Quant Boosters - Gaurav Sharma - Set 2


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


    Number of Questions - 30
    Solved ? - Yes
    Topic - Quant Mixed Bag
    Source - Genius Tutorial Preparation Forum


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


    Q1) In a convex octagon, two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the octagon os
    a) 8/37
    b) 2/31
    c) 7/19
    d) 4/9


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


    Total diagonals = 8C2 – 8 = 20

    Total points of intersection = 20C2

    0_1510551335016_d8ed3e5d-b1fe-40b9-bed4-bc7c9ead7353-qm31.png

    Every combination of 4 vertices (eg: H, A, C, D) of an octagon corresponds to two diagonals intersecting at the interior point.

    Favorable occurrences = 8C4 = 70

    Required probability = 70/20C2 = 7/19


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


    Q2) How many numbers can be formed using all the digits 0, 2, 3, 5 and 6 without repetition such that the hundredth digit is more than the ten’s digit
    a) 72
    b) 56
    c) 48
    d) 60


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


    We have 5 digit numbers using 0, 2, 3, 5 and 6 without repetition.
    Number of cases when 0 is at hundredth place: 4! = 24
    Number of cases when 2 is at hundredth place: 2 x 2 x 1 x 3 x 1 = 12
    Number of cases when 3 is at hundredth place: 2 x 2 x 1 x 2 x 1 = 8
    Number of cases when 5 is at hundredth place: 2 x 2 x 1 x 1 x 1 = 4
    Number of cases when 6 is at hundredth place: 0

    Total case = 24 + 12 + 8 + 4 = 48


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


    Q3) A solution contains milk and water in the ratio 4:3. What part of the solution must be taken out and replaced with water so that the resultant solution contains milk and water in the ratio 2:3?
    a) 25%
    b) 30%
    c) 28%
    d) 23%


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


    Let v be the volume of the solution.

    Given that the ratio of milk and water is 4:3

    Let x litres of the solution be taken out.

    Now the volume v –x will contain milk and water in the ratio 4:3

    Now x litres of water is added to this solution and the ratio of milk and water becomes 2:3 i.e 4:6

    Hence, to v-x litres containing milk and water in the ratio of 4:3, we added x litres of water and the volume v contains milk and water in the ration 4:6

    v - x --- 4 : 3

    v ---- 4 : 6

    Clearly, if v is 10 litres, x is 3 litres of water

    Hence 30% of the solution must be replaced with water.


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


    Q4) If 9^(x^2 – 2) – 3^(x^2 – 2) = 6, then the sum of all possible values of x is
    a) 0
    b) Root(3)
    c) -3
    d) 2root(3)


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


    9^(x^2 – 2) – 3^(x^2 – 2) = 6
    Let 3^(x^2 – 2) = y
    y^2 – y – 6 = 0
    (y-3) (y+2) = 0
    y = 3 and y = -2
    But y cannot be a negative number, so y = 3
    x^2 – 2 = 1 = > x^2 = 3 = > x = root(3) or – root(3)
    Sum = root(3) + (- root(3)) = 0


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


    Q5) log (5x + 6) > log (x^2 + 10) then (log base is 0.6)
    a) -6/5 < x < 1 or x > 4
    b) 1 < x < 4
    c) 0 < x < 0.6
    d) 0 < x < 4


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


    log(5x + 6) > log (x^2 + 10) (base is 0.6)
    Also base 0.6 < 1
    5x + 6 < x^2 + 10
    x^2 – 5x + 4 > 0
    (x – 1)(x – 4) > 0
    x < 1 or x > 4
    Also, log of negative numbers is not possible
    So 5x + 6 > 0 = > x > 6/5
    And x^2 + 10 > 0 for all values
    Hence range will be -6/5 < x < 1 or x > 4


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


    Q6) In an isosceles triangle ABC with AB = AC = 20 cm and BC = 24 cm. Circle with center 0 touches BC at P, CA at Q and AB at R. Find the area of ARPQ


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


    0_1510551965985_5c82ef65-706f-4105-ae14-94ef55819e81-qm41.png

    AC = 20, PC = 12
    So AP = 16 (Pythagoras theorem)
    Now, PC = QC = 12
    AQ = 8ARQ ~ ABC
    AQ/AC = RQ/BC
    8/20 = RQ/24 = > RQ = 48/5
    ARPQ is a Kite (AR = AQ, RP = QP)
    Area = 1/2 x d1 x d2
    = 1/2 x AP x RQ
    = 1/2 x 16 x 48/5
    = 76.8 sq cm


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


    Q7) x, y and z are integers that are sides of an obtuse – angled triangle. If xy = 6, find z
    a) 2
    b) 4
    c) 3
    d) Both (a) & (b)


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


    If xy = 6
    Then possible cases for x, y are: 1 x 6 = 6 and 2 x 3 = 6

    Case 1:
    When the sides of triangle are 1, 6 and z
    Here (6 – 1) < z < (6 + 1) = > 5 < z < 7
    Hence z = 6 but (1, 6, 6) does not form an obtuse triangle.

    Case 2: Sides are 2, 3 and z
    Here (3 – 2) < z < (3 + 2) = > 1 < z < 5
    z = 2, 3, 4

    Triangle 1: (2, 3, 2)
    3^2 > 2^2 + 2^2, hence it is obtuse = > z = 2 is possible

    Triangle 2: (2, 3, 3)
    3^2 # 2^2 + 3^2, hence it is not an obtuse triangle
    z = 3 is not possible

    Triangle 3 (2, 3, 4)
    4^2 > 2^2 + 3^2 hence it is obtuse.
    z = 4 is possible.

    Option d is correct


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


    Q8) The product of digits of a five digit number is 600. How many such numbers are possible?


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


    600 = 2^3 x 3 x 5
    Leaving aside 5^2
    Arrangements of 2^3 x 3 = 24 are
    ( 1 x 8 x 3 ), ( 1 x 4 x 6), (2 x 4 x 3), (2 x 2 x 6)

    Note: We will not consider [1 x 2 x 12] as 12 is a 2 digit number

    Now possible cases for
    (1, 8, 3, 5, 5) are 5!/2! = 60
    (1, 4, 6, 5, 5) are 5!/2! = 60
    (2, 4, 3, 5, 5) are 5!/2! = 60
    (2, 2, 6, 5, 5) are 5!/2!2! = 30

    Total = 60 + 60 + 60 + 30 = 210


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


    Q9) If all the four digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 and 9 without repetition are arranged in ascending order, what will be the rank of number 5293 ?
    a) 897
    b) 898
    c) 908
    d) None of these


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


    For four digit numbers from (1, 2, 3, 4, 5, 6, 7, 9)
    Starting with 1 – 7C3 x 3! = 35 x 6 = 210
    Similarly for 4-digit numbers starting with 2, 3 and 4 we have 210 cases each.
    Total = 210 x 4

    4 digit numbers starting with 5 1 - - = 6C2 x 2! = 30

    4 digit numbers starting with 5 2 1 - = 5C1 = 5

    Similary for 4 digit numbers starting with 5 2 3 - , 5 2 4 - , 5 2 6 - , 5 2 7 - : we will have 5 cases each

    Total 25 cases

    Now the next number will be 5291 – 1 case

    Required number is 5293 – again 1 case

    Total cases = 840 + 30 + 25 + 2 = 897

    Hence rank of 5293 is 897th


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


    Q10) Bells A and B ring 8 times and 38 times in a minute respectively. If they start ringing simultaneously after how much time (in seconds) will B ring exactly 10 times more than A ?
    a) 8
    b) 10
    c) 15
    d) 20


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