Divisibility

Notice that 115 = 5*23. In order to establish is x is divisible by 115, we must establish if it is simultaneously divisible by 5 and 23.

1. tells us that x is divisible by 5 (since 25 = 5^2). But since we do not know if x is divisible by 23 as well, we cannot tell. So it's impossible to say whether x is divisible by 115 or not.

2. tells us that x is divisible by 23 (since 138 = 115 + 23), but this time we don't know if it is divisible by 5. So again, it's insufficient.

Put both stmts together and they complement each other, meaning that x will be divisible by 115. So the answer is C.

Since 115 = 23x5, x will be divisible by 115 provided it is divisible by both 23 and 5.

Statement I: Since it is divisible by 25, it must be divisible by 5. But this alone is not sufficient.

Satement II: Since 138=23x6, if x is divisible by 138, it must be divisible by 23 and 6. But again this alone is not sufficient.

Combining both, x is divisible by 5 as well as 23. Hence x must be divisible by LCM(5,23) i.e 115.

So C