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IBPS Clerk 2018 Aptitude Test - 6
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IBPS Clerk 2018 Aptitude Test - 6
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  • Question 1/10
    1 / -0

    What approximate value should come in the place of the question mark (?) in the following question? (You are not expected to calculate the exact value)

    (8.88 × 5.25) + 54.80 = ? × 4.19 + 27.897

    Solutions

    Follow BODMAS rule to solve this question, as per the order given below,

    Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

    Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,

    Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

    Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

    Given expression is,

    ⇒ (8.88 × 5.25) + 54.80 = ? × 4.19 + 27.897

    We can write the given values as:

    ⇒ 8.88 ≈ 9 and 5.25 ≈ 5 and 54.80 ≈ 55

    ⇒ 4.19 ≈ 4 and 27.897 ≈ 28

    Then,

    ⇒ (9 × 5) + 55 = ? × 4 + 28

    ⇒ 45 + 55 = ? × 4 + 28

    ⇒ 100 = ? × 4 + 28

    ⇒ 100 - 28 = ? × 4

    ⇒ 72 = ? × 4

    ⇒ ? = 72/4

    ∴ ? ≈ 18

     

  • Question 2/10
    1 / -0

    What approximate value should come in the place of the question mark (?) in the following question? (You are not expected to calculate the exact value)

    115.72 ÷ 4.10 × 4.90 = 90.14% of 200 - ?

    Solutions

    Follow BODMAS rule to solve this question, as per the order given below,

    Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

    Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,

    Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

    Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

    Given expression is,

    ⇒ 115.72 ÷ 4.10 × 4.90 = 90.14% of 200 - ?

    We can write the given values as :

    ⇒ 115.72 ≈ 116 and 4.10 ≈ 4

    ⇒ 4.90 ≈ 5 and 90.14 ≈ 90

    Then,

    ⇒ 116 ÷ 4 × 5 = 90% of 200 - ?

    ⇒ 116 ÷ 4 × 5 = (90/100) × 200 - ?

    ⇒ 29 × 5 = 90 × 2 - ?

    ⇒ 145 = 180 - ?

    ⇒ ? = 180 - 145

    ∴ ? ≈ 35

     

  • Question 3/10
    1 / -0

    What should come in place of ‘?’ in the following number series?

    15, 16, 12, 39, 23, ?

    Solutions

    The pattern of the given series is:

    ⇒ 15 + 1= 15 + 1 = 16

    ⇒ 16 - 2= 16 - 4 = 12

    ⇒ 12 + 3= 12 + 27 = 39

    ⇒ 39 - 4= 39 - 16 = 23

    ⇒ 23 + 5= 23 + 125 = 148 = ?

     

  • Question 4/10
    1 / -0

    What should come in place of ‘?’ in the following number series?

    2, 4, 10, 22, ?

    Solutions

    The pattern of the given series is:

    ⇒ 2 + 1+ 1 = 2 + 1 + 1 = 4

    ⇒ 4 + 2+ 2 = 4 + 4 + 2 = 10

    ⇒ 10 + 3+ 3 = 10 + 9 + 3 = 22

    ⇒ 22 + 4+ 4 = 22 + 16 + 4 = 42 = ?

     

  • Question 5/10
    1 / -0

    What should come in place of ‘?’ in the following number series?

    4, 2.5, 3.5, 6.75, ?

    Solutions

    The pattern of the given series is:

    ⇒ 4 × 0.5 + 0.5 = 2 + 0.5 = 2.5

    ⇒ 2.5 × 1 + 1 = 2.5 + 1 = 3.5

    ⇒ 3.5 × 1.5 + 1.5 = 5.25 + 1.5 = 6.75

    ⇒ 6.75 × 2 + 2 = 13.5 + 2 = 15.5 = ?

     

  • Question 6/10
    1 / -0

    What should come in place of ‘?’ in the following number series?

    17, 18, 14, 23, 7, ?

    Solutions

    The pattern of the given series is:

    ⇒ 17 + 1= 17 + 1 = 18

    ⇒ 18 - 2= 18 - 4 = 14

    ⇒ 14 + 3= 14 + 9 = 23

    ⇒ 23 - 4= 23 - 16 = 7

    ⇒ 7 + 5= 7 + 25 = 32 = ?

     

  • Question 7/10
    1 / -0

    Amit lent Rs. 4000 to Raj for a period of 4 years. He lent a part of the sum at 3% simple interest and the rest at 5%. If he received a total of Rs. 640 as interest at the end of 4 years, what was the part of money lent at 5%?

    Solutions

    Let the amount lent at 5% be Rs. x

    ∴ Money lent at 3% = Rs. (4000 – x)

    Simple interest on Rs. x = (Principal × Time × Rate)/100

    ⇒ Simple interest = (x × 4 × 5)/100 = Rs. x/5

    Total interest = Rs. 640

    ∴ Interest on Rs. (4000 – x) = Rs. (640 – x/5)

    ⇒ [(4000 – x) × 4 × 3)]/100 = 640 – x/5

    ⇒ 3(4000 – x)/25 = 640 – x/5

    ⇒ 12000 – 3x = 16000 – 5x

    ⇒ 2x = 4000

    ⇒ x = 2000

    ∴ Money lent at 5% is Rs. 2000

     

  • Question 8/10
    1 / -0

    In an examination, the number of those who passed was 5 times the number of those who failed. If there had been 75 less candidates and 10 more had failed, the ratio of passed to failed candidates would have been 1 : 2, then find the total number of candidates initially?

    Solutions

    Let the number of failed candidates = x

    Number of passed candidates = 5x

    Total number of candidates = 5x + x = 6x

    According to the question

    (5x - 75 - 10)/(x + 10) = 1/2

    ⇒ 10x - 150 - 20 = x + 10

    ⇒ 10x - x = 180

    ⇒ 9x = 180

    ⇒ x = 20

    ∴ Total number of candidates = 20 × 6 = 120

     

  • Question 9/10
    1 / -0

    A train half of the length of a 1 km long bridge covers it in 2 min. A still man sees the train crossing the bridge then what is the speed of train with respect to man?

    Solutions

    ⇒ Length of bridge = 1 km = 1000 m

    ⇒ Length of train = 1000/2 = 500 m

    ⇒ Total distance travelled = (1000 + 500) m = 1500 m

    ⇒ Time taken = 2 min

    ∴ Speed of train = 1500/2 = 750 m/min

     

  • Question 10/10
    1 / -0

    30 litres of milk is mixed N litres of some water. The price of milk is Rs. 40 per litre and price of water is Rs. 1 per litre. Due to the mixing, the cost of mixture became Rs. 24.4 per kg. What is the value of N?

    Solutions

    Total price of milk in mixture = 30 × Rs. 40 = Rs. 1200

    Total price of water in mixture = N × Rs. 1 = Rs. N

    ⇒ Average price of mixture = (1200 + N)/(30 + N) = 24.4

    ⇒ 1200 + N = 732 + 24.4N

    ⇒ N = 468/23.4 = 20

     

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