Metropoolregio rotterdam den haag Interview Questions, Process, and Tips

4

answer :
Let the distance which I walk to the station be x km Then Time needed to reach the station at 3 km/hr = (3x​+602​) hrs and
Time needed to reach the station at 4 km/hr = (4x​+602​) hrs But 3x​+301​=4x​−301​⇒3x​−4x​=−302​=151​
⇒124x−3x​=151​⇒12x​=151​⇒x=1512​=54​ km

the answer is 3:10
explanation:

let the first number be x
and the second number be y
(7/8)((3/7)x)=(25/100)((45/100)y) given
ie.., 21x/56= 9y/80
on solving we get x:y= 3:10
is the answer

the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;

Ans: 60 kph

Suppose Person meets the train everyday at 3 PM at Station A.

His speed is 12kph.

So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.

But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.

Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.

So speed of the train is 5KM/5min = 60 kph.

6 cuts

360 times

A can give B (5 min – 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.

Sum of all numbers in series= 870 (I.e 312+162+132+142+122)
Average=sum of numbers/count of numbers
Average= 870/5
= 174

D

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