Doesn't methyl amine give carbylamine test? A compound $\ce X$ of formula $\ce{C2H5NO2}$ on treatment with $\ce {HNO2}$ 
   $(\ce{NaNO2}+\ce{HCl})$ gives off an inactive gas $\ce Y$ and forms $\ce Z$. When heated with soda lime, $\ce X$ gives $\ce{A(CH5N)}$ soluble in acid. Find the true statements : 
  
  
  $\ce A$ answers carbylamine test.
  $\ce X$ is an amino acid.
  $\ce Y$ is $\ce{CO2}$.
  $\ce Z$ is a hydroxyl acid.
  


My attempt : 

$\ce X$ is glycine.

$\ce A$ is methyl amine.

$\ce Y$ is dinitrogen.

$\ce Z$ is 2-hydroxy ethanoic acid.

Hence, correct options should be : $1,2,4$. But the answer given is $2,4$. 

Can anyone help me to understand if and where I went wrong?
You are indeed correct. 

$\ce{HOOC-CH2-NH2(X) -&gt;[HNO2] HOOC-CH2-OH (Z) + N2(Y)}$

$\ce{HOOC-CH2-NH2(X) -&gt;[soda lime] CH2-NH2(A)}$

Also, $ \ce{(A) + CHCl3 + 3KOH -&gt;[Heat] CH3-NC + 3KCl + 3H2O}$. 

Methylamine does give the carbylamine test.

So clearly, the answer given must be wrong.

The correct options are 1,2 and 4.