Amtrack Express

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Sun Feb 21, 2010 11:32 am

Q: Amtrack train left for New York City ,increasing its speed in each hour.It started its journey from Los Angeles,California but after four hours of its journey it met with an accident .Its speed in the fourth hour of its journey was 7/5 times that of the third hour and the speed in the third hour was 10/7 times that of the second hour and in the second hour it was 7/5 times that of the first hour.If it had travelled with half the speed of that of the third hour ,then it would have gone 160 miles less in the same time ( i.e in four hours).The average speed of the train during the journey of 4 hours was :

a) 60 miles/hr

b)70 miles/hr

c)80 miles/hr

d)90 miles/hr

e)100 miles/hr

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Source: Beat The GMAT — Problem Solving | Sun Feb 21, 2010 11:32 am

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Sun Feb 21, 2010 11:40 am

Aman verma wrote:Q: Amtrack train left for New York City ,increasing its speed in each hour.It started its journey from Los Angeles,California but after four hours of its journey it met with an accident .Its speed in the fourth hour of its journey was 7/5 times that of the third hour and the speed in the third hour was 10/7 times that of the second hour and in the second hour it was 7/5 times that of the first hour.If it had travelled with half the speed of that of the third hour ,then it would have gone 160 miles less in the same time ( i.e in four hours).The average speed of the train during the journey of 4 hours was :

a) 60 miles/hr

b)70 miles/hr

c)80 miles/hr

d)90 miles/hr

e)100 miles/hr

Let the speed in the 4th hour be S4.
S4 = 7/5 x S3 -(1)
=>S3 = 5/7 S4
S3 = 10/7 x S2
=>S2 = 7/10 x S3 =7/10 x 5/7 x S4 = 1/2 x S4 -(2)
S2 = 7/5 x S1
=>S1 = 5/7 x S2 = 5/14 x S4

Avg. speed = (S1 + S2 + S3 + S4)/4 = (36/14) x (S4/4)
=9/14 x S4

Now if train travelled with speed of S3/2(i.e. 5/14 S4),it would have travelled 160 miles less in 4hours.

Now, [(9/14 x S4) - (5/14 x S4)] x 4 =160

Hence,S4 = 140 miles/hr
Average speed = 9/14 X 140 =[spoiler] 90 MILES /HR (d) IS THE ANSWER[/spoiler]

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Sun Feb 21, 2010 11:51 am

Aman verma wrote:Q: Amtrack train left for New York City ,increasing its speed in each hour.It started its journey from Los Angeles,California but after four hours of its journey it met with an accident .Its speed in the fourth hour of its journey was 7/5 times that of the third hour and the speed in the third hour was 10/7 times that of the second hour and in the second hour it was 7/5 times that of the first hour.If it had travelled with half the speed of that of the third hour ,then it would have gone 160 miles less in the same time ( i.e in four hours).The average speed of the train during the journey of 4 hours was :
a) 60 miles/hr
b)70 miles/hr
c)80 miles/hr
d)90 miles/hr
e)100 miles/hr

Let y be the speed of Amtrack train in 1st hour.
so in 2nd hour speed = 7/5 y
in 3rd hour speed= 10/7 x 7/5 y = 2y
in 4th hour speed = 7/5 x 2y = 14/5 y
so if it would have travelled with speed y i.e 4y = 160 --> y = 40
so av speed = (y+7/5y + 2y +14/5y)/4 =(3y +21y/5)/4
(120 +168 )/4 = 288/4 = 68 miles/hr

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firdaus117: Master | Next Rank: 500 Posts; Posts: 102; Joined: Sat Feb 20, 2010 5:38 am; Location: IIM Ahmedabad; Thanked: 10 times

by firdaus117 » Sun Feb 21, 2010 11:52 am

Let the speed for first hour be 'v' miles/hr.
Speed for second hour =7v/5
Speed for third hour=2v
Speed for fourth hour=14/5v
Total distance travelled in this case=(1+ 7/5 + 2 + 14/5)v
=36/5 v miles
In second case,the train will travel with a speed of v miles throghout the 4 hours.
Total distance travelled in this case=4v
Accd to question,
36/5 v - 4v=160
or, v=50miles/hr
Average speed=(36v/5)/4
=90 mph
Option D

Hey Aman,looking at the average speed,I feel you have run an Indian train on NY-LA track.

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Sun Feb 21, 2010 12:02 pm

shashank.ism wrote:
Aman verma wrote:Q: Amtrack train left for New York City ,increasing its speed in each hour.It started its journey from Los Angeles,California but after four hours of its journey it met with an accident .Its speed in the fourth hour of its journey was 7/5 times that of the third hour and the speed in the third hour was 10/7 times that of the second hour and in the second hour it was 7/5 times that of the first hour.If it had travelled with half the speed of that of the third hour ,then it would have gone 160 miles less in the same time ( i.e in four hours).The average speed of the train during the journey of 4 hours was :
a) 60 miles/hr
b)70 miles/hr
c)80 miles/hr
d)90 miles/hr
e)100 miles/hr
Let y be the speed of Amtrack train in 1st hour.
so in 2nd hour speed = 7/5 y
in 3rd hour speed= 10/7 x 7/5 y = 2y
in 4th hour speed = 7/5 x 2y = 14/5 y
so if it would have travelled with speed y i.e 4y = 160 --> y = 40
so av speed = (y+7/5y + 2y +14/5y)/4 =(3y +21y/5)/4
(120 +168 )/4 = 288/4 = 68 miles/hr

Hey shashank,
Look at the bold-faced part above.
Its written 160 miles less not that it will travel "10 miles" in that time-span.
Hence,you have to subtract the two distances and equate that to 4y.

So,in ur case the answer will be [(3y +21y/5)/4]*(4) - 4y =160
THen ,u will get the answer as 90.

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Sun Feb 21, 2010 12:17 pm

harsh.champ wrote:
shashank.ism wrote:
Aman verma wrote:Q: Amtrack train left for New York City ,increasing its speed in each hour.It started its journey from Los Angeles,California but after four hours of its journey it met with an accident .Its speed in the fourth hour of its journey was 7/5 times that of the third hour and the speed in the third hour was 10/7 times that of the second hour and in the second hour it was 7/5 times that of the first hour.If it had travelled with half the speed of that of the third hour ,then it would have gone 160 miles less in the same time ( i.e in four hours).The average speed of the train during the journey of 4 hours was :
a) 60 miles/hr
b)70 miles/hr
c)80 miles/hr
d)90 miles/hr
e)100 miles/hr
Let y be the speed of Amtrack train in 1st hour.
so in 2nd hour speed = 7/5 y
in 3rd hour speed= 10/7 x 7/5 y = 2y
in 4th hour speed = 7/5 x 2y = 14/5 y
so if it would have travelled with speed y i.e 4y = 160 --> y = 40
so av speed = (y+7/5y + 2y +14/5y)/4 =(3y +21y/5)/4
(120 +168 )/4 = 288/4 = 68 miles/hr
Hey shashank,
Look at the bold-faced part above.
Its written 160 miles less not that it will travel "10 miles" in that time-span.
Hence,you have to subtract the two distances and equate that to 4y.

So,in ur case the answer will be [(3y +21y/5)/4]*(4) - 4y =160
THen ,u will get the answer as 90.

oh I missed that I thought it is 4y = 160 ..
I read ur solution , I did a mistake there...thanks...
yeah the answer as 90.

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Sun Feb 21, 2010 1:01 pm

Aman verma wrote:Q: Amtrack train left for New York City ,increasing its speed in each hour.It started its journey from Los Angeles,California but after four hours of its journey it met with an accident .Its speed in the fourth hour of its journey was 7/5 times that of the third hour and the speed in the third hour was 10/7 times that of the second hour and in the second hour it was 7/5 times that of the first hour.If it had travelled with half the speed of that of the third hour ,then it would have gone 160 miles less in the same time ( i.e in four hours).The average speed of the train during the journey of 4 hours was :

a) 60 miles/hr

b)70 miles/hr

c)80 miles/hr

d)90 miles/hr

e)100 miles/hr

Say initial speed was x & distance traveled = x
speed in second hour = 7/5x
speed in third hour = 10/7*7/5x = 2x
speed in the fourth hiyr = 14/5x

Total distance traveled = 3x+21/5x = 36/5x

total distance travelled if the speed were half of 2x = 4x

36/5x -4x = 160
16/5x= 160
x =50

Average speed = 36/(5*4 )*50 = 90 miles per hour

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Wed Feb 24, 2010 6:16 am

firdaus117 wrote:Let the speed for first hour be 'v' miles/hr.
Speed for second hour =7v/5
Speed for third hour=2v
Speed for fourth hour=14/5v
Total distance travelled in this case=(1+ 7/5 + 2 + 14/5)v
=36/5 v miles
In second case,the train will travel with a speed of v miles throghout the 4 hours.
Total distance travelled in this case=4v
Accd to question,
36/5 v - 4v=160
or, v=50miles/hr
Average speed=(36v/5)/4
=90 mph
Option D
Hey Aman,looking at the average speed,I feel you have run an Indian train on NY-LA track.

ANS:[spoiler]d)90[/spoiler] Yes Firdaus117, indeed the train did run very slow in those tracks, maybe that's the reason of accident.

800. Arjun's-Bird-Eye

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