Northgate Interview Questions, Process, and Tips

C

3days

4 years

1/8

877.4

4 boy 3 sis

39 hrs 15 mins..
note:
he will not be having .45 mins for his lunch on saturday..

Let:
1. speed of boat in still water = x
2. speed of stream = y;

Then net speed of boat upstream = x – y = 16km / 4 h = 4kmph
Net speed of boat downstream = x + y = 16 / 2 = 8kmph

Now solve for x to get x = 6kmph in still water.

163

let no of boys in group is = x
then total sum = 30 * x
after joining one more boys with a weight of 35 kg the total sum is = 30x + 35
after joining the new student the weight will increase 1 kg
so total sum
30x + 35 = 31(x+1)
x = 4

540 meters or 0.54 km

130 years

414 Rs
1.2*37.8=45.36=46 sq.feet
46*9=414 RS

The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.