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SSC CHSL 2019-20 Aptitude Test - 46
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SSC CHSL 2019-20 Aptitude Test - 46

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  • Question 1/10
    1 / -0

    If Sin2 30⁰ + Cos4 45⁰ = X then the value of 1/X + 1/x2= ?

    Solutions

    Sin2 30⁰ + Cos4 45⁰ = X

    ⇒ 1/4 + 1/4 = X

    ⇒ 2/4 = X = ½

    So, 2 + 4 = 6

     

  • Question 2/10
    1 / -0

    If Sin θ + Cos θ = 11/9 then find Sin 2θ = ?

    Solutions

    Given,

    (Sin θ + Cos θ)2 = (11/9)2

     ⇒ 1 + 2 Sin θ Cos θ = 121/81

    ⇒ 2 Sin θ Cos θ = 121/81 – 1

    ⇒ 2 Sin θ Cos θ = 40/81 = Sin 2θ

     

  • Question 3/10
    1 / -0

    If (a + b) : (b + c) : (c + a) ∷ 5 : 6 : 8 and a + b + C = 38 then a + c =?

    Solutions

     2 (a + b + C) = 38 K

    (a + b + C) = 19K

    38 = 19K

    K = 2

    So, the correct answer a + c = 26

     

  • Question 4/10
    1 / -0

    In a triangle ABC, ∠A = 120° AB = 8 cm and AC = 12 find BC = ?

    Solutions

    Cos 120° = (82 + 122 – X2)/(2 × 8 × 12) =  (64 + 144 - X2)/96

    ⇒ -1 = (64 + 144 - X2)/96

    ⇒ -96 = 208 - X2

    ⇒ X2 = 304

    ⇒ X = 17.4

     

  • Question 5/10
    1 / -0

    In triangle ABC, ∠B = 150° and ∠C = 100°, AD is angle bisector AE is a perpendicular to BC. Find the angle ∠EAD.

    Solutions

    ∠EAD = ½ |∠B - ∠C| = ½ |150 - 100| = ½ × 50 = 25

     

  • Question 6/10
    1 / -0

    If a person traveling in boat where speed of the boat and stream are 20km/h and 30km/h respectively. How much time will he take to reach a point 600 km away if he travels first hour downstream and next hour upstream in reverse direction and so on?

    Solutions

    Speed of the boat = 20 km/h and speed of the river = 30km/h

    ∴ Speed of the boat in downstream = 20 + 30 = 50 km/h

    And speed of the boat in upstream = 30 – 20 = 10 km/h

    Distance covered by boat in first hour = 50 km

    Distance covered by boat in next hour = 10 km/h

    ∴ Distance covered by boat in 2 hour = 50 – 10 = 40 km

    ∴ Distance covered by boat in 10 hour = 200 km

    Remaining distance = 600 – 200 = 400 km

    And remaining time = (400 km) /(50 km/h) = 8 hours.

     

  • Question 7/10
    1 / -0

    The ratio of milk and water in three samples is 2 ∶ 3, 2 ∶ 5 and 13 ∶ 7. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?

    Solutions

    Let ‘x’ quantity of each of three mixtures to be mixed

    Quantity of milk in the new mixture = (2/5)x + (2/7)x + (13/20)x = 187x/140

    Quantity of water in the new mixture = (3/5)x + (5/7)x + (7/20)x = 233x/140

    ∴ Required ratio = 187 ∶ 233

     

  • Question 8/10
    1 / -0

    In ΔABC, ∠A = 50°. The internal bisector of ∠B and ∠C meet at point G, inside the triangle while external bisector of ∠B and ∠C meet at point Y, outside the triangle. What is the sum of ∠BGC and ∠BYC?

    Solutions

    If the bisectors of ∠B and ∠C meet at point I inside the triangle, then

    ∠BGC = 90° + ∠A/2

    ∠BIC = 90° + 50/2 = 90° + 25° = 115°

    If exterior bisectors of ∠B and ∠C meet at point Y outside of the triangle, then

    ∠BOC = 90° – ∠A/2

    ∠BOC = 90° – 50°/2 = 65°

    Sum of ∠BGC to ∠BYC = 115° + 65° = 180°

     

  • Question 9/10
    1 / -0

    The cost of fencing small square ground is Rs. 15000. The cost of fencing per m is Rs. 20. Find the cost of fencing of large square ground having side is 10% more than the side of small ground.

    Solutions

    Given,

    Cost of fencing small square ground = 15000

    Perimeter of small ground = 18000/20 = 900 m

    Side of small square ground = 900/4 = 225 m

    Side of large square ground = (225 + 225) × 10/100 = 450 + 1/10 m = 450.1

    Perimeter of large square ground = 450.1 × 4 = 1800 m

    Cost of fencing of large square ground = 1800 × 20 = Rs. 36000

     

  • Question 10/10
    1 / -0

    Akash and Bapan are standing on the same side of a wall and observe that the angles of elevation to the top of the wall are 45° and 60° respectively. If the height of the wall is 100 m, the distance between Aakash and Bapan is: (Use √3 = 1.73 and √2 = 1.41)

    Solutions

    Height of the wall PQ = 100m

    In ΔPQB

    tan60° = PQ/BQ

    √3 = 100/BQ

    BQ = 100/√3 m

    In ΔPQA

    Tan 45° = PQ/AQ

    1 = 100/AQ

    AQ = 100m

    AQ = AB + BQ

    100 = AB + 100/√3

    AB = 100 – 100/√3 = 100 – (1 - 1.73)

    AB = 50 – 28.83

    AB = 21.17 M

     

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