Erode sengunthar engineering college Interview Questions, Process, and Tips

S=D/T
S=624KM/6.5HRS
S=96KM/HR

B

11

x and y can be equal to 1..
If so,then he gets 1/1 of Rs.10 which is equal to 10 and
again 1/1 of Rs.10=Rs.10..so,he gets total of 20..and
returns 20..so,no gain and no loss..
If x=1,y=2,he gets, 5+20=25..and returns 20..so he may not
lose..
So,whatever be the values of x and y,only these two answers
are possible..
so its a)He never loses..

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%

1!=1
2!=2×1=2
3!=3x2x1=6
4!=4x3x2x1=24
5!=5x4x3x2x1=120
6!=6x5x4x3x2x1=720
The answer is 120.

i Guess 8 mtrs.

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

The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.

If Vijay gives ‘x’ marbles to Ajay then Vijay and Ajay would have V – x and A + x marbles.
V – x = A + x — (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.

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

EMNXDS

E

TOTAL NUMBER OF BOOKS 40+40=80
ISSUED BOOKS 30
REMAINING BOOKS 80-30=50
RATE=50/80=5/8

500 ways

16

A perfect square is a square number of a digit. eg 64 is a perfect number, a square of 8
Now digits AB9 is a square number of a number.
AB9 Can also be written as A multiply by B multiply by 9
Get the Square root of AB9
Assumption, A=1, B=1
1*1=1=A, 1*1=1=B Therefore,
Square root of A = A, B = B and 9=3
Therefore
An odd number is a number indivisible by 2.
for example 1,3,5,7…….
Therefore Squares A*B*9= AB9
Where a=1, b=1, 3 as digits.
Conclusion
A=1 is an odd number