Global Edge Software Aptitude Placement Papers - Global Edge Software Aptitude Interview Questions and Answers updated on 14.Jun.2024

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.

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.

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.

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.

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.

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.

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.

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.

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.

(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.

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.

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

Adding both side 1

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

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

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

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.

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.

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.

(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.

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.

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.

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.

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.

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.

Speed = (200/24) m/sec = 25/3 m/sec 

convert m/sec to km/hr 

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

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.

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.

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.

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.

3/4 of his share = 75000

so his share = 100000.

2/3 of business value = 100000

so total value = 150000.

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.

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.

Man's row in downstream by speed = 6 kmph

and upstream by speed = 4 kmph

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

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.

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.

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.

Split the series as below

4,2,5 | 9,5,11 | 13,7,16 | 17,9,?4,2,5 -> diff b/w 4,5 = 1

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

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

17,9,21 -> diff b/w 17,21 = 4.

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.

man's upstream speed = 25 kmph

Man's downstream speed = 35 kmph

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

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.

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.

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

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.

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.

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.