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Rate Question, please help!

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axl169
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Rate Question, please help! [#permalink] Rate Question, please help!   07 Oct 2006, 21:02
Hi, got stumped on this one. Any help would be really appreciated it.

2 Trains, X&Y, start simultaneously on opposite ends of a 100-mile route and travled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100 mile trip in 5 hours. Train Y, also traveling at a constant rate completed the 100 mile trip in 3 hours. How many miles had train X traveled when it met Train Y?

A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5



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Re: Rate Question, please help! [#permalink] Rate Question, please help!   07 Oct 2006, 21:59
Rate x Time = Distance
Train X 5 x t = d
Train Y 3 x t = 100-d

The only variable they have in common is that both trains have been travelling for the same amount of time...thus, set each equation equal to each other in terms of d.

so Train X = t = d/5
and Train Y = t = (100-d)/3

Set each equation equal to each other:
d/5 = (100-d)/3
Cross multiply and you get:
3d = 500 - 5d

Solve for d and you get 62.5. Thus the answer is 62.5
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   07 Oct 2006, 22:10
The correct answer was A. 37.5. But I don't know understand how that is.
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   07 Oct 2006, 23:11
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axl169 wrote:
The correct answer was A. 37.5. But I don't know understand how that is.

Yes the answer will be 37.5

Speed of train X = 100/5 mph
Speed of train y = 100/3 mph

Let's say when the 2 trains meet, X had travelled m miles. and Y had traveled 100-m miles.

Since the trains started at the same time, they would have taken same time to cover m and 100-m miles respectively.

m/(100/5) = 100-m/(100/3)
5m = 300-3m
8m = 300
m = 37.5 miles
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 03:12
Alternatively you can solve this this way:

Train X: rate=100/5=20miles/hour; time=t; d=d
Train Y: rate=100/3=33*1/3miles/hour; time=t; d=100-d

construct two equations: r*t=d.... train x: 20*t=d
train y: 33*1/3t=100-d

then solve...

33*1/3t=100-20t
33*1/3t+20t=100
53*1/3t=100
t=300/160
t=1.875

The questions asks how many miles had train x trveled... we know the formula: r*t=d

We go back to the begining and plug in the values for train x....
r=20 miles/hour; t=1.875;d=?

20*1.875=37.5
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 06:27
Clearly the speeds ratio of A and B is 3:5
So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 08:21
cicerone wrote:
Clearly the speeds ratio of A and B is 3:5
So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,


Cicerone
Where do you get 3/8 from? I like your ratio method of doing this.
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 10:06
$uckafr33 wrote:
cicerone wrote:
Clearly the speeds ratio of A and B is 3:5
So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,


Cicerone
Where do you get 3/8 from? I like your ratio method of doing this.


Hey look at the bold part
Since the distance is shared in the raio of 3:5
A should cover 3/8 and in the same time B will cover 5/8 .............
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 10:51
cicerone wrote:
Clearly the speeds ratio of A and B is 3:5
So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,


Yes, it works in this example, but it is better to have a clear concept of solving such problems. This ratio example won't work for other "rate" problems...
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 12:12
SimaQ wrote:
cicerone wrote:
Clearly the speeds ratio of A and B is 3:5
So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,


Yes, it works in this example, but it is better to have a clear concept of solving such problems. This ratio example won't work for other "rate" problems...


Hey SimaQ, please do not say this..............
Any question form Time and Distance out of Speed , Distance and Time if one of them is constant, i bet i will do it using ratios only..............
Why don't u try to test me........
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 13:56
cicerone's method is neater but this is how i did it. trick to remember with this type of question where 2 things travel towards each other is that together they travel the total distance, whether its lifts, trains, cars, water going down a pipe, frogs etc :roll: :roll:

so

where t = time

100/5 t + 100/3 t = 100
300/15 t + 500/15 t = 100
t = 1500/800 = 15/8

15/8 * 100/5 = 75/2 = 37.5
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 18:01
Speed of X is 100/5 = 20.0 m/h.
Speed of Y is 100/3 = 33.3 m/h.

now if X and Y met, that means total distance travelled would be 100 miles (since both start from opposite ends).

adding up the speed gives the consolidated speed which is 20 + 33.3 = 53.3 miles/hour

now, again applying distance-time formula, time taken is 100/53.3 = approx 2 hours.

so, X and Y met after 2 hours of their journey.

X travelled less than 40 miles in approximately 2 hours..

hence A.
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Re: Rate Question, please help! [#permalink] Rate Question, please help!   08 Oct 2006, 19:22
a)37.5

The time after which both trains meet = t
At time t, total distance traveled = 100 miles.

distance traveled by x = (100/5)*t
distance traveled by y = (100/3)*t

Putting these together,

(100/5)*t + (100/3)*t = 100
Simplifying, we get t=15/8

In 15/8 hrs, train X has traveled: 20*(15/8) = 37.5



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Re: Rate Question, please help! [#permalink]
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