$\begin{array}{1 1}(a)\;36666,43333\\(b)\; 43333,36000\\(c)\;36000,43333\\(d)\;43321,36666\end{array}$

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The polymer contains 30% molecules of mass 20,000,40% molecules of molecular mass 30,000 and rest 30% of molecules of molecular mass 60,000

Thus,

$\bar{M_n}=\large\frac{\sum N_iM_i}{\sum N_i}=\frac{(30\times 20000)+(40\times 30000)+(30\times 60000)}{30+40+30}$

$\Rightarrow 36000$

$\bar{M_w}=\large\frac{\sum N_iM_i^2}{\sum N_iM_i}=\frac{30 (20000)^2+40(30000)^2+30( 60000)^2}{30\times 20000+40\times 30000+30\times 60000}$

$\Rightarrow 43333$

Hence (c) is the correct answer.

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