Elements A and B combine to form two compounds  ${\mathrm{A}}_2\mathrm{B}$ and ${\mathrm{A}}_2{\mathrm{B}}_3$ . When 0.2 moles of ${\mathrm{A}}_2\mathrm{B}$  weigh 8.8 grams and 0.05 moles of   ${\mathrm{A}}_2{\mathrm{B}}_3$weigh 3.8 grams then the atomic weights of A and B respectively are

0.2 moles of   ${\mathrm{A}}_2\mathrm{B}$= 8.8 grams
1 mole of  ${\mathrm{A}}_2\mathrm{B}$ = 44 grams
$\therefore 2\mathrm{A}+\mathrm{B}=44--\left(1\right)$
0.05 moles of  ${\mathrm{A}}_2{\mathrm{B}}_3$ = 3.8 grams
1 mole of   ${\mathrm{A}}_2{\mathrm{B}}_3$= 76 grams
$2\mathrm{A}+3\mathrm{B}=76---\left(2\right)$

 Eq(2) –Eq(1)
    B = 16

Substituting the value of ‘B’ in eq(1)
                            2A +16 = 44
                                   2A = 28
                                     A = 14