Quant Sectional Test 1 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.

1. Bhanu has wine and water in ratio 5:2. If he sold 14(2/7)% of mixture and poured 15 litre of water in it so that new mixture becomes 10:9. Find the initial quantity of mixture?
   50 L
   25 L
   49 L
   33 L
   40 L
   Option C
   Let the quantity of wine and water be 5x and 2x resp.
   [5x – 5x/7]/[2x – 2x/7 + 15] = 10/9
   => x = 7
   Initial quantity of mixture = 7x = 7*7 = 49 L
   

    

2. A and B enter into a partnership with investment 3:8 and if ratio of time period investment of A and B is 4:3 resp. B gets Rs.750 more than A as profit share. Find the 40% of the total profit?
   Rs.1000
   Rs.600
   Rs.800
   Rs.700
   Rs.900
   Option E
   Let the investment of A and B be 3x and 8x resp.
   Now,
   8x*3 : 3x*4
   = 2x : x
   Difference
   2x – x = 750
   =>x = 750
   40% of profit = 750*3*2/5 = Rs.900
   

    

3. Mr. Josh sold an article at a loss of 25% . If he has sold the same article Rs.130 more than he would have gained 40%. Find the cost price of the article?
   200
   150
   320
   440
   220
   Option A
   Let the CP be x.
   SP = Rs.0.75x
   Now,
   40 = [0.75x + 130 –x ]/x*100
   => x = 200
   

    

4. When a sphere is cut into two hemisphere the total surface area of two hemisphere is equal to the area of circle whose radius is equal to the side of square whose area is 144 cm^2. Find the radius of the sphere?
   2(7)^1/2 cm
   4(2)^1/2 cm
   5(8)^1/2 cm
   2(6)^1/2 cm
   3(4)^1/2 cm
   Option D
   Let the radius of the sphere be r cm.
   r = (144)^1/2 = 12 cm
   Let the radius of hemisphere be R cm.
   3*pi*R*R + 3*pi*R*R = pi*r*r
   =>R = 2(6)^1/2 cm
   

    

5. Karuna invested some amount in scheme P which offer 20% CI p.a. while some amount in scheme Q which offers 8% SI p.a. After 2 years ratio of interest earn from P to Q is 11:6. Amount invested by Karuna in scheme P is what percent of the amount invested by her in scheme Q.
   
   55(5/7)%
   66(2/3)%
   68(3/5)%
   59(4/3)%
   61(2/5)%
   Option B
   Let the amount invested in P and Q be x and y resp.
   x*[(120/100)^2 – 1]/y*(8*2)/100 = 11/6
   =>x/y = 2/3
   Required% = 2/3*100 = 66(2/3)%
   

    

6. Tap A is 50% more efficient than Tap B. Both Tap together can fill the tank in (96/5) hours. Find the time taken by Tap A to fill the tank alone.
   
   44 hours
   25 hours
   22 hours
   32 hours
   30 hours
   Option D
   Let the time taken by Tap A and B be x and x/1.5 hours resp.
   1/x + 1/(x/1.5) = 1/(96/5)
   =>x = 48
   Time taken by Tap A= 48/1.5 = 32 hours
   

    

7. A vessel contains 540 litres mixture of honey, milk and water in the ratio of 3:4:5 resp. 15 litres of honey and 20 litres of milk is added to this mixture. Find the new ratio of honey, milk and water in the vessel.
   
   3:10:7
   6:8:9
   2:8:5
   6:5:7
   3:6:9
   Option B
   Quantity of honey in the vessel initially = [3/(3+4+5)]*540 = 135 L
   Quantity of milk in the vessel initially = [4/(3+4+5)]*540 = 180 L
   Quantity of water in the vessel initially = [5/(3+4+5)]*540 = 225 L
   Required ratio = (135+15): (180+20): 225 = 6:8:9
   

    

8. The interest earned on Rs.4800 at rate of R% compound annually is Rs.1008 after 2 years. What would be the simple interest earned on the same amount at the rate of (R+5)% per annum after 3 years?
   
   Rs.2874
   Rs.2550
   Rs.2160
   Rs.1880
   Rs.2250
   Option C
   Equivalent interest rate of 2 years = (1008/4800)*100 = 21%
   R% + R% + (R^2/100)% = 21%
   => R = 10
   Required interest = 4800*(10+5)% *3 = Rs.2160
   

    

9. Find the time taken by a boatman to travel a distance of 250 km downstream where downstream speed of boatman is 250% of upstream speed of boatman. Speed of stream is 15 km/hr.
   10 hours
   9 hours
   7 hours
   5 hours
   4 hours
   Option D
   Stream speed = 75/5 = 15 km/hr.
   Speed of boatman be x km/hr.
   Now,
   (x+15) = 2.5(x – 15)
   => x = 52.5/1.5 = 35
   Downstream speed of upstream = 35+15 = 50 km/hr.
   Required time = 250/50 = 5 hours
   

    

10. Find the amount received by Sumit if he invested Rs.18,000 in a scheme offering 13% simple interest for 7 years.
    
    Rs.38951
    Rs.35784
    Rs.34380
    Rs.33201
    Rs.32451
    Option C
    Interest earned by Sumit = 18000*0.13*7 =Rs.16380
    So, the amount received by Sumit after 7 years = 18000+16380 = Rs.34380
    

     

Directions(11-15): Given graph shows percentage of foreigner out of total tourists who visited Taj Mahal in last 5 years. Based on data given, answer the following questions.

Same number of tourists visited Taj Mahal every year .
Total tourists = Indians + Foreigners




11. If Govt. Charges Rs.10 from every Indian tourist and Rs.100 from every foreigner tourist, Find the total fare collected by Govt. In 2017 if number of Indian tourists were 27000?
    Rs.30.5 lakh
    Rs.25.9 lakh
    Rs.34.8 lakh
    Rs.35.7 lakh
    Rs.30.1 lakh
    Option D
    45% -> 27000
    1% -> 600
    55% -> 33000
    Total fare collected = (27000*10) + (33000+10) = Rs.35.7 lakh
    

     

12. The number of Indian tourists in 2013 were what percent more or less than foreigner tourists in 2017?
    14(2/17)%
    11(3/10)%
    17(5/14)%
    18(2/11)%
    15(3/13)%
    Option D
    Required% = (65% – 55%)/55%*100 = 18(2/11)%

     

13. If in year 2012, 77000 foreigners visited Taj Mahal which was 10% more than those visited in year 2013, Find the total number of tourists in 2013?
    
    4 lakh
    2 lakh
    6 lakh
    5 lakh
    3 lakh
    Option B
    Let the total tourists in 2013 be x.
    35/100*x*110/100 = 77000
    => x = 2 lakh
    

     

14. If Indians who visited Taj Mahal in 2015 was 21000, find the difference between number of foreigner tourists in 2015 and 2017 ?
    
    3000
    6000
    7000
    4000
    8000
    Option B
    35% -> 21000
    1% -> 6000
    Required Difference = (65-55)*600 = 6000
    

     

15. If Indian tourists in 2016 were 2250 more than Indian tourists in 2014. Find number of foreigner tourists in 2017?
    9120
    8250
    7854
    6500
    8110
    Option B
    (55% – 40%) -> 2250
    15% -> 150
    55% -> 8250
    Foreigner tourists in 2017 = 8250
    

     


Directions(16-20):Study the following graph carefully and answer the questions that follow :

Three different products (in thousands) produced by a company in five different years :


16. Number of Mouse’s produced in year 2011 was what percentage of total number of Motherboards produced in all the years together ?
    
    30%
    20%
    40%
    60%
    70%
    Option B
    Required% = 30/(30+35+15+25+45)*100 = 20%

     

17. What was the respective ratio between the number of Pen drives produced by company in the year 2011 and number of Mouses produced by the company in the year 2009 ?
    1 : 2
    7 : 9
    5 : 8
    2 : 3
    7 : 5
    Option C
    Req. ratio = 25000/40000 = 5 : 8

     

18. What is the difference between the total number of Motherboards and Pen drives produced by company together in the year 2012 and number of Mouse’s produced by the company in 2008 ?
    
    80
    66
    92
    60
    75
    Option E

    Req. difference = ( 45 + 45 ) – 15 = 75
    

     

19. What was the respective ratio between the number of Mouse’s produced by the company in year 2009, 2011, 2012 ?
    2 : 3 : 5
    3 : 4 : 1
    7 : 8 : 9
    8 : 6 : 7
    7 : 2 : 5
    Option D
    Required ratio = 40000 : 30000 : 35000 = 8 : 6 : 7

     

20. What is the percent increased in production of Motherboards in year 2011 from the previous year ?
    44(5/6)%
    50(7/9)%
    66(2/3)%
    55(5/3)%
    61(2/5)%
    Option C
    Required% increase = (25-15)/15*100 = 66(2/3)%

     

Directions(21-27): What approximate value will come in place of question mark in the following questions.

21. 313.31+116.31+62.03 = ? + 318.78
    172
    165
    180
    192
    160
    Option A
    313.31+116.31+62.03 = ? + 318.78
    => 313+116+62 = ?+319
    =>? = 172
    

     

22. 149.71% of 160 – 60.85* 1.99 = (2)^? + 85.76
    4
    5
    3
    6
    8
    Option B
    149.71% of 160 – 60.85* 1.99 = (2)^? + 85.76
    => 150/100*160 – 61 * 2 = (2)^? + 86
    =>240 – 122 = (2)^? + 86
    =>(2)^? = 32
    => ? = 5
    

     

23. 78% of (6.89)/(5.99) of ? = 83.99% of 1249.81
    660
    845
    740
    600
    720
    Option E

    78% of (6.89)/(5.99) of ? = 83.99% of 1249.81
    =>125/100*7/6*? = 84/100*1250
    =>? = 720
    

     

24. 40% of 540 + 18^2 = ?% of 1200
    33
    45
    40
    51
    37
    Option B
    40% of 540 + 18^2 = ?% of 1200
    =>?*12 = 0.4*540 + 324
    =>?*12 = 216+324
    =>? = 540/12
    =>? = 45

     

25. 34^2 – 26^2 = ?^2 – 30% of 320
    24
    18
    20
    15
    29
    Option A
    34^2 – 26^2 = ?^2 – 30% of 320
    =>?^2 – 0.3*320 = (34+26)(34-26)
    =>?^2 – 96 = 60*8
    =>?^2 576
    =>? = 24
    

     

26. (420/12 * 35 + 45^2 – 15^2) = ?^2
    60
    55
    74
    51
    40
    Option B
    (420/12 * 35 + 45^2 – 15^2) = ?^2
    =>?^2 = 35*35 + 2025 – 225
    =>?^2 = 1225 + 2025 – 225
    =>?^2 = 1225 + 1800
    =>? = (3025)^1/2 = 55
    

     

27. 45% of 480 + 18^2 = ?^2 * 17^2 + 10^2
    20
    27
    10
    15
    18
    Option B
    45% of 480 + 18^2 = ?^2 * 17^2 + 10^2
    =>?^2 – 289 + 100 = 0.45 * 480 + 324
    =>?^2 – 189 = 216 + 324
    =>?^2 = 540 + 189
    =>? = (729)^1/2
    =>? = 27
    

     

Directions(28-30): Solve both the equations individually then choose a correct option.

28. I.2x^2 – 29x + 104 = 0
    II.5y^2 – 87y + 378 = 0
    
    x>=y
    x>y
    y>x
    y>=x
    No relation
    Option C
    I.2x^2 – 29x + 104 = 0
    =>2x^2 – 16x – 13x + 104 = 0
    =>2x(x-8) – 13(x-8) = 0
    =>(x-8)(2x-13) = 0
    => x = 8, 13/2
    II.5y^2 – 87y + 378 = 0
    =>5y^2 – 45y – 42y + 378 = 0
    =>5y(y-9) – 42(y-9) = 0
    =>(y-9)(5y – 42) = 0
    =>y = 9, 42/5
    y>x
    

     

29. I.5x^2 – 47x + 56 = 0
    II.3y^2 – 54y + 240 = 0
    
    No relation
    y>=x
    x>y
    x>=y
    y>x
    Option B
    I.5x^2 – 47x + 56 = 0
    =>5x^2 – 40x – 7x + 56 = 0
    =>(x-8)(5x-7) = 0
    =>x = 8,7/5
    II. 3y^2 – 54y + 240 = 0
    =>3y^2 – 24y – 30y + 240 = 0
    =>(y-8)(3y – 30) = 0
    =>y = 8,10
    y>=x
    

     

30. I.x^2 = 144
    II.3y^2 – 47y + 182 = 0
    
    y>=x
    y>x
    x>=y
    No relation
    x>y
    Option D
    I.x^2 = 144
    =>x = +12,-12
    II.3y^2 – 47y + 182 = 0
    =>3y^2 – 21y – 26y + 182 = 0
    =>(y-7)(3y – 26) = 0
    =>y = 7,26/3
    No relation.
    

     

Directions(31-35): Find the missing term of the following series.

31. 15, 18, 16, 19, 17, 20, ?
    
    22
    25
    18
    30
    10
    Option C
    15 + 3 = 18
    18 – 2 = 16
    16 + 3 = 19
    19 – 2 = 17
    17 + 3 = 20
    20 – 2 = 18
    

     

32. 1050, 420, 168, 67.2, 26.88, 10.752, ?
    
    4.3008
    3.5510
    5.1202
    3.4510
    2.1122
    Option A
    1050 ×2/5 = 420
    420 ×2/5 = 168
    168 ×2/5 = 67.2
    10.752 ×2/5 = 4.3008
    

     

33. 0, 6, 24, 60, 120, 210, ?
    
    550
    400
    482
    300
    336
    Option E
    0 + 1 × 6 = 6
    6 + 2 × 9 = 24
    24 + 3 × 12 = 60
    60 + 4 × 15 = 120
    120 + 5 × 18 = 210
    210 + 6 × 21 = 210 + 126 = 336
    

     

34. 32, 49, 83, 151, 287, 559, ?
    
    1200
    1450
    1220
    1103
    1334
    Option D
    32 + 1 × 17 = 32 + 17 = 49
    49 + 2 × 17 = 49 + 34 = 83
    83 + 4 × 17 = 83 + 68 = 151
    151 + 8 × 17 = 151 + 136 = 287
    287 + 16 × 17 = 287 + 272 = 559
    559 + 32 × 17 = 559 + 544 = 1103
    

     

35. 462, 552, 650, 756, 870, 992, ?
    
    1352
    1440
    1122
    1240
    1650
    Option C
    552 – 462 = 90
    650 – 552 = 98
    756 – 650 = 106
    870 – 756 = 114
    992 – 870 = 122
    ? = 992 + 130 = 1122