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A train left station X at A hour B minutes. It reached station Y a

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A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 10 Aug 2016, 06:15
3
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

39% (02:38) correct 61% (02:38) wrong based on 123 sessions

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A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:

A.0
B.1
C.2
D.3
E.4
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 13 Aug 2016, 14:17
2
msk0657 wrote:
A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:

A.0
B.1
C.2
D.3
E.4



It's been two days that I have posted the question...Since no one dared to write the clear cut solution even after 100+ views..I am giving the solution and expecting alternative solutions..please post them...


Let's try to get hours and minutes values separately..

Starting time + time taken = Reaching time..

A hours + C hours = B hours -------- (i)

A, C and B cannot have values greater than or equal to 24

B minutes + A minutes = C minutes -------- (ii)

Looking at the two equations, we get that no value of A satisfies both the equations.
Hence, option A
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 04 Dec 2016, 03:28
msk0657 wrote:
A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:

A.0
B.1
C.2
D.3
E.4



suppose train left A @ 00:01hrs and reached B @ 01:01hrs then hours of distance travelled =01:00 hrs
where A=00 , B=01 , C=01

similarly other values too can be found..

msk0657
Please clear me??

thanks
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 06 Dec 2016, 11:33
1
msk0657 wrote:
msk0657 wrote:
A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:

A.0
B.1
C.2
D.3
E.4



It's been two days that I have posted the question...Since no one dared to write the clear cut solution even after 100+ views..I am giving the solution and expecting alternative solutions..please post them...


Let's try to get hours and minutes values separately..

Starting time + time taken = Reaching time..

A hours + C hours = B hours -------- (i)

A, C and B cannot have values greater than or equal to 24

B minutes + A minutes = C minutes -------- (ii)

Looking at the two equations, we get that no value of A satisfies both the equations.
Hence, option A



What if A=0 and B and C are the same? In this case the conditions are satisfied
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 06 Dec 2016, 13:10
1
Converting into minutes :

X : 60A+B

Time taken for traveling : 60C+A

Y : 60B+C

so, Y-X = Time taken for traveling

(60B+C) - (60A+B) = 60C+A

59B -59C = 61A

59 (B-C) = 61A

(59/61) (B-C) = A

None of the values from 0-24 can make LHS an integer hence no possible values exist.

Hope this is a valid way to solve this question!
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 07 Dec 2016, 09:11
Anu26 wrote:
Converting into minutes :

X : 60A+B

Time taken for traveling : 60C+A

Y : 60B+C

so, Y-X = Time taken for traveling

(60B+C) - (60A+B) = 60C+A

59B -59C = 61A

59 (B-C) = 61A

(59/61) (B-C) = A

None of the values from 0-24 can make LHS an integer hence no possible values exist.

Hope this is a valid way to solve this question!




A can be 0 right? If A = 0, then for any B = C, the above equation is satisfied. Can you clear this up? Thanks.
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 07 Dec 2016, 21:24
pridhvic wrote:
Anu26 wrote:
Converting into minutes :

X : 60A+B

Time taken for traveling : 60C+A

Y : 60B+C

so, Y-X = Time taken for traveling

(60B+C) - (60A+B) = 60C+A

59B -59C = 61A

59 (B-C) = 61A

(59/61) (B-C) = A

None of the values from 0-24 can make LHS an integer hence no possible values exist.

Hope this is a valid way to solve this question!




A can be 0 right? If A = 0, then for any B = C, the above equation is satisfied. Can you clear this up? Thanks.



Yes that is true. For any B=C A=0 and all those will be possible answers.
Is a) the correct answer? or this approach is wrong?
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 08 Dec 2016, 00:05
convert the minutes to hours by dividing by 60.

then, start time+ travel time= reach time:

A + (B/60) + C + (A/60) = B + (C/60)

Now, with simple simplifications, we get: C= B - (61/59)*A

To note, A, B and C have to be integers, so for C to be an integer, A must be a multiple of 59 or 0.

59 not acceptable=> A=0
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 08 Dec 2016, 10:50
If there is a solution for A, then the answer is B right?

Thanks!
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Re: A train left station X at A hour B minutes. It reached station Y a  [#permalink]

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New post 04 Nov 2017, 11:22
msk0657 wrote:
A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:

A.0
B.1
C.2
D.3
E.4


I solved 2 pairs of equations and found 2 solutions: A=0 and A=59. A<=24 ---> one solution.

Is your OA is correct?

Maybe it has to be B?
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Re: A train left station X at A hour B minutes. It reached station Y a &nbs [#permalink] 04 Nov 2017, 11:22
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