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Extra distance traveled by faster train

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Extra distance traveled by faster train

by makkiemaps » Fri Mar 04, 2011 5:54 pm
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

A. z[(y - x)/x + y]
B. z[(x - y)/x + y]
C.
D.
E.

OA : A

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Instead of telling your way of getting the right answer, I'll be obliged if you can tell what's wrong in my method

Difference in speed = x - y
Time taken to reach the destination : z / (x+y)

Extra distance by faster train = (x - y) * z(x + y)

My answer: "B"
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by srcc25anu » Fri Mar 04, 2011 6:14 pm
difference in speed is not x-y. Rather its z/x - z/y (as x and y are respective hours)
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by srcc25anu » Fri Mar 04, 2011 6:23 pm
let z/x be speed of faster train and z/y be speed of slower train.

time after which faster and slower trains meet = z / (z/x + z/y) that equates to xy / (x+y) hrs

dist traveled by fast train in xy/(x+y) hrs = xy / (x+y) * z/x = zy/(x+Y) --------- 1
dist traveled by fast train in xy/(x+y) hrs = xy / (x+y) * z/y = zx/(x+Y) ------------2

diff in distance = z(y-x) / (x+Y) that is 1 - 2 above
Ans A
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by makkiemaps » Fri Mar 04, 2011 7:43 pm
Thanks guys. My biggest problem in quantitative section is that I don't read the question properly. I read x and y as respective speeds of the train. :)
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by vineeshp » Sat Mar 05, 2011 9:51 am
Very important to read the questions carefully. Especially the units. Sometimes GMAT brings in unconventional units like miles and minutes. You may be inclined to think the speed is miles per hour. Take enough time to read the question. You are not going to lose anything with the extra time you take for that that you wont lose with a wrong answer. :)

All the best! I see your test date is early April. :)
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by makkiemaps » Sat Mar 05, 2011 3:00 pm
vineeshp wrote:Very important to read the questions carefully. Especially the units. Sometimes GMAT brings in unconventional units like miles and minutes. You may be inclined to think the speed is miles per hour. Take enough time to read the question. You are not going to lose anything with the extra time you take for that that you wont lose with a wrong answer. :)

All the best! I see your test date is early April. :)
Thanks for the suggestion Vineesh. Yes, I will be extra careful reading the question from now on. I am also not able to run through entire quant section in 75 mins. In fact in my last two MGMAT practice tests, I was forced to guess & mark the answer after reading half of the question because I was still on Q30 with 5 mins to spare :( .

Scores:

MGMAT 1 - 650 (44Q, 33V)
MGMAT 2 - 670 (46Q, 35V)
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by GMATGuruNY » Sat Mar 05, 2011 4:57 pm
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

(A) z(y - x)/x + y

(B) z(x - y)/x + y

(C) z(x + y)/y - x

(D) xy(x - y)/x + y

(E) xy(y - x)/x + y
We can plug in our own values.

Let z = 200 miles.
Let x = 10 hours.
Rate for high-speed train = d/t = 200/10 = 20 miles per hour.
Let y = 40 hours.
Rate for regular train = d/t = 200/40 = 5 miles per hour.
When elements work together to complete a task, we combine their rates. Since the 2 trains are working together to cover the 200 miles, their combined rate = 20+5 = 25 miles per hour.
Time for the trains to meet = d/r = 200/25 = 8 hours.
Distance traveled by high-speed train = r*t = 20*8 = 160 miles.
Distance traveled by regular train = r*t = 8*5 = 40 miles.
Difference = 160-40 = 120 miles. This is our target.

Now we plug z = 200, x = 10, and y = 40 into the answer choices to see which yields our target of 120.

Only answer choice A works:
z(y - x)/x + y = 200(40-10)/(10+40) = 200*30/50 = 120.

The correct answer is A.
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