G2 Technology Interview Questions, Process, and Tips

The local value of 7 in the number is at 10000 so 7×10000 and face of 7 is 7
So 70000 -7 so the answer is C 69993

7 min clock:|——-7——-|
4 min clock:|—–4—-|—-4—–|—–4—-|—–4—-|
you got 9 min: |—————9————–|

Its 2 times faster than the other train
v1*t=v2
v2*t=4*v1
solving these two,we get
v2/v1=2

Product of two numbers = 1320
HCF = 6
LCM = x
Formula:
Product of two numbers =(HCF *LCM)
1320=(6*x)
x=1320/6
x=220
LCM Of the numbers is220

% error= (new no-actual no)/actual no *100
% error= [(x/7)-7x]/7x * 100
% error= 48x/49x * 100
% error= 0.9795*100=97.95

these components can be used to add specialized
functionality, such as animation or pop-up menus, to Web
pages, desktop applications, and software development tools.

23 Teams

F+V-E =2;
F= faces;V= vertices;E = number of edges

18th

Pfull = 4 Hrs; Qfull = 5 Hrs
Assume velocity Vp = x then Vq = 4/5 x –> avg velocity = (5/5 + 4/5) / 2 = 0.9
If fastest time = 4hrs then the time it would take to fill up tank by alternating is 4/0.9 = 4.44 Hrs

3 groups

NCPQJG

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

Let the firt number be n,
The largest number will be n+1
Their sum
n+n+1=55
2n+1=55
2n=54
n=27
Largest number =27+1=28