6 or -6. Two pairs of values are possible: (6, 0) and (-6, 0)

    When |y| = 2, |x| - 3 = 2, |x| = 5, x can be +5 or --5. There are four possible pairs here: (5, 2) , (-5, 2), (5, -2), (-5, -2)

    When |y| = 8, |x| - 3 = 1, |x| = 4, x can be +4 or --4. There are four possible pairs here: (4, 8) , (-4, 8), (4, -8), (-4, -8)

    Correct Answer: 10 Pairs


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


    Q3) x + |y| = 8, |x| + y = 6. How many pairs of x, y satisfy these two equations?
    a) 2
    b) 4
    c) 0
    d) 1


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


    We start with the knowledge that the modulus of a number can never be negative, though the number itself may be negative.

    The first equation is a pair of lines defined by the equations
    y = 8 – x ------- (i) {when y is positive}
    y = x – 8 ------- (ii) {when y is negative}

    With the condition that x ≤ 8 (because if x becomes more than 8, |y| will be forced to be negative, which is not allowed)

    The second equation is a pair of lines defined by the equations:
    y = 6 – x ------- (iii) {when x is positive}
    y = 6 + x ------- (iv) {when x is negative}

    with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be negative.

    On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and (iv) are parallel (same slope).

    Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv) holds good only when x is negative.

    Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This satisfies both equations.

    Only one solution is possible for this system of equations.


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


    Q4) What is the number of real solutions of the equation x^2 - 7|x| - 18 = 0?
    a) 2
    b) 4
    c) 3
    d) 1


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


    Let us split this into two cases. Case 1, when x is greater than 0 and Case 2, when x is lesser than 0.

    Case 1
    x > 0. Now, |x| = x
    x^2 – 7x – 18 = 0
    (x – 9) (x + 2) = 0
    x is either –2 or +9.

    Case 2
    x < 0. Now, |x| = –x
    x^2 + 7x – 18 = 0
    (x + 9) (x – 2) = 0
    x is either –9 or +2.

    However, in accordance with the initial assumption that x < 0, x can only be –9 (cannot be +2).

    Hence, this equation has two roots: –9 and +9.

    Alternatively, we can treat this as a quadratic in |x|, the equation can be written as |x|2 – 7 |x| – 18 = 0.
    Or, (|x| – 9) (|x| + 2) = 0
    |x| = 9 or –2. |x| cannot be –2.
    |x| = 9, x = 9 or –9.

    Correct Answer: There are 2 solutions.


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


    Q5) x^2 - 9x + |k| = 0 has real roots. How many integer values can 'k' take?
    a) 40
    b) 21
    c) 20
    d) 41


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


    Discriminant, D = 81 – 4|k|
    If roots are real, D > 0
    81 – 4|k| > 0
    4|k| < 81
    |k| < 20.25

    Hence, –20.25 < k < 20.25
    The integer values that k can take are –20, –19, –18 … 0 … 18, 19 and 20.

    41 different values (Remember to include 0. Answer choice 'A' will definitely be 40


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


    Q6) x^2 - 11x + |p| = 0 has integer roots. How many integer values can 'p' take?
    a) 6
    b) 4
    c) 8
    d) More than 8


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


    To start with the discriminant should be a perfect square. Let the dicriminant be 'D'.

    From the quadratic formula: (−11 ± √D)/2, we see that the numerator has to be an even number for the roots to be integers.

    This implies that the discriminant should be a perfect square and be square of an odd number. (Only then we will have odd + odd = even in the numerator)

    D = 121 – 4|p| = 121 – 4|p|
    4|p| cannot be negative => D can take values 121, 81, 49, 25, 9, 1
    |p| can be 0, 10, 18, 24, 28, 30

    p can take 0, ±10, ± 18, ± 24, ± 28, ± 30

    Correct Answer: More than 8 values are possible.


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


    Q7) 2x + 5y = 103. Find the number of pairs of positive integers x and y that satisfy this equation.
    a) 9
    b) 10
    c) 12
    d) 20


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


    Rearranging the equation, we get:
    2x = 103 – 5y

    This says that when you subtract a multiple of 5 from 103, you get an even number. You have to subtract an odd multiple of 5 from 103 in order to get an even number. There are 20 multiples of 5 till 100, ten of which are odd. (Note that you cannot subtract 105, or higher multiples as they result in a negative value for x.)

    So, y can have ten integer values. x also has 10 integer values, each corresponding to a particular value of y.

    y = 1, gives us a potential value for x, so do y = 3, 5, 7….19. y can take 10 values totally.


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


    Q8) Consider three numbers a, b and c. Max (a,b,c) + Min (a,b,c) = 13. Median (a,b,c) - Mean (a,b,c) = 2. Find the median of a, b, and c.
    a) 11.5
    b) 9
    c) 9.5
    d) 12


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


    The numbers can be arranged in ascending order thus: Min < Median < Max
    According to the question, Min + Max = 13
    Median - (Min + Med + Max)/3 = Median - (12 + Med)/3 = 2.
    Solving, we get Median = 9.5.


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


    Q9) a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + c3z = d3.

    Which of the following statements if true would imply that the above system of equations does not have a unique solution?
    i. a1/a2 = b1/b2 = c1/c2 ≠ d1/d2
    ii. a1/a2 = a2/a3 ; b1/b2 = b2/b3
    iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers


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


    If we have three independent equations, we will have a unique solution. In other words, we will not have unique solutions if:

    1. The equations are inconsistent or
    2. Two equations can be combined to give the third

    Now, let us move to the statements.

    i. a1/a2 = b1/a2 = c1/c2 ≠ d1/d2

    This tells us that the first two equations cannot hold good at the same time.

    Think about this:

    x + y + z = 3;
    2x + 2y + 2z = 5.
    Either the first or the second can hold good. Both cannot hold good at the same time.
    So, this will definitely not have any solution.

    ii. a1/a2 = a2/a3 and b1/b2 = b2/b3

    a1, a2, a3 are in GP, b1, b2, b3. This does not prevent the system from having a unique solution.

    For instance, if we have

    x + 9y + 5z = 11
    2x + 3y – 6z = 17
    4x + y – 3z = 15

    This could very well have a unique solution.

    iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers
    This gives us practically nothing. This system of equations can definitely have a unique solution.

    So, only Statement i tells us that a unique solution is impossible.


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


    Q10) Equation x^2 + 5x – 7 = 0 has roots a and b. Equation 2x^2 + px + q = 0 has roots a + 1 and b + 1. Find p + q.
    a) 6
    b) 0
    c) -16
    d) 2


 

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