Globeair ag Interview Questions, Process, and Tips

b

2^a = 2 x 2+a
a^3= 2 x a + 3
=> 4+a= 2a + 3
=> a = 1

Ans : 228
7
(7*1)+1=8
(8*2)+2=18
(18*3)+3=57
(57*4)+4=232 [ given 228 ] odd
(232*5)+5=1165
(1165*6)+6=6996

there wil be two point of intersections

let x be lenghth of candles.

thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.

let he lit for n hrs.

so in n hrs thicker one goes n/6.
and thinner one goes n/4.

so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.

6+8+10+10= 36
6²+8²+10²+12²=344

1 – (-1) = 1 +1 = 2

w, l=2w , area of square =8^2 = 64
area of rectangle = l*b
so w*2w = 8*64
w=16
length = 32

297

Let Suvarna, Tara, Uma and Vibha be S,T,U,V respectively
initially in the beginning each persons share be
V = x U = y T = z
S = w = (x+y+z+32) Reason: She has to double others share, so she should have each and everyone’s share and still should be left out with 32

after 1st Round of game
S loses and is out with 32 and doubles the others share
V = 2x U = 2y T = 2z

After 2nd Round of game
T loses and is out with 32 and doubles the others share
V = 4x U = 4y
This means T had 2z = 2x + 2y + 32
After 3rd round of game
U looses and is out with 32 and doubles others share
V = 8x
This means U initially has 4y = 4x + 32
In the end V = 8x = 32

Solving this we get x = 4, y = 12, z = 32 and w = 80
There fore Suvarna had highest share in the beginning

u c 5 numbers…..
we add the combination of digits…..
a.17->1+7=8(even)
b.23->2+3=5(odd)!!!!!
c.37->3+7=10(even)
d.13->1+3=4(even)
e.73->7+3=10(even)
all combinations are even xept 23
ans:23

CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….

C

10 Camels = 68000

madras

required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)

9 because from 64 stubs 8 cigarettes and again from 8 stubs he’ll get 1 cigarette then his score will be 9