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28 Nov 2018, 19:16
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
67% (03:25) correct
33%
(03:20)
wrong
based on 284
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
A. 20
B. 40
C. 80
D. 100
E. 120
29 Nov 2018, 04:26
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Jazzmin
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 4834 times ]
Archit3110
Major Poster
01 Dec 2018, 01:37
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Jazzmin
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]
Downloaded 109 times
