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SSC CHSL 2025 Aptitude Test - 2
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SSC CHSL 2025 Aptitude Test - 2
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  • Question 1/10
    2 / -0.5

    Which of the following is equivalent to the trigonometric expression given below?

    {sin x × (1 + cos x)}/(1 + cos x - cos2 x - cos3 x)

    Solutions

    The correct answer is option 3 i.e. cosec x.

    {sin x × (1 + cos x)}/(1 + cos x - cos2 x - cos3 x)

    ⇒  {sin x × (1 + cos x)} / (1 + cos x - cos2 x - cos3 x)

    ⇒ {sin x (1 + cos x)}/{(1 - cos2 x) (1 + cos x)}

    ⇒ sin x/(1 - cos2 x)

    ⇒ sin x/ sin2x [∵ 1 - cos2A = sin2A]

    ⇒ 1/sin x = cosec x [∵ 1/ sin A = cosec A]

  • Question 2/10
    2 / -0.5

    Solutions

  • Question 3/10
    2 / -0.5

    If b + 1/b = 4, find the value of (b3 + 1/b3) - 4.

    Solutions

    The correct answer is option 2 i.e. 48.

    Given:

    b + 1/b = 4

    Formula used:

    (a + b)3 = a3 + b3 + 3ab(a + b) ---- (1)

    Calculations:

    Using equation (1), we get

    ⇒ (b + 1/b)3 = (4)3

    ⇒ b3 + 1/b3 + 3 × 4 = 64

    ⇒ b3 + 1/b3 = 64 - 12

    ⇒ b3 + 1/b3 = 52

    Subtracting 4 from both sides we get

    ⇒ b3 + 1/b3 - 4 = 52 - 4

    ⇒ b3 + 1/b3 - 4 = 48

  • Question 4/10
    2 / -0.5

    If x + 1/x = √3, find the value of x24 + x36 + x6 + 3.

    Solutions

    The correct answer is option 1 i.e. 4.

    Concept used:

    If x + 1/x = √3

    x3 + 1/x3 = 3√3 - 3√3

    x6 = -1 ---- (1)

    Calculations:

    ⇒ x24 + x36 + x6 + 3

    It can be written as

    ⇒ (x6)4 + (x6)6 + x6 + 3

    ⇒ (-1)4 + (-1)6 + -1 + 3

    ⇒ 1 + 1 - 1 + 3 = 4

  • Question 5/10
    2 / -0.5

    B purchased 15 kg apples at the rate of ₹180 per kg from a wholesaler who uses a weight of 950 grams for the kg weight. B sold all these apples at ₹180 per kg but used a weight of 750 grams for the kg weight. Find the percentage profit earned by B in the transaction (correct to 2 places of decimal).

    Solutions

    The correct answer is option 1 i.e. 26.67%.

    The cost price and selling price of apple are the same, B first incurs a loss while purchasing and then incurs a gain while selling.

    Profit % = (Selling price- Cost price)/Cost price× 100

    The S.P : C.P ratio is given by

    Percentage profit earned = (95 - 75)/75 × 100 = 2000/75 = 26.67%

  • Question 6/10
    2 / -0.5

    If 30% of x is equal to 5 times of 20% of y. Find the value of x : y.

    Solutions

    The correct answer is option 2 i.e. 10 : 3.

    Given:

    30% of x is equal to 5 times of 20% of y.

    Calculations:

    Hence, the value of x : y = 10 : 3

  • Question 7/10
    2 / -0.5

    Evaluate:

    20% of 300 - {40 ÷ 1/(2)-1}[5.64 + 0.36]2

    Solutions

    The correct answer is option 3 i.e. -660.

    20% of 300 - {40 ÷ 1/(2)-1}[5.64 + 0.36]2

    By using the BODMAS rule:

    = (20/100) × 300 - (40 ÷ 2)[6.00]2

    = 60 - 20 [36]

    = 60 - 720

    = -660

  • Question 8/10
    2 / -0.5

    When a number is divided by 7, it leaves 1 as the remainder. When the cube of this number is divided by 7, what will be the remainder?

    Solutions

    The correct answer is Option 3 i.e. 1.

    Let the number be 8

    When 8 is divided by 7, the remainder is 1

    Thus,

    When the cube of the number is divided by 7

    ⇒ (8)3/7 or 512/7

    The remainder will be 1

  • Question 9/10
    2 / -0.5

    In a group of 7 people of average weight 55 kg, one person is removed from weight X, and a new person of weight Y is added then the average becomes 58 kg.

    What is the absolute difference between X and Y?

    Solutions

    The correct answer is option 4 i.e. 21.

    Average = Sum of all weights/Total number of people

    ⇒ Sum of all weights = 55 × 7

    ⇒ 385kg

    When X is removed from the group the total weight becomes,

    ⇒ 385 - X

    When Y is added,

    ⇒ 385 - X + Y

    New average = 58 × 7

    ⇒ 406

    ⇒ 385 - X + Y = 406

    ⇒ X - Y = -21

    Or |-21| = 21

  • Question 10/10
    2 / -0.5

    A boat goes 60 km upstream in 4 hours and 45 km downstream in 1.5 hours in a river. What is the speed of the boat in still water?

    Solutions

    The correct answer is option 1 i.e. 22.5 km/hr.

    If the speed of the boat in still water  = x km/hr and, the speed of the stream = y km/hr

    The downstream speed = (x + y)

    The upstream speed = (x - y)

    The distance covered in 4 hr upstream = 60 km

    The distance covered in 1.5 hr downstream = 45 km

    Speed = distance/time

    The relative speed in upstream = (x - y)

    ⇒ (x - y) = 60/4 = 15 ....(I)

    The relative speed downstream = (x + y)

    ⇒ (x + y) = 45/1.5 = 30 ....(II)

    Adding (I) and (II) we get,

    ⇒ 2x = (30 + 15) = 45

    ⇒ x = 45/2 = 22.5 km/hr

    So, the speed of the boat = 22.5 km/hr

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