Vision college Interview Questions, Process, and Tips

60000

RABBIT

speed = 72 Kmph
Speed= 72 * 1000 / (60*60) m/s
speed= 72 * 5 / 18
speed= 20 m/s
time = 30 s
distance = 600 m

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

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.

Difference between 2 and 6 is = 4
Difference between 6 and 12 is = 6
Difference between 12 and 20 is = 8
Difference between 20 and 30 is = 10
Difference between 30 and 42 is = 12
Difference between 42 and 56 is = 14
So ne number will be with 16 Difference i .e 72
Therefore Answer will be 72

-Total /n=80
– x/5=40
x=200

Total/n=y/n
y-x/n-5=90
80n-200=90(n-5)
80n-200=90n-450
450-200=10n
n =25 (students) write exam

8/15 left.

1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15

Let the age of the man be x
Then age of his son becomes (x−24)

2 years later from now,

Age of man will be = x+2
and age of his son will be =(x−24+2)=x−22

According to question,

2(x−22)=x+2

i.e., 2x−44=x+2

i.e., x=46

Therefore ,
Present age of man=46 years
And present age of his son =46−24=22 years

Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:

1. Weigh ABC versus DEF.

Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.

2. Weigh G versus H.

Scenario a.i: If these (2) balance, the oddball is I.

Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.

Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.

2. Weigh A versus B.

Scenario b.i: If these (2) balance, the oddball is C.

Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.

answer is maximum of 2.

Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45

Son in law