Charter hall Interview Questions, Process, and Tips

7

i will drop a mail saying that can you pleaae meal me the the originals of the stregic plan

10000

one-legged =5% of 10000=500

remaining=10000-500=9500

barefooted=9500/2=4750

remaining people= 9500-4750=4750

hence required number of shoes= 4750*2+500*1=100005% of 10000 = 500 one legged

9500 / 2 = 4750 bare foot

minium no of shoes = 4750*2 + 500*1 = 10000

1.5 km/hour

225

let the third no be x ,
then the first no is 3x ,
second no is 2*3x ie 6x
average is (3x+6x+x )/3 = 20
10x/3 = 20
x= 60/10 = 6
third no is 6 , second is 6x = 36, first no is 3x = 18
largest no is 36 i.e second no

time (dist covrd by husband) (dist covrd by wife)
0.5hr 20 miles 0 miles
1 hr 40 miles 25 miles
1.5 hr 60 miles 50 miles
2 hrs 80 miles 75 miles
2.5 hrs 100 miles 100 miles
Therefore wife catches up after 2.5 hrs..

75/8 days

Ans. 175
Relative speed in m/s=40-22=18X5/18=5m/s
Total distance=125+x m
T=1min=60 s
D=sXt
125+x=5X60
x=175

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

B

– 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

Pearl.. Head cut then earl.. tail cut then pear.. Head and tail cut then ear

SI = PTR / 100

24

7770-7077=693

GIVEN
80 CARDS IN 30 MINUTES.
Have to prove 7:30 ?
14 * 80 +80.
Answer is =1200.