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by ghacker » Mon Jun 22, 2009 12:06 am
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
A) 5
B) 5(x-y)
C) 20x
D) 20y
E) 35x

The GCD must divide 35x and 20 Y so

35x = 7*5*x and 20y = 2*2*5*y

If y = 7x , 20y = 2*2*5*7*x so 35x can be the GCD
if x = 4y then 35x = 7*5*2*2*y , 20y can be the GCD

5 can be the GCD when x and y are relatively prime
5(x-y) can be the GCD when x-y=1

so 20X cannot be the GCD
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by axat » Wed Jul 22, 2009 2:45 am
Greetings, all. Here's another problem requiring a painfully lengthy solution. Do share your approach and answer, and if possible the time taken by you.


If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0
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by tohellandback » Wed Jul 22, 2009 3:29 am
IMO E (after an eye opener from scoobydooby)

time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
but when b=0,its the opposite
INSUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF

combined we have a<0,b<0 or b=0
so E
Attachments
soln.jpg
Last edited by tohellandback on Thu Jul 23, 2009 3:57 am, edited 1 time in total.
The powers of two are bloody impolite!!
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Cant see how A

by way2kailash » Wed Jul 22, 2009 1:35 pm
tohellandback wrote:IMO A
time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
SUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF
How is it A?

a<0

Ex a= -1, b=3 : a.|b|=-3, a-b =-4 so a.|b|>a-b
a=-1, b=-3: a.|b|=-3, a-b =2 so a.|b|< a-b
Not Suff
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Re: Cant see how A

by tohellandback » Wed Jul 22, 2009 3:50 pm
way2kailash wrote:
tohellandback wrote:IMO A
time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
SUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF
How is it A?

a<0

Ex a= -1, b=3 : a.|b|=-3, a-b =-4 so a.|b|>a-b
a=-1, b=-3: a.|b|=-3, a-b =2 so a.|b|< a-b
Not Suff
you can't take those values. |a| must be > |b|
The powers of two are bloody impolite!!
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Re: Cant see how A

by way2kailash » Thu Jul 23, 2009 3:17 am
tohellandback wrote:
way2kailash wrote:
tohellandback wrote:IMO A
time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
SUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF
How is it A?

a<0

Ex a= -1, b=3 : a.|b|=-3, a-b =-4 so a.|b|>a-b
a=-1, b=-3: a.|b|=-3, a-b =2 so a.|b|< a-b
Not Suff
you can't take those values. |a| must be > |b|
Im a dopehead
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Re: Cant see how A

by scoobydooby » Thu Jul 23, 2009 3:41 am
tohellandback wrote:
way2kailash wrote:
tohellandback wrote:IMO A
time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
SUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF
How is it A?

a<0

Ex a= -1, b=3 : a.|b|=-3, a-b =-4 so a.|b|>a-b
a=-1, b=-3: a.|b|=-3, a-b =2 so a.|b|< a-b
Not Suff
you can't take those values. |a| must be > |b|
for statement 1, if we take b=0, a=-5 say
a.|b|=-5.0=0
a-b=-5-0=-5
=>a.|b|>a-b
so we get both a yes and a no for statment 1. same for statement 2.
after combining, a<0, b<0 or b=0, we still get a yes and no
would go for E on this one.
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by aj5105 » Thu Jul 23, 2009 10:19 am
what was the eye opener?
tohellandback wrote:IMO E (after an eye opener from scoobydooby)

time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
but when b=0,its the opposite
INSUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF

combined we have a<0,b<0 or b=0
so E
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by ketkoag » Thu Jul 23, 2009 3:10 pm
I agree with scooby, we have to consider a case in which b = 0
hence E should be the answer..
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by tohellandback » Thu Jul 23, 2009 6:35 pm
aj5105 wrote:what was the eye opener?
tohellandback wrote:IMO E (after an eye opener from scoobydooby)

time taken 4 minutes
plz check the pic

1) pics 3 and 4 satisfy
now plug in numbers
pic 3: a=-5, b=-2
a.|b|=-10
a-b=-3
a.|b|<a-b

pic4: a=-5, b=4
a.|b|=-20
a-b=-9
but when b=0,its the opposite
INSUFFICIENT

2)pics 1 satisfies
pic 1: a=5,b=2
a.|b|=10
a-b=3


but when b=0, it is the opposite. NOT SUFF

combined we have a<0,b<0 or b=0
so E
I missed b=0, in 1st case when a<0
The powers of two are bloody impolite!!
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