Ardentec Interview Questions, Process, and Tips

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

B. No such pair exists

x/2y=3/2…
therefore
2x=6y
x=3y
therefor
=21y+4y/3y-2y
=25y/y
=25..
there answer=25
so for the above question, we don’t have an option…so answer
is Data Insufficient..

4:3
A—–>B (first train)
B——>A(second train)
A/B=Srureroot((time to B)/(time to A))

10%=394;50%=1970;250%=9850;30%=1182;
250%+30%=9850+1182=11032

10000

38 years

Let Rajan’s present age be x years. Then, his age at the time of marriage = (x – 8) years.

x = 65(x−8)

⇒5x=6x−48

⇒x=48 years

Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = (x – 18) = 30 years

∴ Rajan’s sister’s present age = (30 + 8) years = 38 years

C.1175

yes

assume the ratio 6:4 to be 60:40
boys taking lunch in canteen=60% of 60=36
Girls taking lunch in the canteen=40% of 40=16
Total boys and girls taking lunch in the canteen=52
Total students we assumed is 60+40=100
Therefore percentage of boys and girls taking lunch in the canteen=(52/100)*100=52%

Either 4 or (-)4