TS EAMCET 18-July-2022 Shift 2 Solved Paper

Speed of river =v
Speed of man in still water =2v
∴ Speed upstream =(2v−v)
Speed downstream =(2v+v)
Now, according to the question, total time taken in going upstream and back downstream to the starting point =t
∴‌‌t‌up ‌+t‌down ‌=t‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Let the total distance on one way =d ∴‌‌t‌up ‌=‌

d

2v−v

⇒t‌down ‌=‌

d

2v+v

Putting these values is Eq. (i) we get,
‌

d

2v−v

+‌

d

2v+v

=t
⇒‌‌d[‌

1

v

+‌

1

3v

]=t⇒‌

d

v

[‌

3+1

3

]=t
⇒‌‌t=‌

4d

3v

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
It is given that time taken by the man to cover the same distance in still water =t0
⇒‌‌t0=‌

2d

2v

=‌

d

v

.
∴ From Eq. (ii)
t=‌

4

3

t0⇒‌

t

t0

=‌

4

3