Aenova Interview Questions, Process, and Tips

That will depend heavily on when in the journey the 200 men join.

14

0.18

15km/hr

1/10 + 1/12 -1/20 = 15/2 i.e 7.5 h means 7 h & 30 min

=> 3s = d
=> 5(s-9) = d

therefore
5s-45=3s
2s = 45
s=22.5

hence
d= 67.5

let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer

6+8+10+10= 36
6²+8²+10²+12²=344

18th

FOR 15 = RS.20
FOR 20 = RS.18
FOR 25 = RS.16
FOR 30 = RS.14
FOR 35 = RS.12
FOR 40 = RS.10

let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240