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# The people eating in a certain cafeteria are either faculty members or

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Math Expert
Joined: 02 Sep 2009
Posts: 56244
The people eating in a certain cafeteria are either faculty members or  [#permalink]

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01 Feb 2018, 23:23
00:00

Difficulty:

85% (hard)

Question Stats:

54% (02:26) correct 46% (02:21) wrong based on 118 sessions

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The people eating in a certain cafeteria are either faculty members or students, and the number of faculty members is 15 percent of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50 percent of the original total. The number of students who left is what fractional part of the original number of students?

(A) 17/20
(B) 10/17
(C) 1/2
(D) 7/17
(E) 7/20

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Posts: 59
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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01 Feb 2018, 23:27
C 1/2

85 students +15 faculty =100

Students left and remaining total is 50

X+15=50

So x is 35
Number of students left is 85-35=50

50/100 =1/2

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Manager
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Posts: 210
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The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 00:04
+1 D

Assume,

Total number of Faculty=F
Total number of Students=S
After some students left,New number of students=S1

Now,
we get F=.15(F+S)----(1)
--> F=3/17S

Now After some of the students left, the total number of persons remaining is 50 percent of the original total

i.e, F+S1=.5(F+S)---(2)

Solving 1 and 2 we get

S1=7/17 S

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Joined: 19 Sep 2016
Posts: 13
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 01:31
Let x be the total number of people.therefore number of faculty =0.15x
No of students =0.85x
Let students who left be y then
0.15x+0.85x -y=0.5x
solving we get
x=2y
y/x=1/2
Please let me know if i m wrong somewhere

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Manager
Joined: 26 Sep 2017
Posts: 59
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 01:33
jrk23 wrote:
C 1/2

85 students +15 faculty =100

Students left and remaining total is 50

X+15=50

So x is 35
Number of students left is 85-35=50

50/100 =1/2

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Sorry 50/85 question is asking students left to original students...

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Posts: 13
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 01:38
surbhipuri wrote:
Let x be the total number of people.therefore number of faculty =0.15x
No of students =0.85x
Let students who left be y then
0.15x+0.85x -y=0.5x
solving we get
x=2y
y/x=1/2
Please let me know if i m wrong somewhere

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Sorry didnt readthe question properly i guess
Student left / total students
=x/2×0.85x
=10/17

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Posts: 59
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 01:38
surbhipuri wrote:
Let x be the total number of people.therefore number of faculty =0.15x
No of students =0.85x
Let students who left be y then
0.15x+0.85x -y=0.5x
solving we get
x=2y
y/x=1/2
Please let me know if i m wrong somewhere

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You should not use x in denominator, you should use .85x in denominator

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Intern
Joined: 19 Sep 2016
Posts: 13
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 01:39
Corrected the same sir

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Intern
Joined: 29 Dec 2017
Posts: 11
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 04:02
1
let The Total no of students =100
then no of faculty = 15
no of student =85
let p no of student left
so , 100-p=50*100/100
p=50 students left
so, No of student left /total no of student = 35/85=7/17
Intern
Joined: 29 Dec 2017
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The people eating in a certain cafeteria are either faculty members or  [#permalink]

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Updated on: 02 Feb 2018, 04:41
1
let The Total no of students =100
then no of faculty = 15
no of student =85
let p no of student left
so , 100-p=50*100/100
p=50 students left
so, No of student left /total no of student = 50/85=10/17

Originally posted by CharuSingh26 on 02 Feb 2018, 04:03.
Last edited by CharuSingh26 on 02 Feb 2018, 04:41, edited 1 time in total.
Intern
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Posts: 13
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 04:28
CharuSingh26 wrote:
let The Total no of students =100
then no of faculty = 15
no of student =85
let p no of student left
so , 100-p=50*100/100
p=50 students left
so, No of student left /total no of student = 35/85=7/17

No of student left is 50 as you calculated

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The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 09:05
The people eating in a certain cafeteria are either faculty members or students, and the number of faculty members is 15 percent of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50 percent of the original total. The number of students who left is what fractional part of the original number of students?

(A) 17/20
(B) 10/17
(C) 1/2
(D) 7/17
(E) 7/20

Picking smart numbers start with 85 students and 15 teachers. After some students leave there is half as many people, meaning 50 students have left.
50/85 = 10/17 when simplified.
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Concentration: Finance, Economics
Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 12:47
CanCanuck90 wrote:
The people eating in a certain cafeteria are either faculty members or students, and the number of faculty members is 15 percent of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50 percent of the original total. The number of students who left is what fractional part of the original number of students?

(A) 17/20
(B) 10/17
(C) 1/2
(D) 7/17
(E) 7/20

Picking smart numbers start with 85 students and 15 teachers. After some students leave there is half as many people, meaning 50 students have left.
50/85 = 10/17 when simplified.

50 students do not leave. 50 people leave.
Your scenario has to go from 15/100 people are teachers to 15/50 teachers.
This means you go from 85 students to 35 students.

D. 7/17

It also easier if you pick 30/200 teachers and then after kids leave its 30/100 teachers. 170 students down to 70 students. 70/170 =7/17
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Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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02 Feb 2018, 13:53
IdiomSavant wrote:
CanCanuck90 wrote:
The people eating in a certain cafeteria are either faculty members or students, and the number of faculty members is 15 percent of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50 percent of the original total. The number of students who left is what fractional part of the original number of students?

(A) 17/20
(B) 10/17
(C) 1/2
(D) 7/17
(E) 7/20

Picking smart numbers start with 85 students and 15 teachers. After some students leave there is half as many people, meaning 50 students have left.
50/85 = 10/17 when simplified.

50 students do not leave. 50 people leave.
Your scenario has to go from 15/100 people are teachers to 15/50 teachers.
This means you go from 85 students to 35 students.

D. 7/17

It also easier if you pick 30/200 teachers and then after kids leave its 30/100 teachers. 170 students down to 70 students. 70/170 =7/17

Re-read the question, it says "after some students leave".... I agree with everything else.

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Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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05 Feb 2018, 10:18
Bunuel wrote:
The people eating in a certain cafeteria are either faculty members or students, and the number of faculty members is 15 percent of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50 percent of the original total. The number of students who left is what fractional part of the original number of students?

(A) 17/20
(B) 10/17
(C) 1/2
(D) 7/17
(E) 7/20

If we let the total people = n, then 0.15n = faculty and 0.85n = students. If we let x = the number of students who leave, we can create the equation:

[0.15n + (0.85n - x)]/n = 1/2

(n - x)/n = 1/2

2(n - x) = n

2n - 2x = n

n = 2x

n/2 = x

Thus, the number of students who left is (n/2)/(85n/100) = 100n/170n = 10/17 of the original number of students.

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Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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05 Feb 2018, 10:40
techiesam wrote:
+1 D

Assume,

Total number of Faculty=F
Total number of Students=S
After some students left,New number of students=S1

Now,
we get F=.15(F+S)----(1)
--> F=3/17S

Now After some of the students left, the total number of persons remaining is 50 percent of the original total

i.e, F+S1=.5(F+S)---(2)

Solving 1 and 2 we get

S1=7/17 S

Sorry guys...the answer should be B
I did the calculation right but didnot read the question properly.
ITS 10/17.
So its B
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The people eating in a certain cafeteria are either faculty members or  [#permalink]

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05 Feb 2018, 10:57
1
Let total number of people in cafteria be 100.

Given - 15% of total are faculty members - 15
85 % would be students - 85

2 ] After some students leave the total number of people becomes 50% of what was at the start - 50% of 100 - 50
Number of students who left = 50 - 15 = 35

The number of students who left is what fractional part of the original number of students = 35/85 = 7/17. [D]
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Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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11 Feb 2018, 22:11
this question definitely calls for test it
so draw a table
total student teachers
100 85 15 ........as 15 percent are teacher
50 35 15 ............ as total number reduce to half but teachers are same so only students reduce

required ratio = no of students left/total no of students =85-35/85=50/85=10/17 (B)
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Re: The people eating in a certain cafeteria are either faculty members or  [#permalink]

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19 Feb 2018, 17:53
Really nice trap in answer C (students to overall total, not original students.)
Re: The people eating in a certain cafeteria are either faculty members or   [#permalink] 19 Feb 2018, 17:53
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