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A cyclist covers 44.8 km with a tailwind in 1 hour 36 minutes, and the speed of the wind is 40% of the speed of the cyclist in still air. Determine the ratio of the time taken by the cyclist to cover 108 km against a headwind to the time taken by the cyclist to cover 420 km with a tailwind respectively.

The correct answer is 3 : 5.

Step 1: Understand the given variables and convert units

Let the speed of the cyclist in still air be $v$ km/h.
The speed of the wind is given as 40% of the speed of the cyclist in still air.
Therefore, the speed of the wind $w=0.4v$ km/h.

Step 2: Calculate the effective speeds with tailwind and against headwind

With a tailwind, the effective speed $v_{\text{down}}$ is:
$v_{\text{down}}=v+w=v+0.4v=1.4v$

Against a headwind, the effective speed $v_{\text{up}}$ is:
$v_{\text{up}}=v-w=v-0.4v=0.6v$

Step 3: Determine the value of $v$ from the first scenario

The cyclist covers 44.8 km with a tailwind in 1 hour 36 minutes.
Convert 1 hour 36 minutes into hours:
1 hour 36 minutes = $1+\frac{36}{60}=1+0.6=1.6$ hours.

Using the formula $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$:
$1.4v=\frac{44.8}{1.6}=28$

$v=\frac{28}{1.4}=20$ km/h.

So, the speeds are:
Effective speed against headwind ($v_{\text{up}}$) = $0.6\times 20=12$ km/h.
Effective speed with tailwind ($v_{\text{down}}$) = $1.4\times 20=28$ km/h.

Step 4: Calculate the required times and their ratio

Time taken to cover 108 km against a headwind ($T_1$):
$T_1=\frac{108}{12}=9$ hours.

Time taken to cover 420 km with a tailwind ($T_2$):
$T_2=\frac{420}{28}=15$ hours.

Ratio of $T_1$ to $T_2$:
$\text{Ratio}=\frac{9}{15}=\frac{3}{5}$ or 3 : 5.