Answer :
Answer:
The required option is D) 15(x-y).
Step-by-step 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
[tex]\frac{30x}{GCD}\ \text{and}\ \frac{15y}{GCD}[/tex] 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).