# LIC ADO Mock Test Series 3, LIC ADO Online Test Series 3, LIC ADO Quiz

## LIC ADO Mock Test Series 3, LIC ADO Online Test 3, LIC ADO Quiz

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LIC ADO Mock Test Series 3, LIC ADO Online Test 3, LIC ADO Quiz. LIC ADO Exam Free Online Quiz 2019, 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** 2019. Now Test your self for LIC ADO Exam by using below quiz…

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- Question 1 of 50
##### 1. Question

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?

CorrectD = 540 + 180 = 720

S = 54 * 5/18 = 15 mps

T = 720/15 = 48 secIncorrectD = 540 + 180 = 720

S = 54 * 5/18 = 15 mps

T = 720/15 = 48 sec - Question 2 of 50
##### 2. Question

If 12 men do a work in 80 days, in how many days will 16 men do it?

Correct12 * 80 = 16 * x

x = 60 daysIncorrect12 * 80 = 16 * x

x = 60 days - Question 3 of 50
##### 3. Question

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?

CorrectIncorrect - Question 4 of 50
##### 4. Question

The sum of four consecutive even numbers is 292. What would be the smallest number?

CorrectLet 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.IncorrectLet 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. - Question 5 of 50
##### 5. Question

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?

CorrectLet 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. 72IncorrectLet 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 - Question 6 of 50
##### 6. Question

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

CorrectArea 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 mIncorrectArea 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 - Question 7 of 50
##### 7. Question

What is the sum of all the composite numbers up to 20?

Correct4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 132

Incorrect4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 132

- Question 8 of 50
##### 8. Question

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 one-fourth of his income, find the ratio of their monthly savings?

CorrectLet 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.IncorrectLet 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. - Question 9 of 50
##### 9. Question

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

CorrectL.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.IncorrectL.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. - Question 10 of 50
##### 10. Question

64309 – 8703 + 798 – 437 = ?

Correct64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967

Incorrect64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967

- Question 11 of 50
##### 11. Question

How many figures are required for numbering the pages of a book containing 1000 pages?

Correct1 to 9 = 9 * 1 = 9

10 to 99 = 90 * 2 = 180

100 to 999 = 900 * 3 =2700

1000 = 4

———–

2893Incorrect1 to 9 = 9 * 1 = 9

10 to 99 = 90 * 2 = 180

100 to 999 = 900 * 3 =2700

1000 = 4

———–

2893 - Question 12 of 50
##### 12. Question

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.

CorrectLet 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.IncorrectLet 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. - Question 13 of 50
##### 13. Question

Find the one which does not belong to that group ?

CorrectK

^{+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.IncorrectK

^{+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. - Question 14 of 50
##### 14. Question

Find the one which does not belong to that group ?

CorrectY

^{+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.IncorrectY

^{+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. - Question 15 of 50
##### 15. Question

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?

CorrectIncorrect - Question 16 of 50
##### 16. Question

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?

CorrectThe 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.IncorrectThe 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. - Question 17 of 50
##### 17. Question

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?

CorrectRatio 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.IncorrectRatio 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. - Question 18 of 50
##### 18. Question

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?

CorrectMilk = 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.IncorrectMilk = 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. - Question 19 of 50
##### 19. Question

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?

CorrectLet 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.IncorrectLet 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. - Question 20 of 50
##### 20. Question

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?

CorrectSuppose 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.IncorrectSuppose 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. - Question 21 of 50
##### 21. Question

Find the one which does not belong to that group ?

Correct20 = 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.Incorrect20 = 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. - Question 22 of 50
##### 22. Question

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?

CorrectA 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.IncorrectA 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. - Question 23 of 50
##### 23. Question

The average of first 10 even numbers is?

CorrectSum of 10 even numbers = 10 * 11 = 110

Average = 110/10 = 11IncorrectSum of 10 even numbers = 10 * 11 = 110

Average = 110/10 = 11 - Question 24 of 50
##### 24. Question

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?

CorrectSum = (50 * 100) / (2 * 5) = Rs. 500

Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25

C.I. = (551.25 – 500) = Rs. 51.25.IncorrectSum = (50 * 100) / (2 * 5) = Rs. 500

Amount = [500 * (1 + 5/100)^{2}] = Rs. 551.25

C.I. = (551.25 – 500) = Rs. 51.25. - Question 25 of 50
##### 25. Question

(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?

CorrectAs 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 = 0.1IncorrectAs 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 - Question 26 of 50
##### 26. Question

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?

CorrectLet 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.IncorrectLet 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. - Question 27 of 50
##### 27. Question

An article is bought for Rs.600 and sold for Rs.500, find the loss percent?

Correct600 —- 100

100 —- ? => 16 2/3%Incorrect600 —- 100

100 —- ? => 16 2/3% - Question 28 of 50
##### 28. Question

Find the roots of the quadratic equation: x

^{2}+ 2x – 15 = 0?Correctx

^{2}+ 5x – 3x – 15 = 0

x(x + 5) – 3(x + 5) = 0

(x – 3)(x + 5) = 0

=> x = 3 or x = -5.Incorrectx

^{2}+ 5x – 3x – 15 = 0

x(x + 5) – 3(x + 5) = 0

(x – 3)(x + 5) = 0

=> x = 3 or x = -5. - Question 29 of 50
##### 29. Question

What is the difference between the largest number and the least number written with the figures 3, 4, 7, 0, 3?

Correct74330 Largest

30347 Smallest

————-

43983Incorrect74330 Largest

30347 Smallest

————-

43983 - Question 30 of 50
##### 30. Question

The greatest number of four digits that have 144 for their HCF is?

Correct144) 9999 (69

9936

——–

63

9999 – 63 = 9936Incorrect144) 9999 (69

9936

——–

63

9999 – 63 = 9936 - Question 31 of 50
##### 31. Question

P alone can complete a piece of work in 6 days. Work done by Q alone in one day is equal to one-third 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?

CorrectWork 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.IncorrectWork 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. - Question 32 of 50
##### 32. Question

35 + 15 * 1.5 = ?

CorrectGiven Exp. = 35 + 15 * 3/2

= 35 + 45/2 = 35 + 22.5 = 57.5IncorrectGiven Exp. = 35 + 15 * 3/2

= 35 + 45/2 = 35 + 22.5 = 57.5 - Question 33 of 50
##### 33. Question

36 * 48 ÷ 64 + 36 ÷ 12 = ?

Correct36 * 48 / 64 + 36/12 = 27 + 3 = 30

Incorrect36 * 48 / 64 + 36/12 = 27 + 3 = 30

- Question 34 of 50
##### 34. Question

A number exceeds by 25 from its 3/8 part. Then the number is?

Correctx – 3/8 x = 25

x = 40Incorrectx – 3/8 x = 25

x = 40 - Question 35 of 50
##### 35. Question

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?

Correct1 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.Incorrect1 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. - Question 36 of 50
##### 36. Question

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?

Correct28 * 2 = 56

20 * 3 = 60

———–

4 yearsIncorrect28 * 2 = 56

20 * 3 = 60

———–

4 years - Question 37 of 50
##### 37. Question

Solve the equation for x : 19(x + y) + 17 = 19(-x + y) – 21

Correct19x + 19y + 17 = -19x + 19y – 21

38x = -38 => x = -1Incorrect19x + 19y + 17 = -19x + 19y – 21

38x = -38 => x = -1 - Question 38 of 50
##### 38. Question

Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is -.

CorrectNo 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.IncorrectNo 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. - Question 39 of 50
##### 39. Question

The ratio of two numbers is 2:3 and the sum of their cubes is 945. The difference of number is?

Correct2x 3x

8x cube + 27x cube = 945

35x cube = 945

x cube = 27 => x = 3Incorrect2x 3x

8x cube + 27x cube = 945

35x cube = 945

x cube = 27 => x = 3 - Question 40 of 50
##### 40. Question

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?

CorrectNumber 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.IncorrectNumber 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. - Question 41 of 50
##### 41. Question

Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?

Correct2 * 22/7 * 14 = 88

88 * 1 1/2 = Rs.132Incorrect2 * 22/7 * 14 = 88

88 * 1 1/2 = Rs.132 - Question 42 of 50
##### 42. Question

25 * 25 / 25 + 15 * 40 = ?

Correct(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.

Incorrect(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.

- Question 43 of 50
##### 43. Question

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?

CorrectLet 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 mIncorrectLet 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 - Question 44 of 50
##### 44. Question

Find the one which does not belong to that group ?

CorrectExcept 110, other numbers are divisible by 4.

IncorrectExcept 110, other numbers are divisible by 4.

- Question 45 of 50
##### 45. Question

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?

CorrectLet 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.IncorrectLet 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. - Question 46 of 50
##### 46. Question

The tens digit of a two-digit number is two more than its unit digit. The two-digit number is 7 times the sum of the digits. Find the units digits?

CorrectLet the two-digit 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.IncorrectLet the two-digit 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. - Question 47 of 50
##### 47. Question

In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?

CorrectThe 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.IncorrectThe 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. - Question 48 of 50
##### 48. Question

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?

Correct30 * 33/2 = 5/2 * 2 * x => x = 99

Incorrect30 * 33/2 = 5/2 * 2 * x => x = 99

- Question 49 of 50
##### 49. Question

If 25% of x is 15 less than 15% of 1500, then x is?

Correct25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225

Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840.Incorrect25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225

Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840. - Question 50 of 50
##### 50. Question

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?

CorrectOriginal price = 100

CP = 80

S = 80*(140/100) = 112

100 – 112 = 12%IncorrectOriginal price = 100

CP = 80

S = 80*(140/100) = 112

100 – 112 = 12%