GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Aug 2018, 03:53

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

A train after traveling at 50kmph meets with an accident and then

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 00:43
2
26
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

74% (03:43) correct 26% (03:34) wrong based on 154 sessions

HideShow timer Statistics

A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.

a) 45
b) 33
c) 48
d) 55
e) 61
Most Helpful Community Reply
Senior Manager
Senior Manager
User avatar
Joined: 13 Jun 2013
Posts: 277
Premium Member
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 02:46
14
6
Blackbox wrote:
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.

a) 45
b) 33
c) 48
d) 55
e) 61


here is less mathematical approach.

let the two end points between which this train is running be A and B. Also suppose that accident occurs at C which is at a distance of 50km from A. thus we have

A---------50km----------C----------------------------B 1)

Time taken to travel the distance after the accident i.e. CB is 35 minutes more than the usual time.

now, if the accident were to occur at 24 km farther, then train would have reached only 25 minutes late. i.e.

A--------50 km----------C---------24--------D---------------B 2)

if we look at the situation depicted in 1) and 2) we will see that 10 minute difference in time is occurring because of this 24 Km. which is traveled by its usual speed (=x) in the later case. thus we have

\(\frac{24*4}{3x}\) - \(\frac{24}{x}\) = \(\frac{10}{60}\)
\(\frac{32}{x}\) - \(\frac{24}{x}\) = \(\frac{1}{6}\)
solving this we get x= 48km/h
General Discussion
Manager
Manager
avatar
Joined: 21 Sep 2012
Posts: 218
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 01:43
3
1
Let y be the balance distance to be covered and x be the former speed.
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late
so, y/(3x/4) - y/x = 35/60
4y/3x - y/x = 7/12
y/x(4/3-1)=7/12
y/x*1/3=7/12
y/x=7/4
4y-7x=0 ........ 1

Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late
so, (y-24)/(3x/4) - (y-24)/x = 25/60
4(y-24)/3x - (y-24)/x = 5/12
(y-24)/x (4/3-1) = 5/12
(y-24)/x *1/3 = 5/12
(y-24)*12 = 3x*5
(y-24)*4 = 5x
4y-5x = 96 ....... 2

eq2 - eq1
2x=96
x=48

Ans = C
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 02:13
Chinmay- Thanks for the explanation. I have a follow-up question:
In Eq 1:

You said 'y' is the balance distance . Then in the question -
Quote:
y/(3x/4) - y/x = 35/60

since you are subtracting the latter time (post-accident) from the former time (pre-accident), shouldn't \(\frac{y}{x}\) be \(\frac{50}{x}\)?
Manager
Manager
avatar
Joined: 21 Sep 2012
Posts: 218
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 02:24
1
Blackbox wrote:
Chinmay- Thanks for the explanation. I have a follow-up question:
In Eq 1:

You said 'y' is the balance distance . Then in the question -
Quote:
y/(3x/4) - y/x = 35/60

since you are subtracting the latter time (post-accident) from the former time (pre-accident), shouldn't \(\frac{y}{x}\) be \(\frac{50}{x}\)?


My friend, I am not subtracting the latter time (post accident) from the former time. I am subtracting the time that train would have taken, had there been no accident from the actual time it took. If there had been no accident, the train would have continued at its speed (i.e. x) and would have taken y/x hrs to complete. Since there was an accident, speed was reduced (3x/4) and it actually took y/(3x/4) hrs to cover balance distance. I hope it helps.
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 02:46
Chinmay - Sorry for being thick, but I struggle to understand why you would subtract the times if there were no accident. Is it because they say the train arrived late? What if there was a case when the train arrives early? Would you add times?
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 02:54
Manpreet - Easier solution but I guess I am missing the fundamentals here - "why subtract?"
Senior Manager
Senior Manager
User avatar
Joined: 13 Jun 2013
Posts: 277
Premium Member
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 03:03
1
1
Blackbox wrote:
Manpreet - Easier solution but I guess I am missing the fundamentals here - "why subtract?"


hi, remember the difference of 10 minutes is occurring because the 24km is travelled at two different speed. x is the usual speed and 3/4x is the speed after the accident. and to convert minutes into hour i have divided 10 by 60.
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 07 Dec 2014, 04:22
manpreetsingh86 wrote:
Blackbox wrote:
Manpreet - Easier solution but I guess I am missing the fundamentals here - "why subtract?"


hi, remember the difference of 10 minutes is occurring because the 24km is travelled at two different speed. x is the usual speed and 3/4x is the speed after the accident. and to convert minutes into hour i have divided 10 by 60.


I see it now, so you are only making an equation for that 24 km part equating the difference to 10 minutes. The rest of the question is actually irrelevant. Hmm... interesting. Are there any other problems on GMAT club in line with this concept?

Any other approaches/discussions are welcome.
Intern
Intern
User avatar
Joined: 12 Aug 2014
Posts: 7
Location: Peru
GPA: 3.82
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 08 Dec 2014, 05:55
1
Somebody can explain me why is 4/3x - 24/x and not 24/x- 4/3x????? 24/x is not bigger?
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 08 Dec 2014, 10:44
marcosgensollen wrote:
Somebody can explain me why is 4/3x - 24/x and not 24/x- 4/3x????? 24/x is not bigger?


Not sure if I followed your question. Could you elaborate? Also, I believe it is \(\frac{24*4}{3x} - \frac{24}{x} = \frac{10}{60}\) in which case 24/x is smaller. But please don't take my word for it. I am still brushing up this concept.
Intern
Intern
User avatar
Joined: 12 Aug 2014
Posts: 7
Location: Peru
GPA: 3.82
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 08 Dec 2014, 11:20
Blackbox wrote:
marcosgensollen wrote:
Somebody can explain me why is 4/3x - 24/x and not 24/x- 4/3x????? 24/x is not bigger?


Not sure if I followed your question. Could you elaborate? Also, I believe it is \(\frac{24*4}{3x} - \frac{24}{x} = \frac{10}{60}\) in which case 24/x is smaller. But please don't take my word for it. I am still brushing up this concept.



I understand. If (3x/4)t=24 and (x)(t)= 24 -------> (1) t= 4*24/(3x) and (2) t= 24/x. If you analyze each formula, you are going to conclude that 24/x take LESS time than 4*24/3x to complete the 10 minutes. It's a very confusing exercise because you have to realized that youre substracting time (minutes) and no rates.
SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 09 Dec 2014, 22:17
10
4
Say total distance = d Kms

Speed = s Km/hr

Refer diagram below:

Attachment:
speed.png
speed.png [ 7.74 KiB | Viewed 43909 times ]


Setting up the time equation for scenario I

\(\frac{50}{s} + \frac{d-50}{\frac{3s}{4}} = \frac{d}{s} + \frac{35}{60}\)

4d - 7s = 200 .................. (1)

Setting up the time equation for scenario II

\(\frac{74}{s} + \frac{d-74}{\frac{3s}{4}} = \frac{d}{s} + \frac{25}{60}\)

4d - 5s = 74*4 .................. (2)

(2) - (1)

2s = 74*4 - 200

s = 2(74 - 50) = 48

Answer = C
_________________

Kindly press "+1 Kudos" to appreciate :)

SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 09 Dec 2014, 22:29
1
Blackbox wrote:
Chinmay - Sorry for being thick, but I struggle to understand why you would subtract the times if there were no accident. Is it because they say the train arrived late? What if there was a case when the train arrives early? Would you add times?



That's precisely correct. Please refer my post above.

Usual speed was "s" and the usual time taken would had been "d/s"

However, due to accident, the speed decreased, so negative sign.

Had the question be like "after 50 kms, they added another engine due to which train reached 10 minutes earlier", then it had been positive sign. (Retrospective)

Chinmay & I have approached the same way, however the +ve/-ve signs are taken at different sides

Refer the diagram for more clarity
_________________

Kindly press "+1 Kudos" to appreciate :)

SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 09 Dec 2014, 22:33
1
marcosgensollen wrote:
Somebody can explain me why is 4/3x - 24/x and not 24/x- 4/3x????? 24/x is not bigger?


The equation is setup for time = distance/speed

Less the speed, more time would be required as distance remains constant

Equation setup accordingly
_________________

Kindly press "+1 Kudos" to appreciate :)

SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1835
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 09 Dec 2014, 22:34
1
marcosgensollen wrote:
Blackbox wrote:
marcosgensollen wrote:
Somebody can explain me why is 4/3x - 24/x and not 24/x- 4/3x????? 24/x is not bigger?


Not sure if I followed your question. Could you elaborate? Also, I believe it is \(\frac{24*4}{3x} - \frac{24}{x} = \frac{10}{60}\) in which case 24/x is smaller. But please don't take my word for it. I am still brushing up this concept.



I understand. If (3x/4)t=24 and (x)(t)= 24 -------> (1) t= 4*24/(3x) and (2) t= 24/x. If you analyze each formula, you are going to conclude that 24/x take LESS time than 4*24/3x to complete the 10 minutes. It's a very confusing exercise because you have to realized that youre substracting time (minutes) and no rates.


We cannot add/subtract speeds / rates. However, time can be added/ subtracted.

As mentioned in earlier posts, that's why time equation is created to extract values of "assumed variables" to find correct answer
_________________

Kindly press "+1 Kudos" to appreciate :)

Intern
Intern
avatar
Joined: 23 Sep 2014
Posts: 36
Location: India
Concentration: Marketing, Finance
GMAT 1: 670 Q48 V34
GMAT ToolKit User Reviews Badge
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 10 Dec 2014, 00:32
Blackbox wrote:
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.

a) 45
b) 33
c) 48
d) 55
e) 61



Hi

If the answer choices are such then you should directly use 48 and do backward calculation and you can land up to the answer in less than minute.

Why I am saying for this particular question is

it is given that speed reduces to 3/4th of the orignal

Only 48 is the one which can be divided by 4 and have a realisitc reduced speed i.e 36km/hr.

In rest you get in fractions.

Not full proof, but it works in 90% cases.

:)
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 10 Dec 2014, 08:36
PareshGmat wrote:
...

Setting up the time equation for scenario I
\(\frac{50}{s} + \frac{d-50}{\frac{3s}{4}} = \frac{d}{s} + \frac{35}{60}\)
4d - 7s = 200 .................. (1)
Setting up the time equation for scenario II
\(\frac{74}{s} + \frac{d-74}{\frac{3s}{4}} = \frac{d}{s} + \frac{25}{60}\)
...



This is amazing and exactly how I solved after a lot of digging. Just to add for others, 35/60 is added to d/s (original time taken sans accident) to show the 35 minute delay mentioned. And likewise, for the second equation, 25/60 is added to d/s to show the 25 minute delay. Thanks Paresh!
Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 210
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 10 Dec 2014, 08:47
believer700 wrote:

In rest you get in fractions.

Not full proof, but it works in 90% cases.

:)


You're probably right in this case. The rest of the options were made up by me and I didn't give it enough thought that they would result in a fraction. Back solving isn't a foolproof solution, personally I would do that unless I am in a time crunch.
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1067
A train after traveling at 50kmph meets with an accident and then  [#permalink]

Show Tags

New post 27 Dec 2015, 08:22
1
let d=total distance
divide each case into 3 segments:
1. first 50 km
2. next 24 km
3. last d-74 km
let s=speed of train
for segment 1, in each case, time=50/s
for segment 3, in each case, time=d-74/(3s/4)
therefore, segment 2 must account for the 10 minute total time difference between the two cases
24/(3s/4)-24/s=1/6
s=48 kmh
A train after traveling at 50kmph meets with an accident and then &nbs [#permalink] 27 Dec 2015, 08:22

Go to page    1   2    Next  [ 26 posts ] 

Display posts from previous: Sort by

A train after traveling at 50kmph meets with an accident and then

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.