## Global Edge Software Aptitude Interview Questions & Answers

1. Question 1. A 600m Long Train Is Running At 73 Kmph. How Much Time Train Will Take To Cross An Electric Pole?

Formula Used: Time = ( Distance / Speed)

As all the option given in sec., so convert the train speed (Kmph) in to mps multiply by 5/18

speed (mps) = 73 * 5/18

Time = 600 / (73 * 5/18)

= (600 * 18 )/(73 * 5) sec

= (10800 / 365)

Time take by Train = 29.58Sec.

2. Question 2. A 120 M Long Train Is Running At 72 Kmph. How Much Time Will It Take To Cross A Man Standing On The Platform?

Formula Used: Time = ( Distance / Speed)

As all the option given in sec., so convert the train speed (Kmph) in to mps multiply by 5/18

speed (mps) = 72 * 5/18 = 20 mps

Time = (120 / 20) sec = 6 sec

Time take by Train = 6 Sec.

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4. Question 3. Two Trains 400m And 300m Long Run At The Speeds Of 50 Kmph And 40kmph Respectively In Opposite Directions On Parallel Tracks. The Time Taken To Cross Each Other?

Trains are running in opposite Direction:

So need to find Length of two Trains = 300m + 400m = 700m

and Total Speed = 40 Kmph + 50 Kmph (Opposite Direction)

= 90 Kmph

so speed (m/sec) = 90 * 5/18 m/sec = 25 m/sec

Formula Used: Time = Distance/Speed

Time = 700/ 25 sec

Time = 28 Sec.

5. Question 4. 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?

Let 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.

So Total Distance

=> 20x + 25(x – 1) = 110

=> 45x – 25 = 110 => 45x = 135

=> x = 3.

As They meet x hrs after 7 a.m. so they meet at 10 a.m.

6. Question 5. Two Trains Are Running At 40 Km/hr And 20 Km/hr Respectively In The Same Direction. Fast Train Completely Passes A Man Sitting In The Slower Train In 5 Seconds. What Is The Length Of The Fast Train?

As train are running in same direction

so Relative speed = (40 – 20) km/hr = 20 km/hr

= ( 20 x 5/18 ) m/sec = 50/9 m/sec.

Formula Used: Distance = Speed * Time

Now Length of Faster Train = ( 50/9 x 5 ) m = 250/9 m

= 27 7/9 m.

7. Wipro Aptitude Interview Questions

8. Question 6. A Train Moves Past A Telegraph Post And A Bridge 264 M Long In 8 Seconds And 20 Seconds Respectively. What Is The Speed Of The Train?

Let the length of the train be x meters and its speed by y m/sec.

Then, x/y = 8 => x = 8y ———– (1)

As per the question total distance = (x + 264) meters.

(x + 264)/20 = y

Put the value of x from equation 1.

=> 8y + 264 = 20y

=> y = 22.

Therefore Speed = 22 m/sec = ( 22 x 18 /5) km/hr = 79.2 km/hr.

9. Question 7. An Athlete Runs 200 Metres Race In 24 Seconds. His Speed Is?

Speed = (200/24) m/sec = 25/3 m/sec

convert m/sec to km/hr

(25/3 * 18/5) km/hr = 30 km/hr.

10. ABB Group Aptitude Interview Questions

11. Question 8. Albert Is Travelling On His Cycle And Has Calculated To Reach Point A At 2 P.m. If He Travels At 10 Kmph, He Will Reach There At 12 Noon If He Travels At 15 Kmph. At What Speed Must He Travel To Reach A At 1 P.m.?

Let’s Assume the distance travelled Albert = x km.

Formula Used: Time = Distance/Speed

so (x/10 – x/15) = 2 hrs

=> (3x – 2x)/30 = 2 hrs.

=> (3x – 2x) = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60/10 hrs = 6 hrs.

Formula Used: Speed = Distance/Time

So, Albert started 6 hours before 2 P.M. i.e @ 8 A.M.

Required speed = 60/5 kmph. = 12 kmph.

12. Question 9. A Person Travelled By Train For 1 Hour At A Speed Of 50 Kmph. He Then Travelled By A Taxi For 30 Minutes At A Speed Of 32 Kmph To Complete His Journey. What Is The Average Speed At Which He Travelled During The Journey?

Total distance travelled in 1 hour

= 50 + (&frac; x 32) km

= 66 km

? Average speed = Total distance/Total time

= 66/(3/2) = 44 kmph.

13. Value Labs Aptitude Interview Questions

14. Question 10. A Car Covers Four Successive 6 Km Stretches At Speeds Of 25 Kmph, 50 Kmph, 75 Kmph And 150 Kmph Respectively. Its Average Speed Over This Distance Is?

Time = Distance/Speed

Time taken for each 6 km can be given by

6/25, 6/50, 6/75 and 6/150

Total time = (6/25) + (6/50) + (6/75) + (6/150) = (36 + 18 + 12 + 6)/150 = 72/150

Average speed = Distance/time = (24/72) x 150 = 50 kmph.

15. Question 11. A’ And ‘b’ Complete A Work Togather In 8 Days.if ‘a’ Alone Can Do It In 12 Days.then How Many Day ‘b’ Will Take To Complete The Work?

A & B one day work = 1/8

A alone one day work = 1/12

B alone one day work = (1/8 – 1/12) = ( 3/24 – 2/24)

=> B one day work = 1/24

so B can complete the work in 24 days.

16. Abaxis Aptitude Interview Questions

17. Question 12. If A Alone Can Do A Piece Of Work In 8 Days And B Alone Can Do The Same Work In 12 Days. How Many Days A And B Required To Finish The Same Work If They Work Togather?

A alone one day work = 1/8

B alone one day work = 1/12

Both A and B one day work = (1/8 + 1/12) = (3/24 + 2/24)

= 5/24

so A and B together finish the work in 24/5 day

or 4 4/5 days.

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19. Question 13. A Can Finish A Piece Work In 18 Days And B Can Do The Same Work In Half The Time Taken By A. So If They Working Together, What Part Of The Same Work Can Finished In A Day?

First find the 1 day work of both (A & B)

A’s 1 day’s work = 1/18

and

B’s 1 day’s work = 1/9 (B can do work in half time)

(A + B)’s 1 day’s work = (1/18+1/9)

= (1+2)/18 = 3/18 = 1/6

so A & B together can do 1/6 of work in 1 day.

20. Question 14. A And B Can Together Finish A Work In 30 Days. They Worked Together For 20 Days And Then B Left. After Another 20 Days, A Finished The Remaining Work. In How Many Days A Alone Can Finish The Job ?

(A + B)’s 1 day’s work = 1/30

so (A&B) 20 days work = (20*1/30) = 2/3

so left work = (1?2/3)=1/3

1/3 work is done by A = 20 days.

So whole work will be done by A = (20 x 3) = 60 days.

21. Question 15. A And B Together Can Do A Piece Of Work In 30 Days. A Having Worked For 16 Days, B Finishes The Remaining Work Alone In 44 Days. In How Many Days Shall B Finish The Whole Work Alone ?

A’s 1 day’s work = x

and B’s 1 day’s work = y

So (A & B) 1 day work = 1/30 => x+y =1/30

=> 30x + 30y = 1 ——– (1)

So 16x + 44y = 1 ——– (2)

By Solving above two equations,

x = 1/60 and y = 1/60

B’s 1 day’s work = 1/60

Hence, B alone shall finish the whole work in 60 days.

22. Oracle Aptitude Interview Questions

23. Question 16. Vikas Can Cover A Distance In 1hr 24min By Covering 2/3 Of The Distance At 4 Kmph And The Rest At 5kmph.the Total Distance Is?

Let total distance be S

total time=1hr24min

A to T :: speed=4kmph

diistance=2/3S

T to S :: speed=5km

distance=1-2/3S=1/3S

21/15 hr=2/3 S/4 + 1/3s /5

84=14/3S*3

S=84*3/14*3

= 6km.

24. Question 17. A Cistern Can Be Filled By A Tap In 4 Hours While It Can Be Emptied By Another Tap In 9 Hours. If Both Taps Are Opened Simultaneously, Then After How Much Time Will The Cistern Get Filled ?

Time taken by tap A to fill the cistern=4 hrs

so work done by tap A in 1 hour = 1/4th

Time taken by tap B to empty the full cistern = 9 hours

so work done by tap B in 1 hour = 1/9th

=> Work done by (A + B) in 1 hour=(1/4 – 1/9)=5/36

Therefore, the tank will fill the cistern = 36/5 hours=7.2 hours.

25. TCS Aptitude Interview Questions

26. Question 18. In A Class, There Are 15 Boys And 10 Girls. Three Students Are Selected At Random. The Probability That 1 Girl And 2 Boys Are Selected, Is?

Let’s assume the sample space = S

and Event of selecting 1 girl and 2 boys = E

So, n(S) = Number ways of selecting 3 students out of 25 = 25C3

=> (25 * 24 * 23)/(3 * 2 * 1) = 2300.

n(E) = (10C1* 15C2)

= 10 * [(15 * 14)/(2 * 1)] = 1050.

P(E) = n(E)/n(S) = 1050/2300 = 21/46.

27. Wipro Aptitude Interview Questions

28. Question 19. A Man Swims Downstream 72 Km And Upstream 45 Km Taking 9 Hours Each Time. What Is The Speed Of The Current?

Man’s downstream speed = 72/ 9 kmph => 8kmph

Man’s up stream speed = 45/ 9 => 5 kmph

So speed of current = (8 – 5)/2 = 1.5 kmph.

29. Question 20. A Boat Can Row Upstream At 25 Kmph And Downstream At 35 Kmph, Then The Speed Of The Current Is?

man’s upstream speed = 25 kmph

Man’s downstream speed = 35 kmph

so Speed of current = (35 – 25)/2 = 5 kmph.

30. Infosys Aptitude Interview Questions

31. Question 21. A Man Can Row His Boat With The Stream At 6 Km/h And Against The Stream In 4 Km/h. The Man’s Rate Is?

Man’s row in downstream by speed = 6 kmph

and upstream by speed = 4 kmph

so man rate = (6 – 4)/2 = 1 kmph.

32. Question 22. Two Pipes A And B Can Fill A Tank In 9 Hours And 3 Hours Respectively. If They Are Opened On Alternate Hours And If Pipe A Is Opened First, In How Many Hours Will The Tank Be Full?

Tank part filled by pipe A in 1 hour =1/9

Tank part filled by pipe B in 1 hour =1/3

Given Pipe A and B are opened alternatively.

So Part filled in every 2 hours =(1/9+1/3)=4/9

Tank Part will be filled in 4 hour =2*4/9=8/9

Remaining part = (1-8/9)=1/9

So next is A turn.

So Pipe A will fill remaining 1/9 part in next 1 hour.

Total Time = (4 hrs + 1 hrs) = 5 hrs.

33. Question 23. Pipe A Can Fill A Cistern In 6 Hours Less Than Pipe B. Both The Pipes Together Can Fill The Cistern In 4 Hours. How Much Time Would A Take To Fill The Cistern All By Itself?

Let’s assume time required by Pipe A to fill the cistern = X hours

So Time required by Pipe B to fill the cistern = (X + 6) hours

? Both Pipes (A+B) can fill cistern in 1 hour = [1/X + 1/(X + 6)]

Given Both pipe fill the cistern in 4 hours

=> [1/X + 1/(X + 6)] = 1/4 => [(X+6) + X]/(X+6)*x = 1/4

4X + 24 + 4X = X2 + 6x

X2 – 2X – 24 = 0

(X-6)(X+4) = 0

=> A can fill cistern in 6 hours.

34. IBM Aptitude Interview Questions

35. Question 24. One Pipe Can Fill A Tank Three Times As Fast As Another Pipe. If Together The Two Pipes Can Fill The Tank In 36 Min, Then The Slower Alone Will Be Able To Fill The Tank In?

Lets assume time required by slower pipe alone to fill the tank = x minutes.

Then, faster pipe will fill it in x/3 minutes.

=> 1/x+3/x = 1/36

=>4/x = 1/36 => x = 144 min.

36. ABB Group Aptitude Interview Questions

37. Question 25. 13 Sheeps And 9 Pigs Were Bought For Rs. 1291.85.if The Average Price Of A Sheep Be Rs. 74. What Is The Average Price Of A Pig?

Average price of a sheep = Rs. 74

:Total price of 13 sheeps = (74*13) = Rs. 962

But, total price of 13 sheeps and 9 pigs

= Rs. 1291.85

Total price of 9 pigs

= Rs. (1291.85-962) = Rs. 329.85

Hence, average price of a pig

= (329.85/9) = Rs. 36.65.

38. Question 26. Three Pipes A, B And C Can Fill A Tank In 6 Hours. After Working At It Together For 2 Hours, C Is Closed And A And B Can Fill The Remaining Part In 7 Hours. The Number Of Hours Taken By C Alone To Fill The Tank Is?

Tank part filled by pipes (A+B+C) in 1 hrs = 1/6 ——- (1)

so tank part filled by (A+B+C) in 2 hrs = 2*1/6 = 1/3

Now find the remaining part = (1-1/3) = 2/3

=> (A+B) 7 hs work = 2/3

so (A+B) 1 hrs work = 2/21 —— (2)

To find the C 1 hrs work use eq. 1 & 2

=> 1/6-2/21 = 1/14

so C alone can fill the tank in 14 hrs.

39. Capgemini Aptitude Interview Questions

40. Question 27. A Tap Can Fill A Tank In 6 Hours. After Half The Tank Is Filled, Three More Similar Taps Are Opened. What Is The Total Time Taken To Fill The Tank Completely?

Time taken by one tap to fill half tank = 3hrs.

Part filled by the four taps in 1 hrs = (4*1/6) = 2/3.

Remaining part = (1-1/2) = 1/2.

so 2/3 : 1/2 :: 1:x => x = 1/2*1*3/2) = 3/4 hrs. => 45 min.

=> Total time taken = 3 hrs 45 min.

41. Value Labs Aptitude Interview Questions

42. Question 28. If (x + 1/x) = 2, Then The Value Of (x^100 + 1/x^100) Is?

Quick Approach

for x =1 given eq. will be satisfy. (1+1/1)=2

so (x^100 + 1/x^100) = (1^100 + 1/1^100) = 2.

43. Question 29. If (2x + 2/x) = 1, Then The Value Of (x^3 + 1/x^3) Is?

(2x + 2/x) = 1

=> (x + 1/x) = 1/2

(x^3+1/x^3) = (x+1/x)^3-3x*1/x(x+1/x)

= (1/2)^3-1/2 = (1/8 – 3/2) = (1-12)/8= -11/8.

44. MindTree Aptitude Interview Questions

45. Question 30. 7 Years Ago, The Ages (in Years) Of A And B Were In The Ratio 4:5 And 7 Years Hence They Will Be In The Ratio 5:6. The Present Age Of B Is?

7 years ago, A’s age=4x years

and B’s age=5xyears

so (4x+14)/(5x+l4)=5/6

=> 25x + 70 = 24x + 84

x = (84 – 70) = 14

B’s present age = 5x + 7 = 5*14 + 7 = 77 years.

46. Question 31. The Sum Of Ages Of Family Members (both Children And Parents) Is 360 Years.the Total Ages Of Children And Parents Are In The Ratio 2:1 And The Ages Of Wife And Husband Are In The Ratio 5:7.what Will Be The Age Of Husband?

Given sum of ages is 360 years.

The ratio of children and parents ages is 2:1.

So total age of parents = 360 x 1 / 3 = 120 years

Given Ratio of wife and husband age is 5:7.

So the age of husband = 120 x 7

47. Question 32. The Sum Of The Ages Of A Mother And Her Son Is 45 Years. Five Years Ago, The Product Of Their Ages Was 3 Times The Mother Age At That Time, Then The Present Age Of The Son?

Let’s assume mother age = x years. —– (1)

sum of mother and her son age = 45

so son age will be = (45-x) years. ——- (2)

Five year ago:

mother age will be = (x-5) years

son age will be = (45-x-x) years = (40-x) year.

As per question

(x-5) * (40-x) = 3*(x-5)

=> (40-x) = 3

=> x = 37 year.

so son age will be (45-37) = 8 years.

48. EXL Service Aptitude Interview Questions

49. Question 33. Father Is Aged Three Times More Than His Son Mohit. After 8 Years, He Would Be Two And A Half Times Of Mohit’s Age. After Further 8 Years, How Many Times Would He Be Of Mohit’s Age?

Let’s assume Monit’s present age = X years.

So father’s present age = (X + 3X) years = 4X years.

After 8 years.

(4X + 8) =5/2 * (X + 8)

=> 8X + 16 = 5X + 40

=> 3X = 24 so, X=8

Hence, required ratio = (4X + 16) / (X + 16) = 48/24 = 2.

50. Abaxis Aptitude Interview Questions

51. Question 34. If X^3 + 3x^2 + 3x = 7, Then X Is Equal To?

x^3 + 3x^2 + 3x = 7

=> x^3 + 3x^2 + 3x + 1= 7 +1

=> (x+1)^3=2^3

=> x+1 = 2 => x =1.

52. Question 35. A Man Owns 2/3 Of The Market Research Beauro Business And Sells 3/4 Of His Shares For Rs. 75000. What Is The Value Of Business?

3/4 of his share = 75000

so his share = 100000.

2/3 of business value = 100000

so total value = 150000.

53. Question 36. From Its Total Income, A Sales Company Spent Rs.20,000 For Advertising, Half Of The Remainder On Commissions And Had Rs.6000 Left. What Was Its Total Income?

Let total income is X

X=20,000+(X-20,000/2)+6000

X-X/2=20,000-10,000+6000

X/2=16,000

X=32,000.

54. Oracle Aptitude Interview Questions

55. Question 37. The Sum Of Three Numbers Is 98. If The Ratio Of The First To Second Is 2 :3 And That Of The Second To The Third Is 5 : 8, Then The Second Number Is?

Let the three parts be 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 x 15/49 = 30.

56. Question 38. A Sum Of Money Becomes 2.5 Times Itself At 12.5% Simple Interest P.a. The Period Of Investment Is?

Let the period is ‘T’ and Sum= ‘P’.

As given money become 2.5.

=> 2.5 * P = P + S.I

=> S.I = 1.5 * P ————– (1)

=> S.I = (P * T * 12.5)/100 ———–(2)

By eq. (1) and (2)

=> 1.5 * P = (P * T * 12.5)/100

=> T=12 years.

57. Question 39. Three Partners A, B, C Start A Business. Twice A’s Capital Is Equal To Thrice B’s Capital And B’s Capital Is Four Times C’s Capital. Out Of A Total Profit Of Rs. 16,500 At The End Of The Year, B’s Share Is?

Let C capital = x. so B capital = 4x

2 *(A capital) = 3 * 4x

=> A’s capital = 6x

So A : B : C = 6x : 4x : x = 6 : 4 : 1

So, B’s capital = Rs. [16500 * 4/11] = Rs. 6000.

58. Question 40. If A Man Walks At The Rate Of 5kmph, He Misses A Train By Only 7min. However If He Walks At The Rate Of 6 Kmph He Reaches The Station 5 Minutes Before The Arrival Of The Train.find The Distance Covered By Him To Reach The Station?

Lets assume the required distance = x km.

Difference in the times taken at two speeds=12mins=1/5 hr.

Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.

So required distance = 6 km.

59. TCS Aptitude Interview Questions

60. Question 41. A Batsman In His 18th Innings Makes A Score Of 150 Runs And There By Increasing His Average By 6. Find His Average After 18th Innings?

Let the average for 17 innings is x runs

Total runs in 17 innings = 17x

Total runs in 18 innings = 17x + 150

Average of 18 innings = 17x + 150/18

17x + 150/18 = x + 6 — > x = 42

Thus, average after 18 innings = 42.

61. Question 42. Sum Of Squares Of Two Numbers Is 2754, Their Hcf Is 9, Lcm Is 135, Find The Numbers?

Product of two no. = H.C.F*L.C.M

So,x*y=135*9=1215 —–(1)

and x^2+y^2=2754

So,(x+y)^2=x^2+y^2+2*x*y = 2754+2*1215=5184

So,x+y=72 ———– (2)

By solving eq. (1) & (2)

nos. are 45 and 27.

62. Infosys Aptitude Interview Questions

63. Question 43. Find The Largest 4-digit Number, Which Gives The Remainder 7 And 13 When Divided By 11 And 17?

LCM of 11 and 17 = 187.

When divided by 11 remainder 7,so difference 4.

When devided by 17 remainder 13,so difference 4.

Largest no exactly devide by 11 & 17=9911

The no’s = 9911-4 = 9907.

64. Question 44. What Comes Next In The Sequence?
4, 2, 5, 9, 5, 11, 13, 7, 16, 17, 9