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Need help with these two

by knight247 » Sat Jul 16, 2011 8:43 am
*Is X and integer?
(1)X^3 is an integer
(2)3X is an integer

*If X and Y are two distinct integers such that xy+x+y+1=20, what is the value of x?
(1)x=2n, where n is an integer
(2)x>y

Detailed explanations would be appreciated
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by sandeeptalla450 » Sat Jul 16, 2011 9:59 am
*Is X and integer?
(1)X^3 is an integer
(2)3X is an integer
Consider statement (1):
Let X^3=1, then X=1 => X is an integer.
Let X^3=2, then X=2^(1/3) => X is not an integer.
Hence, statement (1) is not sufficient.

Consider statement (2):
Let 3X=3, then X=1 => X is an integer.
Let 3X=2, then X=2/3 => X is not an integer.
Hence, statement (2) is not sufficient.

Consider both statements together: X^3 and 3X are both integers => X is an integer. (Consider any value.)
Hence, C.
*If X and Y are two distinct integers such that xy+x+y+1=20, what is the value of x?
(1)x=2n, where n is an integer
(2)x>y

The question can be simplified to (x+1)(y+1) = 20
So, the possible values are for (x+1) and (y+1) are
4 5
5 4
-4 -5
-5 -4
So, possible values for x and y are:
x y
3 4
4 3
-5 -6
-6 -5

Consider statement (1):
Let x=2n, where n is an integer => x is an EVEN integer which is true for (x,y) values of (4,5) and (-4,-5).
Hence, statement (1) is not sufficient.

Consider statement (2):
x>y which is true for (x,y) values of (5,4) and (-4,-5).
Hence, statement (2) is not sufficient.

Considering both statements together: There is only one possible value for (x,y) = (-4,-5)
Hence, C.
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by Frankenstein » Sat Jul 16, 2011 10:01 am
Hi,
Q1)From(1):
if x^3 = 8, x=2(integer)
if x^3 = 7, x is not integer
Not sufficient
From(2):
if 3x =2, xis not integer
if 3x =3, x is integer
Not sufficient
Both (1) and (2):
x = n/3 where n is an integer
x^3 = n^3/27 is an integer. So, n^3 must be multiple of 27.
As n is already an integer n should be of the form 3*p
So, x is an integer
Hence, C

Q2)xy+x+y+1=20
So, (1+x)(1+y) = 20
From(1): x is even
x=4, y=3 satisfies
x= -2, y= -21 satisfies
In fact, there are more values which satisfy. Using this case because these will satisfy the combined case as well.
Not sufficient
From(2):
x=4, y=3
x= -2, y= -21
Not sufficient
Both(1) and (2):
x=4, y=3
x= -2, y= -21
Not sufficient

Hence, E
Cheers!

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by gmat25 » Sat Jul 16, 2011 10:18 am
*Is X and integer?
(1)X^3 is an integer
(2)3X is an integer
IMO C

(1) Insufficient
if x = cube root of any int. then x^3 = int ---> means x is not an int still x^3 is int
if x = int then x^3 = int ---> means x is an int and x^3 is int

so, x can be an int or non-int, hence Insufficient

(2) Insufficient

if X = non int = 1/3 still 3x = int
if x = int = 1,2 or 3 etc 3x = int

COMBINE THEM, if x = cube root of any int. then 3x will not be int that means u cannot take non int value of x hence sufficient.
*If X and Y are two distinct integers such that xy+x+y+1=20, what is the value of x?
(1)x=2n, where n is an integer
(2)x>y
IMO A

xy+x+y+1=20 = (x+1)(y+1) = 20

Op A

x=2n, where n is an integer ---> putting this value in above eq'n

(2n+1)(y+1) = 20

Now if u notice, (2n+1) = odd, why because, 2n = even so even + 1 = odd

now only possible even * odd = 20 combination is

5*4 = 20 no other combination is possible

so X + 1 = 5 and Y + 1 = 4, hence u get the value of X

Op B

x>y

(x+1)(y+1) = 20

now combinations can be

10 * 2 , here X = 9 and Y = 1

5 * 4, here x = 4 and Y = 3

so as u can see more than one possible value of X is possible hence insufficient
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by Frankenstein » Sat Jul 16, 2011 10:28 am
gmat25 wrote:
x=2n, where n is an integer ---> putting this value in above eq'n

(2n+1)(y+1) = 20

Now if u notice, (2n+1) = odd, why because, 2n = even so even + 1 = odd

now only possible even * odd = 20 combination is

5*4 = 20 no other combination is possible

so X + 1 = 5 and Y + 1 = 4, hence u get the value of X
Hi,
How about 1*20, (-1)*(-20) and (-5)*(-4)?
These are possibilities as well right?
Cheers!

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by knight247 » Sat Jul 16, 2011 10:34 am
Thanks Frank!! As always...Ur the most reliable...lol
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by gmat25 » Sun Jul 17, 2011 12:17 am
Frankenstein wrote:
gmat25 wrote:
x=2n, where n is an integer ---> putting this value in above eq'n

(2n+1)(y+1) = 20

Now if u notice, (2n+1) = odd, why because, 2n = even so even + 1 = odd

now only possible even * odd = 20 combination is

5*4 = 20 no other combination is possible

so X + 1 = 5 and Y + 1 = 4, hence u get the value of X
Hi,
How about 1*20, (-1)*(-20) and (-5)*(-4)?
These are possibilities as well right?
Yes u are right...i missed this point. Thanks
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