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SSC CGL 2019 Aptitude Test - 38
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SSC CGL 2019 Aptitude Test - 38
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  • Question 1/10
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    If in a triangle, angles are in the ratio 1:1:2 and the length of its longest side is L cm, then what is the perimeter of the triangle?

    Solutions

    Let the angles of the triangle are A, A and 2A respectively.

    4A = 180 0 ...........................(Angle sum property)

    A = 45 0

    Angles are 45 0 , 45 0 and 90 0 respectively.

    It is a right angle isosceles tringle,

    Other 2 sides = L/ (√2)

    Perimeter = L + (L/√2) + (L/√2)

    = L +L√2 = L (1 +√2) cm.

     

  • Question 2/10
    1 / -0

    If d is the side of a triangle and 2-(1/d) - (15/d 2 ) = 0, then what is the value of ((d 4 -4)/(d 3 -3))?

    Solutions

    2-(1/d) - (15/d 2 ) = 0

    2d 2 -d - 15 = 0

    (d-3) x (2d +5) = 0

    d = 3 or d = - 5/2

    But 'd' is the side of a triangle .

    so  d will be positive .

    ((d 4 -4)/(d 3 -3)) = (81-4)/(27-3) = (77/24)

     

  • Question 3/10
    1 / -0

    In a trapezium MNOP, MN || OP, diagonal PN and OM are intersect at point A and ratio of area of △MNA to the area of △OPA is 4 : 9, then find OP : MN.

    Solutions

    In △MNA and △OPA, ∠MAN = ∠PAO (∴ Opposite angles)

    ∠MNA = ∠APO (∴ Alternate angles)

    ∠NMA = ∠AOP (∴ Alternate angles)

    So, △MNA and △OPA are similar.

    In similar triangle, Ratio of their area = Square of ratio of their corresponding sides

    => Area of △MNA : Area of △OPA = (MN) 2 : (OP) 2

    => 4 : 9 = (MN) 2 : (OP) 2

    => MN : OP = 2 : 3

    => OP : MN = 3 : 2

     

  • Question 4/10
    1 / -0

    If 4x 2 + 16y 2 -  12x  -  40y + 34 = 0, then find the value of x/y.

    Solutions

    4x 2 + 16y 2 -  12x  -  40y + 34= 0

    (4x 2 -  12x + 9) + (16y 2 -  40y + 25) = 0

    (2x  -  3) 2 + (4y  -  5) 2 = 0

    2x  -  3 = 0 => x = 3/2

    4y  -  5 = 0 => y = 5/4

    (x/y) = x * (1/y) = (3/2) * (4/5) = 6/5

     

  • Question 5/10
    1 / -0

    A chord of length 14 cm is drawn perpendicular to the diameter of the circle having radius 25 cm. Find the distance of point of intersection of chord and diameter from the centre of circle.

    Solutions

    In triangle OPA, OA = 25 cm (Radius)

    AP = AB/2 = 14/2 = 7cm

    OP = √(OA 2 -  AP 2 ) = √(25 2 - 7 2 ) = √(625 - 49) = √576 = 24 cm

    Hence, distance of point of intersection of chord and diameter from the centre of circle = 24 cm

     

  • Question 6/10
    1 / -0

    If P is m% more than Q, then Q is?

    Solutions

    As P is m% more than Q

    P = Q + ((m/100) x Q)

    P = ((100 +m)/100) x Q

    Q = (100/ (100 +m)) x P

     

  • Question 7/10
    1 / -0

    If 2x 3 + x 2 -mx -4 is divided by x -2, then remainder is zero, then what is the value of m?

    Solutions

    When 2x 3 + x -mx -4 is divided by x -2 and remainder is zero, it means

    For x = 2, 2x 3 + x -mx -4 = 0

    2 x (2) 3 +(2) 2 -  (2m) -4 =0

    -2m =-16

    m = 8

     

  • Question 8/10
    1 / -0

    How many numbers from 1 to 901 are there which are divisible by either 3 or 9 but not by both?

    Solutions

    Total numbers from 1 to 901 which are divisible by 3 = (900/3) = 300

    Total numbers from 1 to 901 which are divisible by 9 = (900/9) = 100

    All numbers which are divisible by 9 are divisible by 3 also

    Total numbers divisible by either 3 or 9 but not by both = 300-100 = 200

     

  • Question 9/10
    1 / -0

    Directions For Questions

    Following table shows number of T-20 matches played, runs scored,50s and 100s by 4 Indian batsmen in a year.

    ...view full instructions


    What is the average of the total number of runs scored by all 4 batsmen together?

    Solutions

    Required average = (900 +840 +1050 +450)/4 = 3240/4 = 810

     

  • Question 10/10
    1 / -0

    Directions For Questions

    Following table shows number of T-20 matches played, runs scored,50s and 100s by 4 Indian batsmen in a year.

    ...view full instructions


    What is the difference between average runs per match scored by Virat and average runs per match scored by Suresh?

    Solutions

    Average runs per match scored by Virat = (900/16) = (225/4)

    Average runs per match scored by Suresh = (450/12) = (150/4)

    Required difference = (225/4)  -  (150/4) = (75/4) = 18.75

     

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