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SSC CPO 2018 Aptitude Test 33
Result
SSC CPO 2018 Aptitude Test 33

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  • Question 1/10
    1 / -0

    The value of sin A sin (60 – A)  sin (60 + A) :- 

    Solutions

     

  • Question 2/10
    1 / -0

    C twice efficient as A, B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e. AB, BC, CA)  started with AB on first day and BC on second and AC on third and so on, then how many days are required to finish the work?

    Solutions

    A does in 12 days.

    C does in 6 days.

    B does in 18 days.

    Let total work be 36 units

     

  • Question 3/10
    1 / -0

    If  cosθ + secθ = √3    then, the value of   cos3θ + sec3θ

    Solutions

     

  • Question 4/10
    1 / -0

     The ratio of the areas of the circumcircle and the incircle of a square is

    Solutions

    Let the side of the square = a

     

  • Question 5/10
    1 / -0

    ABC is a right triangle and BD is perpendicular to AC. Find the length of AD + DC + BD.

    Solutions

    Let AD = x

    And DC = y, BD = z

    Now,

    62 + 82 = (x + y)2

    x + y = 10 …..(i)

    Also,

    z2+x2 = 36 ….(ii)

    and z2 + y2 = 64……..(iii)

    Subtracting (ii) from (iii), we get

    y2 –x2 = 28

    (y-x)(y+x)=28

    y-x=2.8 ……(iv) 

    From (i) and (iv)

    y = 6.4

    And so, x = 10 – 6.4 = 3.6

    z2 = 82 – y2 = 82 – (6.4)2

    z = 4.8

    AD + DC + BD  = 3.6 + 6.4 + 4.8 = 14.8

     

  • Question 6/10
    1 / -0

    Given PQRS is a square. A circle is inscribed in the square. AB is the diameter of the circle and C is the mid-point of PQ. Given PQ = 8 m. What is the area of shaded portion? 

    Solutions

    Area of square PQRS = 8 × 8 = 64 m2

    AB = PQ =8 m, radius of circle = 4 m

    Area of circle = π r2 =  22//7   × 4 × 4 = 50.28 m2

    Remaining area= (64 – 50.28) m2 = 13.72 m2

     

  • Question 7/10
    1 / -0

     O is the centre of circle and  ∠DAB    = 50°. Calculate the value of x and y.

    Solutions

    ∎    ABCD is a cyclic quadrilateral

    y + 500 = 1800

    y = 1300

    In,  ∆    OAB

    OA = OB = Radius

    ∠   OAB =  ∠   OBA

    ∠   AOB = 1800 – (50 + 50) = 800

    x = 180 – 80 = 1000

     

  • Question 8/10
    1 / -0

    What is the value of  cos20º + cos23º + cos26º + …. + cos2 90º ?

    Solutions

    2 0º + cos2 3º + cos2 6º + ……. + cos2 90º

    cos2 0º = 1, cos2 90º = 0

    cos2 3º + cos2 6º + ……… + cos2 84º + cos2 87º = cos2 3º + cos2 6º + ……….. + sin2 6º + sin2 3º

    = 1 + 1 x 14 +  1/2   = 15.5

     

  • Question 9/10
    1 / -0

    ABCD is a cyclic quadrilateral. The side AB is extended to E in such a way that BE = BC. If  ∠   ADC = 70º,  ∠   BAD = 95º, then  ∠   DCE is equal to – 

    Solutions

     

  • Question 10/10
    1 / -0

    Solutions

     

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