# Quant Sectional Test 1 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.

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 â€“ 5x/7]/[2x â€“ 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 â€“ 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 â€“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

11. 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

12. 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

13. 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)%

14. 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

15. 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

16. 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

17. 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 :

18. 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%

19. 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

20. 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 ) â€“ 15 = 75

21. 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

22. 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)%

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

24. 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

25. 149.71% of 160 â€“ 60.85* 1.99 = (2)^? + 85.76
4
5
3
6
8
Option B
149.71% of 160 â€“ 60.85* 1.99 = (2)^? + 85.76
=> 150/100*160 â€“ 61 * 2 = (2)^? + 86
=>240 â€“ 122 = (2)^? + 86
=>(2)^? = 32
=> ? = 5

26. 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

27. 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

28. 34^2 â€“ 26^2 = ?^2 â€“ 30% of 320
24
18
20
15
29
Option A
34^2 â€“ 26^2 = ?^2 â€“ 30% of 320
=>?^2 â€“ 0.3*320 = (34+26)(34-26)
=>?^2 â€“ 96 = 60*8
=>?^2 576
=>? = 24

29. (420/12 * 35 + 45^2 â€“ 15^2) = ?^2
60
55
74
51
40
Option B
(420/12 * 35 + 45^2 â€“ 15^2) = ?^2
=>?^2 = 35*35 + 2025 â€“ 225
=>?^2 = 1225 + 2025 â€“ 225
=>?^2 = 1225 + 1800
=>? = (3025)^1/2 = 55

30. 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 â€“ 289 + 100 = 0.45 * 480 + 324
=>?^2 â€“ 189 = 216 + 324
=>?^2 = 540 + 189
=>? = (729)^1/2
=>? = 27

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

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

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

34. I.x^2 = 144
II.3y^2 â€“ 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 â€“ 47y + 182 = 0
=>3y^2 â€“ 21y â€“ 26y + 182 = 0
=>(y-7)(3y – 26) = 0
=>y = 7,26/3
No relation.

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

36. 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

37. 1050, 420, 168, 67.2, 26.88, 10.752, ?
4.3008
3.5510
5.1202
3.4510
2.1122
Option A
1050 Ã—2/5 = 420
420 Ã—2/5 = 168
168 Ã—2/5 = 67.2
10.752 Ã—2/5 = 4.3008

38. 0, 6, 24, 60, 120, 210, ?
550
400
482
300
336
Option E
0 + 1 Ã— 6 = 6
6 + 2 Ã— 9 = 24
24 + 3 Ã— 12 = 60
60 + 4 Ã— 15 = 120
120 + 5 Ã— 18 = 210
210 + 6 Ã— 21 = 210 + 126 = 336

39. 32, 49, 83, 151, 287, 559, ?
1200
1450
1220
1103
1334
Option D
32 + 1 Ã— 17 = 32 + 17 = 49
49 + 2 Ã— 17 = 49 + 34 = 83
83 + 4 Ã— 17 = 83 + 68 = 151
151 + 8 Ã— 17 = 151 + 136 = 287
287 + 16 Ã— 17 = 287 + 272 = 559
559 + 32 Ã— 17 = 559 + 544 = 1103

40. 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

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