A red car and a blue car started racing towards each other at the same time from point $A$ and $B$ respectively, as shown above, where $AB$ was $600$ km. long.

Then $6$ hours later, both cars met up at point $C$ with constant speeds.

Afterwards, when the blue car reached the destination at point $A$ , the red car was passing point $D$ .

If $CD$ was $160$ km. long, how many more hours would the red car take to travel from point $D$ to the destination $B$ ?

The answer is 5.

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Let $u$ be the red car's speed and $v$ be the blue car's speed. The distance traveled will equal to speed multiplied by time.

Since both cars met up at point $C$ , their combined speed towards each other could be set up in the equation: $600 = 6\times (u+v)$ ; $u+v = 100$ .

Now, for the rest of the next journey with time $t$ , the red car had traveled $160$ km.:

$ut = 160$

This time $t$ was the same time the blue car used to finish up the final distance, which the red car had first traveled. In other words, the first distance = $6u = vt$ . Therefore, $t = \dfrac{6u}{v}$ .

Substituting this term in the previous equation, we'll get: $ut = \dfrac{6u^2}{v} = 160$ .

$3u^2 = 80v = 80(100-u)$

$3u^2 + 80u - 8000 = 0$

$(u-40)(3u+200) = 0$

Thus, $u=40$ ; $v=60$ ; $t= \dfrac{6\times 40}{60} = 4$ .

In other words, in $10$ hours, the blue car had traveled $60\times 10 = 600$ km. while the red car had traveled $40\times 10 = 400$ km. and would have $200$ km. to go.

As a result, it would take $\dfrac{200}{40} = \boxed{5}$ hours more for the red car to travel from point $D$ to the destination $B$ .