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RRB NTPC Aptitude Test - 22
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RRB NTPC Aptitude Test - 22

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  • Question 1/10
    1 / -0

    Calculate the volume of a prism which has a base of square of side 9 cm and has a height of 27 cm.

    Solutions

    Volume of a prism = Area of the base × height

    = 81× 27

    = 2187

     

  • Question 2/10
    1 / -0

    Each edge of a regular tetrahedron is 6cm.Find the total surface area of the tetrahedron in (sq.cm)?

    Solutions

    Total surface area of a regular tetrahedron = area of 4 equilateral triangles

     

  • Question 3/10
    1 / -0

    Sin 22.5 + cos 112.5 =?

    Solutions

    Sin 22.5 + cos (90 + 22.5) 

    ∵Cos (90 + θ) = - sin θ

    Sin 22.5 - sin22.5 = 0

     

  • Question 4/10
    1 / -0

    If p/q + q/p = 1, then p3+ q3 = ?

    Solutions

     

  • Question 5/10
    1 / -0

    O is the circumcentre of triangle ABC. Find the value of x if angle BAC = 60° and angle BOC = 4x+52

    Solutions

    If O is the circumcentre then BOC = 2BAC= 120

    Then 4x + 52 = 120

    4x = 68

    x = 17

     

  • Question 6/10
    1 / -0

    What is the mean of the marks scored by students in a science exam?

    42, 40, 53, 49, 55

    Solutions

    Mean = (42 + 40 + 53 + 49 + 55) ÷ 5 = 47.8

     

  • Question 7/10
    1 / -0

    Directions For Questions

    The following table represents the percentage marks of five students in six subjects. Consider the table and answer the following questions.

    ...view full instructions

    The average marks obtained by the students in English and Hindi respectively

    Solutions

    Average in English = (72 + 70 + 71 + 75 + 67)/5 = 71

    Average in Hindi = (65 + 70 + 71 + 70 + 74)/5 = 70

     

  • Question 8/10
    1 / -0

    Directions For Questions

    The following table represents the percentage marks of five students in six subjects. Consider the table and answer the following questions.

    ...view full instructions

    Who has the highest total marks?

    Solutions

    Binod = 75 + 80 + 65 + 68 + 72 + 65 = 425

    Bindu = 80 + 85 + 75 + 65 + 70 + 70 = 445

    Baibhav = 82 + 88 + 70 + 69 + 71 + 71 = 451

    Bimal = 78 + 87 + 65 + 70 + 75 + 70 = 445

    Binu = 76 + 82 + 55 + 72 + 67 + 74 = 426

    ∵ Baibhav secured the highest total with 451 marks.

     

  • Question 9/10
    1 / -0

    Find the difference of average marks obtained by all the students in Science and History

    Solutions

    Total marks obtained by all the students in Science = 80 + 85 + 88 + 87 + 82 = 422

    Average = 422 ÷ 5 = 83.4

    Total marks obtained by all the students in History = 65 + 75 + 70 + 65 + 55 = 330

    Average = 330 ÷ 5 = 66

    Difference of average of science and history = 83.4 – 66 = 17.4

     

  • Question 10/10
    1 / -0

    If b = (2a+3)/(3a – 2) , find a when b = 5

    Solutions

    b = (2a+3)/(3a-2)

    b = 5

    5 = (2a+3)/(3a-2)

    5(3a – 2) = 2a + 3

    15a – 10 = 2a + 3

    15a – 2a = 3 + 10

    13a = 13

    a = 1

     

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