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# One day at a restaurant there were a total of m workers, consisting of

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Manager
Joined: 07 Jun 2018
Posts: 51
Location: United States
One day at a restaurant there were a total of m workers, consisting of  [#permalink]

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28 Nov 2018, 19:16
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Difficulty:

55% (hard)

Question Stats:

69% (03:21) correct 31% (03:18) wrong based on 69 sessions

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One day at a restaurant there were a total of m workers, consisting of only waiters and cooks, some of whom had college degrees and others did not. Of these workers n were cooks and 5n/4 had college degrees. If 3n/4 cooks had degrees, m/4 of the waiters did not have degrees and 30 of the workers did not have degrees, how many workers were at the restaurant?

A. 20
B. 40
C. 80
D. 100
E. 120
Math Expert
Joined: 02 Aug 2009
Posts: 7960
One day at a restaurant there were a total of m workers, consisting of  [#permalink]

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29 Nov 2018, 04:26
Jazzmin wrote:
One day at a restaurant there were a total of m workers, consisting of only waiters and cooks, some of whom had college degrees and others did not. Of these workers n were cooks and 5n/4 had college degrees. If 3n/4 cooks had degrees, m/4 of the waiters did not have degrees and 30 of the workers did not have degrees, how many workers were at the restaurant?

A. 20
B. 40
C. 80
D. 100
E. 120

m workers, n cooks and m-n waiters...
(I) 3n/4 cooks had degrees, so n/4 cooks do not have degrees AND m/4 of the waiters did not have degrees.
so n/4+m/4=30....n+m=120....(i)
(II) 5n/4 had college degrees, so m-30 had degrees
5n/4=m-30....5n=4m-120....(ii)

substitute value of (i)...
5n=4(120-n)-120......5n=480-4n-120......9n=360.....n=40
so m = 120 - 40 = 80
C

let us solve it logically..
30 did not have degrees and from wordings we can say that there are more with degrees than without degrees..
so our answer has to be > 30*2 or >60.
Now from choices available take m as 80..
so m/4 or 80/4 or 20 cooks do not have degree.
Total 30 did not have degrees, so 30-20 or 10 waiters did not have degrees..
a) 3n/4 cooks had degrees, so n/4 did not have degree, so n/4=10...n=40
b) 5n/4 had college degrees, so 5*40/4=50 had degrees

Total degrees + no degrees = 50+30=80

C

Ofcourse it can be easily done by 2 by 2 matrix.
Attachments

set.png [ 35.1 KiB | Viewed 675 times ]

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Re: One day at a restaurant there were a total of m workers, consisting of  [#permalink]

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01 Dec 2018, 01:37
Jazzmin wrote:
One day at a restaurant there were a total of m workers, consisting of only waiters and cooks, some of whom had college degrees and others did not. Of these workers n were cooks and 5n/4 had college degrees. If 3n/4 cooks had degrees, m/4 of the waiters did not have degrees and 30 of the workers did not have degrees, how many workers were at the restaurant?

A. 20
B. 40
C. 80
D. 100
E. 120

really an amazing question , I solved it using 2x2 matrix..

from the given matrix we can say
m/4+n/4=30
m+n=120
and 2n/4+m/4=m-n
3m=6n
m=2n

solve for m & n we get m= 80 and n=40

IMO C 80 is correct
Attachments

File comment: 2x2 matrix
cooks and waiter 2x2.xlsx [30.37 KiB]

Duke & Cornell Moderator
Joined: 29 May 2018
Posts: 94
Location: India
GMAT 1: 700 Q48 V38
GMAT 2: 720 Q49 V39
GPA: 4
Re: One day at a restaurant there were a total of m workers, consisting of  [#permalink]

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24 Dec 2018, 05:43
This is a good but poorly worded question. M/4 of the waiters implies m/4*(m-n) whereas you want to say m/4 waiters. Kindly make this essential correction
Re: One day at a restaurant there were a total of m workers, consisting of   [#permalink] 24 Dec 2018, 05:43
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# One day at a restaurant there were a total of m workers, consisting of

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