RackWare Interview Questions, Process, and Tips

Ans is 10%

Printer price = x;
Computer price = 3x
Total overall price = 20x + 60 x 3x = 200x

Printer percent = (20x/200x ) x 100 = 10%

Fuel to go = x + (x/4)=5x/4
Fuel to come = x
now,
x+(5x/4) = 4.5
9x/4 = 4.5
9x = 18
x = 2 (Fuel to comeup)
Fuel to go will be:
2+(2/4) = 2 + (0.5) = 2.5

The probability will be
white ball = 5/9
red ball = 4/9
The answer can be 5/9 and 4/9 or 1

427 is the answer
Because
385 means (3+5=8) or (8-3=5) ,
462 means (4+2=6) or (6-2=4) ,
572 means (5+2=7) or (7-5=2) ,
396 means (6+3=9) or (9-3=6) ,
671 means (6+1=7) or (7-6=1) ,
264 means (4+2=6) or (6-2=4) ,
but only 427 is the number in which if we do addition of any of its two digits (e.g 4+2 or 4+7) then its answer doesn’t come 3rd digit
like if we do 4+2 then answer will be 6 so the number should be 426 but it is not
or if we do 7-4 then answer will be 3 (not 2)
so the right answer is 427

M
M. L
L

360

Article price=200
After 25%increase =250
After 25% decrease =187.5

West

90 hours

1200*5 : 750*4
6000:3000
2:1

– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.

– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m

To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec

In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m

at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m

If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)

Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.

For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.

This will take an additional (9.6 + 2) seconds.

Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds

150 m

I’m the capable of what u expect from me and you should say what is the tallent u have which is used for that company…🥰

Because from there i can do more and learned more too