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# There was a cycle race going on. 1/5th of those in front of a person a

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Intern
Joined: 19 May 2019
Posts: 3
There was a cycle race going on. 1/5th of those in front of a person a  [#permalink]

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17 Jul 2019, 20:54
3
00:00

Difficulty:

75% (hard)

Question Stats:

42% (01:56) correct 58% (01:56) wrong based on 62 sessions

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There was a cycle race going on. 1/5th of those in front of a person and 5/6 of those behind him gives the total number of participants. How many people took part in the race?

1. 35
2. 61
3. 31
4. 30
5. 91
Intern
Joined: 16 Aug 2016
Posts: 2
Re: There was a cycle race going on. 1/5th of those in front of a person a  [#permalink]

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18 Jul 2019, 05:16
2
eenu5 wrote:
There was a cycle race going on. 1/5th of those in front of a person and 5/6 of those behind him gives the total number of participants. How many people took part in the race?

1. 35
2. 61
3. 31
4. 30
5. 91

let number in front are x and behind him are y. According to the quest:

X/5= 5/6*(Y)...
and total participants become X+Y -----> (ii)

Expressing X in terms of Y the eq. (i) gives ----> X= (25/6)*Y

therefore, the total number of participants are: (25/6)*Y +Y = (31/6)*Y

Since the above number needs to be an integer, Y can be 6, 12, 18.... and hence the participants can be 31, 62, 93...

hence the answer is [C].

-Abhinav
Intern
Joined: 01 Oct 2018
Posts: 14
Location: India
Concentration: Technology, Social Entrepreneurship
GPA: 3.88
There was a cycle race going on. 1/5th of those in front of a person a  [#permalink]

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18 Jul 2019, 06:15
4
We can approach to this question in back track by eliminating options
as we are given there is one person and people are standing either in front or in back
and total including that person should satisfy our equation

take option A 35 (subtract person from that )
1/5 * 34 ahead
5/6 * 34 behind
Not possible as no of persons should be integer

take option B 61 (subtract person from that )
1/5 * 60 = 12
5/6 * 60 = 50
Total comes as 12 + 50 = 62
so this option not possible

Take option C 31 (subtract person from that)
1/5 * 30 = 6
5/6 * 30 = 25
Total comes as 6 + 25 = 31
Bingo this is correct

Lets try other 2 options also
30
(subtract person from that)
1/5 * 29 and 5/6 * 29
Not an integer so eliminated straightforward.

91
(subtract person from that)
1/5*90 = 18 and 5/6*90 = 75
Total comes 18 + 75 = 93 which is wrong

so correct option is C
Intern
Joined: 14 Jun 2015
Posts: 14
Re: There was a cycle race going on. 1/5th of those in front of a person a  [#permalink]

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22 Jul 2019, 18:36
eenu5 Pls mention the source of this question..
Intern
Joined: 24 Mar 2018
Posts: 7
Re: There was a cycle race going on. 1/5th of those in front of a person a  [#permalink]

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03 Aug 2019, 04:36
Let us assume x as the total number of participants.

Number of people ahead of that one person is x-1 and similarly number of people behind that one person is x-1.

Hence, as per the question stem:

1/5(x-1) + 5/6(x-1) = x
Hence, x = 31.

Posted from my mobile device
Re: There was a cycle race going on. 1/5th of those in front of a person a   [#permalink] 03 Aug 2019, 04:36
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# There was a cycle race going on. 1/5th of those in front of a person a

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