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At 10:30 am car $A$ starts from point $A$ towards point $B$ at the speed of $65$ km/hr, at the same time another car left from point $B$ towards point $A$ at the speed of $70$ km/hr, the total distance between two points is $810$ km, at what time does these two cars meet ?

This is what I have tried,

$t_1 = \frac{810}{65} \,$ km/hr $= 12.46 \, $hr
$t_2 = \frac{810}{70} \,$ km/hr $= 11.57\,$ hr
$t_1-t_2 = 0.89 \,$ hr
$t_1-t_2 = 0.89\cdot 60 = 53.4 \,$ minutes

but this couldn't be the answer because how could these two cars could meet after $53.4$ minutes ?

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@Joe, when they start they are 810km apart; after 1 hour the person leaving from point A would have traveled 65km, so if the other person had not left from point B, then the distance between them would be 810-65=745km. But the other person also helps close the distance between them by traveling 70km in the 1st hour from point B towards A, so the distance between them is 810-(65+75)=810-135=675km. So they are effectively subtracting 135km worth of distance per hour from the original distance between them.

So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.

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    $\begingroup$ it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? $\endgroup$ – Joe Jan 19 '15 at 18:52
  • $\begingroup$ Yes, this is exactly right. $\endgroup$ – Acemanhattan Jan 19 '15 at 18:56
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    $\begingroup$ But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. $\endgroup$ – turkeyhundt Jan 19 '15 at 19:00
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    $\begingroup$ Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. $\endgroup$ – Acemanhattan Jan 19 '15 at 19:02
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    $\begingroup$ Thanks a lot for great explanation. =) $\endgroup$ – Joe Jan 19 '15 at 19:03
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Hint: Think about how much closer the cars get each hour.

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They are approaching each other at an effective speed of 135 km/hr...

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  • $\begingroup$ but at what time does these two cars meet ? would that be 135/2 ? $\endgroup$ – Joe Jan 19 '15 at 18:27
  • $\begingroup$ like 135/2 = 67.5 and then 810/67.5 = 12 ? $\endgroup$ – Joe Jan 19 '15 at 18:29
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    $\begingroup$ no. it would be how long it would take to travel 810km at a rate of 135km/hr. Because every hour, they get 135km closer to each other. $\endgroup$ – turkeyhundt Jan 19 '15 at 18:50
  • $\begingroup$ it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? $\endgroup$ – Joe Jan 19 '15 at 18:54
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    $\begingroup$ Yes. very good. $\endgroup$ – turkeyhundt Jan 19 '15 at 18:59
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Hint: The total distance between the two cars is initially $810~\text{km}$. Thus, in order to meet, the combined distance they travel is $810~\text{km}$. Since they are travelling toward each other, they get $65~\text{km} + 70~\text{km} = 135~\text{km}$ closer to each other each hour.

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  • $\begingroup$ so then I have two divide 135 by 2 and then divide the answer by 810, like this, 135/2 = 67.5 then 810/67.5 = 12hours right ? so they will meet at 10:30 pm right ? $\endgroup$ – Joe Jan 19 '15 at 18:48
  • $\begingroup$ No. What you did is compute the time it would take for the average distance travelled by the cars to be $810~\text{km}$, by which time they would have long passed each other. The cars get $135~\text{km}$ closer to each other each hour. You want to reduce the distance between them, which is initially $810~\text{km}$, to $0~\text{km}$. $\endgroup$ – N. F. Taussig Jan 19 '15 at 18:51
  • $\begingroup$ it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? $\endgroup$ – Joe Jan 19 '15 at 18:53
  • $\begingroup$ That is correct. $\endgroup$ – N. F. Taussig Jan 19 '15 at 18:54
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    $\begingroup$ They do not have to cover $810~\text{km}$ individually. After six hours, the slower car has travelled $$65~\frac{\text{km}}{\text{h}} \cdot 6~\text{h} = 390~\text{km}$$ so it is now $810~\text{km} - 390~\text{km} = 420~\text{km}$ from where the faster car began. Since the faster car has travelled $$70~\frac{\text{km}}{\text{h}} \cdot 6~\text{h} = 420~\text{km}$$ toward the slower car, they meet at 4:30 pm. Draw a picture. $\endgroup$ – N. F. Taussig Jan 19 '15 at 19:03

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