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which is the greatest?

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Director
Joined: 25 Oct 2008
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11 Aug 2009, 07:39
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80% (01:34) correct 20% (00:00) wrong based on 8 sessions

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

[Reveal] Spoiler:
OA:A

Guys pls help me with this..took a long time to type!!!
Allrite i went with C in a jiffy..we want the greatest so d and E are out as they have some values which are to be substrcted.B is out cos C is greater than b.Remained with A and C..C has squares in it so that has to be greatest..
Okay..messed up logic I know dont kill me help me!!;)
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Re: which is the greatest? [#permalink]

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11 Aug 2009, 08:29
1
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tejal777 wrote:
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.$$1-\frac{3}{4}+\frac{4}{5}+\frac{1}{3}$$

[Reveal] Spoiler:
OA:A

Guys pls help me with this..took a long time to type!!!
Allrite i went with C in a jiffy..we want the greatest so d and E are out as they have some values which are to be substrcted.B is out cos C is greater than b.Remained with A and C..C has squares in it so that has to be greatest..
Okay..messed up logic I know dont kill me help me!!;)

you are right here that D and E are out because some values needs to be subtracted.

C is also out because we are squaring the fractions ( For e.g 1/2 > 1/4 and 1/4 is a square of 1/2 )

So now we are left with A and B.

Now, if we take square root of any fraction the value will be higher than the fraction itself. ( Lets consider the same example sqrt of 1/4 = 1/2 )

So B out and hence the answer is A.

I hope this helps. !

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Current Student
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Re: which is the greatest? [#permalink]

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11 Aug 2009, 08:34
tejal777 wrote:
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.$$1-\frac{3}{4}+\frac{4}{5}+\frac{1}{3}$$

[Reveal] Spoiler:
OA:A

Guys pls help me with this..took a long time to type!!!
Allrite i went with C in a jiffy..we want the greatest so d and E are out as they have some values which are to be substrcted.B is out cos C is greater than b.Remained with A and C..C has squares in it so that has to be greatest..
Okay..messed up logic I know dont kill me help me!!;)

well for C when you raise a fraction to a power it will always go smaller (imagine fraction as decimal like .2 if you raise it to 2 it will be smaller). For A I just multiplied the entire equation by sqrt(60)/(sqrt(60)) which is legal since it is just 1. since all the values share a factor in sqrt(60) you will see that the result is basically b with each fraction multiplying by a sqrt term like:
2/3 *(sqrt(30)/sqrt(20)). since the sqrt terms are greater than 1 (just by looking at them) we know the fractional term wont be smaller so A is good.
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Director
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Re: which is the greatest? [#permalink]

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11 Aug 2009, 15:58
Quote:
So now we are left with A and B.

Now, if we take square root of any fraction the value will be higher than the fraction itself. ( Lets consider the same example sqrt of 1/4 = 1/2 )

So B out and hence the answer is A.

could somebody explain between A and B more clearly?
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Kudos [?]: 1143 [0], given: 100

Current Student
Joined: 12 Jun 2009
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Kudos [?]: 274 [1], given: 52

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Re: which is the greatest? [#permalink]

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11 Aug 2009, 19:59
1
KUDOS
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!!!!
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Re: which is the greatest? [#permalink]

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12 Aug 2009, 23:50
nitishmahajan wrote:
tejal777 wrote:
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.$$1-\frac{3}{4}+\frac{4}{5}+\frac{1}{3}$$

[Reveal] Spoiler:
OA:A

Guys pls help me with this..took a long time to type!!!
Allrite i went with C in a jiffy..we want the greatest so d and E are out as they have some values which are to be substrcted.B is out cos C is greater than b.Remained with A and C..C has squares in it so that has to be greatest..
Okay..messed up logic I know dont kill me help me!!;)

you are right here that D and E are out because some values needs to be subtracted.

C is also out because we are squaring the fractions ( For e.g 1/2 > 1/4 and 1/4 is a square of 1/2 )

So now we are left with A and B.

Now, if we take square root of any fraction the value will be higher than the fraction itself. ( Lets consider the same example sqrt of 1/4 = 1/2 )

So B out and hence the answer is A.

I hope this helps. !

This approach is better and faster. Thanks!

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Manager
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Re: which is the greatest? [#permalink]

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25 Aug 2009, 22:45
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....

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Senior Manager
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Re: which is the greatest? [#permalink]

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26 Aug 2009, 01:50

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Director
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Re: which is the greatest? [#permalink]

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25 Sep 2009, 22:33
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!
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Re: which is the greatest? [#permalink]

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25 Sep 2009, 22:41
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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

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Re: which is the greatest? [#permalink]

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25 Sep 2009, 23:06
tejal777 wrote:
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!

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

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Re: which is the greatest?   [#permalink] 25 Sep 2009, 23:06
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