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# I always have trouble with these... does any one have

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Intern
Joined: 20 May 2008
Posts: 7
I always have trouble with these... does any one have [#permalink]

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31 May 2008, 12:00
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I always have trouble with these... does any one have cheatsheets for how to solve these kinds?

1. Is x less than y?
(1) x-y+1<0
(2) x-y-1<0

OA is A.

2. If b is an integer, is an integer?
(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0

OA is B

3. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

OA is D
Director
Joined: 23 Sep 2007
Posts: 783

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31 May 2008, 12:13
Manager
Joined: 28 May 2008
Posts: 94

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31 May 2008, 12:27
1
KUDOS
sportypk wrote:
I always have trouble with these... does any one have cheatsheets for how to solve these kinds?

1. Is x less than y?
(1) x-y+1<0
(2) x-y-1<0

OA is A.

2. If b is an integer, is an integer?
(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0

OA is B

3. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

OA is D

Hey there..
1) x-y+1<0
which means x < y-1
Which means x is clearly less than y.
for eg y=5
this means x<4 which is less than 5. right? SO, A is sufficient !
Option B says
x-y-1<0
which means x<y+1
Now in this case x could be less than y or greater than Y. I'll show you how
Lets say Y= 5. Now x<(5+1) ie x<6
Now x could be 4 which is less than Y. But, x could also be 5.5. right?
Hence B is not sufficient !
Therefore A !

2) A says
a^2 + b^2 is an integer.
since b is an integer , b^2 is obviously an integer.
If a is an integer, a^2 will be an integer.
Even if a is not an integer(lets say a is sqrt(3) , then a^2 will be 3) a^2 could end up an integer.
Hence A is not sufficient.
coming to B
a^2 – 3b^2 = 0
which means a^2=3b^2
this means a=sqrt(3)b (+ or -)
which means a cant be an integer .as we r multiplying by sqrt(3). right?
Hence B is sufficient !

3)x^2<1
means -1<x<1
which means |x| <1 (this expression means -1<x<1)
Hence A is sufficient
Now B
|x| < 1/x
since x is an integer (excluding zero)
(1/any integer) would be less than 1
Hence B sufficient
Therefor D would be the answer !!

BTW: whats a cheat sheet.m pretty new here

Please shoot any questions or doubts you might have !

Cheers!!
Director
Joined: 14 Oct 2007
Posts: 753
Location: Oxford
Schools: Oxford'10

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31 May 2008, 15:05
zeenie wrote:
sportypk wrote:
I always have trouble with these... does any one have cheatsheets for how to solve these kinds?

1. Is x less than y?
(1) x-y+1<0
(2) x-y-1<0

OA is A.

2. If b is an integer, is an integer?
(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0

OA is B

3. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

OA is D

Hey there..
1) x-y+1<0
which means x < y-1
Which means x is clearly less than y.
for eg y=5
this means x<4 which is less than 5. right? SO, A is sufficient !
Option B says
x-y-1<0
which means x<y+1
Now in this case x could be less than y or greater than Y. I'll show you how
Lets say Y= 5. Now x<(5+1) ie x<6
Now x could be 4 which is less than Y. But, x could also be 5.5. right?
Hence B is not sufficient !
Therefore A !

2) A says
a^2 + b^2 is an integer.
since b is an integer , b^2 is obviously an integer.
If a is an integer, a^2 will be an integer.
Even if a is not an integer(lets say a is sqrt(3) , then a^2 will be 3) a^2 could end up an integer.
Hence A is not sufficient.
coming to B
a^2 – 3b^2 = 0
which means a^2=3b^2
this means a=sqrt(3)b (+ or -)
which means a cant be an integer .as we r multiplying by sqrt(3). right?
Hence B is sufficient !

3)x^2<1
means -1<x<1
which means |x| <1 (this expression means -1<x<1)
Hence A is sufficient
Now B
|x| < 1/x
since x is an integer (excluding zero)
(1/any integer) would be less than 1
Hence B sufficient
Therefor D would be the answer !!

BTW: whats a cheat sheet.m pretty new here

Please shoot any questions or doubts you might have !

Cheers!!

Hi Zeenie, welcome to the club.

I agree with the OAs for 1 and 3 as per your working (this is what I got)

but for OA for 3 is wrong since if a=0, you get a^2 = sqrt(3)*b^2
which makes a an integer. As per your working a can also be non integer.

Manager
Joined: 28 May 2008
Posts: 94

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01 Jun 2008, 02:17
zeenie wrote:
sportypk wrote:
I always have trouble with these... does any one have cheatsheets for how to solve these kinds?

1. Is x less than y?
(1) x-y+1<0
(2) x-y-1<0

OA is A.

2. If b is an integer, is an integer?
(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0

OA is B

3. If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x

OA is D

Hey there..
1) x-y+1<0
which means x < y-1
Which means x is clearly less than y.
for eg y=5
this means x<4 which is less than 5. right? SO, A is sufficient !
Option B says
x-y-1<0
which means x<y+1
Now in this case x could be less than y or greater than Y. I'll show you how
Lets say Y= 5. Now x<(5+1) ie x<6
Now x could be 4 which is less than Y. But, x could also be 5.5. right?
Hence B is not sufficient !
Therefore A !

2) A says
a^2 + b^2 is an integer.
since b is an integer , b^2 is obviously an integer.
If a is an integer, a^2 will be an integer.
Even if a is not an integer(lets say a is sqrt(3) , then a^2 will be 3) a^2 could end up an integer.
Hence A is not sufficient.
coming to B
a^2 – 3b^2 = 0
which means a^2=3b^2
this means a=sqrt(3)b (+ or -)
which means a cant be an integer .as we r multiplying by sqrt(3). right?
Hence B is sufficient !

3)x^2<1
means -1<x<1
which means |x| <1 (this expression means -1<x<1)
Hence A is sufficient
Now B
|x| < 1/x
since x is an integer (excluding zero)
(1/any integer) would be less than 1
Hence B sufficient
Therefor D would be the answer !!

BTW: whats a cheat sheet.m pretty new here

Please shoot any questions or doubts you might have !

Cheers!!

Hi Zeenie, welcome to the club.

I agree with the OAs for 1 and 3 as per your working (this is what I got)

but for OA for 3 is wrong since if a=0, you get a^2 = sqrt(3)*b^2
which makes a an integer. As per your working a can also be non integer.

Heyy there Buff
Thank you for the welcome! :D
I am sure I'll get to learn loads from all of you members.
Ans yes you r right
a=sqrt(3)*b
if b=0, a will be 0 which is an integer
and if b is any other integer, other than 0
then a will be irrational.
Hence insufficient!
Thank you for pointing that out !

cheers !!
Re: Few problems   [#permalink] 01 Jun 2008, 02:17
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