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6/(2+3)-(2+3)/6
=6/5-5/6
=(36-25)/30
=11/30
ans is 11/30
In order to enhance my career, I asked my colleagues about my strength and weaknesses, the areas where I need to improve more, and acquired more knowledge from my superiors. I worked on my weaknesses to convert into a positive response.
In 30 mins- Thief – 20km, Owner – 0km
In 1hr – Thief – 40km, Owner – 25km
In 1.5hrs – Thief – 60km, Owner – 50km
In 2hrs – Thief – 80km, Owner – 75km
In 2.5hrs – Thief – 100km, Owner – 100km
Answer: In 2.5 hours the owner would have overtaken the Theif
24
6620
4/5
Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.
First write equations from info:
(A) (Mon + Tue + Wed)/3 = 111 Rearrange as ——–> Tue + Wed = 111 – Mon
(B) (Tue + Wed + Thu)/3 =102 Rearrange as ——–> Tue + Wed = 102 – Thu
(C) Thu = 0.8(Mon)
Substitute equation C into B:
(B) Tue + Wed = 102 – 0.8(Mon)
At this point I changed the values for clearer algebra:
Mon = x
Tue + Wed = y
Re-write equations A & B with new values:
(A) y = 111 – x
(B) y = 102 – 0.8x
Solve simultaneous equations:
111 – x = 102 – 0.8x
111 – 102 = x – 0.8x (Re-arraged)
9 = 0.2x
x = 45
Thus, Mon = 45C
Thu = 0.8(45)
Thu = 36C
So the answer is it was 36C on Thursday
Ravi do a work in 30 days
Prakash do a work in 40days
then both work together 1/30+1/40=70/1200
then solve values 120/7 is 17(1/7) days
both a and b is correct
NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission
2 and half day.
Three man’s on day work = 1/6+1/7+1/8=73/168
three man’s can complete the work in 1/73/168 days
= 168/73 days
Now if they works together fro the alternet days they will
complete the worksin 2*168/73 days
(If three mans working for the alternate days then work
completion time will be doubbled)
10%
Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
=>800+x=653∗60=>800+x=20∗65=1300=>x=1300−800=500