DevSquare Quantitative Aptitude Speed Time and Distance / Boats and Streams Questions

Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph

Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18)*(18/5) kmph
= 5 kmph

It is very important to check, if the boat speed given is in still water
or with water or against water. Because if we neglect it
we will not reach on right answer. I just mentioned here because
mostly mistakes in this chapter are of this kind only.

Lets see the question now.
Speed downstream = (16 + 5) = 21 kmph

Time = distance/speed = 84/21 = 4 hours

3 rotates 3,6,9,12,15
5 rotates 5,10,15
because ans is 15

First train takes 4 hours and the second train takes 3.5 hours.

Time ratio is 8:7. Therefore, the speed ratio will be 7:8.

Let the speeds be 7x and 8x, and distance be 28x ( 4x7 or 3.5x8).

At 7 AM, the first train must have covered a distance of 14x.
Therefore, at 7 A.M. the distance between the two trains is 28x-14x=14x.

Time taken to meet = 14x/(7x+8x)=14/15 hour or 56 minutes.

Hence, the two trains meet at 7.56 AM.

Let train A take x hours to reach the point of meeting(let it be P)
Then train B will reach the same point in
x-2 hours.
Distance covered to reach P by train A in x hours=Distance covered by B to reach P in x-2 hrs.
80*x=100(x-2)
80x=100x-200
x=10 hrs.
Distance from Nampali=x*speed of train A
=10*80
=800 km

Here the Distance is constant. So, let the actual time is x.
Then,
600/x-600/(x+(1/2))=200
After we solve this we will get 1hr.
So, the actual time taken for the journey is 1hr.

Let the two opposite ends of the river be X and Y and the distance
between them be D meters.(i.e., width = D meters)
Let P and Q be the two men starting from the opposite banks
(i.e., from X and Y respectively).
Let the speed of P and Q be A and B m/hr .

I meet :
During I meet, P travels 340m from X while Q travels
(D - 340)m from Y.
Therefore, Time taken for P to travel 340m = Time taken for
Q to travel (D - 340)
Or 340 / A = (D - 340) / B
Or 340 / (D - 340) = A / B ...(1)

II meet :
After crossing spot I, both of them proceed in their respective directions,
reach banks and return back to cross each other at
Spot II which is 170m from Y.
From Spot I to Spot II, P would had travelled a distance of
(D - 340) + 170 m
From Spot I to Spot II, Q would had travelled a distance of
340 + (D - 170) m
Time taken by P to travel from Spot I to Spot II will be the same
as that of Q from Spot I to Spot II

Therefore, A / (D - 340) + 170 = B / 340 + (D - 170)
Or
(D - 340) + 170 / 340 + (D - 170) = A / B ...(2)
From equations I and II, we get,
340 / (D - 340) = (D - 340) + 170 / 340 + (D - 170)
340 / (D - 340) =
D - 170 / D + 170
By Cross- Multiplying,
340 (D + 170) = (D - 170) (D - 340)
340D + 57800 =
D2 - 170D - 340D + 57800
D2 - 850D = 0
By Factorizing,
D(D - 850) = 0
D = 850
Hence the width of the river = 850 m

Let d be the length of the journey,

time taken to travel distance d at speed of 50 kmph = d/50

time taken to travel distance d at speed of 60 kmph = d/60

difference in time = 15 min = 15/60 hr

d/50 - d/60 = 15/60

d/5 - d/6 = 15/6

d/5 =15

d = 15*5

d = 75 km

 

Let time travelled in train is x min then in cycle (x+20) min.
(30/60)x + (15/60)(x + 20) = 50 or x = 60
Total time taken 60 + (60 + 20) = 2 hr 20 min