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M12-16

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Intern
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Re: M12-16  [#permalink]

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New post 29 Nov 2018, 13:50
Oh such a confusion i have created for applying much shortcuts.
a^2-b^2= (a+b)(a-b)
and i applied for
(a+b)^2
I think i have to take some rest. Exam pressure force me to do this type of 3rd quality of mistakes.

Thanks bunuel.
You are great seriously.
1 kudos for you.
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Re: M12-16  [#permalink]

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New post 21 Mar 2019, 20:00
Dear Bunnel

,I solved this sum as follows

I assumed x/y as a.Therefore a+1/a>2 which becomes (a-1)^2>0.which implies a>1 therefore x/y>1.

a)In Option a since x not equal to y.either x can be greater or y can be greater therefore not sufficient.

b)x and y can have any number therefore not sufficient.

Due to this i choose option E.What mistake am I making
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Re: M12-16  [#permalink]

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New post 08 Jul 2019, 17:24
If \(x\) and \(y\) are positive integers, is \(\frac{x}{y}+ \frac{y}{x} \gt 2\)?

I think it's easier to NOT manipulate the prompt here, at least for me.
Since they are +ints and reciprocal terms it's basically asking if x=y because that is the only way to get 1+1 = 2 (NO answer), and anything else will give us a number > 2 (YES answer).

(1) \(x\) does not equal \(y\)
Directly answers the prompt, sufficient.
To test plug in some random numbers like x=1,y=2 or x=9,y=10

(2) \(x\) and \(y\) do not share any common divisors except 1
The trap here is that we might assume this means they are prime numbers. However, that's not the case.
For example, 9 is not a prime number but does not share any divisors except 1 with 10. If x=9, y=10 then the above case works and gives a YES.
However the main point here is to note that 1 only has a divisor of 1, so if x & y = 1 then this fulfils the prompt and gives a NO.
Therefore, not sufficient.
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M12-16  [#permalink]

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New post Updated on: 22 Oct 2019, 10:33
Sum of a positive number with its reciprocal is always greater than or equal to 2(when number is equal to 1).

x + 1/x = [(x)^1/2 - (1/x)^1/2]^2 +2

Originally posted by gauradhya on 22 Oct 2019, 10:28.
Last edited by gauradhya on 22 Oct 2019, 10:33, edited 1 time in total.
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M12-16   [#permalink] 22 Oct 2019, 10:28

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