Quant Sectional Test 2 for LIC AAO 2019 Prelim Exam

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

Directions(1-4): Compare both the equations and choose a correct option.

  1. I.6x^2 – 25x + 14 = 0
    II.2y^2 – 13y + 6 = 0

    x>=y
    x > y
    y>=x
    y > x
    No relation
    Option E
    I.6x^2 – 25x + 14 = 0
    =>6x^2 -21x – 4x + 14 = 0
    =>(2x-7)(3x – 2) = 0
    => x = 7/2, 2/3
    II.2y^2 – 13y + 6 = 0
    =>2y^2 – 12y – y + 6 = 0
    =>(y-6)(2y – 1) = 0
    => y = 6, ½
    No relation.

     

  2. I.5x^2 – 27x + 10 = 0
    II.y^2 – 14y + 48 = 0
    y > x
    No relation
    y>=x
    x > y
    x>=y
    Option A
    I.5x^2 – 27x + 10 = 0
    =>5x^2 – 25x – 2x + 10 = 0
    =>(x-5)(5x-2) = 0
    =>x = 5,2/5
    II.y^2 – 14y + 48 = 0
    =>y^2 – 8y – 6y + 48 = 0
    =>(y-8)(y-6) = 0
    => y = 8,6
    y > x

     

  3. I.x^2 + 12x + 32 = 0
    II.y^2 + 10y + 16 = 0
    x > y
    y>=x
    No relation
    x>=y
    y > x
    Option C
    I.x^2 + 12x + 32 = 0
    =>x^2 + 8x + 4x + 32 = 0
    =>(x+4)(x+8) = 0
    =>x = -4,-8
    II.y^2 + 10y + 16 = 0
    =>y^2 + 2y + 8 y + 16 = 0
    =>(y+2)(y+8) = 0
    => y = -2,-8
    No relation.

     

  4. I.2x^2 – 18x + 36 = 0
    II.2y^2 – 30y + 108 = 0
    No relation
    x > y
    y > x
    y>=x
    x>=y
    Option D
    I.2x^2 – 18x + 36 = 0
    =>2x^2 – 12x – 6x + 36 = 0
    =>(2x-6)(x – 6) = 0
    => x = 3,6
    II.2y^2 – 30y + 108 = 0
    =>2y^2 – 12y – 18u + 108 = 0
    =>(2y – 18)(y-6) = 0
    => y = 6,9
    y>=x

     

  5. Directions(5-20): What will be the approximate value come in place of question mark ‘?’ in the following questions.

  6. 750 *(10.24)^1/2 = ?% of 480
    500
    440
    320
    400
    510
    Option A
    750 *(10.24)^1/2 = ?% of 480
    =>750*3.2 = ?*4.8
    =>? = 500

     

  7. (75% of 1820) – 25^2 = ?^2 -25% of 176
    20
    18
    11
    28
    17
    Option D
    75% of 1820) – 25^2 = ?^2 -25% of 176
    =>?^2 – ¼*176 = 0.75*1820 – 625
    =>?^2 – 44 = 1365- 625
    =>?^2 = 740 + 44
    =>? = 28

     

  8. {(484)^1/2 + 224 / 2^3 }% of 960 = ?% of 640
    66
    81
    70
    75
    90
    Option D
    {(484)^1/2 + 224 / 2^3 }% of 960 = ?% of 640
    =>?% of 640 = (22+28)% of 960
    =>? * 6.4 = 480
    =>? = 75

     

  9. (600/24)*25 + 25^2 + 125% of 280 = ?^2
    33
    40
    25
    41
    53
    Option B
    (600/24)*25 + 25^2 + 125% of 280 = ?^2
    =>?^ 2 = 25*25 +625 + 350
    =>?^2 = 1250 + 350
    => ? = 40

     

  10. 15*8 + ? – 9 * 11 = 12*18
    195
    152
    170
    164
    175
    Option A
    15*8 + ? – 9 * 11 = 12*18
    => 120 + ? – 99 = 216
    =>? + 21 = 216
    =>? = 195

     

  11. 16% of 1750 -20% of 1150 =25% of ?
    110
    100
    200
    324
    220
    Option C
    16% of 1750 -20% of 1150 =25% of ?
    =>0.25*? = 0.16*1750 – 0.20*1150
    =>0.25*? = 280 – 230
    =>? = 200

     

  12. 244 – 156 + ?^2 -13^2 = 15*14 + 25*22
    20
    16
    14
    29
    19
    Option D
    244 – 156 + ?^2 -13^2 = 15*14 + 25*22
    =>88 + ?^2 – 169 = 210 + 550
    =>?^2 – 81 = 760
    =>?^2 = 841
    => ? = 29

     

  13. {(3072/16)/8)/4}+ 14^2 = ?
    195
    174
    166
    202
    188
    Option D
    {(3072/16)/8)/4}+ 14^2 = ?
    =>? = {(192/8)/4} + 14^2
    => ? = 24/4 +196
    =>? = 202

     

  14. 648/9 + 18^2 – 20% of 630 = ?*6
    41
    22
    30
    35
    45
    Option E
    648/9 + 18^2 – 20% of 630 = ?*6
    =>?*6 = 72 + 324 – 0.20*630
    =>?*6 = 396 – 126
    =>?*6 = 270
    => ? = 45

     

  15. 613 + 762 + ? – 981 = 549 – 375 + 817
    660
    597
    500
    495
    510
    Option B
    613 + 762 + ? – 981 = 549 – 375 + 817
    =>1375 + ? – 981 = 991
    =>394 + ? = 991
    =>? = 597

     

  16. 84^2 – 79^2 + (1936)^1/2 – 20% of 980 = ?
    740
    600
    548
    663
    875
    Option D
    84^2 – 79^2 + (1936)^1/2 – 20% of 980 = ?
    =>815 + 44 – 196 = ?
    =>? = 663

     

  17. 958, 833, 733, 658, 608, (?)
    448
    415
    583
    574
    545
    Option C
    958, 833, 733, 658, 608, (?)
    958 – 833 = 125
    833 – 733 = 100
    733 – 658 = 75
    658 – 608 = 50
    ? = 608 – 25 = 583

     

  18. 11, 10, 18, 51, 200, (?)
    900
    880
    995
    745
    717
    Option C
    11 × 1 – 1 = 10
    10 × 2 – 2 = 18
    18 × 3 – 3 = 51
    51 × 4 – 4 = 200
    200 × 5 – 5 = 995

     

  19. 25, 48, 94, 186, 370, (?)
    695
    687
    770
    712
    738
    Option E
    25 × 2 – 2 = 50 – 2 = 48
    48 × 2 – 2 = 96 – 2 = 94
    94 × 2 – 2 = 188 – 2 = 186
    186 × 2 – 2 = 372 – 2 = 370
    370 × 2 – 2 = 740 – 2 = 738

     

  20. 14, 24, 43, 71, 108, (?)
    154
    140
    125
    136
    118
    Option A
    14 + 10 = 24
    24 + 19 (=10 + 9) = 43
    43 + 28 (= 19 + 9) = 71
    71 + 37 (= 28 + 9) = 108
    108 + 46 (=37 + 9) = 154

     

  21. 144, 173, 140, 169, 136, (?)
    140
    165
    157
    125
    133
    Option B
    144 + 29 = 173
    173 – 33 = 140
    140 + 29 = 169
    169 – 33 = 136
    136 + 29 = 165

     

  22. Directions(21-25): Answer the following questions based on the following graph.
    The given chart shows the number of Passenger trains and Express trains passing through different platforms of a railway station on a particular day.

  23. The respective ratio of number of Passenger trains to the number of Express trains passing through platform number 7 is 5:7. If the total number of trains passing through platform number 7 is 50% more than the total number of trains passing through platform number 2. Find the number of Passenger trains passing through platform number 7.
    71
    52
    45
    60
    84
    Option D
    Total number of trains passing through platform number 2 = 56+40 = 96
    Total number of trains passing through platform number 7 = 96*1.5 = 144
    Total number of Passenger trains passing through platform number 7 = 5/12*144 = 60

     

  24. The number of passenger trains passing through platform number 6 to the number of passenger trains passing through platform number 5 are in the ratio 4:5 resp. and the number of Express trains passing through platform number 6 are 25% more than number of Express trains passing through platform number 5 . Find the total number of trains passing through platform number 6.
    125
    111
    108
    114
    117
    Option A
    Total number of Passenger trains passing through platform number 6 = 50/5*4 = 40
    Total number of Express trains passing through platform number 6 = 68*1.25 = 85
    Total number of trains passing through platform number 6 = 40+85 = 125

     

  25. Find the ratio of number of trains passing through platform number 2 to the total number of trains passing through platform number 3?
    1:3
    3:2
    4:5
    5:7
    3:4
    Option C
    Total number of trains passing through platform number 2 = 56+40 = 96
    Total number of trains passing through platform number 3 = 48+72 = 120
    Required ratio = 4:5

     

  26. Total number of trains passing through platform number 4 is what percentage more or less than the number of Express trains passing through platform number 3?
    32%
    10%
    18%
    25%
    20%
    Option D
    Total number of trains passing through platform number 4 = 36+54 = 90
    Required% = (90-72)/72*100 = 25%

     

  27. Find the average of the number of passenger trains passing through platform number 1,2 and 3 taken together.
    30
    38
    44
    56
    47
    Option D
    Required average = (64+56+48)/3 = 56

     

  28. Prakash borrowed some money at the rate of 3 percent per annum for the first 4 years, 7 percent per annum for the next 5 years and at the rate of 9 percent per annum for the next 7 years. If the total interest paid by him at the end of 16 years is 4400. Find the money borrowed by Prakash?
    Rs.3000
    Rs.2000
    Rs.4000
    Rs.6000
    Rs.5000
    Option C
    Let the sum be Rs.x.
    (3*4*x)/100 + (7*5*x)/100 + (9*7*x)/100 = 4400
    =>x = 440000/110 = Rs.4000

     

  29. Prabir invests an amount of Rs.15860 in the names of her three daughters A,B and C in such a way that they get the same interest after 2,3 and 4 years resp. If the rate of interest is 5% p.a. Find the ratio of the amounts invested among A,B and C.
    1:4:2
    5:3:7
    3:7:2
    5:1:4
    6:4:3
    Option E
    (P1*2*5)/100 : (P2*3*5)/100 : (P3*4*5)/100
    => 10P1 : 15P2 : 20P3
    =>P1:P2:P3 =30:20:15 = 6:4:3

     

  30. If numerical value of percentage markup price, marked price per kg and percentage discount offered on marked price of a product are in the ratio 20:12:9. Find the percentage profit on the product.
    7.7%
    9.2%
    5.8%
    6.5%
    8.3%
    Option B
    MP = 12
    SP = (100-9)/100*12 = 10.92
    Let CP be x.
    (100+20)/100*x = 12
    => x= 10
    Profit% = 0.92/10*100 = 9.2%

     

  31. Sahil buys wheat at Rs.10 per kg and puts a price tag on it so as to earn a profit of 30%. However his faculty shows 1 kg when it is actually 700 gm. What is his actual profit percent?
    613/2%
    600/7%
    515/4%
    623/8%
    551/5%
    Option B
    CP of 1 kg = Rs.10
    CP of 1 gm = 1/100
    SP of 700 gm = Rs.13
    SP of 1 gm = 13/700
    Profit% = {[13/700 – 1/100]/[1/100]}*100 = 600/7%

     

  32. To do a certain work A is 100/3% more efficient than B. They started to work alternatively on the same work starting from A on one day and B on day two and finished 60% of work in 30 days. Find the time taken by A alone to finish the whole work?
    163/5 days
    157/2 days
    166/5 days
    175/4 days
    171/3 days
    Option D
    Ratio of efficiency of A and B is 4:3.
    Let in one day A does 4x unit of work and in one day B does 3x unit of work.
    In 30 days 60% work is completed.
    So, in 50 days whole work will be completed.
    In 2 days 7x unit is completed so in 50 days.
    = 50*7x/2 unit done
    Whole work is completed by A in = (50*7x)/(4x*2) = 175/4 days

     

  33. A car covered 48 km less in T hours than a train, if speed of train is 25% more than the speed of car, which is 64 km/hr. and speed of an aircraft is 62(1/2)% more than the sum of speed of car and train. Find how much distance will be covered by aircraft in (T-1/4) hour?
    873.7 km
    741.6 km
    555.5 km
    643.5 km
    612.4 km
    Option D
    Speed of train = 64*5/4
    = 80 km/hr.
    Now,
    80T – 64T = 48
    =>T = 3 hours
    Speed of Aircraft = (144*13/8) = 234 km/hr.
    Required distance = 234*(3 – 1/4) = 643.5 km

     

  34. Sam invested some amount in scheme P which offers 20% C.I. p.a. while some amount in scheme Q which offers 8% S.I. p.a. After 2 years, ratio of interest earn from P to Q is 11:6. Amount invested by Sam in scheme P is what percent of the amount invested by him in scheme Q.
    53(4/5)%
    66(2/3)%
    44(2/7)%
    50(1/2)%
    61(3/5)%
    Option B
    Let the amount invested in scheme P and Q be x and y resp.
    x*[(120/100)^2 – 1]/(y*8*2/100) = 11/6
    =>(x*0.44)/(y*0.16) = 11/6
    =>x / y = 2/3
    Required% = 2/3*100 = 66(2/3)%

     

  35. There are two vessels A and B which contains mixture of sulphuric acid and nitrous oxide in the ratio of 7:2 and 3:4 resp. Mixture of both vessels are mixed to obtain a mixture of 390 ml, in which quantity of nitrous oxide is 160 ml. Find the ratio of quantity of mixture in vessel A to quantity of mixture in vessel B ?
    3:4
    2:5
    6:7
    7:8
    3:5
    Option C
    Let the total mixture of sulphuric acid and nitorus oxide in vessel A and vessel B be x and y resp.
    x + y = 390 – (1)
    2x/9 + 4y/7 = 160
    14x+36y = 10080 –(2)
    On solving these two equations, we get
    x = 180 ml
    y = 210 ml
    Required ratio = 6:7

     

  36. The surface of sphere is 423.5 cm^2 less than total surface area of a hemisphere. If ratio between radius of hemisphere and sphere is 3:2. Find the radius of hemisphere?
    8.4 cm
    12.3 cm
    13.7 cm
    7.4 cm
    10.5 cm
    Option E
    3*pi*r*r*r – 4*pi*r*r = 423.5
    =>x = 3.5 cm
    Radius of hemisphere = 21/2 = 10.5 cm

     

  37. In first bucket there are five apples, Y oranges and 7 bananas. Probability of drawing one orange from bucket is 2/5. In second bucket there are (Y-3) apples, (Y-4) oranges and 6 bananas, if two fruits are drawn from second bucket. Find the probability that both are apples?
    1/17
    4/27
    2/21
    4/23
    3/19
    Option C
    Probability of drawing one orange = Y/(12+Y) = 2/5
    => Y = 8
    Required Probability = 5C2/15C2 = 2/21

     


Related posts

Leave a Comment