Fivium Interview Questions, Process, and Tips

s1(1+x/100)+s2(1+x/100)+s3(1+x/100)

It is assumed to be constant. Now, to cross past the pole, the train should cover a distance of x meters. Now, the time taken by the train to cross a platform of length 100 m is 25 seconds. Hence, the length of the train is 150 m.

7days

3days

Circular Track sis of 11 km

Speed of Mens are

4 , 5.5 , 8 km/hr

Time at Which they are at starting Points again

11/4 , 11/5 , 11/8

11/4 , 2.2 , 11/8

We need to find LCM of these

to find at what time they Meet again at starting point

LCM 11 * 2 = 22

After 22 Hrs they will meet at starting point

Answer: z and u

Explanation:

Y is to the right of U and exactly in front of V. Therefore,

U Y
V

Z is behind W and W and X are at extreme ends. So, W has to be to the right of Y. The final arrangement is as follows.
U Y W
X V Z Therefore, Z and U are at extreme ends is true.

share= 5:7
i.e.,
=9600*5/12
= 4000

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

1.5

let convert meter to kilometer by multiply 18/5 then answer be 14*(18/5)=50.4 km/s