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
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