which is the greatest?

could somebody explain between A and B more clearly?

ok i will try explain more thoroughly.

For b it is obvious what the result is right?
For a:
What i did is trying to get into the "same format" as b as in:
2/3+3/4+4/5+5/6.
it is pretty obvious that A has similar fractions except they are all square rooted. So to get them into same format multiply each fraction by "1" to clear out the square root. So we do:
sqrt(2)/sqrt(3) *(sqrt(6)/sqrt(6)) ..... + sqrt(5)/sqrt(6) *(sqrt(30)/sqrt(30)). you notice that the sqrt multipliers are all "1s" and that they are the products of the fraction values in question - sqrt(6) = sqrt(2)*sqrt(3) for example.
This way you get something like:
2/3 *(sqrt(3)/sqrt(2)) + .... 5/6 *(sqrt(6)/sqrt(5)).
So you have the basic pattern like B but each fraction has a multiple... you should notice that those multiples (sqrt(3)/sqrt(2)) etc.. result in a value GREATER than 1. so basically each similar fraction like 2/3 is multiplying by something greater than 1. With that you can pretty much deduce that the result will be greater than B's. Hope it helps because there's no doubt you will see something like this on the exam!!!!

This approach is better and faster. Thanks!

A..it is..making the denominator same is best approach in such questions...clearly D & E are out....

We will comapre remainign three...by making denominator same....

nice post..helpful

So basically the takeaway is when one squares fractions the value reduces..and when you take the square root the value is greater than the fraction.Hmm..gotta be careful!

a.\frac{\sqrt{2}}{\sqrt{3}} +\frac{\sqrt{3}}{\sqrt{4}}+\frac{\sqrt{4}}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt{6}}

b.\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}

c.\frac{2^2}{3^2}+\frac{3^2}{4^2}+\frac{4^2}{5^2}+\frac{5^2}{6^2}

d.1-\frac{1}{3}+\frac{4}{5}-\frac{3}{4}

e.1-\frac{3}{4}+\frac{4}{5}+\frac{1}{3}


Soln: I too agree with Soln A

First considering the expressions in A,B,C
sqrt(fraction) > fraction>square(fraction)
so A > B > C

now considering D and E
D has two negatives while E has one negative
and since both have a common negative number of (-3/4)
therefore E ( which has +1/3) is greater than D(which has -1/3)
E > D

now comparison is only if A or E is greater
Comparing E with B, we can see that B > E
but since A > B, thus A > E also

therefore A is the largest fraction

Yes right.
This is the point to be observed in this question.

Answer A