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LIC AAO & AE 2020 Aptitude Test - 1
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LIC AAO & AE 2020 Aptitude Test - 1
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  • Question 1/10
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    From a vessel filled with milk, 1/4 of its content is removed, and the vessel is then filled up with water. If this be done four times in succession, what proportion of the milk originally contained in the vessel will have been removed from it?

    Solutions

     

  • Question 2/10
    1 / -0

    Rahul got married 5 years ago. His age is  6/5 times of his age at the time of his marriage. Two years ago, his son was 2 years old. The sum of their present ages (Rahul & his son) is.

    Solutions

    Let Rahul’s present age be x years
    ∴ His age during marriage = (x — 5) years
    Acc. to ques,
    x = 6/5 × (x − 5)
    x = 30 years
    ∴  present age of son = 2 + 2 = 4 years
    Sum of Rahul & his son present ages = 30 + 4 = 34 years

     

  • Question 3/10
    1 / -0

    Three persons P, Q and R can do a job in 10 days, 15 days and 20 days respectively. They started working together and after working 4 days P and R leave the Job. Find in how many days Q will complete the remaining job?

    Solutions

     

  • Question 4/10
    1 / -0

    The fare of a OLA taxi is Rs 36 for the first five kilometers and Rs 13 per kilometre thereafter. If a passenger pays Rs 2,402/- for a journey of how many distance in kilometres?

    Solutions

    Let the total distance travelled by taxi be x km
    2402 = 36 + 13 × (x − 5)
    2402 = 36 + 13x − 65
    2431 = 13x
    x = 187 km

     

  • Question 5/10
    1 / -0

    Mahesh borrows Rs. 3904 and promises to pay back with compound interest at 25% p.a. in 3 early equal installments at the end of first, second and third year. Find the amount of each installment.

    Solutions

    The value of 3904 after 3 years = the value of the first installment after 2 year + the value of second installment after 1 year + the value of third installment Let the amount of installment be x

     

  • Question 6/10
    1 / -0

    A cuboidal tank measuring 1m × 2.1 m × 22.5 m is dug in one corner of a field measuring 67.5 m × 0.5m. The earth dug out is spread evenly over the remaining portion of the field. How much is the level of the field raised?

    Solutions

    We know that, volume of cuboid = length × breadth × height
    ∴  Volume of earth dug out from the tank = 1m × 2.1 m × 22.5m = 47.25  m3
    Also, Volume of the cuboidal tank = volume of earth dug out from the tank.
    Now,
    the exposed area of the field upon which the earth dug out is spread evenly = area of the total field – area of the top face of the tank
    Area of the total field = 67.5 m × 0.5m = 33.75 sq. m.
    Area of the top face of the tank = 1m × 2.1 m = 2.1 sq. m.
    The exposed area of the field = (33.75 – 2.1) sq. ft. = 31.65 sq. m.
    Now, the earth dug out is spread evenly over 31.65 sq. m. area.
    If level of field raises by l m. then from volume constancy we can say,
    31.65 × l = 47.25
    ⇒  l =  1.492  ≈ 1.5 m

     

  • Question 7/10
    1 / -0

    What should come in place of question mark (?) in the following questions?

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  • Question 8/10
    1 / -0

    What should come in place of question mark (?) in the following questions?

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  • Question 9/10
    1 / -0

    What should come in place of question mark (?) in the following questions?

    Solutions

     

  • Question 10/10
    1 / -0

    What should come in place of question mark (?) in the following questions?

    (29.8% of 260) – (56% of 225) + (60.01% of 510) – 103.57 = ?

    Solutions

     

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