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Re: Of the 15 employees in a company, the mean of the salaries of seven em
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18 Nov 2024, 13:57
Given that the mean of the 7 employees who are the least paid is X, the sum of their salaries is \(7X\)
That leaves 8 employees. Given that the mean of those 8 employees who are the highest paid is Y, the sum of their salaries is \(8Y\)
The total sum of all 15 employees is \(7X + 8Y\)
Since the mean of a set = total sum / # of values, the mean of the salaries of all 15 employees \(= \dfrac{7X+8Y}{15}\)
To compare Quantity A and B, multiply both by 30 (their least common denominator) and simplify.
Quantity A = \(\dfrac{7X+8Y}{15} * 30 = 14X+16Y\)
Quantity B = \(\dfrac{X+Y}{2} * 30 = 15X+15Y\)
Subtract (14X + 15Y) from both.
Quantity A = \(14X+16Y - (14X + 15Y) = Y\)
Quantity B = \(15X+15Y - (14X + 15Y) = X\)
Now we're left comparing Y and X. Since we're given that X is the mean of the least paid employees, Y is greater than X.
Thus, Quantity A is greater.