Quant Sectional Test 3 for LIC AAO 2019 Prelim Exam

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We have come up with Sectional Tests for upcoming LIC AAO 2019 Prelim Exam. Practice the questions to ace the exam.

Directions(1-10): What approximate value will come in place of question mark ‘?’ in the following questions.

  1. 32% of 250 + 18% of 350 + (8*9) = ?
    220
    195
    200
    222
    215
    Option E
    32% of 250 + 18% of 350 + (8*9) = ?
    =>80+ 63 + 72 = ?
    => ? = 215

     

  2. 427 + ? – 732 = 29^2 – 456
    622
    690
    740
    800
    770
    Option B
    427 + ? – 732 = 29^2 – 456
    =>? – 305 = 841 – 456
    =>? -305 =385
    =>? = 690

     

  3. 84^2 – 79^2 + (1936)^1/2 – 20% of 980 = ?
    693
    555
    663
    744
    721
    Option C
    84^2 – 79^2 + (1936)^1/2 – 20% of 980 = ?
    =>(84+79)(84 -79) + (1936)^1/2 – 20% of 980 = ?
    =>815 + 44 – 196 = ?
    =>? = 663

     

  4. 3.2% of 500 * 2.4 % of ? = 288
    815
    700
    666
    750
    781
    Option D
    3.2% of 500 * 2.4 % of ? = 288
    =>16*2.4/100 * ? =288
    => ? = 750

     

  5. 3024 / 12 + 14^2 – 25% of 820 = ? – 20% of 650
    320
    373
    400
    410
    375
    Option B 3024 / 12 + 14^2 – 25% of 820 = ? – 20% of 650
    =>? – 0.20 * 650 = 252 + 196 -0.25* 820
    =>? – 130 =448 – 205
    =>? = 373

     

  6. 32*15 + ? – 25*16 = 45 * 20
    740
    815
    820
    723
    700
    Option C
    32*15 + ? – 25*16 = 45 * 20
    =>480 + ? – 400 = 900
    =>? = 820

     

  7. 25*13 + 148 – 28 – 42 = ? – 224
    715
    620
    600
    627
    750
    Option D
    25*13 + 148 – 28 – 42 = ? – 224
    => ? = 325 + 148 – 70 + 224
    =>? = 627

     

  8. (784)^1/2 * 29 – ?^3 =180 + 17^2
    5
    9
    8
    7
    6
    Option D
    (784)^1/2 * 29 – ?^3 =180 + 17^2
    =>28*29 – ?^3 = 180 + 289
    =>?^3 = 343
    => ? = 7

     

  9. (300/11)% of 4620 – 840 = ? * 7
    60
    55
    42
    65
    50
    Option A
    (300/11)% of 4620 – 840 = ? * 7
    =>(3/11) of 4620 – 840 = ? * 7
    =>1260 – 840 = ? * 7
    =>? = 60

     

  10. 2620% of 40 + 40% of 825 = ?
    1220
    1424
    1378
    1300
    1175
    Option C
    2620% of 40 + 40% of 825 = ?
    => ? = 40% of 2620 + 40% of 825
    =>? = 1048 + 330
    =>? = 1378

     

  11. Directions(11-15): Following the graph shows the percentage profit gained by two companies A and B over the year 2007 to 2012.

  12. In which of the following years is the percentage of expenditure with respect to income is 80% for Company A?
    90%
    60%
    80%
    50%
    70%
    Option C
    Let the expenditure be x.
    Income = x*(100+25)/100 = 1.25x
    % = x/1.25x*100 = 80%

     

  13. The income of Company A in the year 2012 and the expenditure of Company B in the year 2009 was the same, that is Rs.90 lakh. What will be the ratio of the income of Company B in 2009 to the expenditure of Company A in the year 2012?
    6:7
    2:3
    7:6
    9:5
    8:7
    Option D
    % PA = 20%
    ExpenditureA = I/1.2 = 90/1.2 = 75 lakhs
    % PB = 35%
    IncomeB = 90 × 1.35 = 135 lakhs
    Ratio = 135/75 = 9/5

     

  14. If the expenditure of Company B in the year 2010 was Rs.40 lakh, what was its income (in Rs) in the year 2013?
    70,000
    80,000
    50,000
    60,000
    Data Inadequate
    Option E
    Data Inadequate

     

  15. If the expenditure of Company A in the year 2011 and 2012 was in the ratio 6 : 5, what was the ratio of its incomes?
    15 : 14
    13 : 10
    8 : 9
    15 : 11
    17 : 12
    Option B
    E1 = 6
    E2 = 5
    Now,I1 = E1 *(100+30)/100 = E1 * 1.3
    I1/I2 = (E1/E2) * (1.3/1.2) = 78/6
    I1 : I2 = 13 : 10

     

  16. If the income of Company B in year 2009 was Rs91.8 lakh, what was its expenditure (in Rs) in that year?
    Rs. 80 lakh
    Rs. 74 lakh
    Rs. 55 lakh
    Rs. 68 lakh
    Rs. 60 lakh
    Option D
    % profit = 35%
    Expenditure = Income × 100/(100 %P)
    91.8*(100/135) = Rs. 68 lakh

     

  17. I. 4x^2 – 13x + 10 = 0
    II.4y^2 + 11y – 3 = 0

    y >= x
    y > x
    No relation
    x > y
    x >= y
    Option D
    I. 4x^2 – 13x + 10 = 0
    =>4x^2 – 8x – 5x + 10 = 0
    =>(x-2)(4x- 5) = 0
    =>x = 2,5/4
    II.4y^2 + 11y – 3 = 0
    =>4y^2 + 12y – y – 3 = 0
    =>(y+3)(4y – 1) = 0
    =>y = -3,1/4
    x > y

     

  18. I. 2x^2 – 15x + 25 = 0
    II. 3y^2 = 4y + 15
    y > x
    x > y
    x >= y
    y >= x
    No relation
    Option E
    I. 2x^2 – 15x + 25 = 0
    =>2x^2 – 10x – 5x + 25 = 0
    =>(2x-5)(x-5) = 0
    =>x = 5,5/2
    II. 3y^2 = 4y + 15
    =>3y^2 – 4y – 15 = 0
    =>3y^2 – 9y + 5y – 15 = 0
    =>(3y + 5)(y – 3) = 0
    =>y = 3, – 5/3
    No relation

     

  19. I.x^2 + 20x + 96 = 0
    II.y^2 + 25y + 156 = 0
    y > x
    No relation
    x >= y
    y >= x
    x > y
    Option C
    I.x^2 + 20x + 96 = 0
    =>x^2 + 12x + 8x + 96 = 0
    =>(x+8)(x+12) = 0
    =>x= -8, – 12
    II.y^2 + 25y + 156 = 0
    =>y^2 + 13y + 12y + 156 = 0
    =>(y+13)(y+12) = 0
    =>y = -13,-12
    x >= y

     

  20. I. 2x + 5y = 51
    II.9x – 4y = 44
    x > y
    No relation
    y >= x
    y > x
    x >= y
    Option A
    On solving both the equations, we get
    x = 8
    y = 7
    x > y

     

  21. I.3x^2 – 29x + 66 = 0
    II. 4y^2 – 57y + 198 = 0
    x >= y
    No relation
    x > y
    y >= x
    y > x
    Option D
    I.3x^2 – 29x + 66 = 0
    =>3x^2 – 18x – 11x + 66 = 0
    =>(x – 6)(3x – 11) = 0
    =>x = 6,11/3
    II. 4y^2 – 57y + 198 = 0
    =>4y^2 – 24y – 33y + 198 = 0
    =>(y-6)(4y – 33) = 0
    =>y = 6, 33/4
    y >= x

     

  22. 8, 11, 17, 47, 128, 371, 1100
    11
    8
    17
    47
    371
    Option C
    8 + 3 =11
    11 + 3^2 = 11 + 9 = 20 == 17
    20 + 3^3 = 20 + 27 = 47
    47 + 3^4 = 47 + 81 = 128
    128 + 3^5 = 128 + 243 = 371

     

  23. 1, 5, 13, 31, 61, 125, 253
    31
    13
    5
    1
    61
    Option A
    1 + 2^2 = 1 + 4 = 5
    5 + 2^3 = 5 + 8 = 13
    13 + 2^4 = 13 + 16 = 29 == 31
    29 + 2^5 = 29 + 32 = 61
    61 + 2^6 = 61 + 64 = 125

     

  24. 150, 290, 560, 1120, 2140, 4230
    150
    1120
    290
    2140
    4230
    Option B
    150 × 2 – 1 × 10
    = 300 – 10 = 290
    290 × 2 – 2 × 10
    = 580 – 20 = 560
    560 × 2 – 3 × 10 = 1120 – 30
    = 1090 == 1120
    1090 × 2 – 4 × 10 = 2180 – 40 = 2140
    2140 × 2 – 5 × 10 = 4280 – 50 = 4230

     

  25. 10, 8, 13, 35, 135, 671, 4007
    135
    35
    10
    671
    8
    Option D
    10 × 1 – 2 = 8
    8 × 2 – 3 = 13
    13 × 3 – 4 = 35
    35 × 4 – 5 = 135
    135 × 5 – 6 = 675 – 6
    = 669 == 671
    669 × 6 – 7 = 4014 – 7 = 4007

     

  26. 29, 37, 21, 43, 13, 53, 5
    29
    37
    21
    53
    43
    Option E
    29 + 1 × 8 = 37
    37 – 2 × 8 = 37 – 16 = 21
    21 + 3 × 8 = 21 + 24 = 45 == 43
    45 – 4 × 8 = 45 – 32 = 13
    13 + 5 × 8 = 13 + 40 = 53
    53 – 6 × 8 = 53 – 48 = 5

     

  27. C is 40% more efficient than B who is 25% more efficient than A. If A,B and C together can complete a work in 35 days. How many days B alone can complete 75% of the work?
    80 days
    77 days
    60 days
    74 days
    84 days
    Option E
    Let the workdone by A in one day = x units
    workdone by B in one day = 1.25x units
    workdone by C in one day = 1.25x *1.40 = 1.75x units
    Total work = 35(x + 1.25x + 1.75x) = 140x units
    Time taken by B alone to complete 75% work = (0.75*140x)/1.25x = 84 days

     

  28. The ratio of speed of Anil and Sunil is 4:3. Ravi can travel a distance of 840 km in 14 hours. If the speed of Sunil is 20% less than the speed of Ravi. What will be the time taken by Anil to travel to a distance of 512 km?
    6 hours
    7 hours
    5 hours
    8 hours
    9 hours
    Option D
    Speed of Ravi = 840/14 = 60 km/hr.
    Speed of Sunil = 0.8*60 = 48 km/hr.
    Speed of Anil = (4/3)*48 = 64 km/hr.
    Required time = 512/64 = 8 hours

     

  29. Varun deposited Rs.12000 in a bank for 2 years at 15% simple interest per annum. After 2 years, the total amount was transferred to a scheme for 2 years at a compound interest of 10% per annum. Find the compound interest earned from the scheme.
    Rs.4124
    Rs.3276
    Rs.2850
    Rs.2152
    Rs.3200
    Option B
    Amount deposited in the scheme = 12000+(12000*15*2)/100 = Rs.15600
    Required compound interest earned from the scheme = 15600*{(1.1)^2 -1} = Rs.3276

     

  30. The length and breadth of a rectangular plot are in the ratio of 3:2 resp. If the cost incurred for fencing the plot at the rate of Rs.12.5/m is Rs.1000. Find the area of the equilateral triangle whose side is equal to the breadth of the reactangle.
    50(3)^1/2 m^2
    48(3)^1/2 m^2
    64(3)^1/2 m^2
    55(3)^1/2 m^2
    60(3)^1/2 m^2
    Option C
    2(3x+2x) = 1000/12.5
    =>x = 8
    Length =24 m
    Breadth = 16 m
    Area of equilateral triangle = (3)^1/2 /4 * 16*16 = 64(3)^1/2 m^2

     

  31. A boatman can cover 192 km upstream and 260 km downstream in 12.5 hours. If the ratio of upstream speed to downstream speed is 4:5 resp. Find the length of the train which takes 24 seconds to cross the tree at the speed equal to the speed of boat in still water.
    280 m
    240 m
    250 m
    180 m
    200 m
    Option B
    Let the upstream and downstream speed be 4x and 5x km/hr resp.
    192/4x + 260/5x = 12.5
    =>x = 8
    Speed in still water =( 4x+5x )/2 = 9x/2 = 36 km/hr = 10 m/s
    Length of train = 10*24 = 240 m

     

  32. A man sold two books, one at a loss of 10% and other at a gain of 25%. If the cost price of each book is same , find the overall profit or loss percentage in this transaction.
    5.8%
    7.5%
    8.1%
    6.6%
    4.5%
    Option B
    Let the CP of each book be Rs.x.
    SP of both books = 0.9x + 1.25x = Rs.2.15x
    CP of both books = Rs.2x
    Required profit% = (0.15x/2x)*100 = 7.5%

     

  33. A and B together started a business with investment of Rs.580 and Rs.x resp. If the profit share of B is 75% more than that of A. Find the value of x.
    1000
    1015
    880
    928
    950
    Option B
    Let the profit share of A be Rs.x
    Profit share of B = Rs.1.75y
    Ratio = y:1.75y = 4:7
    Now,
    580/x = 4/7
    =>x = 1015

     

  34. Pipe A and Pipe B can fill a tank together in (60/7) minutes. If pipe A is 33.33% more efficient than pipe B. Find the time taken by pipe A alone to fill the tank.
    18 minutes
    15 minutes
    8 minutes
    5 minutes
    10 minutes
    Option B
    Let the efficiency of pipe B be 3x.
    Efficiency of pipe A = 4x
    Now,
    1/3x + 1/4x = 7/60
    =>x = 5
    Time taken by pipe A alone to fill the tank = 3x = 15 minutes

     

  35. A solution contains 48 ml of acid and 52 ml of water is completely mixed with some quantity of 25% acid solution. If the ratio of acid to water in the final mixture is 3:5. Find the quantity of 25% acid solution mixed to the initial solution.
    84
    70
    74
    81
    68
    Option A
    Let the quantity of second solution be x ml.
    Amount of acid in final mixture = (48+0.25x) ml
    Amount of water in final mixture = (52+0.75x) ml
    (48+0.25x)/ (52+0.75x) = 3/5
    =>x= 84

     

  36. A bag contains 8 red balls and 3 orange balls. Three balls are randomly selected. Find the number of ways in which atleast 2 orange balls are selected.
    22 ways
    10 ways
    25 ways
    18 ways
    20 ways
    Option C
    Required number of ways = 3C2*8C1 + 3C3 = 25 ways

     


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