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

Answer :

Formula Used: Time = ( Distance / Speed)

As all the choice given in sec., so convert the educate pace (Kmph) in to mps multiply by using 5/18

pace (mps) = seventy three * 5/18

Time = 600 / (73 * five/18)

= (six hundred * 18 )/(seventy three * five) sec

= (10800 / 365)

Time take by Train = 29.58Sec.

Question 2. A a hundred and twenty M Long Train Is Running At seventy two Kmph. How Much Time Will It Take To Cross A Man Standing On The Platform?

Answer :

Formula Used: Time = ( Distance / Speed)

As all the choice given in sec., so convert the teach pace (Kmph) in to mps multiply with the aid of 5/18

pace (mps) = 72 * five/18 = 20 mps

Time = (120 / 20) sec = 6 sec

Time take with the aid of Train = 6 Sec.

Aptitude Interview Questions

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?

Answer :

Trains are jogging in contrary Direction:

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

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

= 90 Kmph

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

Formula Used: Time = Distance/Speed

Time = 700/ 25 sec

Time = 28 Sec.

Question four. 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?

Answer :

Let they meet x hours after 7 a.M.

Distance included by way of A in x hours = 20x km.

Distance protected by means of B in (x - 1) hours = 25(x - 1) km.

So Total Distance

=> 20x + 25(x - 1) = one hundred ten

=> 45x - 25 = a hundred and ten => 45x = 135

=> x = three.

As They meet x hrs after 7 a.M. So they meet at 10 a.M.

Question 5. Two Trains Are Running At forty 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?

Answer :

As educate are walking in same route

so Relative velocity = (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/nine x five ) m = 250/9 m

= 27 7/9 m.

Wipro Aptitude Interview Questions

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

Answer :

Let the duration of the teach be x meters and its velocity by way of y m/sec.

Then, x/y = eight => x = 8y ----------- (1)

As according to the query overall 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.

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

Answer :

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

convert m/sec to km/hr

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

ABB Group Aptitude Interview Questions

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.?

Answer :

Let's Assume the space 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 tour 60 km at 10 km/hr = 60/10 hrs = 6 hrs.

Formula Used: Speed = Distance/Time

So, Albert began 6 hours before 2 P.M. I.E @ eight A.M.

Required velocity = 60/five kmph. = 12 kmph.

Question nine. 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?

Answer :

Total distance travelled in 1 hour

= 50 + (&frac; x 32) km

= sixty six km

? Average velocity = Total distance/Total time

= sixty six/(3/2) = 44 kmph.

Value Labs Aptitude Interview Questions

Question 10. A Car Covers Four Successive 6 Km Stretches At Speeds Of 25 Kmph, 50 Kmph, seventy five Kmph And a hundred and fifty Kmph Respectively. Its Average Speed Over This Distance Is?

Answer :

Time = Distance/Speed

Time taken for every 6 km can be given by using

6/25, 6/50, 6/75 and 6/one hundred fifty

Total time = (6/25) + (6/50) + (6/seventy five) + (6/a hundred and fifty) = (36 + 18 + 12 + 6)/150 = seventy two/a hundred and fifty

Average speed = Distance/time = (24/72) x a hundred and fifty = 50 kmph.

Question eleven. A' And 'b' Complete A Work Togather In eight Days.If 'a' Alone Can Do It In 12 Days.Then How Many Day 'b' Will Take To Complete The Work?

Answer :

A & B in the future paintings = 1/8

A alone in the future paintings = 1/12

B by myself at some point work = (1/eight - 1/12) = ( 3/24 - 2/24)

=> B sooner or later work = 1/24

so B can entire the work in 24 days.

Abaxis Aptitude Interview Questions

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?

Answer :

A on my own at some point work = 1/8

B on my own at some point paintings = 1/12

Both A and B sooner or later paintings = (1/eight + 1/12) = (three/24 + 2/24)

= five/24

so A and B collectively finish the paintings in 24/five day

or 4 4/five days.

Aptitude Interview Questions

Question thirteen. 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?

Answer :

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

A's 1 day's paintings = 1/18

and

B's 1 day's paintings = 1/9 (B can do paintings in 1/2 time)

(A + B)'s 1 day's paintings = (1/18+1/nine)

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

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

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 ?

Answer :

(A + B)'s 1 day's work = 1/30

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

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

1/3 work is accomplished by means of A = 20 days.

So entire paintings will be carried out through A = (20 x 3) = 60 days.

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

Answer :

A's 1 day's paintings = x

and B's 1 day's paintings = y

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

=> 30x + 30y = 1 -------- (1)

So 16x + 44y = 1 -------- (2)

By Solving above equations,

x = 1/60 and y = 1/60

B's 1 day's work = 1/60

Hence, B on my own shall end the entire paintings in 60 days.

Oracle Aptitude Interview Questions

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

Answer :

Let overall distance be S

overall time=1hr24min

A to T :: speed=4kmph

diistance=2/3S

T to S :: velocity=5km

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

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

84=14/3S*three

S=eighty four*three/14*three

= 6km.

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

Answer :

Time taken by using faucet A to fill the cistern=4 hrs

so work performed by means of faucet A in 1 hour = 1/4th

Time taken through tap B to drain the whole cistern = 9 hours

so paintings performed with the aid of tap B in 1 hour = 1/9th

=> Work done by way of (A + B) in 1 hour=(1/four - 1/nine)=five/36

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

TCS Aptitude Interview Questions

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?

Answer :

Let's anticipate the pattern area = S

and Event of choosing 1 woman and a pair of boys = E

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

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

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

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

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

Wipro Aptitude Interview Questions

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

Answer :

Man's downstream velocity = seventy two/ 9 kmph => 8kmph

Man's up flow speed = forty five/ 9 => five kmph

So pace of current = (8 - five)/2 = 1.Five kmph.

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

Answer :

man's upstream pace = 25 kmph

Man's downstream pace = 35 kmph

so Speed of present day = (35 - 25)/2 = 5 kmph.

Infosys Aptitude Interview Questions

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?

Answer :

Man's row in downstream with the aid of pace = 6 kmph

and upstream by using velocity = 4 kmph

so man fee = (6 - four)/2 = 1 kmph.

Question 22. Two Pipes A And B Can Fill A Tank In 9 Hours And three 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?

Answer :

Tank part crammed through pipe A in 1 hour =1/9

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

Given Pipe A and B are opened instead.

So Part stuffed in every 2 hours =(1/9+1/three)=4/9

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

Remaining component = (1-8/nine)=1/9

So subsequent is A flip.

So Pipe A will fill last 1/9 element in subsequent 1 hour.

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

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

Answer :

Let's expect time required with the aid of Pipe A to fill the cistern = X hours

So Time required through 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/four => [(X+6) + X]/(X+6)*x = 1/four

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

X2 - 2X - 24 = 0

(X-6)(X+four) = zero

=> A can fill cistern in 6 hours.

IBM Aptitude Interview Questions

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?

Answer :

Lets anticipate time required by using slower pipe by myself to fill the tank = x minutes.

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

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

=>four/x = 1/36 => x = a hundred and forty four min.

ABB Group Aptitude Interview Questions

Question 25. Thirteen Sheeps And 9 Pigs Were Bought For Rs. 1291.Eighty five.If The Average Price Of A Sheep Be Rs. 74. What Is The Average Price Of A Pig?

Answer :

Average fee of a sheep = Rs. Seventy four

:Total charge of thirteen sheeps = (74*thirteen) = Rs. 962

But, total charge of thirteen sheeps and 9 pigs

= Rs. 1291.85

Total price of nine pigs

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

Hence, common fee of a pig

= (329.85/nine) = Rs. 36.Sixty five.

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?

Answer :

Tank element stuffed through pipes (A+B+C) in 1 hrs = 1/6 ------- (1)

so tank part crammed with the aid of (A+B+C) in 2 hrs = 2*1/6 = 1/three

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

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

so (A+B) 1 hrs paintings = 2/21 ------ (2)

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

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

so C by myself can fill the tank in 14 hrs.

Capgemini Aptitude Interview Questions

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?

Answer :

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

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

Remaining component = (1-half) = 1/2.

So 2/three : half :: 1:x => x = 1/2*1*three/2) = three/four hrs. => forty five min.

=> Total time taken = three hrs 45 min.

Value Labs Aptitude Interview Questions

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

Answer :

Quick Approach

for x =1 given eq. Will be fulfill. (1+1/1)=2

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

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

Answer :

(2x + 2/x) = 1

=> (x + 1/x) = half

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

= (half of)^three-half = (1/eight - three/2) = (1-12)/eight= -11/eight.

MindTree Aptitude Interview Questions

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

Answer :

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 = (eighty four - 70) = 14

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

Question 31. The Sum Of Ages Of Family Members (each 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?

Answer :

Given sum of a while is 360 years.

The ratio of kids and parents a long time is 2:1.

So total age of dad and mom = 360 x 1 / 3 = a hundred and twenty years

Given Ratio of spouse and husband age is five:7.

So the age of husband = a hundred and twenty x 7

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

Answer :

Let's anticipate mother age = x years. ----- (1)

sum of mother and her son age = 45

so son age can be = (forty five-x) years. ------- (2)

Five yr in the past:

mom age will be = (x-5) years

son age might be = (45-x-x) years = (forty-x) 12 months.

As in line with query

(x-five) * (forty-x) = 3*(x-five)

=> (forty-x) = 3

=> x = 37 yr.

So son age may be (45-37) = eight years.

EXL Service Aptitude Interview Questions

Question 33. Father Is Aged Three Times More Than His Son Mohit. After eight 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?

Answer :

Let's anticipate Monit's gift age = X years.

So father's gift age = (X + 3X) years = 4X years.

After eight years.

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

=> 8X + 16 = 5X + 40

=> 3X = 24 so, X=8

Hence, required ratio = (4X + sixteen) / (X + 16) = forty eight/24 = 2.

Abaxis Aptitude Interview Questions

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

Answer :

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

Adding each aspect 1

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

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

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

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?

Answer :

three/4 of his share = 75000

so his share = 100000.

2/3 of enterprise value = one hundred thousand

so general value = 150000.

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?

Answer :

Let general profits 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.

Oracle Aptitude Interview Questions

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

Answer :

Let the 3 components be A, B, C. Then,

A : B = 2 : three and B : C = five : eight = five * 3/5 : eight * three/5 = 3 : 24/five

=> A : B : C = 2 : 3 : 24/five = 10 : 15 : 24

=> B = 98 x 15/49 = 30.

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

Answer :

Let the period is 'T' and Sum= 'P'.

As given cash turn out to be 2.Five.

=> 2.Five * P = P + S.I

=> S.I = 1.5 * P -------------- (1)

=> S.I = (P * T * 12.Five)/a hundred -----------(2)

By eq. (1) and (2)

=> 1.Five * P = (P * T * 12.5)/100

=> T=12 years.

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. Sixteen,500 At The End Of The Year, B's Share Is?

Answer :

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.

Question forty. 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.Locate The Distance Covered By Him To Reach The Station?

Answer :

Lets assume the required distance = x km.

Difference inside the instances taken at two speeds=12mins=1/five hr.

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

So required distance = 6 km.

TCS Aptitude Interview Questions

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?

Answer :

Let the common for 17 innings is x runs

Total runs in 17 innings = 17x

Total runs in 18 innings = 17x + a hundred and fifty

Average of 18 innings = 17x + a hundred and fifty/18

17x + a hundred and fifty/18 = x + 6 -- > x = forty two

Thus, common after 18 innings = 42.

Question 42. Sum Of Squares Of Two Numbers Is 2754, Their Hcf Is nine, Lcm Is a hundred thirty five, Find The Numbers?

Answer :

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

So,x*y=a hundred thirty five*nine=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=seventy two ----------- (2)

By fixing eq. (1) & (2)

nos. Are 45 and 27.

Infosys Aptitude Interview Questions

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

Answer :

LCM of 11 and 17 = 187.

When divided by eleven the rest 7,so difference 4.

When devided through 17 the rest 13,so distinction 4.

Largest no precisely devide through 11 & 17=9911

The no's = 9911-4 = 9907.

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

Answer :

Split the series as underneath

four,2,five nine,5,eleven thirteen,7,16 17,9,?Four,2,5 -> diff b/w four,five = 1

nine,5,eleven -> diff b/w nine,11 = 2

thirteen,7,16 -> diff b/w thirteen,sixteen = three

17,9,21 -> diff b/w 17,21 = four.