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From city A, car a sets on it's road towards city B. 9 Hours later, car b, sets from city B towards city A. The two cars met along the way. At the point of the meeting, car a, passed 240 km more than car b had. car a arrived at city B 10 hours after the meeting. car b arrived at city A 9 hours after said meeting. Velocities of both cars were constant.

What is the distance that car b passed until the meeting? What are the velocities of both cars?

I got 3 equalities here, in 4 unknowns, which didn't allow me to reach a finite solution to the problem; \begin{cases} Va (t+9) = Vb(t) + 240 \\ Va (t+9) + Va (10) = S \\ Vb(9)+Vb(t) = S \end{cases} where Va, Vb, velocities of a and b, concordantly, and S the total distance between the cities. t the time b traveled until the meeting.

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..........What? – Bak1139 Jan 16 '14 at 20:35
We have a total distance from $A$ to $B$ : say $d$ km. When they meet, car $a$ has travelled a distance $d_1$ that is his velocity (say $v_a$) times $t_1$ hours; but car $b$ has travelled for $t_1-9$ hours with velocity $v_b$ traversing a distance $d_2=d-d_1$. But you know also that $d_1=d_2+240$ km. In this way, you can remove $d$. – Mauro ALLEGRANZA Jan 16 '14 at 20:39

You have enough information for the unknowns.

Let $t_{m}$ be the meeting time, in hours. Let $d_A, d_B$ be the distances in kilometers that cars $A, B$ travel before they meet. Let $v_A, v_B$ be the speeds of the cars in km/hr. That's five unknowns.

Before the meeting, we know that $d_A = v_A t_m$ and $d_B = v_B (t_m - 9 \text{ hr}).$

We know also that $d_A - d_B = 240 \text{ km}.$

We know after the meeting that $d_B/v_A = 10 \text{ hr}$ and $d_A/v_B = 9 \text{ hr}.$

Five equations, five unknowns. Now solve for $d_B, v_A, v_B.$

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