# Quant Sectional Test 2 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(1-4): 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