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IBPS PO Aptitude Test - 1
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IBPS PO Aptitude Test - 1
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  • Question 1/10
    1 / -0

    In each of these questions, one term in the given number series is wrong. Find out the wrong term.

    5, 15, 81, 479, 3360, 26897

    Solutions

    5 x 4 – 5 = 15

    15 x 5 + 6 = 81

    81 x 6 – 7 = 479

    479 x 7 – 8 = 3361 (not 3360)

    3361 x 8 + 9 = 26897

     

  • Question 2/10
    1 / -0

    In each of these questions, one term in the given number series is wrong. Find out the wrong term.

    23, 46, 76, 110, 152, 192

    Solutions

    4 x 6 – 1 = 23

    6 x 8 – 2 = 46

    8 x 10 – 4 = 76

    10 x 12 – 8 = 112 (Not 110)

    12 x 14 – 16 = 152

    14 x 16 – 32 = 192

     

  • Question 3/10
    1 / -0

    What value should come at the place of question mark in the following number series?

    10, 15, 30, 65, 155 ?

    Solutions

    10 x 0.5 + 10 = 15

    15 x 1 + 15 = 30

    30 x 1.5 + 20 = 65

    65 x 2 + 25 = 155

    155 x 2.5 + 30 = 417.5

     

  • Question 4/10
    1 / -0

    What value should come at the place of question mark in the following number series?

    215, 335, 485, 665, 875, ?

    Solutions

    63 – 13 = 215

    73 – 23 = 335

    83 – 33 = 485

    93 – 43 = 665

    103 – 53 = 875

    113 – 63 = 1115

     

  • Question 5/10
    1 / -0

    What value should come at the place of question mark in the following number series?

    44, 72, 104, 140, 180, ?

    Solutions

    4 x 11 = 44

    6 x 12 = 72

    8 x 13 = 104

    10 x 14 = 140

    12 x 15 = 180

    14 x 16 = 224

     

  • Question 6/10
    1 / -0

    In each of the following questions, two equations are given. You have to solve these equations to find the relation between x and y.

    1. 8x2 – 15x + 7 = 0

    2. 2y2 – 15y + 28 = 0

    Solutions

    I.8x2 – 15x + 7 = 0

    => 8x2 – 8x – 7x + 7 = 0

    => 8x(x – 1) – 7(x – 1) = 0

    => (8x – 7)(x – 1) = 0

    => x = 7/8, 1

    II.2y2 – 15y + 28 = 0

    => 2y2 – 8y – 7y + 28 = 0

    => 2y(y – 4) – 7(y – 4) = 0

    => (2y – 7)(y – 4) = 0

    => y = 7/2, 4

    Hence, x < y

     

  • Question 7/10
    1 / -0

    In each of the following questions, two equations are given. You have to solve these equations to find the relation between x and y.

    1. 2x2 – 7x + 6 = 0

    2. y2 + 13y + 42 = 0

    Solutions

    I.2x2 – 7x + 6 = 0

    => 2x2 – 4x – 3x + 6 = 0

    => 2x(x – 2) – 3(x – 2) = 0

    => (2x – 3)(x – 2) = 0

    => x = 3/2, 2

    II.y2 + 13y + 42 = 0

    => y2 + 6y + 7y + 42 = 0

    => y(y + 6) + 7(y + 6) = 0

    => (y + 7)(y + 6) = 0

    => y = -7, -6

    Hence, x > y

     

  • Question 8/10
    1 / -0

    In each of the following questions, two equations are given. You have to solve these equations to find the relation between x and y.

    1. 2x2– 9x + 9 = 0

    2. 8y2+ 34y + 21 = 0

    Solutions

    I.2x2– 9x + 9 = 0

    => 2x2 – 6x – 3x + 9 = 0

    => 2x(x – 3) – 3(x – 3) = 0

    => (2x – 3)(x – 3) = 0

    => x = 3/2, 3

    II.8y2+ 34y + 21 = 0

    => 8y2 + 28y + 6y + 21 = 0

    => 4y(2y + 7) + 3(2y + 7) = 0

    => (4y + 3)(2y + 7) = 0

    => y = -3/4, -7/2

    Hence, x > y

     

  • Question 9/10
    1 / -0

    In each of the following questions, two equations are given. You have to solve these equations to find the relation between x and y.

    1. 2x2 – 13x + 15 = 0

    2. 3y2 + 28y + 65 = 0

    Solutions

    I.2x2 – 13x + 15 = 0

    => 2x2 – 10x – 3x + 15 = 0

    => 2x(x – 5) – 3(x – 5) = 0

    => (2x – 3)(x – 5) = 0

    => x = 3/2, 5

    II.3y2 + 28y + 65 = 0

    => 3y2 + 15y + 13y + 65 = 0

    => 3y(y + 5) + 13(y + 5) = 0

    => (3y + 13)(y + 5) = 0

    => y = -13/3, -5

    Hence, x > y

     

  • Question 10/10
    1 / -0

    In each of the following questions, two equations are given. You have to solve these equations to find the relation between x and y.

    1. x2 = 2401

    2. y3 = 117649

    Solutions

    I.x2 = 2401

    => x = ± √2401

    => x = ± 49

    II.y3 = 117649

    => y =  3√117649

    => y = 49

    Hence, x ≤ y

     

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