Answer:
The required option is D) 15(x-y).
Explanation:
Consider the provided information.
If x and y are positive integers, each of the following could be the greatest common divisor of 30x and 15y
That means GCD must divide 30x and 15y
must be an integer.
Now consider the provided option.
Option A) 30x
Let x = 1 and y = 2 Then the numbers are 30x = 30, 15y = 30 and 30x = 30
GCD(30,30) = 30 thus the option A can be greatest common divisor.
Option B) 15y
Let x = 1 and y = 1 Then the numbers are 30x = 30, 15y = 15 and 15y = 15
GCD(30,15) = 15 thus the option B can be greatest common divisor.
Option C) 15(x+y)
Let x = 1 and y = 1 Then the numbers are 30x = 30, 15y = 15 and 15(x+y) = 30
GCD(30,15) = 15 ≠ 30 Thus, the option C cannot be greatest common divisor because 15(x+y) > 15y for any positive integer.
Option D) 15(x-y)
Let x = 2 and y = 1 Then the numbers are 30x = 60, 15y = 15 and 15(x-y) = 15
GCD(60,15) = 15 thus the option D can be greatest common divisor.
Option E) 15000
Let x = 500 and y = 1000 Then the numbers are 30x = 15000, 15y = 15000 and 15000
GCD(15000,15000) = 15000 thus the option D can be greatest common divisor.
Hence, the required option is D) 15(x-y).