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.
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