Smart recruit online Interview Questions, Process, and Tips

1

4

G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15

– 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

5! = 120

39/50

56.25%

Blue

3×3=9

X-340 = X x 4 x 8/ 100

100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.

132

5A=3B,
*3 SO,
15A=9B

30mins

Let the total time taken to fill the tank be x minutes.

Then, work done by B in half hour + work done by (A+B) in half hour =1.

(
40
1
​
×
2
x
​
)+(
60
1
​
+
40
1
​
)×
2
x
​
=1

⇒
80
x
​
+
48
x
​
=1

⇒
240
3x+5x
​
=1

⇒8x=240

⇒x=30 min

C-2896