Quant Marathon by Gaurav Sharma - Set 9


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    How many values can natural number x take so that (4x + 7)^2/(x + 5) is an integer?

    1. 1
    2. 2
    3. 3
    4. 4

    (4x + 7)^2/(x + 5) = [ (x + 5) (16x – 24) + 169) / (x + 5) ]

    For the equation to have integer values, 169 should be divisible by (x + 5)

    So can take 2 values, 8 and 164

    Option (b) is correct.

    If the equation |x^2 + bx + c| = k has four real roots, then

    1. b^2 – 4c > 0 and 0 < k < (4c – b^2)/4
    2. b^2 – 4c < 0 and 0 < k < (4c – b^2)/4
    3. b^2 – 4c > 0 and k < (4c – b^2)/4
    4. None of these

    For the equation to have four real roots, the line y = k must intersect y = |x^2 + bx + c| at four points.

    D > 0 and k e (0, -D/4)

    Find the missing term: 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ?

    4 ( 3 and 5 are prime )

    6 ( 5 and 7 are prime )

    12 ( 11 and 13 are prime )

    18 ( 17 and 19 are prime )

    30 ( 29 and 31 are prime )

    42 ( 41 and 43 are prime )

    108 ( 107 and 109 are prime )

    138 ( 137 and 139 are prime )

    If a, b, c, d, e, f are non negative real numbers such that a + b + c + d + e + f = 1, then the maximum value of ab + bc + cd + de + ef is

    1. 1/6
    2. 1
    3. 6
    4. 1/4

    If we take a = b = 12 and c = d = e = f = 0 then the given equation holds so the maximum value is ¼

    Option b and c  do not hold as they are integers and option a i.e 1/6 is less than ¼ so option (a) is also rejected. Hence option d is correct.

    The number of integral values of x satisfying root(-x^2 + 10x – 16) < x – 2 is

    1. 0
    2. 1
    3. 2
    4. 3

    root(-x^2 + 10x – 16) < x – 2

    -x^2 + 10x – 16 ≥ 0

    x^2 - 10x + 16 ≤ 0

    2 ≤ x ≤ 8 … (1)

    Also, -x^2 + 10x – 16 < x^2 – 4x + 4

    2x^2 – 14x + 20 > 0

    x^2 – 7x + 10 > 0

    x > 5 or x < 2 … (2)

    From (1) and (2), 5 < x ≤ 8

    x = 6, 7, 8

    A is the father of C and D is son of B and E is brother of A. if C is sister of D, how is B related to E ?

    1. Sister in law
    2. Brother in law
    3. Brother
    4. Can’t say

    A is the father of C and C is sister of D. So A is father of D but D is son of B

    So, B is the mother of D and wife of A

    E is the brother of A so B is the sister in law of E.

    The sum of the following series is

    (1 x 2 x 3)/5^1 + (2 x 3 x 5)/5^2 + (3 x 4 x 5)/5^3 + (4 x 5 x 6)/5^4 + …

    1. 375/128
    2. 125/64
    3. 625/128
    4. 875/64

    S = (1 x 2 x 3)/5^1 + (2 x 3 x 5)/5^2 + (3 x 4 x 5)/5^3 + (4 x 5 x 6)/5^4 + …

    S = 6/5 + 24/5^2 + 60/5^3 + 120/5^4 + …

    S/5 = 6/5^2 + 24/5^3 + 60/5^4 + 120/5^5 + …

    S – S/5 = 4S/5 = 6/5 + 18/5^2 + 36/5^3 + 60/5^4 + 90/5^5 + …

    4S/25 = 6/5^2 + 18/5^3 + 36/5^4 + 60/5^5 + …

    4S/5 – 4S/25 = 16S/25 = 6/5 + 12/5^2 + 18/5^3 + 24/5^4 + 30/5^5 + …

    16S/125 = 6/5^2 + 12/5^3 + 18/5^4 + 24/5^5 + 30/5^6 + …

    64S/125 = 6/5 + 6/5^2 + 6/5^3 + …

    S = (6/5 / ( 1 – 1/5) ) X 125/64 = 375/128

    Using the vertices of a regular hexagon as vertices, how many triangles can be formed with distinct areas?

    1. 24
    2. 20
    3. 18
    4. 28

    Let us number the vertices from 1 – 6, there are three different types of triangles that can be formed.

    Type 1 : (1, 2, 3) [ this is same as (2, 3, 4), (3, 4, 5) … etc)

    Type 2 : (1, 2, 4) [ this is same as (1, 2, 5) (2, 3, 5) (3, 4, 6)]

    Type 3 : (1, 3, 5) and (2, 4, 6) have same shape and area

    So there will be three triangles of different areas that can be formed from a regular hexagon.

    There are 6 triangles of Type 1, 12 of type 2 and 2 of type 3.

    Adding these we get 6 + 12 +  = 20 triangles.

    Total number of triangles possible = 6C3 = 20

    There are 11 numbers of the blackboard – six zeroes and five ones. You have to perform the following operation 10 times : cross out any two numbers. If they were equal write a zero on the blackboard and if they were not equal, write a one. This operation can be performed in any order as you wish. What is the number you have at the end.

    1. 11
    2. 00
    3. 1
    4. Cant say

    The operation could be performed in a number of different ways, but 1 thing is always the same : After each operation the sum of the numbers on the blackboard is always odd.

    Also, we are losing 1 digit in each operation. So, finally we must have a single digit – either 0 or 1

    But it should be odd so our result must be 1.

    A natural number is chosen at random from the first 100 natural numbers. The probability that x + 100/x > 50 is

    1. 1/2
    2. 7/18
    3. 5/12
    4. 11/20

    We have x + 100/x > 50

    x^2 + 100 > 50x

    (x – 25)^2 > 525

    x – 25 < root(525) or x – 25 > root(525)

    x < 25 – root(525) or 25 + root(525)

    As x is positive and root(525) = 22.9

    We must have x ≤ 2 or x ≥ 48

    Thus 2 + 53 = 55 favorable cases.

    Probability = 55/100 = 11/20

     


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