Henkel loctite Interview Questions, Process, and Tips

12

Number of persons between Vijay and Jack = 48 – (14 + 17) = 17.
Now, Mary lies in middle of these 17 persons i.e. at the eight position.
So, number of persons between Viji and mary = 7.

– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.

– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m

To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec

In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m

at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m

If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)

Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.

For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.

This will take an additional (9.6 + 2) seconds.

Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds

324:400:576?
18×18:20×20:24×24:26×26
324:400:576:676

The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392

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….

pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240

Active listening,Effective communication,Confidence.

1km = 1000m
Xkm = 300m
X=0.30km

1hr = 3,600s
X = 40.5s
0.01125hr
Therefore; 0.3km/0.1125hr= 26.6666=26.67km/hr

9

9/25

let the total work to be done =1
so a+b+c=1 (say)
then a and b are supposed to do 7/11 th of work
so c= 1- (7/11)
c=4/11
therefore c gets (4/11)*550
=200 is the answer

24.
exp: 0 is formed by multiplying 5 with 2. so first we find
how many 5’s and 2’s are there in 100!.
No of 5’s : 100/5=20/5=4/5.
20 +4 =24.
No of 2’s : 100/2=50/2=25/2=12/2=6/2=3/2=1/2.
50 +25 +12 +6 +3 +1.
It has 24 5’s and 2’s.
so the no. of zeros=24.

one cat kill one rat six minutes
so 1 cat kill 100 rat willbe needed 6*100=600minutes
then 100 rates kill 50minutes means then 600/50=12
so the answer is 12

8:20

C. 96

sample space=36
prob of one of the dice getting face 6
i.e n={(1,6)(2,6)(3,6)(4,6)(5,6)(6,1)(6,2)(6,3)(6,4)
(6,5)(6,6)}=11
p=n/s
p=11/36
ans:11/36