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how to solve this! What is the concept!

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Manager
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how to solve this! What is the concept! [#permalink]

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how to solve this! What is the concept!


\(\frac{1}{x} + \frac{1}{y} = \frac{36}{323}\)
find the value of x-y

(1) x and y are prime.
(2) x>y>1


From the \(\frac{1}{x} + \frac{1}{y} = \frac{36}{323}\) we get that
x+y=36 and x*y = 323

this comes down to an equation where
-y*y+36y-323=0

where D<0 thus cannot be solved,

Does this mean that OA is E

Artuurs

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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 07:21
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x+y=36
xy=323

19 and 17 satisfy this condition
but again either of x or y could be 17 hence x-y can be 2 or -2

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Manager
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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 07:24
iwillwin wrote:
x+y=36
xy=323

19 and 17 satisfy this condition
but again either of x or y could be 17 hence x-y can be 2 or -2


Was it guessing or what!

what is the concept being tested

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Senior Manager
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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 07:25
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1/x +1/y is same as (x+y)/xy

so we know that xy = 323 or a multiple of it

to find out x and y, factorize 323

You will find that 17*19 = 323
also 17+19 = 36

so we know that x and y are 17 and 19, but not exactly which one is what.

Now look at the options.

For stmt 1: doesnt tell us the values of x and y, insufficient
Stmt 2: tells u that 1<y<x, so you know y=17 and x=19 and both are greater than 1, so no scope for decimals. Sufficient.

ans is B

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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 07:30
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Note that question doesn't tell us whether x and y are integers or not. So decimals too could satisfy this condition. If it were not given in stmt 2 that x and y are greater than 1, then ans would have been C.

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Director
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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 12:44
rashminet84 wrote:
Note that question doesn't tell us whether x and y are integers or not. So decimals too could satisfy this condition. If it were not given in stmt 2 that x and y are greater than 1, then ans would have been C.

Didn't get this one..!?
Even if x and y are greater than 1, they can be decimals and can satisfy the proportion.

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Director
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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 13:01
imo C!

Economist is right!!!

lets say, we just have x>y>1

if we take x as 18.9

then value of Y will be given from

\(1/y = 36/323 - 1/18.9\)

Do you think we can't solve for y. I guess it'll be less than 18.9.
so there are multiple set possible on real number line for this equation.

Need 1 and 2 ==> C
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Senior Manager
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Re: how to solve this! What is the concept! [#permalink]

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New post 22 Jul 2009, 13:19
sudeep wrote:
imo C!

Economist is right!!!

lets say, we just have x>y>1

if we take x as 18.9

then value of Y will be given from

\(1/y = 36/323 - 1/18.9\)

Do you think we can't solve for y. I guess it'll be less than 18.9.
so there are multiple set possible on real number line for this equation.


Need 1 and 2 ==> C

Yes both of u r right. It should be C.

why didnt i think of decimals greater than 1 when i could think of those less than 1 :wall

Kudos [?]: 150 [0], given: 15

Re: how to solve this! What is the concept!   [#permalink] 22 Jul 2009, 13:19
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