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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.

- 32% of 250 + 18% of 350 + (8*9) = ?

220195200222215Option E

32% of 250 + 18% of 350 + (8*9) = ?

=>80+ 63 + 72 = ?

=> ? = 215 - 427 + ? – 732 = 29^2 – 456
622690740800770Option B

427 + ? – 732 = 29^2 – 456

=>? – 305 = 841 – 456

=>? -305 =385

=>? = 690 - 84^2 – 79^2 + (1936)^1/2 – 20% of 980 = ?
693555663744721Option 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 - 3.2% of 500 * 2.4 % of ? = 288

815700666750781Option D

3.2% of 500 * 2.4 % of ? = 288

=>16*2.4/100 * ? =288

=> ? = 750 - 3024 / 12 + 14^2 – 25% of 820 = ? – 20% of 650

320373400410375Option B 3024 / 12 + 14^2 – 25% of 820 = ? – 20% of 650

=>? – 0.20 * 650 = 252 + 196 -0.25* 820

=>? – 130 =448 – 205

=>? = 373 - 32*15 + ? – 25*16 = 45 * 20

740815820723700Option C

32*15 + ? – 25*16 = 45 * 20

=>480 + ? – 400 = 900

=>? = 820 - 25*13 + 148 – 28 – 42 = ? – 224
715620600627750Option D

25*13 + 148 – 28 – 42 = ? – 224

=> ? = 325 + 148 – 70 + 224

=>? = 627 - (784)^1/2 * 29 – ?^3 =180 + 17^2

59876Option D

(784)^1/2 * 29 – ?^3 =180 + 17^2

=>28*29 – ?^3 = 180 + 289

=>?^3 = 343

=> ? = 7 - (300/11)% of 4620 – 840 = ? * 7

6055426550Option A

(300/11)% of 4620 – 840 = ? * 7

=>(3/11) of 4620 – 840 = ? * 7

=>1260 – 840 = ? * 7

=>? = 60 - 2620% of 40 + 40% of 825 = ?
12201424137813001175Option C

2620% of 40 + 40% of 825 = ?

=> ? = 40% of 2620 + 40% of 825

=>? = 1048 + 330

=>? = 1378 - 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% - 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?

6:72:37:69:58:7Option 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 - If the expenditure of Company B in the year 2010 was Rs.40 lakh, what was its income (in Rs) in the year 2013?
70,00080,00050,00060,000Data InadequateOption E

Data Inadequate - 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 : 1413 : 108 : 915 : 1117 : 12Option 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

- If the income of Company B in year 2009 was Rs91.8 lakh, what was its expenditure (in Rs) in that year?

Rs. 80 lakhRs. 74 lakhRs. 55 lakhRs. 68 lakhRs. 60 lakhOption D

% profit = 35%

Expenditure = Income × 100/(100 %P)

91.8*(100/135) = Rs. 68 lakh - I. 4x^2 – 13x + 10 = 0

II.4y^2 + 11y – 3 = 0

y >= xy > xNo relationx > yx >= yOption 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 - I. 2x^2 – 15x + 25 = 0

II. 3y^2 = 4y + 15

y > xx > yx >= yy >= xNo relationOption 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 - I.x^2 + 20x + 96 = 0

II.y^2 + 25y + 156 = 0

y > xNo relationx >= yy >= xx > yOption 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 - I. 2x + 5y = 51

II.9x – 4y = 44

x > yNo relationy >= xy > xx >= yOption A

On solving both the equations, we get

x = 8

y = 7

x > y - I.3x^2 – 29x + 66 = 0

II. 4y^2 – 57y + 198 = 0

x >= yNo relationx > yy >= xy > xOption 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 - 8, 11, 17, 47, 128, 371, 1100
1181747371Option 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 - 1, 5, 13, 31, 61, 125, 253
31135161Option 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 - 150, 290, 560, 1120, 2140, 4230
150112029021404230Option 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 - 10, 8, 13, 35, 135, 671, 4007
13535106718Option 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 - 29, 37, 21, 43, 13, 53, 5
2937215343Option 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 - 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 days77 days60 days74 days84 daysOption 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 - 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 hours7 hours5 hours8 hours9 hoursOption 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 - 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.4124Rs.3276Rs.2850Rs.2152Rs.3200Option 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 - 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^248(3)^1/2 m^264(3)^1/2 m^255(3)^1/2 m^260(3)^1/2 m^2Option 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 - 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 m240 m250 m180 m200 mOption 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 - 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% - 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.

10001015880928950Option 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 - 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 minutes15 minutes8 minutes5 minutes10 minutesOption 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 - 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.
8470748168Option 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 - 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 ways10 ways25 ways18 ways20 waysOption C

Required number of ways = 3C2*8C1 + 3C3 = 25 ways

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