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

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

81% (01:19) correct
19% (00:18) wrong based on 26 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?

Re: Simple Problem..But small doubt [#permalink]
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 _________________

Re: Simple Problem..But small doubt [#permalink]
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 _________________

Re: Simple Problem..But small doubt [#permalink]
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 !!...

Re: Simple Problem..But small doubt [#permalink]
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

Re: Simple Problem..But small doubt [#permalink]
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

Re: Simple Problem..But small doubt [#permalink]
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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