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Global Edge Software Aptitude Placement Papers - Global Edge Software Aptitude Interview Questions and Answers

Q1. 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 expect the pattern area = S

and Event of choosing 1 woman and 2 boys = E

So, n(S) = Number approaches of selecting three college 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/forty six.

Q2. 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 44 Days. In How Many Days Shall B Finish The Whole Work Alone ?

A's 1 day's paintings = x

and B's 1 day's work = 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 by myself shall finish the whole paintings in 60 days.

Q3. Vikas Can Cover A Distance In 1hr 24min By Covering 2/three Of The Distance At four Kmph And The Rest At 5kmph.The Total Distance Is?

Let general distance be S

overall time=1hr24min

A to T :: velocity=4kmph

diistance=2/3S

T to S :: velocity=5km

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

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

84=14/3S*3

S=84*3/14*3

= 6km.

Q4. Two Stations A And B Are one hundred ten 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 eight A.M. And Travels Towards A At A Speed Of

Let they meet x hours after 7 a.M.

Distance included by using A in x hours = 20x km.

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

So Total Distance

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

=> 45x - 25 = 110 => 45x = 135

=> x = three.

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

Q5. 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 Sha

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

2 *(A capital) = three * 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.

Q6. The Sum Of Three Numbers Is nine

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

A : B = 2 : three and B : C = five : 8 = five * 3/5 : 8 * three/five = 3 : 24/5

=> A : B : C = 2 : 3 : 24/5 = 10 : 15 : 24

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

Q7. Find The Largest 4-digit Number, Which Gives The Remainder 7 And 13 When Divided By eleven And 17?

LCM of 11 and 17 = 187.

When divided by way of eleven remainder 7,so difference 4.

When devided by using 17 the rest 13,so difference four.

Largest no exactly devide through eleven & 17=9911

The no's = 9911-4 = 9907.

Q8. 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?

Let the period of the train be x meters and its velocity with the aid of y m/sec.

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

As in step with the question overall distance = (x + 264) meters.

(x + 264)/20 = y

Put the fee of x from equation 1.

=> 8y + 264 = 20y

=> y = 22.

Therefore Speed = 22 m/sec = ( 22 x 18 /5) km/hr = seventy nine.2 km/hr.

Q9. 13 Sheeps And 9 Pigs Were Bought For Rs. 1291.Eighty five.If The Average Price Of A Sheep Be Rs. 7

Average fee of a sheep = Rs. 74

:Total rate of 13 sheeps = (seventy four*13) = Rs. 962

But, general charge of 13 sheeps and 9 pigs

= Rs. 1291.Eighty five

Total rate of 9 pigs

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

Hence, average price of a pig

= (329.85/9) = Rs. 36.Sixty five.

Q10. If (2x + 2/x) = 1, Then The Value Of (x^3 + 1/x^three) Is?

(2x + 2/x) = 1

=> (x + 1/x) = half

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

= (half)^three-half = (1/eight - 3/2) = (1-12)/8= -eleven/8.

Q11. 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 every 6 km may be given by using

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

Total time = (6/25) + (6/50) + (6/seventy five) + (6/150) = (36 + 18 + 12 + 6)/one hundred fifty = 72/a hundred and fifty

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

Q12. If X^3 + 3x^2 + 3x = 7, Then X Is Equal To?

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

Adding each aspect 1

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

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

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

Q13. 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 Fil

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

so tank element stuffed by (A+B+C) in 2 hrs = 2*1/6 = 1/three

Now locate the closing element = (1-1/3) = 2/3

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

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

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

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

so C on my own can fill the tank in 14 hrs.

Q14. 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 count on mom age = x years. ----- (1)

sum of mom and her son age = 45

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

Five year ago:

mom age could be = (x-5) years

son age could be = (forty five-x-x) years = (forty-x) yr.

As in line with query

(x-five) * (40-x) = three*(x-five)

=> (40-x) = three

=> x = 37 12 months.

So son age can be (45-37) = 8 years.

Q15. A Sum Of Money Becomes 2.Five Times Itself At 12.5% Simple Interest P.A. The Period Of Investment Is?

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

As given cash turn out to be 2.5.

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

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

=> S.I = (P * T * 12.5)/one hundred -----------(2)

By eq. (1) and (2)

=> 1.Five * P = (P * T * 12.Five)/100

=> T=12 years.

Q16. 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 overall income is X

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

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

X/2=sixteen,000

X=32,000.

Q17. 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 paintings = (20*1/30) = 2/three

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

1/three paintings is completed by means of A = 20 days.

So entire work may be carried out through A = (20 x three) = 60 days.

Q18. If (x + 1/x) = 2, Then The Value Of (x^100 + 1/x^one hundred) Is?

Quick Approach

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

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

Q19. 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 by myself someday paintings = 1/eight

B alone someday work = 1/12

Both A and B sooner or later paintings = (1/8 + 1/12) = (three/24 + 2/24)

= five/24

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

or four four/five days.

Q20. 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 ?

Time taken by using tap A to fill the cistern=four hrs

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

Time taken with the aid of faucet B to drain the entire cistern = 9 hours

so paintings carried out by faucet B in 1 hour = 1/ninth

=> Work carried out by way of (A + B) in 1 hour=(1/four - 1/9)=five/36

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

Q21. 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

Let's Assume the gap 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 began 6 hours earlier than 2 P.M. I.E @ eight A.M.

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

Q22. 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 means of slower pipe by myself to fill the tank = x minutes.

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

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

=>four/x = 1/36 => x = one hundred forty four min.

Q23. An Athlete Runs 200 Metres Race In 24 Seconds. His Speed Is?

Speed = (2 hundred/24) m/sec = 25/three m/sec

convert m/sec to km/hr

(25/three * 18/five) km/hr = 30 km/hr.

Q24. 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 five Minutes Before The Arrival Of The Train.Discover The Distance Covere

Lets expect the desired distance = x km.

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

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

So required distance = 6 km.

Q25. Sum Of Squares Of Two Numbers Is 2754, Their Hcf Is nine, Lcm Is 135, Find The Numbers?

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

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

By fixing eq. (1) & (2)

nos. Are forty five and 27.

Q26. 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

Total distance travelled in 1 hour

= 50 + (&frac; x 32) km

= sixty six km

? Average speed = Total distance/Total time

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

Q27. 7 Years Ago, The Ages (in Years) Of A And B Were In The Ratio four:5 And 7 Years Hence They Will Be In The Ratio 5:

7 years ago, A's age=4x years

and B's age=5xyears

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

=> 25x + 70 = 24x + eighty four

x = (eighty four - 70) = 14

B's present age = 5x + 7 = 5*14 + 7 = seventy seven years.

Q28. A Man Swims Downstream 72 Km And Upstream 45 Km Taking nine Hours Each Time. What Is The Speed Of The Current?

Man's downstream pace = 72/ 9 kmph => 8kmph

Man's up circulation pace = forty five/ nine => 5 kmph

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

Q29. A Man Owns 2/3 Of The Market Research Beauro Business And Sells 3/4 Of His Shares For Rs. 750

three/four of his proportion = 75000

so his percentage = one hundred thousand.

2/3 of enterprise cost = a hundred thousand

so overall price = 150000.

Q30. A 600m Long Train Is Running At seventy three 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 via five/18

pace (mps) = 73 * five/18

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

= (six hundred * 18 )/(73 * 5) sec

= (10800 / 365)

Time take via Train = 29.58Sec.

Q31. Two Pipes A And B Can Fill A Tank In nine 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 stuffed through pipe A in 1 hour =1/nine

Tank component crammed via pipe B in 1 hour =1/three

Given Pipe A and B are opened instead.

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

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

Remaining element = (1-8/nine)=1/nine

So next is A flip.

So Pipe A will fill closing 1/nine part in next 1 hour.

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

Q32. 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 through speed = 6 kmph

and upstream by means of pace = 4 kmph

so man price = (6 - 4)/2 = 1 kmph.

Q33. 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 strolling in opposite Direction:

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

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

= ninety 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.

Q34. 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?

As teach are jogging in equal route

so Relative speed = (forty - 20) km/hr = 20 km/hr

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

Formula Used: Distance = Speed * Time

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

= 27 7/9 m.

Q35. 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 discover the 1 day paintings of each (A & B)

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

and

B's 1 day's work = 1/nine (B can do work in half of time)

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

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

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

Q36. What Comes Next In The Sequence? Four, 2, five, 9, 5, eleven, thirteen, 7, sixteen, 17, 9

Split the collection as below

4,2,five nine,5,11 thirteen,7,16 nine,?4,2,5 -> diff b/w four,five = 1

9,5,eleven -> diff b/w nine,11 = 2

13,7,sixteen -> diff b/w thirteen,16 = 3

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

Q37. A Batsman In His 18th Innings Makes A Score Of a hundred and fifty Runs And There By Increasing His Average By

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, common after 18 innings = forty two.

Q38. A Boat Can Row Upstream At 25 Kmph And Downstream At 35 Kmph, Then The Speed Of The Current Is?

Guy's upstream velocity = 25 kmph

Man's downstream speed = 35 kmph

so Speed of modern = (35 - 25)/2 = five kmph.

Q39. 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 via one tap to fill half tank = 3hrs.

Part crammed by the four faucets in 1 hrs = (four*1/6) = 2/3.

Remaining part = (1-half of) = half of.

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

=> Total time taken = three hrs forty five min.

Q40. 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 anticipate time required by means 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 = 1/4

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

X2 - 2X - 24 = zero

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

=> A can fill cistern in 6 hours.

Q41. 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 B

Given sum of a while is 360 years.

The ratio of children and mother and father a while is two:1.

So total age of mother and father = 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

Q42. 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 eight Years, How Many Times Would He Be Of Mohit's Age?

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

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

After 8 years.

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

=> 8X + 16 = 5X + forty

=> 3X = 24 so, X=eight

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

Q43. 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 in the future paintings = 1/eight

A by myself in the future work = 1/12

B on my own sooner or later work = (1/eight - 1/12) = ( three/24 - 2/24)

=> B one day work = 1/24