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A train left station X at A hour B minutes. It reached station Y a [#permalink]
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10 Aug 2016, 06:15
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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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13 Aug 2016, 14:17
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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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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.. msk0657Please 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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06 Dec 2016, 11:33
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 AWhat 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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06 Dec 2016, 13:10
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Converting into minutes :
X : 60A+B
Time taken for traveling : 60C+A
Y : 60B+C
so, YX = Time taken for traveling
(60B+C)  (60A+B) = 60C+A
59B 59C = 61A
59 (BC) = 61A
(59/61) (BC) = A
None of the values from 024 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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07 Dec 2016, 09:11
Anu26 wrote: Converting into minutes :
X : 60A+B
Time taken for traveling : 60C+A
Y : 60B+C
so, YX = Time taken for traveling
(60B+C)  (60A+B) = 60C+A
59B 59C = 61A
59 (BC) = 61A
(59/61) (BC) = A
None of the values from 024 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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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, YX = Time taken for traveling
(60B+C)  (60A+B) = 60C+A
59B 59C = 61A
59 (BC) = 61A
(59/61) (BC) = A
None of the values from 024 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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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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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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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?




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