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14 Apr 2021, 07:45
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Difficulty:
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Question Stats:
74% (01:43) correct
26%
(02:08)
wrong
based on 68
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The average (arithmetic mean) annual salary of the employees of a company was $70,000. If the male employees’ annual salary average was $65,000 and that of female employees’ annual salary was $80,000, what could be the number of male employees and female employees, respectively, in the company?
(A) 6; 7
(B) 7; 15
(C) 7; 14
(D) 14; 7
(E) 15; 7
(A) 6; 7
(B) 7; 15
(C) 7; 14
(D) 14; 7
(E) 15; 7
Bunuel
Let the number of males be m and females be f
Total salary of males be x and females be y
Given:\(\frac{ x}{m }= 65000\) (As, \(Average = \frac{Sum}{Number}\))
x = 65000m
Similarly, y = 80000f
Now, \(\frac{65000m + 80000f}{m + f} = 70000\) (As, average salary of employees = 70000)
Reducing the equation:
65m + 80f = 70m + 70f
10f = 5m
2f = m
Number of males and females = 14, 7 (Plug in the options if you are finding it hard to see)
14 Apr 2021, 23:28
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Let number of males be X
Let number of females be Y
total is x+y
Average :
(65000x+80,000y)/x+y = 70,000(x+y)
Solving the above give x:y =2:1
Option D Satisfies the above 2:1 criteria
Let number of females be Y
total is x+y
Average :
(65000x+80,000y)/x+y = 70,000(x+y)
Solving the above give x:y =2:1
Option D Satisfies the above 2:1 criteria
