Question

A man crosses a 320m wide river perpendicular to the current in 4 min. If in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in m/min is

A. 30
B. 40
C. 50
D. 60

Answer 1

If in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in m/min is (A.) 30.

How to find speed?

Denote the speed of the man in still water as
`\(V_m\)` and the speed of the current as
`\(V_c\)`.

The effective speed is the width of the river divided by the time taken:

`\[ \text{Effective speed} = \frac{\text{Width of the river}}{\text{Time taken}} \]`

Given that width is 320 m and the time taken is 4 min:

`\[ V_m + V_c = \frac{320 \, \text{m}}{4 \, \text{min}} \]`

Since the man swims with a speed
`\(V_m\)` that is 5/3 times that of the current:

`\[ V_m = (5)/(3) V_c \]`

Now, substitute this relationship into the effective speed equation:

`\[ (5)/(3) V_c + V_c = \frac{320 \, \text{m}}{4 \, \text{min}} \]`

Combine the terms on the left side:

`\[ (8)/(3) V_c = 80 \, \text{m/min} \]`

Now, solve for
`\(V_c\)`:

`\[ V_c = (3)/(8) * 80 \, \text{m/min} \]`

`\[ V_c = 30 \, \text{m/min} \]`

So, the speed of the current is 30 m/min, and the correct answer is A. 30.