tough question

Expert replies

sana.noor: Legendary Member; Posts: 512; Joined: Mon Jun 18, 2012 11:31 pmFollowed by: 20 members; Thanked: 42 times

by sana.noor » Mon May 13, 2013 4:53 am

At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E

Work hard in Silence, Let Success make the noise.

If you found my Post really helpful, then don't forget to click the Thank/follow me button.

Like this post

Source: Beat The GMAT — Problem Solving | Mon May 13, 2013 4:53 am

mkdureja: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Fri Jan 04, 2013 12:55 am; Thanked: 11 times

by mkdureja » Mon May 13, 2013 5:16 am

From 1PM to 1.30 PM, only Train X covers the distance between the two trains.
Its speed = r miles/hr. So, distance it covers in 1/2 hr. is r/2 miles.

Now, from then on, both the trains cover the remaining distance between them 'p-r/2' miles(subtracting the distance covered by Train X in the first 1/2 hr. from the total distance), with the speed of 'r+s' miles/hr (combined speed).

So, time it will take them to reduce their distance to zero will be =
(p-r/2)/(r+s).

Add initial 1/2 hr. to get the answer = 1/2 + (p-r/2)/(r+s).

Hope its clear.

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon May 13, 2013 5:43 am

sana.noor wrote:At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E

Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Like this post

subhakam: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Mon Jun 21, 2010 12:37 pm

by subhakam » Tue May 14, 2013 6:03 pm

GMATGuruNY wrote:
sana.noor wrote:At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E
Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

Hello GMAT GuruNY - can you please comment whether what my algebraic approach is correct ? (Need to know where I am going wrong)

Let us say we need to fine time (t) when the two trains meet after 1 pm

For Train X - speed is r miles per hour
since it leaves at 1 pm it has 30 minutes head start over train y , therefore total time it gets to travel is (t+1/2)
For Train Y - speed is s miles per hour and time it travels is t
Now , both of them cover the given distance p , therefore equation is
r(t+1/2) + st = p => [r{(2t+1)}/]2 + st = p
=> 2rt+ r + 2st = 2p
=> 2rt + 2st = 2p-r => 2t(r+s)= 2p-r
=> t = [p-r/2] / (r+s) = option D
However do we need to add 1/2 more since travel by x is actually at 1 pm and total time traveled by train X = actually t +1/2 ? in other words
1/2 + ([p-r/2]/(r+s)) ? Let me know
In the same way could one of the right options ever be t = ([p+s/2]/(r+s))- 1/2 ?? (if we say from point of X time traveled is t and from point of Y time traveled is (t-1/2)i.e. train Y travels 30 minutes less than that traveled by train X ?
Please let me know - i want to get this right conceptually and algebraically
Thanks
Subhakam

Like this post

Atekihcan: Master | Next Rank: 500 Posts; Posts: 149; Joined: Wed May 01, 2013 10:37 pmFollowed by: 9 members; Thanked: 54 times

by Atekihcan » Wed May 15, 2013 12:11 am

subhakam wrote:Let us say we need to fine time (t) when the two trains meet after 1 pm

For Train X - speed is r miles per hour
since it leaves at 1 pm it has 30 minutes head start over train y , therefore total time it gets to travel is (t+1/2)
For Train Y - speed is s miles per hour and time it travels is t

As you are assuming that t is the time in hours after 1 PM when they meet, train X has traveled for t hours and train Y has traveled for (t - 1/2) hours.

You are assuming that t is the time in hours after 1 PM when they meet, but you are doing the calculation assuming that t is the time in hours after 1:30 PM when they meet.

So, in our final answer you need to add 1/2.

Like this post

[email protected]: Master | Next Rank: 500 Posts; Posts: 429; Joined: Wed Sep 19, 2012 11:38 pmFollowed by: 4 members; Thanked: 6 times

by [email protected] » Wed May 15, 2013 2:57 am

GMATGuruNY wrote:
sana.noor wrote:At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

Why have we taken 2 as the rate of Y can you please help me understand!

OA is E
Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed May 15, 2013 3:16 am

[email protected] wrote:
GMATGuruNY wrote:
sana.noor wrote:At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E
Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.
Why have we taken 2 as the rate of Y can you please help me understand!

When p=10 and r=20, train Y does not travel any portion of the distance.
Thus, we can plug in ANY VALUE for s (train Y's speed).
I chose s=2 because small values typically are easier to evaluate.
If we plug in p=10, r=20, and s=1000, the result is the same: only C and E yield the target value (1/2).

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review