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# Is x > 1/5 ? (1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27

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Is x > 1/5 ? (1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27 [#permalink]

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08 Jul 2008, 14:56
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Is x > 1/5 ?

(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26

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08 Jul 2008, 15:07
kevincan wrote:
Is x > 1/5 ?

(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26

from 1, Since "1/23 + 1/24 + 1/25 + 1/26 + 1/27" is > 1/5 and x > "1/23 + 1/24 + 1/25 + 1/26 + 1/27". So suff.

from 2, "1/22 + 1/23 + 1/24 + 1/25 + 1/26" is > 1/5 but x is less than "1/22 + 1/23 + 1/24 + 1/25 + 1/26". so nsf.

A.
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08 Jul 2008, 19:48
GMAT TIGER wrote:
kevincan wrote:
Is x > 1/5 ?

(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26

from 1, Since "1/23 + 1/24 + 1/25 + 1/26 + 1/27" is > 1/5 and x > "1/23 + 1/24 + 1/25 + 1/26 + 1/27". So suff.

from 2, "1/22 + 1/23 + 1/24 + 1/25 + 1/26" is > 1/5 but x is less than "1/22 + 1/23 + 1/24 + 1/25 + 1/26". so nsf.

A.

How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?

1/5 = 5/25

1/23 + 1/24 > 2/25

1/26 + 1/27 < 2/25

???

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08 Jul 2008, 20:47
gmatnub wrote:
GMAT TIGER wrote:
kevincan wrote:
Is x > 1/5 ?

(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26

from 1, Since "1/23 + 1/24 + 1/25 + 1/26 + 1/27" is > 1/5 and x > "1/23 + 1/24 + 1/25 + 1/26 + 1/27". So suff.

from 2, "1/22 + 1/23 + 1/24 + 1/25 + 1/26" is > 1/5 but x is less than "1/22 + 1/23 + 1/24 + 1/25 + 1/26". so nsf.

A.

How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?

1/5 = 5/25

1/23 + 1/24 > 2/25

1/26 + 1/27 < 2/25

???

= 1/23 + 1/24
= (24+23)/(23x24)
= (47)/(23x24)
= (47)/[(20+3) (25-1)]
= (47)/(500 + 75 - 20 - 3)
= 47/552

= 1/26 + 1/27
= (27+26)/(26x27)
= (53)/(26x27)
= (53)/[(25+1) (25+2)]
= (53)/(625+25+50+2)]
= 53/702

clearly 47/552 > 53/702
and also 47/552 > 2/25

so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5
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08 Jul 2008, 20:53
gmatnub wrote:

How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?

1/5 = 5/25

1/23 + 1/24 > 2/25

1/26 + 1/27 < 2/25

???

you dont need to do complex calculation. Just compare

A = 1/23-1/25 and B = 1/25-1/27
A = 2/(23*25) and B = 2/(25*27)

clearly A > B

So ((1/23 + 1/24) - 2/25 ) is more than (2/25 - (1/26 + 1/27))

First two terms are increasing the sum from 1/5 and last two terms are decreasing it.
But Contribution of first two terms is more than that of last two terms

We can say that 1/23 + 1/24 + 1/25 + 1/26 + 1/27 > 1/5

Last edited by durgesh79 on 08 Jul 2008, 23:54, edited 2 times in total.

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08 Jul 2008, 21:11
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A

The numbers look to be in perfect progression so I Just took similar example with:

Is x > 3/6 ? (3 times the middle number of condition (1))

(1) x > 1/5 + 1/6 + 1/7
(2) x < 1/4 + 1/5 + 1/6

May not be fool proof but it worked.

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08 Jul 2008, 22:38
1
KUDOS
indy123 wrote:
A

The numbers look to be in perfect progression so I Just took similar example with:

Is x > 3/6 ? (3 times the middle number of condition (1))

(1) x > 1/5 + 1/6 + 1/7
(2) x < 1/4 + 1/5 + 1/6

May not be fool proof but it worked.

This is great!

3 numbers: question asks 3 times the middle number
5 numbers: question asks 5 times the middle number

+1 for you with the hope that this will work for all similar problems

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09 Jul 2008, 01:18
GMAT TIGER wrote:
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?

[...]

clearly 47/552 > 53/702
and also 47/552 > 2/25

so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5

I don't agree with your conclusion, sorry .

From 47/552 > 53/702 and 47/552 > 2/25, how do you know that 53/702 > 2/25? Because you're assuming that, and it is :
- not prooved
- and more importantly: not true

Last edited by Oski on 09 Jul 2008, 01:49, edited 2 times in total.

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09 Jul 2008, 01:24
I think a good way to compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 and 1/5 is to look at the following:

$$\frac{1}{23} + \frac{1}{27} = \frac{27+23}{23*27} = \frac{50}{23*25 + 23*2} > \frac{50}{23*25 + 2*25} = \frac{2*25}{25*25} = \frac{2}{25}$$

you can do exactly the same to proove that $$\frac{1}{24} + \frac{1}{26} > \frac{2}{25}$$

And then, yes, we have $$\frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} > \frac{1}{5}$$

Edit: the hint here is "when you are faced a symmetrical expression, always think of symmetry"

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09 Jul 2008, 01:58
I am impressed by the solution proposed by indy123 as its very fast.
Hope it works for other set of values also.

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09 Jul 2008, 03:25
RayOfLight wrote:
I am impressed by the solution proposed by indy123 as its very fast.
Hope it works for other set of values also.

This is not a solution per se.

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09 Jul 2008, 08:45
Oski wrote:
RayOfLight wrote:
I am impressed by the solution proposed by indy123 as its very fast.
Hope it works for other set of values also.

This is not a solution per se.

I disagree there with you. Its a non sense solution but very accurate for these questions.

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09 Jul 2008, 09:16
indy123 wrote:
A

The numbers look to be in perfect progression so I Just took similar example with:

Is x > 3/6 ? (3 times the middle number of condition (1))

(1) x > 1/5 + 1/6 + 1/7
(2) x < 1/4 + 1/5 + 1/6

May not be fool proof but it worked.

nice but wanna clarify they are not in progression and to be precise you are talking about arithmatic progression

1/6 -1/7 is not equal to 1/5 -1/6

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09 Jul 2008, 13:40
Oski wrote:
GMAT TIGER wrote:
How do you compare 1/23 + 1/24 + 1/25 + 1/26 + 1/27 to 1/5 without long calculation?

[...]

clearly 47/552 > 53/702
and also 47/552 > 2/25

so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5

I don't agree with your conclusion, sorry .

From 47/552 > 53/702 and 47/552 > 2/25, how do you know that 53/702 > 2/25? Because you're assuming that, and it is :
- not prooved
- and more importantly: not true

Did I assume 53/702 > 2/25 ? Can you show me? hmm..........
Of course I assumed 47/552 > 2/25.
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09 Jul 2008, 20:52
In situations like this where you have a progression of the sume of fractions like

$$\frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27}$$

When we are comparing the sum against the middle fraction * the number of fractions, the sum will always be greater than the middle fraction * the number of fractions, provided the numerators are the same and the denominators are consecutive.

The reason behind this is if you start out with the middle, here $$\frac{1}{25}$$ as the "average" we need to figure out which way the two fractions on either side tip the scales.

Attachment:
FractionSumComparison.xls [17.5 KiB]

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Last edited by jallenmorris on 10 Jul 2008, 04:44, edited 1 time in total.

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10 Jul 2008, 01:11
GMAT TIGER wrote:
Did I assume 53/702 > 2/25 ? Can you show me? hmm..........
Of course I assumed 47/552 > 2/25.

I really think you did (it is not because you didn't explicitly wrote it that you didn't and you apparently used it).

Otherwise tell me then (I don't get it), I do you go from there
GMAT TIGER wrote:
1/23 + 1/24 = 47/552
1/26 + 1/27 = 53/702

clearly 47/552 > 53/702
and also 47/552 > 2/25
to there :
GMAT TIGER wrote:
so "1/23 + 1/24 + 1/25 + 1/26 + 1/27" > 1/5
?

Cause with the equations you wrote, we can say : 1/23 + 1/24 + 1/25 + 1/26 + 1/27 > 2/25 + 1/25 + 53/702

But how do you go from there to > 1/5 if you don't assume that 53/702 > 2/25 too ?

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10 Jul 2008, 01:37
kevincan wrote:
Is x > 1/5 ?

(1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27
(2) x < 1/22 + 1/23 + 1/24 + 1/25 + 1/26

Note that for 0 < k < 25, 1/(25 - k) + 1/(25 + k) = 50/($$25^2 - k^2)$$ > 50/$$25^2$$= 2/25

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10 Jul 2008, 01:41
kevincan wrote:
Note that for 0 < k < 25, 1/(25 - k) + 1/(25 + k) = 50/($$25^2 - k^2)$$ > 50/$$25^2$$= 2/25

Exactly. Symmetry, symmetry

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Re: DS Fractions   [#permalink] 10 Jul 2008, 01:41
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