What is the maximum mass of a bug to not tip the straw?

[Nathanael]

Homework Statement

"A drinking straw of length $1.5a$ and mass $2m$ is placed on a square table of side $a$ parallel to one of its sides such that one third of its length extends beyond the table. An insect of mass $\frac{1}{2}m$ lands on the inner end of the straw (ie, the end which lies on the table) and walks along the straw until it reaches the outer end. It does not topple even when another insect lands on top of the first one. Find the largest mass of the second insect that can have without toppling the straw. Neglect friction.
Take $a$=1 meter, and $m$=0.02kg"

2. The attempt at a solution

To summarize my method:

I figured that "Neglect friction" was a sneaky way of saying that the center of mass of the straw and bug does not move.

So I found where the center of mass of the bug+straw is (0.35 meters from the edge of the table) and then I found how far out from the edge of the table the bug will be when he walks all the way along the straw (0.25 meters).

Then I used an equation for the torque (using the edge of the table as the pivot point) to find the maximum mass of the second bug, which I found to be 0.07kg, which is wrong.Is my method wrong, am I overlooking something? Or did I just make mistakes in my calculations? (I went through them twice)
Edited because I accidentally wrote "0.55" instead of "0.35"

 

[haruspex]

So I found where the center of mass of the bug+straw is (0.35 meters from the edge of the table)

That's not what I get. Please post your working. Your method looks fine.

 

[Nathanael]

The center of mass of the straw is 0.25 meters from the edge of the table, and the mass of the straw is 2m

The bug is 1 meter from the edge of the table, with a mass of m/2

The center of mass (as measured from the edge of the table) would then be:

$\frac{0.25*2m+1*m/2}{2.5m}=\frac{0.5m+0.5m}{2.5m}=\frac{1+1}{5}=0.4$

Silly me, I must've made the same mistake twice, and I don't even know what that mistake was!

Anyway, thanks haruspex, and sorry for making you go through the calculation.
(I was hoping it was a conceptual error, not a computational one.)

Next time I will triple check my answers! P.S.
I now get the correct answer.

 

[haruspex]

Anyway, thanks haruspex, and sorry for making you go through the calculation.

No problem.