What smallest fraction should be added to $$3\frac{2}{3}+6\frac{7}{12}+4\frac{9}{36}+5+7\frac{1}{12}$$​ to make the sum a whole number?

Correct option is B

Given:
$$3\frac{2}{3}+6\frac{7}{12}+4\frac{9}{36}+5+7\frac{1}{12}$$​
Solution: 
Let x be added to make it whole number equal to 1.
​$$3\frac{2}{3}+6\frac{7}{12}+4\frac{9}{36}+5+7\frac{1}{12}\\$$ + x = 1​

​$$\frac{11}{3} + \frac{79}{12} + \frac{17}{4} + 5 + \frac{85}{12} + x = 1\\\Rightarrow \frac{319}{12} - 1 = x$$​​
$$\therefore 26+ \frac 7{12} - 1 = x\\$$

To make it whole x should be $$\frac 5{12}$$​​