#### Answer

% abundance of $^{107}Ag = 51.835\%$
% abundance of $^{109}Ag = 48.165\%$

#### Work Step by Step

Atomic mass of Silver $= 107.8682 amu$
Mass of $^{107}Ag = 106.90509 amu$
Mass of $^{109}Ag =108.9047 amu$
Let the fraction of $^{107}Ag = x$
Then the fraction of $^{109}Ag = (1-x)$
The atomic mass of an element is the weighted average of the masses of its constituent isotopes.
Therefore,
$107.8682amu = x(106.90509amu) + (1-x)(108.9047amu)$
$107.8682=106.90509x+108.9047-108.9047x$
$-1.0365=-1.99961x$
$x = 0.51835$
The fraction of $^{107}Ag = 0.51835$
% abundance of $^{107}Ag = 51.835\%$
The fraction of $^{109}Ag=0.48165$
% abundance of $^{109}Ag = 48.165\%$