If x^2 + 5y = 49. Is y an integer?

Note that y to be an integer 49-x^2 should be a multiple of 5.

(1) 1<x<4, well if x is an integer 2 or 3 then the answer will be YES but x may as well be some fraction say 3/2 and in this case the answer will be NO. Not sufficient.

(2) x^2 is an integer --> clearly insufficient.

(1)+(2) Now, \(x^2\) is an integer DOES NOT mean that \(x\) is an integer it also can be some irrational number: \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), ... Note that as these values are in the range \(1<x<4\) then they are valid values for \(x\) and for them \(y\) is not an integer. So we still have Yes (\(x=2\) for example) and No (for \(x=\sqrt{2}\)) answers. Not sufficient.

Answer E.

Side note: \(x\) can be an irrational number and \(y\) still be an integer for example \(x=\sqrt{14}<4\) --> \(14+5y=49\) --> \(y=7\).

Hope it's clear.
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x^2+5y=49 .Is y an integer?
1. 1<x<4
2. X^2 is an integer

Statement 1 alone insufficient: As possible values of 'x' includes fractions- which means a value for 'y' could be an integer or non-integer.

Statement 2 alone insufficient: As 'x^2' could have any positive value. Which means 'y' could be a fraction or an integer (either +ve / or -ve).

Together:
'x^2' is an integer,and and 'x' = 2 or 3. Thus, the equation boils down to 5y = 45 (if x=2) and 5y = 40 (if x=3). So, Y= 9 or 8. Answers the question whether Y is an integer or not.

So, I will go with 'C' as the answer choice.

State 1: Clearly insufficient
State 2: Clearly insufficient

Combined,
Considering \(x^2\) is an integer, \(x = [2, \sqrt{2}, \sqrt{3}, 3]\)
If \(x = [2, 3]\),then Y is Integer
If \(x = [\sqrt{2}, \sqrt{3}]\), then Y is NOT Integer

Combined Insufficient.

Answer is E.
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Consider KUDOS for good posts

This is in addition to the previous post and to reply 'Bunnel'.

If x^2 is an integer,then x is an integer - WRONG.
If x^2=5, x is not an integer.

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I agree with the above statements. But together statement 1 & 2 are STILL sufficient to answer the question asked. Even if x^2 = 5, 'x' falls between 1<x<5. If we replace 'x^2' = 5 in the original equation, we can decisively say whether or not 'y' is an anteger.

Other possible values for 'x' satisfying "Bunnel's" proposition (also satisfying statement 1) ranges from x^2 = 2 to x^2 = 24.

If we replace 'x^2' in the original statement "x^2+5y=49" with any value from "x^2 = 2 to x^2+5y= 24"; We are in a position to say whether or not 'y' is an integer. Because, the question asked "is 'y' an integer?". The answer could be "yes" or "no". From my analysis, I would be able to say either "yes" or "no" with the combination of the two statements.

Therefore, I will still go for choice 'C'.

I take back my two earlier posts and seek sincere apology for my misunderstanding. Yes now, I do agree that together the statements are not syfficient, as there are ranges of values. If "x^2 = an even number" than 'y' is an integer and if "x^2 = an odd number", then 'y' is not an integer. Thus, I can't decisively say whether or not 'y' is an integer.

Thus, the answer choice is E.

Please accept my apology.

No need to apologies buddy. Everybody in this forum makes mistake. We all are here with a common goal. It is always better to make mistakes here than to make those in real test

If X^2+ 5Y = 49, is Y an integer.

1. 1 < X < 4

2. \(X^2\) is an integer

Even I get C..

Given \(y=\frac{49-x^2}{5}\)

S1. 1<x<4 means x could be an number integer or fraction between this range. If a fraction, Y is not an integer. If an integer, Y is an integer. Hence NOT SUFF

S2: \(x^2\) is an integer doesn't give any definite ans as for some values of x, Y is an integer and for some, it is not. Hence NOT SUFF.

S1 + S2: SUFF.... as now x is only 2 or 3.. since only then \(x^2\) would be an integer. Therefore SUFF.

What's the source?

stmnt 1) let X = 2 then X^2 +5Y = 4 + 5Y = 49 => Y = 9 integer

let X = \(\sqrt{2}\) then X^2 + 5Y = 2 +5Y = 49 => Y = 47/5 not an integer

hence insuff

stmnt 2) let X = 2 then X^2 = 4 (integer) => X^2 +5Y = 4 + 5Y = 49 => Y = 9 integer

let X = \(\sqrt{2}\) then X^2 = 2 (integer) => X^2 + 5Y = 2 +5Y = 49 => Y = 47/5 not an integer

hence insuff

taking together as well for value of X as \(\sqrt{2}\) and 2 we get Y as both integer and non integer. Hence E

I got this wrong... Thanks for your help bunuel and hgp2k...

if x^2+5y=49, is Y an integer?

1) 1<x<4
2) x^2 is an integer


I have an issue with the kaplan answer here. My answer would be c because stmt one tells you x is between integers 1 and 4, but it could be a non-integer so not sufficient, and 2 is not sufficient because x could be an integer. However, together the stmts satisfy as x could only be 2 or 3 and 2^2 and 3^2 both agree with the equation making y an integer. Please advise.

Hi!

You've misinterpreted statement (2).

From (2), we know that x^2 is an integer. Based on this, you've erroneously concluded that x must be an integer as well.

However, that's not the case. If x = root2 or root3, x^2 will still be an integer. Moreover, root2 and root3 both satisfy statement 1 as well. In fact, all we know is that:

root1 < x < root16,

so x could be root2, root3, root4, root5, ...., root 15. root4 and root9 are both integers, but none of the others are.

So, even combined, we can't determine whether y is an integer.

May be E i think if you mean x2 = x^2

A. not sufficient ... x can be 2 in which case y is an integer..
x can be 3.3 .. and y is not integer..

B. not sufficient.. x^2 is an integer.. if x^2 = 4 .. y is an integer
but if x^2 = 3... y is not an integer..

considering both statements, again, since x is between 1 and 4.. x can be 1.73 so x^2 = 3 then y is not an integer..
or x can be 2 and x^2=4 ..so y is an integer..

\(y=\frac{49-x^2}{5}\)

If unit's digit of x^2=4,9. y will be an integer

(1) 1<x<4
\(x=\sqrt{2}\)
y=Not Integer
\(x=2\)
y=Integer
Not Sufficient.

(2) x^2 is integer
If x=1; x^2=1; unit digit is neither 9 nor 4. y=integer
If x=2; x^2=4; unit digit is 4. y=integer
Not Sufficient.

Together;
x can be either 3 or 2; square of both; 9 and 4; unit digit either 9 or 4.
y must be integer
Sufficient.

Ans: "C"

Is approximation allowed in such problems? I thought we need to ensure that X^2+ 5Y= 49. This equation will never satisfy a non integer, unless we approximate.

fluke,

C is incorrect.

I see how you have assumed x to be an integer when it is stated that x^2 is an integer. What if x is square root of 3? x^2 will still be an integer, and still agree with 1 < x < 4. However, if you solve x^2 + 5y = 49, y will be a non-integer.

If x is an integer, x^2 is always an integer, but not the other way around.

Hence, answer is E.

Hope this helps you.

If x^2 + 5y = 49, is y an integer?

(1) 1 < x < 4

(2) x^2 is an integer.


Ans: Stmt1: x can be 2 , then y is an integer
x = 1.5 then y non integer Insuff

Stmt 2: x = 2 then x^2=4 and y is an integer
x=1 then x^2 = 1 and y is a non integer . Hence insuff

Combining 1 and 2

x can take only two values either 2 or 3 to result in x^2 as an integer. Both values will give an integer value for y. Suff. (C)

it is nowhere mentioned that X is an integer
Hope we do not miss such points actual GMAT
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\(5y=49-x^2…y=\frac{49-x^2}{5}…49-x^2=multiple.5…x^2=(4,9,49)…x=(2,3,7)\)

(1) 1 < x < 4: \(x=(integer,non-integer,irrational)\) insufic.
(2) x^2 is an integer: \(x=(integer,irrational)…x^2=(\sqrt{2}^2)=2=integer\) insufic.
(1&2) \(x=(\sqrt{2},2,3,non-integer)…x^2=(2,4,9…)\) insufic.

Answer (E)