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Is √(x + y) an integer? [#permalink]
2
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To find: √x+y = an integer?

Statement 1: X^3 = 64
=> X = 4 {what about Y?}
(Insufficient)

Statement 2: X^2 = Y-3
There can be many values for both X and Y here ranging from positive to negative (Insufficient).

Statement 1&2 : X^2 = Y-3 & X = 4
(4)^2 = Y-3
16+3 = Y
19 = Y
Putting Value of X and Y in √x+y
√4+19 = √23 (not an integer)
(Sufficient)

IMO C

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Re: Is √(x + y) an integer? [#permalink]
1
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Is \(\sqrt{x + y}\) an integer?


(1) \(x^3 = 64\)--> insuff: x=4, if y=0, then \(\sqrt{x + y}\) = \(\sqrt{4 + 0}\)=2 --> yes, but if y=1, then \(\sqrt{x + y}\) = \(\sqrt{4 + 1}\)=\(\sqrt{5}\) --> no

(2) \(x^2 = y – 3\) --> insuff: y=x^2+3, \(\sqrt{x + y}\) = \(\sqrt{x + x^2+3}\) =\(\sqrt{(x+1/2)^2+3-1/2^2}\) =\(\sqrt{(x+1/2)^2+11/4}\), if x=2, \(\sqrt{(x+1/2)^2+11/4}\) = 3, if x =4, \(\sqrt{(x+1/2)^2+11/4}\) not equant to integer

combining (1) & (2)=>
from (1) x=4
y=4^2+3=19
so \(\sqrt{x + y}\) = \(\sqrt{4 + 19}\)= \(\sqrt{23}\) --> yes, \(\sqrt{x + y}\) is not an integer
Answer: C
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Re: Is √(x + y) an integer? [#permalink]
sqrt( x+y) = integer ?
x+y = int^2
it means x+y must be a integer who is perfect square.
x=? ; y=?

Statement 1: x^3 = 64
=> x = 4
what about y =?
Not Sufficient


Statement 2: x^2 = y - 3
Not Sufficient.

Statement 1 & Statement 2 :

From (1) ; x=4
From (2) ; y = 4^2 + 3 = 16+3 = 19
sqrt(4+19) is not an integer
Sufficient

Answer(C)
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Re: Is √(x + y) an integer? [#permalink]
1
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From statement 1,
We get x=4,
Nothing is mentioned about y
So, simply discarding it.
Hence, insufficient.

From statement 2,
x^2=y-3
Let, y=12, we get x= 3 and -3
Again putting values of x and y, 3 and 12 respectively we get sqrt(x+y) = sqrt(15) (not an integer)
Putting x and y, -3 and 12 respectively we get sqrt(x+y) = 3 (integer)
Hence, insufficient.

Combining,
Putting x as 4 on statement 2 we get y= 19
Thus sqrt(x+y) = sqrt(19) not an integer.

Combining works and we get no the sqrt(x+y) is not an integer.

Hence, IMO ans is C.

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Re: Is √(x + y) an integer? [#permalink]
1
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Bunuel wrote:
Is \(\sqrt{x + y}\) an integer?


(1) \(x^3 = 64\)

(2) \(x^2 = y – 3\)



imo C .

(1) \(x^3 = 64\)

No info about y .. so insufficient .

(2) \(x^2 = y – 3\)

2 variables unknown.. No info .. so insufficient .

Combining 1 and 2 .. we can find x and y and can find whether \(\sqrt{x + y}\) is integer or not ..

So ans is C
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Re: Is √(x + y) an integer? [#permalink]
What if we substitute y=x^2-3 in the question ?
Then we know for sure that sq.root(x+(x^2)-3)!=integer
Then answer would be B
Can someone help me out here please ?
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Is √(x + y) an integer? [#permalink]
1
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Anush56 wrote:
What if we substitute y=x^2-3 in the question ?
Then we know for sure that sq.root(x+(x^2)-3)!=integer
Then answer would be B
Can someone help me out here please ?


Anush56

Even if we substitute \((y=x^2 + 3)\) in \( \sqrt{(x+y)}\)

K = \(\sqrt{x + (x^2 + 3) }\)

if x = 2 ,then K = 3 = integer .

if x = 0.5 ,then K = \(\frac{ 6}{4 }\) is not an integer

So option B is insufficient

Hope it helps..
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Re: Is √(x + y) an integer? [#permalink]
Kindly see the attachment.
Combined statements are sufficient.
C
Attachments

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Re: Is (x + y) an integer? [#permalink]
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Re: Is (x + y) an integer? [#permalink]
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