$\begin{array}{1 1}(a)\;86.0&(b)\;40.0 \\(c)\;85.5&(d)\;75.5\end{array} $

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Average atomic weight =$\large\frac{R.A(1)\times\text{at.mass}(1)+R.A(2)\times\text{at.mass}(2)}{R.A(1)+R.A(2)}$

Average atomic weight=$\large\frac{85\times 75\times 87\times 25}{100}$

$\Rightarrow \large\frac{6375+2175}{100}$

$\Rightarrow \large\frac{8550}{100}$$=85.5$

Hence (c) is the correct answer.

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