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I was asked the following in an interview:

B is 50% faster than A. A starts at 9:00 and B at 10:00. They are now 300km apart. The speed of A is 50kmph. At what time will they meet if both move in opposite direction?

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Are they on a circular track? Or are we supposed to think of them on the surface of the Earth? Otherwise they'll never meet moving in opposite directions... –  Zev Chonoles Jul 25 '12 at 15:33
    
Do they start at the same point? –  copper.hat Jul 25 '12 at 15:33
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4 Answers

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I think there is an ambiguity about when 'now' is in the question :-). There is nothing that indicates that now is before or after 10:00, for example, so one has to be careful about assuming that the position of $A$ and $B$ can be modeled by a linear (well, affine) equation.

All of the other solutions assume that 'now' is 9:00. I believe the answer should be that they meet no earlier than 12:00.

I am assuming that we are dealing with a 'flat' surface here, with no spherical trickiness.

Suppose $A$ starts at $x=0$, and $B$ at $x=L$, where I assume $L>0$. Also, the question asks what time 'will' they meet, so we are looking for a time in the future when they meet.

Then their positions are given by: $A(t) = 50 \max (t-9,0)$, $B(t) = L-75 \max(t-10,0)$.

Let 'now' be $t_0$. Then $t_0$ must satisfy the constraint $|A(t_0)-B(t_0)| = 300$. If $A(t_0)>B(t_0)$ they will never meet, so in order to meet we have the constraint $B(t_0)-A(t_0) = 300$. Suppose they meet at $t_1$, then we require $t_1 \geq t_0$ and $B(t_1) - A(t_1) = 0$.

If $t_0<9$, then we must have $B(t_0)-A(t_0) = L = 300$. From this we see that $A$ covers 50km by 10:00, and the remaining 250km in $\frac{250}{125} = 2$, hence they meet at $t_1 = $ 12:00.

Now suppose that $t_0 \in [9,10)$. Then $B(t_0)-A(t_0) = L-50(t_0-9) = 300$, from which we get $t_0 = \frac{L+150}{50}$. Since $t_0 \in [9,10)$, we must have $L \in 50[9,10)-150 = [300,350)$. It is clear that $t_1>10$, so then $B(t_1) - A(t_1) = L-75(t_1-10)-50(t_1-9) = 0$, from which we get $t_1 = \frac{L+1200}{125}$. Thus we have $t_1 \in [\mbox{12:00},\mbox{12:24}]$.

Now suppose $t_0 \geq 10$. Then $B(t_0)-A(t_0) = L-75(t_0-10)-50(t_0-9) = 300$, from which we get $t_0 = \frac{L+900}{125}$. (Since $t_0 \geq 10$, this implies that $L \geq 350$.) From $B(t_1) - A(t_1) = L-75(t_1-10)-50(t_1-9) = 0$, we get $t_1 = \frac{L+1200}{125}$, as above. Since $L \geq 350$, we have $t_1 \geq \mbox{12:24}$.

So, tell your interviewer that the question is ambiguous :-).

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At 10:00 they would be $300-(50*1) = 250 km$. The relative speed of B and A then is $50+(1.5*50) = 125 kmph$. The time taken for them to meet would then be $250/125 = 2 hrs$. So they meet at $12:00$.

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First substract the 50 km A walks to get the distance between them at 10:00 is 250.

Say X is the speed of A and y is the speed of B. Then speed of $\frac{3X}{2}=b$ so their speed toward each other is $\frac{5X}{2}$=$\frac{250}{2}$ so $250t=500$ so $t=2 hours So they will meet at 12:00

Here is the link to a similar problem that was showed in the tv series numbers: http://numb3rs.wolfram.com/504/puzzle.html

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At $10:00$ they would be only $250\ km$ apart.

The speed of $A$ is $50$kph and the speed of $B$ is $75$kph. let $d_A$ and $d_B$ represent the distances they each travel after $10:00$. When they hit:

$d_A+d_B=250$

remember, $d=v\times t$. Denote $v_A$ and $v_B$ accordingly, noting that they'll both be travelling for the same amount of time. so:

$$v_At+v_Bt=250\implies t=\frac{250}{v_A+v_B}=\frac{250}{125}=2\ hrs$$ They meet at $12:00$.

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