Standing Waves on a String: Finding Mass per unit Length

Two strings (string A and string B) are set up on a benchtop alongside each other. Each string has two bridges along its length the same distance apart. The
two strings have equal tension. A nearby loudspeaker produces sound at 440 Hz.
String A shows three nodes between the bridges, string B shows two nodes
between the bridges.
Calculate µA/µB where µA and µB are the linear mass densities of string*A*and*string*B.

Well, as the question tells us how many nodes are between the bridges in both cases, can you work out what harmonic string A and string B are oscillating at? We know that T, L and f are the same for both of the strings, so can you think of any equations that link those pieces of information?

String A is in its second harmonic, while string B is in its first harmonic.
The equation relating those quantities are v = √(T/μ ), v=fλ and λ=2L/n.
I have done this question using what I mentioned above and the values provided by the question but my answer is incorrect.

Not sure if I’m interpreting the question incorrectly, but isn’t string A in its 4th harmonic and string B in its 3rd harmonic? The question says that the nodes are between the bridges, so do you also need to include the two additional nodes at the bridges?

That could be why my answer is wrong, I'll try it out and let you know!

What you said was correct, thank you for your help!
P.S. To anyone who still can't get the correct answer, be careful when simplifying/dividing. To make it easier, you can enter the unsolved ratio (no simplification done) of μA/μB in the calculator and substitute the value of T by 1 and L by 2 (since they are the same for the two strings).