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if x2 + y2 = 29, what is the value of (x+y)2 1. xy = 10 2. x

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if x2 + y2 = 29, what is the value of (x+y)2 1. xy = 10 2. x [#permalink] New post 19 Sep 2010, 05:48
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A
B
C
D
E

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71% (01:12) correct 29% (00:18) wrong based on 17 sessions
if x2 + y2 = 29, what is the value of (x+y)2

1. xy = 10
2. x = 5

simple problem..just have to bring an equation for xy here..A is sufficient..as xy value is given..but i feel B is also sufficient..as i can solve the value for y with the value of x.. i.e x2 + y2 = 29 - 2xy

I feel it should be D? or am i assuming sumthing rong here?
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Re: Simple Problem..But small doubt [#permalink] New post 19 Sep 2010, 06:56
If x^2+y^2=29, and even if you know x=5, it is not sufficient to solve for y, since the first equation is quadratic and you will get two answers for y, namely \sqrt{24} and -\sqrt{24}. So its impossible to know the sign of y, and hence the sign of xy. This is why (2) is not sufficient
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Re: Simple Problem..But small doubt [#permalink] New post 19 Sep 2010, 15:36
Jayanth2689 wrote:
if x2 + y2 = 29, what is the value of (x+y)2

1. xy = 10
2. x = 5

simple problem..just have to bring an equation for xy here..A is sufficient..as xy value is given..but i feel B is also sufficient..as i can solve the value for y with the value of x.. i.e x2 + y2 = 29 - 2xy

I feel it should be D? or am i assuming sumthing rong here?


Option A is enough.
Option B: you definitely need to know the value of y otherwise ur equation will be like (X+Y)^2=29+10y. You cant proceed forward without knowing the value of Y.
Hence the ans: A
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Re: Simple Problem..But small doubt [#permalink] New post 21 Sep 2010, 08:43
shrouded1 wrote:
If x^2+y^2=29, and even if you know x=5, it is not sufficient to solve for y, since the first equation is quadratic and you will get two answers for y, namely \sqrt{24} and -\sqrt{24}. So its impossible to know the sign of y, and hence the sign of xy. This is why (2) is not sufficient


Man I liked your status ... I hopeyou are just kidding by asking this question .... pls do app !!...

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Re: Simple Problem..But small doubt [#permalink] New post 21 Sep 2010, 08:48
lol!! tx for ur replies! and i have an uncanny knack to make simple problems complicated ;) thats y the status!
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Re: Simple Problem..But small doubt [#permalink] New post 16 Oct 2010, 07:48
shd be A...easy one...
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Re: Simple Problem..But small doubt [#permalink] New post 17 Oct 2010, 04:47
Given X^2 + Y^2 = 29, from a) XY = 10
Therefore (x & y) can be (5 & 2) or (-5 & -2)
Therefore We cannot come to conclusion from a) right
Please clarify
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Re: Simple Problem..But small doubt [#permalink] New post 17 Oct 2010, 23:29
praba1130 wrote:
Given X^2 + Y^2 = 29, from a) XY = 10
Therefore (x & y) can be (5 & 2) or (-5 & -2)
Therefore We cannot come to conclusion from a) right
Please clarify


x^2+y^2=29
xy=10
x^2+y^2+2xy=49
(x+y)^2=49

Hence, it is sufficient
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Re: Simple Problem..But small doubt [#permalink] New post 20 Oct 2010, 09:24
Jayanth2689 wrote:
if x2 + y2 = 29, what is the value of (x+y)2

1. xy = 10
2. x = 5

simple problem..just have to bring an equation for xy here..A is sufficient..as xy value is given..but i feel B is also sufficient..as i can solve the value for y with the value of x.. i.e x2 + y2 = 29 - 2xy

I feel it should be D? or am i assuming sumthing rong here?


1 Alone sufficient....

(x+y)^2 = x^2 + y^2 + 2xy ----> 29 + xy

but from 1 , xy = 10 ...so i can get the value of (x+y)^2

from 2 it is not sufficient very clearly...
so ans is : 1
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Re: Simple Problem..But small doubt [#permalink] New post 20 Oct 2010, 12:59
A.

1) X^2+Y^2=29 and 2XY=20 sufficient

2) 25+y^2=29 Y=+2 or -2 thus giving two answers. insufficient
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Re: Simple Problem..But small doubt   [#permalink] 20 Oct 2010, 12:59
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