'Though this be madness yet there is method in it.'

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

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

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

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

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|

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

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.

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 agree with scooby, we have to consider a case in which b = 0
hence E should be the answer..

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