Data Sufficiency

Bus no. X420 leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Bus no.Y007, which is making the trip from Boston to New York at a constant speed. If Bus no.Y007 left Boston at 3:50 PM and if the combined travel time of the two buses is 2 hours, what time did Bus no.Y007 arrive in New York?
(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.

OA:[spoiler]later[/spoiler]

gmatrix wrote:Bus no. X420 leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Bus no.Y007, which is making the trip from Boston to New York at a constant speed. If Bus no.Y007 left Boston at 3:50 PM and if the combined travel time of the two buses is 2 hours, what time did Bus no.Y007 arrive in New York?
(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.

OA:[spoiler]later[/spoiler]

S (distance from New York to Boston) >140
S =1(hour)*X420(speed) + t1*X420, t1 is time of X420
S =1/6(10 minutes)*Y007(speed) + t2*Y007, t2 is time of Y007

X420 (speed)=100 m/h, S > 140, 1*100 + t1*X420 >140, t1*X420 >40, t1*100 >40, t1 >0.4 (24 minutes)
Y007 (speed)=?

We are given combined time 2 hours, therefore t2 =~ 2 hours - 1 hour 24 minutes (X420 travel time) =~ 36
However, t2 < 36 minutes

3 pm + 1 hour 24 minutes [4 pm 24 minutes] < t1, t2 < 3 pm 50 minutes + 36 minutes [4 pm 26 minutes]
Statement 1 is NOT SUF. alone

Statement 2 says Y007 arrives before X420, so
3 pm + 1 hour 24 minutes [4 pm 24 minutes] < t2 -- interval -- t1 < 3 pm 50 minutes + 36 minutes [4 pm 26 minutes]

Statement 2 is NOT SUF. alone

Statements 1 and 2 are SUF.

Pick C.

gmatrix wrote:Bus no. X420 leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Bus no.Y007, which is making the trip from Boston to New York at a constant speed. If Bus no.Y007 left Boston at 3:50 PM and if the combined travel time of the two buses is 2 hours, what time did Bus no.Y007 arrive in New York?
(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.

OA:[spoiler]later[/spoiler]

X -> NY to Boston at 3PM -> 100mph
In one hour, it covers 100 miles. So at 4 PM the bus met Y. Y travelled a distance of 100 from there.

Y-> Boston to Newyork -> S speed -> left at 3.50 PM
So Bus Y travelled for10 minutes, then travelled for 100/S hours.
So total time = 1/6 + 100/S hours.

For the initial 10 minutes that Y travelled, Distance = D-100. Hence,
(D-100)/S = 1/6
D-100 = S/6
D = S/6 +100 = (600+S)/6

Let the total distnace be D. then
D/100 + D/S = 2
1/100 + 1/S = 12/(600+S)
6 + S/100 + 600/S + 1 = 12
S/100 + 600/S = 5
S^2 + 60000 = 500S
S^2 -500S +60000 = 0
S = 300,100


(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
it says D/S < D/100
implies S > 100
From this we get that S = 300.
So, we can get the Time.
Hence SUFF

(2) The distance between New York and Boston is greater than 140 miles.
D>140
(600+S)/6 > 140
S>240
Hence S = 300
We can get the time.
SUFF

pick D.

Hope this helps!!

kvcpk


In your explanation cited below

>>>
(2) The distance between New York and Boston is greater than 140 miles.
D>140
(600+S)/6 > 140
S>240
Hence S = 300
We can get the time.
SUFF
>>>

you infer S>240 Hence S = 300; S = 300 is derived from Statement (1). How you come to assign S=300 for the Statement (2) and count this as SUF. alone.

I challenge your D.

Night reader wrote:kvcpk

you infer S>240 Hence S = 300; S = 300 is derived from Statement (1). How you come to assign S=300 for the Statement (2) and count this as SUF. alone.

I challenge your D.

Hi Nightreader,

S=300 or 100 is derived from the information in the problem. This did not include Statement1.

If S>240, then S has to be 300. bcos 240>100

Hope this helps!!

kvcpk wrote:

Night reader wrote:kvcpk

you infer S>240 Hence S = 300; S = 300 is derived from Statement (1). How you come to assign S=300 for the Statement (2) and count this as SUF. alone.

I challenge your D.

Hi Nightreader,

S=300 or 100 is derived from the information in the problem. This did not include Statement1.

If S>240, then S has to be 300. bcos 240>100

Hope this helps!!

yes but S is assigned as the speed for Y007 ... and 100 is the speed of X...

Night reader wrote: yes but S is assigned as the speed for Y007 ... and 100 is the speed of X...

I am sorry NightReader, I didnt get you.

S is the speed for Y.
Using the Inofmration in th eproblem, without using any of the statements, we got that S can be 100 or 300.
But, we are not sure whether S is 100 or 300.

From Stmt1, we get S>100. Hence S should be 300.
From Stmt2, we get S>240, Hence should be 300.

Hence D.

Can you explain what your concern is again?

@gmatrix - what is OA?

kvcpk wrote:

Night reader wrote: yes but S is assigned as the speed for Y007 ... and 100 is the speed of X...

I am sorry NightReader, I didnt get you.

S is the speed for Y.
Using the Inofmration in th eproblem, without using any of the statements, we got that S can be 100 or 300.
But, we are not sure whether S is 100 or 300.

From Stmt1, we get S>100. Hence S should be 300.
From Stmt2, we get S>240, Hence should be 300.

Hence D.

Can you explain what your concern is again?

@gmatrix - what is OA?

Exactly after solving for S from problem we are not sure for 100 or 300, only two statements together let us abandon 100 and leave 300 as S for Y007

gmatrix wrote:Bus no. X420 leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Bus no.Y007, which is making the trip from Boston to New York at a constant speed. If Bus no.Y007 left Boston at 3:50 PM and if the combined travel time of the two buses is 2 hours, what time did Bus no.Y007 arrive in New York?
(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.

OA:[spoiler]later[/spoiler]

What is OA?

Night reader wrote: Exactly after solving for S from problem we are not sure for 100 or 300, only two statements together let us abandon 100 and leave 300 as S for Y007

Ok.. So you agree with me till the point that S can be 100 or 300 based ont he problem.

Using Stmt1:
(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
Distance is the same. Time taken by Y is less than that of Time taken by X.
Which means the speed of Y is greater than speed of X.
Speed of X is 100. So, Speed of Y should be greater than 100.
So speed of Y shud be 300.

Using stmt2:
(2) The distance between New York and Boston is greater than 140 miles.
Distance is greater than 140.
We know the relation between speed and Distance. So we get speed > 240.
So speed should be 300.

think this is ok?

my logic precludes the combination of 2 statements; without checking on either statement we may not abandon one of the quadratics values of S.

hi......OA is [spoiler]D[/spoiler]

kvcpk wrote:

gmatrix wrote:Bus no. X420 leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Bus no.Y007, which is making the trip from Boston to New York at a constant speed. If Bus no.Y007 left Boston at 3:50 PM and if the combined travel time of the two buses is 2 hours, what time did Bus no.Y007 arrive in New York?
(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.

OA:[spoiler]later[/spoiler]

X -> NY to Boston at 3PM -> 100mph
In one hour, it covers 100 miles. So at 4 PM the bus met Y. Y travelled a distance of 100 from there.

Y-> Boston to Newyork -> S speed -> left at 3.50 PM
So Bus Y travelled for10 minutes, then travelled for 100/S hours.
So total time = 1/6 + 100/S hours.

For the initial 10 minutes that Y travelled, Distance = D-100. Hence,
(D-100)/S = 1/6
D-100 = S/6
D = S/6 +100 = (600+S)/6

Let the total distnace be D. then
D/100 + D/S = 2
1/100 + 1/S = 12/(600+S)
6 + S/100 + 600/S + 1 = 12
S/100 + 600/S = 5
S^2 + 60000 = 500S
S^2 -500S +60000 = 0
S = 300,100

S = 200 & 300 and not 300,100 when u factorize S^2-500S+60000=0

(1) Bus no.Y007 arrived in New York before Bus no. X420 arrived in Boston.
it says D/S < D/100
implies S > 100
From this we get that S = 300.
So, we can get the Time.
Hence SUFF

S can be either 200 or 300... Putting this back into the equation gives us two values for time... Hence A, should not be sufficient...

(2) The distance between New York and Boston is greater than 140 miles.
D>140
(600+S)/6 > 140
S>240
Hence S = 300
We can get the time.
SUFF

pick D.

Statement B is sufficient, since S = 300...

After we get speed of Y =200 or 300

we can also get total distance= 133.33 or 150 miles

Total time for Y to reach = 133.33/ 200 or 150/300= 40 mins or 30mins
Actual time Y reaching NY= 4.30 or 4.20

otal time for X to reach = 133.33/ 100 or 150/100= 80 mins or 90 min
Actual time X reaching Boston= 4.20 or 4.30

first statement- It is saying the Y reaches before X - so it is sufficient.
second statement - Distance greater than 140- so it it is sufficient

vijchid wrote:After we get speed of Y =200 or 300

we can also get total distance= 133.33 or 150 miles

Total time for Y to reach = 133.33/ 200 or 150/300= 40 mins or 30mins
Actual time Y reaching NY= 4.30 or 4.20

otal time for X to reach = 133.33/ 100 or 150/100= 80 mins or 90 min
Actual time X reaching Boston= 4.20 or 4.30

first statement- It is saying the Y reaches before X - so it is sufficient.
second statement - Distance greater than 140- so it it is sufficient

The question explicitly asks: what time did Bus no.Y007 arrive in New York?

Hence we can have only one value for time... We cannot say the bus arrives at 4:20 or 4:30...