If a sound increases 5 db, the sound becomes times louder.
a. five
b. ten
c. three
d. fifteen

Well, I'll try to write the formula in a way that's not confusing,
but I'm afraid it might be slightly confusing anyway.

When you're working with dB, the basic rule is

       A change of 10 dB means either multiplying or dividing by 10 .

     Multiply something by 10  ==>  it increases by 10 dB.
     Divide something by 10   ==>  it decreases by 10 dB.

It turns out that another way to write all of this is . . .

     An increase of 10 dB ===> multiply the original amount by 10¹ 
     An increase of 20 dB ===> multiply the original amount by 10²
     An increase of, say, 7 dB ===> multiply the original amount by 10⁰·⁷

     A decrease of 10 dB ===> multiply the original amount by 10⁻¹
     A decrease of 30 dB ===> multiply the original amount by 10⁻³
     A decrease of, say, 13 dB ===> multiply the original amount by 10⁻¹·³

This question says:  The sound increases by  5 dB .

That means the original 'intensity' or 'power' of the sound
is multiplied by
                           10⁰·⁵  =  √10  =  about 3.162 (rounded) . 

From the choices listed, the closest one is  (c).

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way