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## Kaplan DS

This topic has 2 member replies
gmatme Junior | Next Rank: 30 Posts
Joined
06 Mar 2007
Posted:
18 messages

#### Kaplan DS

Sun Apr 29, 2007 10:02 am
if X^2 + 5Y = 49, is Y an Integer?

1) 1 < X < 4
2) X^2 is an integer.

The answer is E? I think it should be C, can some one explain why E?

The following is my reason for C)

From 1) X could be 2, 3, 3/2 etc. If X is 2 or 3 then Y is an integer but if 3/2 then it is not so, 1) is not sufficient.

From 2) not sufficient

From 1 and 2. X^2 is integer then X should be integer so possible values are 2,3. For these 2 values Y is integer - so sufficient.

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bww Senior | Next Rank: 100 Posts
Joined
29 Apr 2007
Posted:
55 messages
1
Mon Apr 30, 2007 7:04 pm
here's how i look at it:

1) X can be any value between 1 and 4, not inclusive. let's assume X=1.1. plugging that into the question stem gives us 1.21+5Y=49. pretty obvious that Y is not an integer. let's say X=2. then we have 4+5Y=49. Y then is an integer. so we have determined both "yes" and "no" to the question. 1) is not sufficient. rule out A and D.

2) cover 1) up momentarily and just consider 2). X^2 is an integer, but X is not necessarily an integer. in the example given by jayhawk2001, the sqrt of 3 is roughly 1.73. point is, unless we know what X is, X^2 can be any integer (1, 5, 85, etc.), which gives us both "yes" and "no" to the question if Y is an integer. thus, 2) is not sufficient, so we rule out B. but 2) also indirectly addresses combining statements 1) and 2). since X^2 is part of the question and we know that X^2 IS an integer, we get different values for Y when X^2 takes on different values. thus, 1) and 2) are both insufficient. the answer is E.

jayhawk2001 Community Manager
Joined
28 Jan 2007
Posted:
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Sun Apr 29, 2007 1:16 pm
gmatme wrote:
From 1 and 2. X^2 is integer then X should be integer so possible values are 2,3. For these 2 values Y is integer - so sufficient.
If X^2 is an integer, X need not necessarily be an integer. Take X^2 = 3.
X = sqrt(3) [ positive root ] which is not an integer.

Hence E

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