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LIC ADO Online Test Series 1, LIC ADO Mock Test, LIC ADO Online Quiz. LIC ADO Free Mock Test Exam 2020. LIC ADO Exam Free Online Quiz 2020, LIC ADO Full Online Mock Test Series 1st in English. Practice Online LIC ADO Sample Test papers and learn which areas need improvement to score better in actual LIC ADO Exam. Start preparing by solving objective questions based on previous years as well as this years new format. Compare scores and work on your weak areas thoroughly. LIC ADO Question and Answers in English and Hindi Series 1. Here we are providing LIC ADO Full Mock Test Paper in English. LIC ADO Mock Test Series 1st 2020. Now Test your self for LIC ADO Exam by using below quiz…
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45 men working 8 hours per day dig 30 m deep. How many extra men should be put to dig to a depth of 50 m working 6 hours per day?
(45 * 8)/30 = (x * 6)/50 => x =100
100 – 45 = 55
(45 * 8)/30 = (x * 6)/50 => x =100
100 – 45 = 55
The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is .
x Not younger_______ ↑
The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.
The required number of ways = 6(6!) = 4320
x Not younger_______ ↑
The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.
The required number of ways = 6(6!) = 4320
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LIC ADO Online Test Series 2, LIC ADO Mock Test Series 2, LIC ADO Quiz.Take this LIC ADO Mock Test  Free Online Test and find out how much you score before you appear for your next interview and written test. LIC ADO English Language Online Test in English Series 2nd. Take LIC ADO Online Quiz. The LIC ADO Full online mock test paper is free for all students. LIC ADO Question and Answers in English and Hindi Series 2.Most of the organizations are opting for computer based online mode of examination. So, a student needs to practice the question papers in online mod before appearing for the exam. Here we are providing LIC ADO Full Mock Test Paper in English. LIC ADO Mock Test Series 2nd 2020. Now Test your self for LIC ADO Exam by using below quiz…
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The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?
Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l^{2} = 324 => l = 18.
Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l^{2} = 324 => l = 18.
Find the least number with which 16200 should be multiplied, to make it a perfect cube.
16200 = 2^{3} * 3^{4} * 5^{2}
A perfect cube has a property of having the indices of all its prime factors divisible by 3.
Required number = 3^{2} * 5 = 45.
16200 = 2^{3} * 3^{4} * 5^{2}
A perfect cube has a property of having the indices of all its prime factors divisible by 3.
Required number = 3^{2} * 5 = 45.
Calculate the number of bricks, each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar?
10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x
10 * 20 * 90 = 15 * 2 * x => x = 6000
10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x
10 * 20 * 90 = 15 * 2 * x => x = 6000
Two cylinders are of the same height. Their radii are in the ratio 1: 3. If the volume of the first cylinder is 40 cc. Find the volume of the second cylinder?
r_{1} = x r_{2} = 3x
Π * x^{2} * h = 40
Π9 x^{2}h = 40 * 9 = 360
r_{1} = x r_{2} = 3x
Π * x^{2} * h = 40
Π9 x^{2}h = 40 * 9 = 360
Two pipes can fill a tank in 18 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?
Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes.
Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes.
(64 + 9 + 9) / (2 * 20 + 1) = ?
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
The difference between simple and compound interest on Rs. 1200 for one year at 10% per annum reckoned halfyearly is?
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)2 – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)2 – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.
(24 * 5 * 7 * 9) / ? = 216
? = (24 * 5 * 7 * 9) / 216 = 35
? = (24 * 5 * 7 * 9) / 216 = 35
A goods train runs at the speed of 72 km/hr and crosses a 250 m long platform in 26 sec. What is the length of the goods train?
Speed = 72 * 5/18 = 20 m/sec.
Time = 26 sec.
Let the length of the train be x meters.
Then, (x + 250)/26 = 20
x = 270 m.
Speed = 72 * 5/18 = 20 m/sec.
Time = 26 sec.
Let the length of the train be x meters.
Then, (x + 250)/26 = 20
x = 270 m.
The average height of 35 boys in a class was calculated as 180cm. It has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm. Find the actual average height of the boys in the class (Round off your answer to two decimal places).
Calculated average height of 35 boys = 180 cm.
Wrong total height of 35 boys = 180 * 35 cm. This was as a result of an actual height of 106 cm being wrongly written as 166 cm. Correct total height of 35 boys = 180 * 35 cm – 166 cm + 106 cm
= 180 * 35 cm – 166 cm + 106 cm/35 = 180 cm – 60 /35 cm
= 180 cm – 1.71 cm = 178.29 cm.
Calculated average height of 35 boys = 180 cm.
Wrong total height of 35 boys = 180 * 35 cm. This was as a result of an actual height of 106 cm being wrongly written as 166 cm. Correct total height of 35 boys = 180 * 35 cm – 166 cm + 106 cm
= 180 * 35 cm – 166 cm + 106 cm/35 = 180 cm – 60 /35 cm
= 180 cm – 1.71 cm = 178.29 cm.
In a colony of 70 residents, the ratio of the number of men and women is 4:3. Among the women, the ratio of the educated to the uneducated is 1:4. If the ratio of the number of education to uneducated persons is 8:27, then find the ratio of the number of educated and uneducated men in the colony?
Number of men in the colony = 4/7 (70) = 40
Number of women in the colony = 3/7 (70) = 30
Number of educated women in the colony = 1/5 (30) = 6
Number of uneducated women in the colony = 4/5 (30) = 24
Number of educated persons in the colony = 8/35 (70) = 16
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 – 10 = 30
Number of educated men and uneducated men are in the ratio 10:30 => 1:3
Number of men in the colony = 4/7 (70) = 40
Number of women in the colony = 3/7 (70) = 30
Number of educated women in the colony = 1/5 (30) = 6
Number of uneducated women in the colony = 4/5 (30) = 24
Number of educated persons in the colony = 8/35 (70) = 16
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 – 10 = 30
Number of educated men and uneducated men are in the ratio 10:30 => 1:3
64309 – 8703 + 798 – 437 = ?
64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967
64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967
When the numerator of a fraction is decreased by 25% and its denominator is decreased by 20%, the new fraction obtained is 3/4. Find the original fraction?
Let the fraction be x/y.
When the numerator decreased by 25% and the denominator decreased by 20%.
(x * 3/4)/(y * 4/5) = 3/4 => x/y = 3/4 * (4/5)/(3/4) = 4/5
Let the fraction be x/y.
When the numerator decreased by 25% and the denominator decreased by 20%.
(x * 3/4)/(y * 4/5) = 3/4 => x/y = 3/4 * (4/5)/(3/4) = 4/5
Find the one which does not belong to that group ?
Star, Moon, Comet and Planet are natural celestial bodies, while Rocket is man made.
Star, Moon, Comet and Planet are natural celestial bodies, while Rocket is man made.
(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = ?
(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = (22 * 33 * 15 + 9 * 99 * 2 * 22) / (33 * 22 * 8 + 5 * 11 * 22 * 12)
= (22 * 33 * 15 + 22 * 33 * 54) / (33 * 22 * 8 + 33 * 22 * 20) = [22 * 33(15 + 54)] / [33 * 22(8 + 20)] = 69/28
(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = (22 * 33 * 15 + 9 * 99 * 2 * 22) / (33 * 22 * 8 + 5 * 11 * 22 * 12)
= (22 * 33 * 15 + 22 * 33 * 54) / (33 * 22 * 8 + 33 * 22 * 20) = [22 * 33(15 + 54)] / [33 * 22(8 + 20)] = 69/28
A train passes a man standing on the platform. If the train is 170 meters long and its speed is 72 kmph, how much time it took in doing so?
D = 170
S = 72 * 5/18 = 20 mps
T = 170/20 = 8 ½ sec
D = 170
S = 72 * 5/18 = 20 mps
T = 170/20 = 8 ½ sec
On dividing a number by 68, we get 269 as quotient and 0 as remainder. On dividing the same number by 67, what will be the remainder?
Number = 269 * 68 + 0 = 18292
67) 18292 (273
18291
——–
1
Required number = 1.
Number = 269 * 68 + 0 = 18292
67) 18292 (273
18291
——–
1
Required number = 1.
The least number of four digits which is divisible by 15, 25, 40 and 75 is:
Greatest number of 4 digits is 9999. L.C.M of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399. Required number = 9999 – 399 = 9600
Greatest number of 4 digits is 9999. L.C.M of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399. Required number = 9999 – 399 = 9600
A man swims downstream 30 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water?
30 — 3 DS = 10
? — 1
18 — 3 US = 6
? — 1 M = ?
M = (10 + 6)/2 = 8
30 — 3 DS = 10
? — 1
18 — 3 US = 6
? — 1 M = ?
M = (10 + 6)/2 = 8
Express 25 mps in kmph?
25 * 18/5 = 90 kmph
25 * 18/5 = 90 kmph
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, which A and C together can do it in 2 hours. How long will B alone take to do it?
A’s 1 hour work = 1/4;
(B + C)’s 1 hour work = 1/3;
(A + C)’s 1 hour work = 1/2
(A + B + C)’s 1 hour work = (1/4 + 1/3) = 7/12
B’s 1 hour work = (7/12 + 1/2) = 1/12
B alone will take 12 hours to do the work.
A’s 1 hour work = 1/4;
(B + C)’s 1 hour work = 1/3;
(A + C)’s 1 hour work = 1/2
(A + B + C)’s 1 hour work = (1/4 + 1/3) = 7/12
B’s 1 hour work = (7/12 + 1/2) = 1/12
B alone will take 12 hours to do the work.
Rs.4500 amounts to Rs.5544 in two years at compound interest, compounded annually. If the rate of the interest for the first year is 12%, find the rate of interest for the second year?
Let the rate of interest during the second year be R%. Given,
4500 * {(100 + 12)/100} * {(100 + R)/100} = 5544
R = 10%
Let the rate of interest during the second year be R%. Given,
4500 * {(100 + 12)/100} * {(100 + R)/100} = 5544
R = 10%
A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023
If a : b = 4 : 1, then find (a – 3b)/(2a – b).
a : b = 4 : 1
a/b = 4/1 => a = 4b
(a – 3b)/(2a – b) = (4b – 3b)/[2(4b) – b]
= b/(8b – b) = b/7b => 1/7
a : b = 4 : 1
a/b = 4/1 => a = 4b
(a – 3b)/(2a – b) = (4b – 3b)/[2(4b) – b]
= b/(8b – b) = b/7b => 1/7
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.
Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126
Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126
The average of 35 students in a class is 16 years. The average age of 21 students is 14. What is the average age of remaining 14 students?
Sum of the ages of 14 students
= (16 * 35) – (14 * 21) = 560 – 294 = 266
Required average = (266/14) = 19 years.
Sum of the ages of 14 students
= (16 * 35) – (14 * 21) = 560 – 294 = 266
Required average = (266/14) = 19 years.
The salary of a worker is first increased by 30% and afterwards reduced by 30%. What is net change in his salary?
(30 * 30)/100 = 9% decrease
(30 * 30)/100 = 9% decrease
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream?
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.
A is twice as good a workman as B and they took 7 days together to do the work B alone can do it in.
WC = 2:1
2x + x = 1/7
x = 1/21 => 21 days
WC = 2:1
2x + x = 1/7
x = 1/21 => 21 days
After decreasing 24% in the price of an article costs Rs.912. Find the actual cost of an article?
CP* (76/100) = 912
CP= 12 * 100 => CP = 1200
CP* (76/100) = 912
CP= 12 * 100 => CP = 1200
The smallest ratio out of 1:1, 2:1, 1:3 and 3:1 is?
Sachin is younger than Rahul by 4 years. If their ages are in the respective ratio of 7:9, how old is Sachin?
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.
A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then, A’s speed is equal to?
Let A’s speed = x km/hr. Then, B’s speed = (7 – x) km/ hr.
So, 24/x + 24/(7 – x) = 14
x^{2} – 98x + 168 = 0
(x – 3)(x – 4) = 0 => x = 3 or 4.
Since, A is faster than B, so A’s speed = 4 km/hr and B’s speed = 3 km/hr.
Let A’s speed = x km/hr. Then, B’s speed = (7 – x) km/ hr.
So, 24/x + 24/(7 – x) = 14
x^{2} – 98x + 168 = 0
(x – 3)(x – 4) = 0 => x = 3 or 4.
Since, A is faster than B, so A’s speed = 4 km/hr and B’s speed = 3 km/hr.
A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
A man can do a job in 15 days. His father takes 20 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together?
1 day work of the three persons = (1/15 + 1/20 + 1/25) = 47/300
So, all three together will complete the work in 300/47 = 6.4 days.
1 day work of the three persons = (1/15 + 1/20 + 1/25) = 47/300
So, all three together will complete the work in 300/47 = 6.4 days.
A person purchases 90 clocks and sells 40 clocks at a gain of 10% and 50 clocks at a gain of 20%. If he sold all of them at a uniform profit of 15%, then he would have got Rs. 40 less. The cost price of each clock is:
Let C.P. of clock be Rs. x.
Then, C.P. of 90 clocks = Rs. 90x.
[(110% of 40x) + (120% of 50x)] – (115% of 90x) = 40
44x + 60x – 103.5x = 40
0.5x = 40 => x = 80
Let C.P. of clock be Rs. x.
Then, C.P. of 90 clocks = Rs. 90x.
[(110% of 40x) + (120% of 50x)] – (115% of 90x) = 40
44x + 60x – 103.5x = 40
0.5x = 40 => x = 80
Income and expenditure of a person are in the ratio 5 : 4. If the income of the person is Rs. 18000, then find his savings.
Let the income and the expenditure of the person be Rs. 5x and Rs. 4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs. 3600.
Let the income and the expenditure of the person be Rs. 5x and Rs. 4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs. 3600.
What is the unit digit in 7^{105} ?
Unit digit in 7^{105} = unit digit in [(7^{4})^{26} * 7]
But, unit digit in (7^{4})^{26} = 1
unit digit in 7^{105} = (1 * 7) = 7
Unit digit in 7^{105} = unit digit in [(7^{4})^{26} * 7]
But, unit digit in (7^{4})^{26} = 1
unit digit in 7^{105} = (1 * 7) = 7
What percent of 125 is 16?
Let the required percentage be x%.
x% of 125 = 16 => x/100 * 125 = 16
=> x = 1600/125 = 64/5 = 12 4/5
Let the required percentage be x%.
x% of 125 = 16 => x/100 * 125 = 16
=> x = 1600/125 = 64/5 = 12 4/5
Rs. 6000 is lent out in two parts. One part is lent at 7% p.a simple interest and the other is lent at 10% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest?
Let the amount lent at 7% be Rs. x
Amount lent at 10% is Rs. (6000 – x)
Total interest for one year on the two sums lent
= 7/100 x + 10/100 (6000 – x) = 600 – 3x/100
=> 600 – 3/100 x = 450 => x = 5000
Amount lent at 10% = 1000
Required ratio = 5000 : 1000 = 5 : 1
Let the amount lent at 7% be Rs. x
Amount lent at 10% is Rs. (6000 – x)
Total interest for one year on the two sums lent
= 7/100 x + 10/100 (6000 – x) = 600 – 3x/100
=> 600 – 3/100 x = 450 => x = 5000
Amount lent at 10% = 1000
Required ratio = 5000 : 1000 = 5 : 1
√3584 * 52.034 = ?
√3600 * 52 = ?
=> ? = 60 * 52 = 3120
√3600 * 52 = ?
=> ? = 60 * 52 = 3120
Sum of two consecutive even terms lacks by 98 from their product. Find the sum of these numbers?
x(x + 2) – (x + x + 2) = 98
x(x + 2) – (x + x + 2) = 98
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.
Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.
Find the one which does not belong to that group ?
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.
21345 * 45 = 4270 * ?
? = (21345 * 45) / 4270
? = 225
? = (21345 * 45) / 4270
? = 225
The sum of three numbers is 98. If the ratio of the first to the second is 2:3. And that of the second to the third is 5:8, then the second number is:
Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30.
Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30.
Rahim bought 65 books for Rs.1150 from one shop and 50 books for Rs.920 from another. What is the average price he paid per book ?
Average price per book = (1150 + 920) / (65 + 50) = 2070 / 115 = Rs.18
Average price per book = (1150 + 920) / (65 + 50) = 2070 / 115 = Rs.18
The radius of a cylindrical vessel is 7cm and height is 3cm. Find the whole surface of the cylinder?
r = 7 h = 3
2πr(h + r) = 2 * 22/7 * 7(10) = 440
r = 7 h = 3
2πr(h + r) = 2 * 22/7 * 7(10) = 440
7 2/5 of 110 / ? = 844
? = 844 – 7 2/5 of 110 = 844 – 37 * 22 = 30
? = 844 – 7 2/5 of 110 = 844 – 37 * 22 = 30
A can do a piece of work in 12 days. When he had worked for 2 days B joins him. If the complete work was finished in 8 days. In how many days B alone can finish the work?
8/12 + 6/x = 1
X = 18 days
8/12 + 6/x = 1
X = 18 days
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LIC ADO Online Test Series 2  50  Go to Test 
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Life Insurance Corporation of India (LIC) one of the biggest stateowned insurance group and investment company in India. It was established in 1956. It offers various services to its customer like insurance plans, pension plans, unitlinked plans, special plans and group schemes. Every year LIC professes a notification regarding upcoming vacancies in LIC to invite all those candidates who aspire to be a part of Life Insurance Corporation and want to pursue it as a career option. By following the same path LIC has come up with a new notification for various posts like Assistant Administrative Officers, Apprentice Development Officer, Associates posts, LIC Insurance Advisor, Urban Career Agents, CCA, LIC Agents etc.
LIC ADO exam pattern and syllabus comprises of General Knowledge and Awareness, Verbal and Nonverbal Reasoning, Numerical Ability, English Language and a part of Insurance and Financial Marketing Awareness. The examination will comprise of total 150 questions of 1 mark each. Three paper will be conducted and the candidate should obtain minimum passing marks in each paper. Total of all the three paper will be counted to make the merit list and negative marking will be provided for every wrong answer.
Duration – 2 hours
Paper  Topics  No. of Questions  Total Marks 
Paper I  Reasoning Ability  25  25 
Numerical Ability  25  25  
Paper II  GK & Current Affairs  25  25 
English Language  25  25  
Paper III  Insurance and Financial Marketing  50  50 
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LIC ADO Mock Test Series 3, LIC ADO Online Test 3, LIC ADO Quiz. LIC ADO Exam Free Online Quiz 2020, LIC ADO Full Online Mock Test Series 3rd in English.The LIC regularly conducted exams for the LIC ADO over the years. All old last year question papers for the above post are available in ebook (PDF) format which you can download from the link given below. LIC ADO Free Mock Test Series in English. LIC ADO Free Mock Test Series 3. LIC ADO English Language Online Test in English Series 3rd. Take LIC ADO Online Quiz. The LIC ADO Full online mock test paper is free for all students. Here we are providing LIC ADO Full Mock Test Paper in English. LIC ADO Mock Test Series 3rd 2020. Now Test your self for LIC ADO Exam by using below quiz…
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A train 540 meters long is running with a speed of 54 kmph. The time taken by it to cross a tunnel 180 meters long is?
D = 540 + 180 = 720
S = 54 * 5/18 = 15 mps
T = 720/15 = 48 sec
D = 540 + 180 = 720
S = 54 * 5/18 = 15 mps
T = 720/15 = 48 sec
If 12 men do a work in 80 days, in how many days will 16 men do it?
12 * 80 = 16 * x
x = 60 days
12 * 80 = 16 * x
x = 60 days
On dividing number by 5, we get 3 as remainder. What will be the remainder when the square of this number is divided by 5?
The sum of four consecutive even numbers is 292. What would be the smallest number?
Let the four consecutive even numbers be 2(x – 2), 2(x – 1), 2x, 2(x + 1)
Their sum = 8x – 4 = 292 => x = 37
Smallest number is: 2(x – 2) = 70.
Let the four consecutive even numbers be 2(x – 2), 2(x – 1), 2x, 2(x + 1)
Their sum = 8x – 4 = 292 => x = 37
Smallest number is: 2(x – 2) = 70.
An article was sold after a discount of 20% and there was a gain of 20%. If the profit made on it was Rs. 6 less than the discount offered on it, find its selling price?
Let CP = Rs. 100x
SP = Rs. 120x
MP = 120x/80 * 100 = Rs. 150x
D = Rs. 150x – Rs. 120x = Rs. 30x
D – P = 30x – 20x = Rs. 6, 10x = Rs. 6
120x = 120/10 * 6 = Rs. 72
Let CP = Rs. 100x
SP = Rs. 120x
MP = 120x/80 * 100 = Rs. 150x
D = Rs. 150x – Rs. 120x = Rs. 30x
D – P = 30x – 20x = Rs. 6, 10x = Rs. 6
120x = 120/10 * 6 = Rs. 72
A 25 cm wide path is to be made around a circular garden having a diameter of 4 meters. Approximate area of the path is square meters is
Area of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}^{2} – ∏[4/2]^{2}
= ∏[2.25^{2} – 2^{2}] = ∏(0.25)(4.25) { (a^{2} – b^{2} = (a – b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m
Area of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}^{2} – ∏[4/2]^{2}
= ∏[2.25^{2} – 2^{2}] = ∏(0.25)(4.25) { (a^{2} – b^{2} = (a – b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m
What is the sum of all the composite numbers up to 20?
4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 132
4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 132
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves onefourth of his income, find the ratio of their monthly savings?
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
L.C.M of 5, 6, 4 and 3 = 60.
On dividing 2496 by 60, the remainder is 37. Number to be added = 60 – 37 = 23.
L.C.M of 5, 6, 4 and 3 = 60.
On dividing 2496 by 60, the remainder is 37. Number to be added = 60 – 37 = 23.
64309 – 8703 + 798 – 437 = ?
64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967
64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967
How many figures are required for numbering the pages of a book containing 1000 pages?
1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 999 = 900 * 3 =2700
1000 = 4
———–
2893
1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 999 = 900 * 3 =2700
1000 = 4
———–
2893
The total marks obtained by a student in Mathematics and Physics is 60 and his score in Chemistry is 20 marks more than that in Physics. Find the average marks scored in Mathamatics and Chemistry together.
Let the marks obtained by the student in Mathematics, Physics and Chemistry be M, P and C respectively.
Given , M + C = 60 and C – P = 20 M + C / 2 = [(M + P) + (C – P)] / 2 = (60 + 20) / 2 = 40.
Let the marks obtained by the student in Mathematics, Physics and Chemistry be M, P and C respectively.
Given , M + C = 60 and C – P = 20 M + C / 2 = [(M + P) + (C – P)] / 2 = (60 + 20) / 2 = 40.
Find the one which does not belong to that group ?
K^{+2}M^{+1}N^{2}L, P^{+2}R^{+1}S^{2}Q, V^{+1}W^{+2}Y^{+1}Z, J^{+2}L^{+1}M^{2}K and W^{+2}Y^{+1}Z^{2}X.
Except VWYZ, all the other groups follow similar pattern.
K^{+2}M^{+1}N^{2}L, P^{+2}R^{+1}S^{2}Q, V^{+1}W^{+2}Y^{+1}Z, J^{+2}L^{+1}M^{2}K and W^{+2}Y^{+1}Z^{2}X.
Except VWYZ, all the other groups follow similar pattern.
Find the one which does not belong to that group ?
Y^{+4}C^{2}A^{+4}E^{2}C, K^{+4}O^{2}M^{+4}Q^{2}O, P^{+4}T^{2}R^{+3}U^{1}T, G^{+4}K^{2}I^{+4}M^{2}K and D^{+4}H^{2}F^{+4}J^{2}H.
Except PTRUT, all other groups follow similar pattern.
Y^{+4}C^{2}A^{+4}E^{2}C, K^{+4}O^{2}M^{+4}Q^{2}O, P^{+4}T^{2}R^{+3}U^{1}T, G^{+4}K^{2}I^{+4}M^{2}K and D^{+4}H^{2}F^{+4}J^{2}H.
Except PTRUT, all other groups follow similar pattern.
A sum amount to Rs.1344 in two years at simple interest. What will be the compound interest on the same sum at the same rate of interest for the same period?
There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in ²⁰P₂ ways.
²⁰P₂ = 20 * 19 = 380.
The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in ²⁰P₂ ways.
²⁰P₂ = 20 * 19 = 380.
In a covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is?
Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x – 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.
Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x – 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
The weights of three boys are in the ratio 4:5:6. If the sum of the weights of the heaviest and the lightest boy is 45 kg more than the weight of the third boy, what is the weight of the lightest boy?
Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy = 4k = 4 * 9 = 36 kg.
Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy = 4k = 4 * 9 = 36 kg.
Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25(x – 1) km.
Therefore 20x + 25(x – 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25(x – 1) km.
Therefore 20x + 25(x – 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.
Find the one which does not belong to that group ?
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.
A can do a piece of work in 10 days and B can do the same work in 12 days. A and B worked together for 2 days. How many more days are required to complete the remaining work if they work together?
A can do 1/10 of the work in a day.
B can do 1/12 of the work in a 1 day.
Both of them together can do (1/10 + 1/12) part of work in 1 day = (6 + 5)/60 = 11/60
They take 60/11 days to complete the work together.
Given that they already worked for 2 days.
The number of days required to complete remaining work => 60/11 – 2 = 38/11 = 3 (5/11) days.
A can do 1/10 of the work in a day.
B can do 1/12 of the work in a 1 day.
Both of them together can do (1/10 + 1/12) part of work in 1 day = (6 + 5)/60 = 11/60
They take 60/11 days to complete the work together.
Given that they already worked for 2 days.
The number of days required to complete remaining work => 60/11 – 2 = 38/11 = 3 (5/11) days.
The average of first 10 even numbers is?
Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11
Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11
If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same sum at the rate and for the same time?
Sum = (50 * 100) / (2 * 5) = Rs. 500
Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25
C.I. = (551.25 – 500) = Rs. 51.25.
Sum = (50 * 100) / (2 * 5) = Rs. 500
Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25
C.I. = (551.25 – 500) = Rs. 51.25.
(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
Eight years ago, Ajay’s age was 4/3 times that of Vijay. Eight years hence, Ajay’s age will be 6/5 times that of Vijay. What is the present age of Ajay?
Let the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.
A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)
3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8
V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8
=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8
=> 10 – 20/3 = 10/12 A – 9/12 A
=> 10/3 = A/12 => A = 40.
Let the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.
A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)
3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8
V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8
=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8
=> 10 – 20/3 = 10/12 A – 9/12 A
=> 10/3 = A/12 => A = 40.
An article is bought for Rs.600 and sold for Rs.500, find the loss percent?
600 — 100
100 — ? => 16 2/3%
600 — 100
100 — ? => 16 2/3%
Find the roots of the quadratic equation: x^{2} + 2x – 15 = 0?
x^{2} + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = 5.
x^{2} + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = 5.
What is the difference between the largest number and the least number written with the figures 3, 4, 7, 0, 3?
74330 Largest
30347 Smallest
————
43983
74330 Largest
30347 Smallest
————
43983
The greatest number of four digits that have 144 for their HCF is?
144) 9999 (69
9936
——–
63
9999 – 63 = 9936
144) 9999 (69
9936
——–
63
9999 – 63 = 9936
P alone can complete a piece of work in 6 days. Work done by Q alone in one day is equal to onethird of the work done by P alone in one day. In how many days can the work be completed if P and Q work together?
Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.
Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.
35 + 15 * 1.5 = ?
Given Exp. = 35 + 15 * 3/2
= 35 + 45/2 = 35 + 22.5 = 57.5
Given Exp. = 35 + 15 * 3/2
= 35 + 45/2 = 35 + 22.5 = 57.5
36 * 48 ÷ 64 + 36 ÷ 12 = ?
36 * 48 / 64 + 36/12 = 27 + 3 = 30
36 * 48 / 64 + 36/12 = 27 + 3 = 30
A number exceeds by 25 from its 3/8 part. Then the number is?
x – 3/8 x = 25
x = 40
x – 3/8 x = 25
x = 40
12 men complete a work in 9 days. After they have worked for 6 days, 6 more men join them. How many days will they take to complete the remaining work?
1 man’s 1 day work = 1/108
12 men’s 6 day’s work = 1/9 * 6 = 2/3
Remaining work = 1 – 2/3 = 1/3
18 men’s 1 day work = 1/108 * 18 = 1/6
1/6 work is done by them in 1 day.
1/3 work is done by them in 6 * 1/3 = 2 days.
1 man’s 1 day work = 1/108
12 men’s 6 day’s work = 1/9 * 6 = 2/3
Remaining work = 1 – 2/3 = 1/3
18 men’s 1 day work = 1/108 * 18 = 1/6
1/6 work is done by them in 1 day.
1/3 work is done by them in 6 * 1/3 = 2 days.
The average age of a husband and a wife is 23 years when they were married five years ago but now the average age of the husband, wife and child is 20 years(the child was born during the interval). What is the present age of the child?
28 * 2 = 56
20 * 3 = 60
———–
4 years
28 * 2 = 56
20 * 3 = 60
———–
4 years
Solve the equation for x : 19(x + y) + 17 = 19(x + y) – 21
19x + 19y + 17 = 19x + 19y – 21
38x = 38 => x = 1
19x + 19y + 17 = 19x + 19y – 21
38x = 38 => x = 1
Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is .
No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.
No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.
The ratio of two numbers is 2:3 and the sum of their cubes is 945. The difference of number is?
2x 3x
8x cube + 27x cube = 945
35x cube = 945
x cube = 27 => x = 3
2x 3x
8x cube + 27x cube = 945
35x cube = 945
x cube = 27 => x = 3
In a colony of 70 residents, the ratio of the number of men and women is 4 : 3. Among the women, the ratio of the educated to the uneducated is 1 : 4. If the ratio of the number of educated to uneducated persons is 8 : 27, then find the ratio of the number of educated to uneducated men in the colony?
Number of men in the colony = 4/7 * 70 = 40.
Number of women in the colony = 3/7 * 70 = 40.
Number educated women in the colony = 1/5 * 30 = 6.
Number of uneducated women in the colony = 4/5 * 50 = 24.
Number of educated persons in the colony = 8 /35 * 70 = 16.
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 – 10 = 30.
Number of educated men and uneducated men are in the ratio 10 : 30 i.e., 1 : 3.
Number of men in the colony = 4/7 * 70 = 40.
Number of women in the colony = 3/7 * 70 = 40.
Number educated women in the colony = 1/5 * 30 = 6.
Number of uneducated women in the colony = 4/5 * 50 = 24.
Number of educated persons in the colony = 8 /35 * 70 = 16.
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 – 10 = 30.
Number of educated men and uneducated men are in the ratio 10 : 30 i.e., 1 : 3.
Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132
2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132
25 * 25 / 25 + 15 * 40 = ?
(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.
(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.
Train X crosses a stationary train Y in 60 seconds and a pole in 25 seconds with the same speed. The length of the train X is 300 m. What is the length of the stationary train Y?
Let the length of the stationary train Y be L_{Y}
Given that length of train X, L_{X} = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 —> ( 1 )
(300 + L_{Y}) / V = 60 —> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + L_{Y})/12 = 60
=> 300 + L_{Y} = 60 (12) = 720
=> L_{Y} = 720 – 300 = 420 m
Length of the stationary train = 420 m
Let the length of the stationary train Y be L_{Y}
Given that length of train X, L_{X} = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 —> ( 1 )
(300 + L_{Y}) / V = 60 —> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + L_{Y})/12 = 60
=> 300 + L_{Y} = 60 (12) = 720
=> L_{Y} = 720 – 300 = 420 m
Length of the stationary train = 420 m
Find the one which does not belong to that group ?
Except 110, other numbers are divisible by 4.
Except 110, other numbers are divisible by 4.
The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2} + b^{2} = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l^{2} + b^{2})/lb = 25/12
12(l^{2} + b^{2}) = 25lb
Adding 24lb on both sides
12l^{2} + 12b^{2} + 24lb = 49lb
12(l^{2} + b^{2} + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)^{2} = 49lb
=> 12(14)^{2} = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2} + b^{2} = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l^{2} + b^{2})/lb = 25/12
12(l^{2} + b^{2}) = 25lb
Adding 24lb on both sides
12l^{2} + 12b^{2} + 24lb = 49lb
12(l^{2} + b^{2} + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)^{2} = 49lb
=> 12(14)^{2} = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.
The tens digit of a twodigit number is two more than its unit digit. The twodigit number is 7 times the sum of the digits. Find the units digits?
Let the twodigit number be 10a + b
a = b + 2 — (1)
10a + b = 7(a + b) => a = 2b
Substituting a = 2b in equation (1), we get
2b = b + 2 => b = 2
Hence the units digit is: 2.
Let the twodigit number be 10a + b
a = b + 2 — (1)
10a + b = 7(a + b) => a = 2b
Substituting a = 2b in equation (1), we get
2b = b + 2 => b = 2
Hence the units digit is: 2.
In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?
The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.
The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.
How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
30 * 33/2 = 5/2 * 2 * x => x = 99
30 * 33/2 = 5/2 * 2 * x => x = 99
If 25% of x is 15 less than 15% of 1500, then x is?
25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225
Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840.
25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225
Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840.
A trader bought a car at 20% discount on its original price. He sold it at a 40% increase on the price he bought it. What percent of profit did he make on the original price?
Original price = 100
CP = 80
S = 80*(140/100) = 112
100 – 112 = 12%
Original price = 100
CP = 80
S = 80*(140/100) = 112
100 – 112 = 12%
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