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There are 101 students in a school. The students are split [#permalink]
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Updated on: 17 Aug 2013, 10:58
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There are 101 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams? A.70 B.65 C.63 D.62 E.61
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Originally posted by pjagadish27 on 17 Aug 2013, 09:43.
Last edited by pjagadish27 on 17 Aug 2013, 10:58, edited 1 time in total.



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Re: There are 100 students in a school. The students are split [#permalink]
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17 Aug 2013, 10:09
pjagadish27 wrote: There are 100 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams?
A.70 B.65 C.63 D.62 E.61 wrong question. There should be 101 students in order to support the official Answer
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Re: There are 100 students in a school. The students are split [#permalink]
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17 Aug 2013, 10:59
Asifpirlo wrote: pjagadish27 wrote: There are 100 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams?
A.70 B.65 C.63 D.62 E.61 wrong question. There should be 101 students in order to support the official Answer Though I have changed the total to 101, Why can't the total be 100? My Logic: Number the students from 1 to 100 A,B and C have students 162 in common. B,C have student 63 in common. Now there are 100(62+1)=37 students to be split. A can have 8 students, B can have 12 students and C can have 17 students. So in total, there are 62 students common to A,B,C.



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Re: There are 100 students in a school. The students are split [#permalink]
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17 Aug 2013, 11:15
pjagadish27 wrote: Asifpirlo wrote: pjagadish27 wrote: There are 100 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams?
A.70 B.65 C.63 D.62 E.61 wrong question. There should be 101 students in order to support the official Answer Though I have changed the total to 101, Why can't the total be 100? My Logic: Number the students from 1 to 100 A,B and C have students 162 in common. B,C have student 63 in common. Now there are 100(62+1)=37 students to be split. A can have 8 students, B can have 12 students and C can have 17 students. So in total, there are 62 students common to A,B,C. The problem is there must be 70 stu in a , 75 in B and 80 in C. you can't change their number
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Re: There are 100 students in a school. The students are split [#permalink]
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17 Aug 2013, 11:25
Asifpirlo wrote: pjagadish27 wrote: Asifpirlo wrote: wrong question. There should be 101 students in order to support the official Answer Though I have changed the total to 101, Why can't the total be 100? My Logic: Number the students from 1 to 100 A,B and C have students 162 in common. B,C have student 63 in common. Now there are 100(62+1)=37 students to be split. A can have 8 students, B can have 12 students and C can have 17 students. So in total, there are 62 students common to A,B,C. The problem is there must be 70 stu in a , 75 in B and 80 in C. you can't change their number Even now, that count does exist. A =62+8=70 B=62+1+12=75 C=62+1+17=80 I don't see any mistake.



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Re: There are 101 students in a school. The students are split [#permalink]
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17 Aug 2013, 11:29
Attachment:
101 students.jpg [ 15.12 KiB  Viewed 2068 times ]
We Have, a + d + e + g = 70 b + d + f + g = 75 c + e + f + g = 80 Adding them we get, a + b + c + 2(d + e + f) + 3g = 225 also, a + b + c + d + e + f + g = 101 Subtracting them, (d+e+f) + 2g = 124 To maximize g, d + e + f = 0 > 2g = 124 Hence, g = 62 Correct Ans: D
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Re: There are 101 students in a school. The students are split [#permalink]
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17 Aug 2013, 11:36
rockstar23 wrote: Attachment: 101 students.jpg We Have, a + d + e + g = 70 b + d + f + g = 75 c + e + f + g = 80 Adding them we get, a + b + c + 2(d + e + f) + 3g = 225 also, a + b + c + d + e + f + g = 101 Subtracting them, (d+e+f) + 2g = 124 To maximize g, d + e + f = 0 > 2g = 124 Hence, g = 62 Correct Ans: D... yes thats the Answer. Reply: 62 common students in three groups and different students are in A , B and C 8,13 and 17. so the total students = 62+8+13+17 = 101
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Re: There are 101 students in a school. The students are split [#permalink]
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13 Mar 2016, 07:01
pjagadish27 wrote: There are 101 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams?
A.70 B.65 C.63 D.62 E.61 Supposing 'x' is the number of students who can be present in all three teams, we've x+a+b+c = 101, also x+a=70 x+b=75 x+c=80 =>3*x+a+b+c=225 Therefore, 2*x = 124 x = 62



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Re: There are 101 students in a school. The students are split [#permalink]
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26 Mar 2017, 10:11
pjagadish27 wrote: There are 101 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams?
A.70 B.65 C.63 D.62 E.61 My Approach:Without the condition of total number of students, max number present in all three teams would be  70. Now given this, we are left with team B  5 extra and team C  10 extra. But this does not address the total number condition. We are left with 16 extra (unaccounted) students (101  70(common)  5(B)  10(C)) To address the total number condition, we can divide the extra 16 students into B and C groups and remove 8 students from common to make additions perfect. 708 = 62..



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Re: There are 101 students in a school. The students are split [#permalink]
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23 May 2018, 15:03
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