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Hi good day
I would just like to verify if my answer is correct on this one. I do apologize if some of you might think that "oh this is just very simple how could she be so unsure with the answer?"  hahaha. I am, you see, very conscious that MAYBE there might be some hidden question between the question.
So the question is this:

 The pressure drop in a 6-in. pipe (I.D. = 6.065 in.) of length L is 1 psi with the flow in the highly
turbulent region. If the same liquid flows at the same volumetric flow rate in a 3-in. pipe (I.D. =
3.068 in.) with length L, the pressure drop (psi) is most nearly:

what i did is the fact that P1/p2=(d2^5)/(d1^5) as i equate it from the original pressure drop formula.
so 1/x=(3.068^5)/(6.065^5). that is 30.

thank you very much everyone!