Techextensor Interview Questions, Process, and Tips

1(100-1)+ 2(100-2)+…..99(100-99)
N=100
solution: N(N-1)
100(99)= 9900

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….

the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;

(p*r*t)/100 = I ………………..(p*3*10)/100 = 840 …..p = 84000/30 = 2800

17.9

60000

c=a/b
c=a-1
=>b=a/c
=>b=a/a-1

c

1) time =20mins,speed=15kmph=15*(5/18)=25/6 m/sec
distance=20*(25/6)=250/3 m
then 2) given as time =15min
speed=250/3*15=(50/9)*(18/5)kmph=20kmph

mausi