Nuvasive Interview Questions, Process, and Tips

Speed = 13+4 km/hr
Distance = 68 km/hr
Time=distance/speed
68/17=4hrs

In plan A 7500/- and other amounts in plan B

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

I agree with @TUMWINE PETER

I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps

The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.

The time to travel 400m at 20mps is 20secs

11

11 14
12 15
13 16
=======
14 17
the answers is 14

E non of those

b

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

divisible by 8 – 124
” 12 – 83
” both – 41
124 + 83 – 41 = 166
999 – 166 = 833
ans is 833…..

(130+x)/30=(45×5/18)
245 meters

required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)

2:3

let the speed limit be x.
speed by first rider S1 and 2nd rider be S2.
S1=x+10;
ans S2=x+2*10; and given S2=35.
solving we get x=15mph.

d

11 times in 12hours

there wil be two point of intersections