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# If x and y are integers and 2*x^(1/2) + y^2 < 5, then x*y can take how

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Math Expert
Joined: 02 Sep 2009
Posts: 64111
If x and y are integers and 2*x^(1/2) + y^2 < 5, then x*y can take how  [#permalink]

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30 Sep 2019, 04:50
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95% (hard)

Question Stats:

34% (02:31) correct 66% (02:15) wrong based on 64 sessions

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If x and y are integers and $$2\sqrt{x} + y^2 < 5$$, then x*y can take how many different values ?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

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Joined: 02 Aug 2009
Posts: 8587
Re: If x and y are integers and 2*x^(1/2) + y^2 < 5, then x*y can take how  [#permalink]

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30 Sep 2019, 05:47
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Bunuel wrote:
If x and y are integers and $$2\sqrt{x} + y^2 < 5$$, then x*y can take how many different values ?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

$$2\sqrt{x} + y^2 < 5$$
x has to be non-negative...
1) when x=0, $$0+y^2<5$$...irrespective of the value of y, xy will be 0.
2) when x=1, $$2*1+y^2<5.....y^2<3$$, so y=+1, -1, 0... xy will be 0, 1 and -1.
3) when x=4, $$4+y^2<5$$, so y will be 0... xy will be 0.
4) when x is 2 or 3 similarly, y can be 1 or -1..possible value of xy = 2, -2, 3 and -3

Different values of xy = 0, 1, -1, 2, -2, 3, -3..so 7 values

D
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Joined: 24 Nov 2016
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If x and y are integers and 2*x^(1/2) + y^2 < 5, then x*y can take how  [#permalink]

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29 Mar 2020, 14:18
Bunuel wrote:
If x and y are integers and $$2\sqrt{x} + y^2 < 5$$, then x*y can take how many different values ?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

$$2\sqrt{x}+y^2<5…y=(negative,zero,positive)$$
$$\sqrt{anything.in.gmat}≥0: x=non.negative.integer$$

$$y=0:2\sqrt{x}+y^2<5…2\sqrt{x}<5…\sqrt{x}<2.5$$
$$x={0,1,2,3}…xy=x*0={0}…(always 0)$$

$$y=1:2\sqrt{x}+y^2<5…2\sqrt{x}<4…\sqrt{x}<2$$
$$x={0,1,2,3}…xy=x*1={0,1,2,3}$$

$$y=-1:2\sqrt{x}+y^2<5…2\sqrt{x}<4…\sqrt{x}<2$$
$$x={0,1,2,3}…xy=x*(-1)={0,-1,-2,-3}$$

xy={-1,-2,-3,0,1,2,3}=7

Ans (D)
If x and y are integers and 2*x^(1/2) + y^2 < 5, then x*y can take how   [#permalink] 29 Mar 2020, 14:18