Mystifly Interview Questions, Process, and Tips

30+32+34+36+38+40+42+44+46+48=390..

Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s

Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs

#Train length = 125Mtrs

Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s

Sp = D/t
=139M/10s
=13.9M/s

0

1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0

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

red

9

9/25

I am Srilaxmi. I born and bought-up in WARANGAL. My father Agriculture cum politician and my Mom homemaker. I blessed with three brothers and one sister. My elder brother was married and working as a TA in MPDO office at Warangal. My first younger brother was also married and working with Sushee Infra Pvt Ltd as a Asset Manager at Hyderabad. Younger brother and sister are into their studies. Coming to me I completed my MBA from ICFAI university in 2008 and worked as a Audit assistant in accounting firm for two years. I had actively participated in bank audits i.e. is in CBI (Concurrent audit) and SBI(Statutory audit) during my services. Later I pursued CS Executive and attempted for five times then I give up to clear my M.Com. At present I was studying postal studies of CS Executive programme. My hobbies are reading books and listening to music.

Let 10’splace digit is x and unit’s place digit y

First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y

Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
After solving, we get —-> y=6x but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 KM/HR

If there are n bikes then the no. of people are 2n+1. n are dropped in firs trip, in second trip 3 bikes are unused, so 2n-3 + n covers 2n+1 people. i.e. n + 2n – 3 = 2n +1
n = 4
No. of people = 9

ans 810 i think

The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.