Rajsriya Automotive Industries Interview Questions, Process, and Tips

100 steps(1step/sec)

pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240

30.5

TWICE

8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8

121

A die is thrown twice n(s)=36
The probability of getting same number is (1,1)(2,2)(3,3)
(4,4)(5,5)(6,6) i.e n(A)=6
Ans. P(A)=n(A)/n(s)
= 6/36
=1/6.
So, The Answer is 1/6

The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392

15a=45

340

Correct option is C)
Let be students are consider as child
Let the age of child added later be x years.
average age of 12 children =20 years
∴ Total age of 12 children =20×12=240 years
After one more child is added-
Average of 13 children =20−1=19 years
Average age of 13 children =
13
sum of 12 children+x
​

19=
13
240+x
​

⇒240+x=247
⇒x=247−240=7 years
Hence the age of child added later is 7 years.

b=15,l=10 or b=10,l=15