What does "a metallic oxide has 40 % oxygen" mean?
Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily.  However, you can find the equivalent weight of the metal without any assumptions about its formula.  

The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen.  We know that the compound contains 40% oxygen, so it must contain 60% metal.  Suppose we had 100g of this compound; that would be 40g O and 60g M.  

$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)
$8 g\ \ce{O}:12g\ \ce{M}$

The equivalent weight of the metal must be 12g.  

If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)

Given an equivalent weight for M of 12g:
$\ce{M2O-&gt; 12g/mol, carbon}$ (not a metal)
$\ce{MO-&gt; 24g/mol, magnesium}$
$\ce{M2O3-&gt; 36g/mol, chlorine}$ (not a metal)
$\ce{MO2-&gt;48g/mol, titanium}$

Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.