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RTD or RTW problems - specific to time

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RTD or RTW problems - specific to time

by priya_kapoor » Sun Jul 18, 2010 1:28 pm
So I've been doing these RTD RTW problems for a while following solutions but I really want to understand the concept. Here is a good question to ask my questions with:

" Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather walked when they two meet?"

So I get the question. I guessed I am confused as to why the time would be the following:

Stacy = 6 X (T +1/4); Heather = 5 X T

rather than it being:

Stacy= 6 X T; Heather= 5 X (T-1/4)

Heather's time is indeed 1/4 less than Stacy's time since she started 1/4 after her... I guess I am wondering why is it this way when the problem states "Heather starts her journey 24 minutes after Stacy..."

Any help provided would be so helpful. Thanks in advance!
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by Stuart@KaplanGMAT » Sun Jul 18, 2010 1:43 pm
priya_kapoor wrote:So I've been doing these RTD RTW problems for a while following solutions but I really want to understand the concept. Here is a good question to ask my questions with:

" Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather walked when they two meet?"

So I get the question. I guessed I am confused as to why the time would be the following:

Stacy = 6 X (T +1/4); Heather = 5 X T

rather than it being:

Stacy= 6 X T; Heather= 5 X (T-1/4)

Heather's time is indeed 1/4 less than Stacy's time since she started 1/4 after her... I guess I am wondering why is it this way when the problem states "Heather starts her journey 24 minutes after Stacy..."

Any help provided would be so helpful. Thanks in advance!
Hi!

Neither of those equations is correct, since the difference in times is 24 minutes, which is 2/5 of an hour, not 1/4.

If T is the total time of travel from when Stacy started walking, then you'd use:

t(stacy) = T
t(heather) = T - 2/5

However, this question is asking for Heather's distance travelled. Since we want to solve for d(heather), it's going to make the math simpler to set:

t(heather) = T
t(stacy) = T + 2/5

We know the total distance travelled is 20 miles, and that:

r(stacy) = 6
r(heather) = 5

So:

Distance = rate * time

20 = 6(T + 2/5) + 5T
20 = 6T + 12/5 + 5T

100/5 - 12/5 = 11T
88/5 = 11T

88/55 = T
8/5 = T

Finally, solving for d(heather)

d(heather) = r(heather) * t(heather)
d(heather) = 5 * (8/5) = 8

So, Heather has travelled 8 miles total when they meet.

(As an aside, I'd solve the question without all the math by drawing a diagram, plotting how far apart they are after 24 minutes and either solving a simpler equation or using weighted averages.)
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by jaskaran » Sun Jul 18, 2010 1:47 pm
Hi Priya,

Both the equations are the same.

'If Heather starts her journey 24 minutes after Stacy' - This statement means that Stacey has a head start and has traveled for 24 mins more than Heather, so if you represent Stacey's time as 't', then you would have to represent Heather with (t-0.4) because you have take into consideration that Heather 24 mins lesser than Stacey, which make the equation right.

Similarly, if you choose to represent Heather to have traveled 't' for hours, then Stacey would have traveled (t+0.4)

Hope this clarified the confusion to some extent!

PS- 24 mins converted to hours would 24/60=2/5-0.4
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by Stuart@KaplanGMAT » Sun Jul 18, 2010 1:47 pm
priya_kapoor wrote:Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather walked when they two meet?
The way I'd solve:

Stacy starts 24 minutes (or 2/5 of an hour early), so let's calculate how far Stacy has travelled before Heather gets moving.

d = rt
d = 6(2/5) = 12/5 miles

So, when Heather starts moving they're 20 - 12/5 = 88/5 miles apart.

When two objects move directly toward each other, you find the combined rate by ADDING their individual rates.

So, each hour Heather and Stacy get 5 + 6 = 11 miles closer together (combined rate of 11mph).

Since Heather is travelling at 5mph, Heather will cover 5/11 of the distance before they meet.

(5/11) * (88/5) = 88/11 = 8 miles.

Done!
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by kmittal82 » Sun Jul 18, 2010 2:46 pm
There's a pretty cool trick you can use when you get problems involving 2 objects moving towards/away from each other. You can add/subtract their speeds and treat it as one object (you can prove this by simple arithmetic). Lets apply it here.

Firstly, lets see how much distance did Stacy travel in 24 minutes.

Stacy's speed = 6 miles per 60 mins, thus in 24 minutes she has already travelled 2.4miles

Hence, total distance left is 20 - 2.4 = 17.6

Now, for the trickery :)

Since heather and stacy are travelling towards each other, you can add their speeds. Thus, speed = 5+6 = 11mph
Distance = 17.6
Time when they meet = 17.6/11

In that much time, Heather travels => 17.6/11 x 5 = 88/11 = 8

For other problems, maybe when 2 objects are moving in the same direction, you should subtract their velocities and use a similar method. Its the best shortcut I've ever found! :)
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