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

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

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New post 28 Nov 2018, 19:16
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Question Stats:

68% (03:21) correct 32% (03:14) wrong based on 65 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
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One day at a restaurant there were a total of m workers, consisting of  [#permalink]

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New post 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
set.png [ 35.1 KiB | Viewed 568 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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New post 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
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cooks and waiter 2x2.xlsx [30.37 KiB]
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Re: One day at a restaurant there were a total of m workers, consisting of  [#permalink]

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New post 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
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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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