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WHile on a straight road, car X and Y are travelling at [#permalink]
 13 Oct 2008, 05:36
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The question is:
WHile on a straight road, car X and Y are travelling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of Car Y?
1. car X is travelling at 50 mph and car Y is travelling at 40 mph
2. 3 minutes ago, car X was 1/2 mile ahead of car Y.
My problem is, while it makes logical sense in my head that both will be individually be enough to answer the question, I am not sure, because I can not reach the answer using the second statement.
Couldl somebody please help explain it to me.
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Re: DS- Cars on a road! [#permalink]
13 Oct 2008, 11:11
Let's say after certain time T, the two cars are 2 miles apart.
In this time T, the distance travelled by car X = D1
and in the same Time T, distance travelled by car Y = D2
We know D(x) - D(y) = 2 => D(y) = D(x) + 2
Lets take case 1)
Car X is speeding at 50MPH
Car Y is speeding at 40MPH
So, (D(x) / 50) = (D(y) / 40) => (D(x) / 5) = ((D(x) + 2) / 4) => D(x) = 10 & D(y) = 8
So Time taken = 8/40 (or) 10/50 = 0.2 hrs.
So I) is sufficient
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Re: DS- Cars on a road! [#permalink]
13 Oct 2008, 12:38
Stmt2 is also sufficient. It essentially gives an idea of relative speed of car X compared to car Y.
If relative speed of car X to car Y is x miles per minute then in 3 minutes, it travelled 1/2 miles and hence, in 6 minutes it will travel 1 mile.
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Re: DS- Cars on a road! [#permalink]
13 Oct 2008, 22:35
Thanks Smahmkl and shahthakur.
The correct answer is actually D. Both statements are individually correct.
Intuitively, that's what I thought as well..
But could somebody shed mroe light on it, by arriving at the answer mathematically, from the second statement.
Please taht would hlep me understand it better.
he question is:
WHile on a straight road, car X and Y are travelling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of Car Y?
1. car X is travelling at 50 mph and car Y is travelling at 40 mph
2. 3 minutes ago, car X was 1/2 mile ahead of car Y.
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Re: DS- Cars on a road! [#permalink]
13 Oct 2008, 22:42
hbs2012 wrote: Thanks Smahmkl and shahthakur.
The correct answer is actually D. Both statements are individually correct.
Intuitively, that's what I thought as well..
But could somebody shed mroe light on it, by arriving at the answer mathematically, from the second statement.
Please taht would hlep me understand it better.
he question is:
WHile on a straight road, car X and Y are travelling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of Car Y?
1. car X is travelling at 50 mph and car Y is travelling at 40 mph
2. 3 minutes ago, car X was 1/2 mile ahead of car Y. Is my explanation above of stmt2 not clear?
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Re: DS- Cars on a road! [#permalink]
14 Oct 2008, 04:41
Thanks SCthakur...
I just went through ur explanation once more.. and giving it better thought.. it does seem to make more sense now.
thanks for your help.
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Re: DS- Cars on a road!
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14 Oct 2008, 04:41
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