Quant Test for LIC AAO Prelims Exam Set – 17

Directions(1-5): What approximate value will come in place of question mark ‘?’ in the following questions.

  1. 25% of 499.85 – 12.12% of 675.12 = 19.89% of ?
    205
    220
    222
    230
    241
    Option B
    25% of 499.85 – 12.12% of 675.12
    = 19.89% of ?
    =>0.20* ? = 0.25*500 – 0.12*675
    =>0.20*? = 125 – 81
    =>? = 220

     

  2. 34.99% of 980 – 6.11^3 = ?^2 – (289.2)^1/2
    9
    12
    5
    10
    18
    Option B
    34.99% of 980 – 6.11^3 = ?^2 – (289.2)^1/2
    =>35% of 980 – 6^3 = ?^2 – 216
    =>?^2 – 17 = 0.35*980 – 216
    =>?^2 = 343 – 216 +17
    =>?^2 = 144
    =>? = 12

     

  3. ¾ % of 2400.31 + 23(1/2)% of 800 + 45^2 = 1125/25+4 * ?
    700
    686
    650
    776
    712
    Option D
    ¾ % of 2400.31 + 23(1/2)% of 800 + 45^2 = 1125/25 + 4* ?
    => 51/400*2400 + 47/200*800 + 2025 = (1125/25)+ 4*?
    =>306+818 + 2025 – 45 = 4*?
    =>? = 776

     

  4. (913 + 329 + 522 + 343 )/(18 + 24 – ? + 18) = 43
    18
    11
    20
    15
    13
    Option B
    (913 + 329 + 522 + 343 )/(18 + 24 – ? + 18)
    = 43
    =>2107/(60 – ?) = 43
    =>2107/43= 60 –?
    =>? = 11

     

  5. [(528.8)^1/2 + (729.15)^1/2 ]* 18.01 = ?% of 14.99^2
    600
    500
    300
    200
    400
    Option E
    [(528.8)^1/2 + (729.15)^1/2 ]* 18.01 = ?% of 14.99^2
    =>(23 + 27)*18 = ?% of 225
    =>?*2.25 = 50* 18
    =>? = 900/2.25
    =>? = 400

     

  6. The average marks of 12 students are 58. If two more students with average marks 60 are added and one student is removed. The new average becomes 59. Find the marks scored by the student who was removed.
    38 marks
    49 marks
    55 marks
    40 marks
    52 marks
    Option B
    Total marks of 12 students = 58*12 = 696
    Total marks of 14 students = 696 + 2*60 = 816
    Let the marks of the students who was removed be x. 816 – x = 59*13
    =>x = 49 marks

     

  7. The radius of the cone is 12 cm and slant height is 3 cm longer than the height of the cone. Find the total surface area of the cone. Take pi = 3.
    1850 cm^2
    1440 cm^2
    1350 cm^2
    1470 cm^2
    1120 cm^2
    Option C
    Let the height of the cone be x cm and slant height
    = (x+3) cm
    Now, (12)^2 + x^2 = (x+3)^2
    =>x = 22.5 cm
    Height = 22.5 cm and slant height = 22.5+3 = 25.5 cm
    T.S.A of the cone = 3*12*(25.5+12) = 1350 cm^2

     

  8. A shopkeeper sold a bicycle for Rs.7200 and earned a profit percent equal to the profit/loss percent earned when 32 pens are sold at the cost price of 48 pens. Find the cost price of the cycle.
    Rs.4700
    Rs.4500
    Rs.4000
    Rs.4100
    Rs.4800
    Option E
    Let the CP of 1 pen = Rs.10
    CP of 32 pens = Rs.320
    SP of 32 pens = CP of 48 pens = Rs.480
    Profit% = (480-320)/320*100 = 50%
    CP = 7200/1.5 = Rs.4800

     

  9. Area of right – angled isosceles triangle is 1/4th the area of the square having perimeter 64cm. Find the hypotenuse of the triangle.
    13 cm
    19 cm
    16 cm
    12 cm
    10 cm
    Option C
    Side of the square = 64/4 = 16 cm
    Area of the square = 256 cm^2
    Area of the triangle = 256/4 = 64 cm
    Base of the triangle = (64*2)^1/2 = 8(2)^1/2 cm
    Hypotenuse = 16 cm

     

  10. A committee of 3 members is to be made out of 6 men and 5 women. What is the probability that the committee has at least two women?
    12/37
    11/30
    13/32
    14/33
    19/35
    Option D
    Number of possible combination of 3 persons
    = 5C2*6C1 + 5C3
    Total possible outcomes = 11C3
    Required Probability = 14/33

     


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