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Man Cat 4 #12-High speed train v. Regular Train

Expert replies
by joyseychow » Tue May 19, 2009 10:42 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)
Last edited by joyseychow on Thu May 21, 2009 5:14 am, edited 1 time in total.
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Source: — Problem Solving |

by dumb.doofus » Wed May 20, 2009 12:07 am
Answer Choice A

Speed of highspeed train = z/x
Speed of lowspeed train = z/y

Suppose they meet after time T..

Distance covered by HST = zT/x
Distance covered by LST = zT/y

zT/x + zT/y = z

or T = xy/x+y

So now distance covered by HST - Distance covered by LST is what the question is asking..

=> zy/(x+y) - zx/(x+y)

= z(y-x)/(x+y) --------- Answer Choice A
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by lunarpower » Wed May 20, 2009 3:10 am
NOTE: this problem is missing ALL the parentheses in denominators. this makes the answer choices misleading.

every one of the answer choices is supposed to be a single fraction.
for instance, (a) and (b) are fractions with denominator (x + y), etc.

--

the above poster's solution is good.

here are some other things about this problem:

--

* you can PLUG IN NUMBERS.
this is not entirely easy to do; you have to select an x, y, z that don't make the numbers too ugly. however, it can certainly be done.
if you want to see a specific example of plug-in-numbers for this problem, go ahead and post back.

--

* you can use PROCESS OF ELIMINATION. here are easy ways to eliminate three of the choices:

since the "regular" train is slower than the high-speed train, y must be greater than x.
since y is greater than x, (b) and (d) are negative numbers, and are thus wrong.

if x = y, then the answer to the problem must be zero (because the two trains would then be moving at the same speed, despite the designation of one of them as "fast" and the other as "slow").
choice (c) is not zero if x = y (in fact it's infinity!), so it's wrong.

that leaves (a) vs. (e).

you can choose between these two if you take a slightly more complicated tack: use UNITS.

the units of choice (a) are (miles * hours) / hours, which works out to miles. this is as required.

the units of choice (e) are (hours * hours * hours) / hours, which works out to hours squared. that makes no sense.
you can also eliminate choice (d) on this basis, too.

we're left with (a).
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by joyseychow » Thu May 21, 2009 8:09 pm
lunarpower wrote:NOTE: this problem is missing ALL the parentheses in denominators. this makes the answer choices misleading.

every one of the answer choices is supposed to be a single fraction.
for instance, (a) and (b) are fractions with denominator (x + y), etc.

Sorry. I've already edited that.
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by joyseychow » Thu May 21, 2009 8:16 pm
lunarpower wrote:NOTE: this problem is missing ALL the parentheses in denominators. this makes the answer choices misleading.

every one of the answer choices is supposed to be a single fraction.
for instance, (a) and (b) are fractions with denominator (x + y), etc.

--

the above poster's solution is good.

here are some other things about this problem:

--

* you can PLUG IN NUMBERS.
this is not entirely easy to do; you have to select an x, y, z that don't make the numbers too ugly. however, it can certainly be done.
if you want to see a specific example of plug-in-numbers for this problem, go ahead and post back.

--

* you can use PROCESS OF ELIMINATION. here are easy ways to eliminate three of the choices:

since the "regular" train is slower than the high-speed train, y must be greater than x.
since y is greater than x, (b) and (d) are negative numbers, and are thus wrong.

if x = y, then the answer to the problem must be zero (because the two trains would then be moving at the same speed, despite the designation of one of them as "fast" and the other as "slow").
choice (c) is not zero if x = y (in fact it's infinity!), so it's wrong.

that leaves (a) vs. (e).

you can choose between these two if you take a slightly more complicated tack: use UNITS.

the units of choice (a) are (miles * hours) / hours, which works out to miles. this is as required.

the units of choice (e) are (hours * hours * hours) / hours, which works out to hours squared. that makes no sense.
you can also eliminate choice (d) on this basis, too.

we're left with (a).
I didn't thought of method of elimination. Brilliant!
Appreciate if you could demonstrate picking number method too. At times, this could be faster that dealing with just variables. Thanks!
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