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by rn4gmat » Tue Apr 20, 2010 8:19 pm
If we simply do a square for all these three it can be seen that

c : 16 = 8 + 8
b : 8 + 2 (15)^1/2 .Here 2 (15)^1/2 will be less than 2(16)^1/2 which is equal to 8 hence this will be less than c.
a: 8 + 2 (12)^1/2.Here 2 (12)^1/2 will be less than 2(15)^1/2 hence this will be less than b.

hence the correct order will be : c,b,a.
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by DeepthiRajan » Tue Apr 20, 2010 9:19 pm
IMO A as the ques asks for the order from least to greatest
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by Rahul@gurome » Tue Apr 20, 2010 9:44 pm
Solution

It is given that a = sqrt(2) + sqrt(6) and b = sqrt(3) + sqrt(5).
Consider sqrt(6) - sqrt(5) which is same as {sqrt(6) -sqrt(5)} * {sqrt(6) + sqrt(5)}/ {sqrt(6) + sqrt(5)}.
In the above step we are both multiplying and dividing by the same quantity {sqrt(6) + sqrt(5)}.
So sqrt(6) - sqrt(5) = (6-5)/ {sqrt(6) + sqrt(5)} = 1/{sqrt(6) + sqrt(5)}.
Similarly sqrt(3) - sqrt(2) = 1/{sqrt(3) + sqrt(2)}.
Obviously {sqrt(6) + sqrt(5)} > {sqrt(3) + sqrt(2)} since both 6 and 5 are greater than 3 and 2 and so the sum of the roots of 6 and 5 will also be greater than the sum of roots of 3 and 2.
So 1/ {sqrt(6) + sqrt(5)}< 1/{sqrt(3) + sqrt(2)}.
Or sqrt(6) - sqrt(5) < sqrt(3) - sqrt(2).
Or sqrt(2) + sqrt(6) < sqrt(5) + sqrt(3).
So a < b.
Now 4 = sqrt(4) + sqrt (4).
Next consider b = sqrt(3) + sqrt(5) and c = sqrt(4) + sqrt (4).
So sqrt(5) - sqrt(4) = {sqrt(5) - sqrt(4)}* {sqrt(5) + sqrt(4)}/ {sqrt(5) + sqrt(4)}
= (5 - 4)/ {sqrt(5)+sqrt(4)}=1/{sqrt(5)+sqrt(4)}.
Similarly sqrt(4) - sqrt(3) = 1/{sqrt(4) + sqrt(3)}
Now 5 > 4 and 4 > 3.
So sqrt(5) > sqrt(4) and sqrt(4) > sqrt(3).
Hence sqrt(5) + sqrt(4) > sqrt(4) + sqrt(3).
Or 1/ {sqrt(5) + sqrt(4)} < 1/ {sqrt(4) + sqrt(3)}.
Or {sqrt(5) - sqrt(4)} < {sqrt(4) - sqrt(3)}.
Or sqrt(5) + sqrt(3) < sqrt(4) + sqrt(4)
Or sqrt(5) + sqrt(3) < 4.
Or b < c.
So a < b < c.
The correct answer is (a).
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by gmatmachoman » Wed Apr 21, 2010 12:39 am
rn4gmat wrote:If we simply do a square for all these three it can be seen that

c : 16 = 8 + 8
b : 8 + 2 (15)^1/2 .Here 2 (15)^1/2 will be less than 2(16)^1/2 which is equal to 8 hence this will be less than c.
a: 8 + 2 (12)^1/2.Here 2 (12)^1/2 will be less than 2(15)^1/2 hence this will be less than b.

hence the correct order will be : c,b,a.
Nice Approach!!
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by kstv » Wed Apr 21, 2010 2:01 am
rn4gmat wrote:If we simply do a square for all these three it can be seen that
c : 16 = 8 + 8
b : 8 + 2 (15)^1/2 .Here 2 (15)^1/2 will be less than 2(16)^1/2 which is equal to 8 hence this will be less than c.
a: 8 + 2 (12)^1/2.Here 2 (12)^1/2 will be less than 2(15)^1/2 hence this will be less than b.
hence the correct order will be : c,b,a.
c² = 8 + 2√16
b² = 8 + 2√15 8 = 3+5
a² = 8 + 2√12 8 = 2+6
that will make it clearer
so ascending order
a, b & c as they are +ve nos.
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by pkw209 » Wed Apr 21, 2010 4:10 pm
c² = 8 + 2√16
b² = 8 + 2√15 8 = 3+5
a² = 8 + 2√12 8 = 2+6
Thanks, KSTV. Would it be possible for you to clarify that portion just a little more? I kind of follow it, but also a little lost.
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