College of the sequoias Interview Questions, Process, and Tips

X = 1/2 * b*h
base incresed by 4 times & height
is devided by 2
X = 1/2 * 4b * h/2
X = h/4 * 4b
X = hb

Niece

Circular Track sis of 11 km

Speed of Mens are

4 , 5.5 , 8 km/hr

Time at Which they are at starting Points again

11/4 , 11/5 , 11/8

11/4 , 2.2 , 11/8

We need to find LCM of these

to find at what time they Meet again at starting point

LCM 11 * 2 = 22

After 22 Hrs they will meet at starting point

The correct ANS: option C

The answer for this question is very simple: it is called Ceaser Cipher.
It consists of replacing a character by another located in (current character position + (key-1)), where key = how many positions you want to skip.
For Example: VICTORY -> YLFWRUB with the key = 3.
Then SUCCESS -> QXFFHQQ

Note: it should work like a circle(Z+1 = A).

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

64

s=a+(a+10)+(a+10+10)+(a+10+10+10)
s=4a+60
a=(s-60)/4

I am working institute

8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here

( A ) 124