$\begin {array} {1 1} (A)\;\large\frac{1}{4} & \quad (B)\;\large\frac{1}{7} \\ (C)\;\large\frac{1}{8} & \quad (D)\;\large\frac{1}{49} \end {array}$

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Total number of cases = $ 100 \times 100$

Now $ 7^m + 7^n$ is divisible by 5 if one of the term has to end with 9 and other with 1.

$ 7^x$ cannot be divisible by 9.

$ \therefore $ m can be 2,6,10,14....98 (25 values ) and n can be 4, 8, 12....100 ( 25 values ).

Since m and n can interchange

$ \therefore $ Required probability = $ \large\frac{ 2 \times 25 \times 25}{100 \times 100}$

= $ \large\frac{1}{8}$

Ans : (C)

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