The descending order of the numbers $$5\sqrt{6}, 10\sqrt{3}, 7\sqrt{5}, 6\sqrt{7}$$ is

This question can be resolved by knowing approximate values of the square roots of 3, 5, 7. 
Sqrt(3) = 1.73 approx., sqrt(5) = 2.23 approx., sqrt(7) = 2.6 approx.

Now, we can see that $$10\sqrt{\ 3}$$ will be larger than 17 while $$7\sqrt{\ 5}$$ would be a little over 15. 
While in contrast, $$5\sqrt{\ 6}$$ would not even be 15. 
Also, $$6\sqrt{\ 7}$$ would be greater than 15. 

Hence, we can say that $$10\sqrt{\ 3}$$ is the largest of them all and $$5\sqrt{\ 6}$$ would be the smallest. 
$$6\sqrt{\ 7}\ >\ 7\sqrt{\ 5}\ $$. 
Hence, answer is B