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08 Jul 2014, 07:29
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A theater box office sold an average (arithmetic mean) of 65 tickets per staff member to a particular movie. Among the daytime staff, the average number sold per member was 76, and among the evening staff, the average number sold was 60. If there are no other employees, what was the ratio of the number of daytime staff members to the number of evening staff members?
A) 1:1
B) 1:4
C) 5:11
D) 7:11
E) 65:76
A) 1:1
B) 1:4
C) 5:11
D) 7:11
E) 65:76
08 Jul 2014, 08:12
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Gnpth
A theater box office sold an average (arithmetic mean) of 65 tickets per staff member to a particular movie. Among the daytime staff, the average number sold per member was 76, and among the evening staff, the average number sold was 60. If there are no other employees, what was the ratio of the number of daytime staff members to the number of evening staff members?
A) 1:1
B) 1:4
C) 5:11
D) 7:11
E) 65:76
A) 1:1
B) 1:4
C) 5:11
D) 7:11
E) 65:76
Deviation from the mean for the daytime staff = 76 - 65 = 11.
Deviation from the mean for the evening staff = 65 - 60 = 5.
Thus, the ratio of the number of daytime staff members to the number of evening staff members = 5:11.
Answer: C.
Or: (76x + 60y)/(x + y) = 65 --> x/y = 5/11.
Answer: C.
08 Jul 2014, 20:58
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Let "d" is the total no. of day staff
So, total collection in day = 76d .............. (1)
Let "n" is the total no. of night staff
So, total collection in night = 60n ................. (2)
Total Day & Night staff = (d+n) & overall average = 65
So, total collection (Day & Night) = 65(d+n) ........... (3)
(1) + (2) = (3)
76d + 60n = 65d + 65n
11d = 5n
\(\frac{d}{n} = \frac{5}{11}\)
Answer = C
So, total collection in day = 76d .............. (1)
Let "n" is the total no. of night staff
So, total collection in night = 60n ................. (2)
Total Day & Night staff = (d+n) & overall average = 65
So, total collection (Day & Night) = 65(d+n) ........... (3)
(1) + (2) = (3)
76d + 60n = 65d + 65n
11d = 5n
\(\frac{d}{n} = \frac{5}{11}\)
Answer = C
