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.