I maximized B so I choose B = 9 hence to hold the inequality true A has to between 6 to 9 . Now looking at quantity A for min. value of A i.e. 6 Qty A becomes 90 --> 15 x 6 . Now , whatever values we will choose for B between 0-9 Qty A will always be greater.

I chose the numbers 200 and 100 and So first case a=2 and b=1 which makes 5(3.0A) = 30 and 3(5.0B) = 15. Hence on plugging the values we find 15A=30 and 9B=1 . Hence A is bigger

For the second case i chose numbers -100 and -200 so a = -1 and b=-2 which makes 5(3.0A) = -15 and 3(5.0B) = -30 which satisfies the condition. Hence on plugging the values we get 15A = -15 and 9B = -18. Hence again A is bigger. Hence it should be **option A**

Correct, that’s a fundamental PEMDAS rule

first we solve the head:

So, 15.0A> 15.0B–> A must be ( GREATER) than B.

Now Quantity A and B both divided by 3

So, Quantity A= 5 A and Quantity B= 3 B

Since 5 > 3 and A > B … Then A is greater .