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Bus started at 8:00
it travelled with 18mph…
distance of destination is 27 miles…
we know velocity=(distance)/time
i.e., time=27/18hours=3/2hours=90 min…
i.e., the bus reached destination at 9:30
the bus stayed for 30 min…
so the bus started return journey at 10:00
now the bus returned with velocity 18 + (18/2)=27mph
the taken to the bus to travel = 27/27=1 hour = 60 min…
so bus would be returned on 11:00…
Sachin is 14 years old whereas Rahu is 18 years old.
10m/sec
21
40
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
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
3, 7, 15, 39, 63, 127, 255, 511
C. 39
3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511
2*3*5*7*11*13*17*19=9699690
The answer cannot be determined as there is a particular formula where the consecutive numbers start.