EBIW Info Analytics Interview Questions, Process, and Tips

6/(2+3)-(2+3)/6
=6/5-5/6
=(36-25)/30
=11/30
ans is 11/30

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

30/70*100=42.87

( b ) 40 sq cm

23. 15s squre is 225and 8 squre is 64 addition is 289 product also 120 addition 15+8=23

10m

B

GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC

Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.

FASHION= FOIHSAN
It is F-ASHIO-N to F-OIHSA-N.
So PROBLEM will be P-ROBLE-M,
the answer is PELBORM.

Time =9 seconds

Speed =(60×518) m/sec=(503)m/sec

∴ Speed =DistanceTime

Length of the train (Distance) = (Speed × Time) = (503×9) m=150 m.

3days

I’m interested in this job because I can see that, in this role, my skills could help solve this problem within your company. I also see an opportunity for me to learn and grow these skills, so we both would benefit personally, professionally, and financially.

2/3 * 15 miles = 10 miles then

time for 10 miles travel is t1 = 10/40 = 1/4

then remaning distance is 5 miles

time for 5 miles travel is t2 = 5/60 = 1/12

t1+ t2= 1/3 in seconds * 60 = 20 min

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