Three more Power Prep questions... interesting..Explinations plzz...

[givinggmat]

1. If X and Y are positive integers such that X = 8Y + 12, what is the greatest common divisor of X and Y?
a. X = 12U, where U is an integer
b. Y = 12Z, where Z is an integer


2. In 1995 Division A of company X had 4,850 customers. If there were 86 service errors in division A that year, what was the service-error rate, in number of service errors per customers, for Division B of Company X in 1995?
a. In 1995 the overall service-error rate for Divisions A and B combined was 1.5 service errors per 100 custmers
b. In 1995 Division B had 9,359 customers, none of whom were customers of Division A.


3. In the XY-Coordinate plane, line l and line k intersect at the point (4,3), Is the product of their slopes negative?
a. The product of the x-intercepts of lines l and k is positive
b. The product of the Y-intercepts of lines l and k is negative

[madboy]

# 1:

From a
X=12U
Y=(3/2)(U-1) [By substituting X=12U in X=8Y+12 and solving for Y]
Since Y is an integer (U-1)/2 has to be an integer (U should be odd in this case). Since the greatest common divisor of U and (U-1)/2 is 1, the GCD of X and Y has to be 3. Hence SUFFICIENT

From b
X=12(2Z+1) [By substituting Y=12Z in X=8Y+12 and solving for X]
Y=12Z
Again, the GCD of z and 2z+1 is 1, so the GCD of X and Y has to be 12.Hence SUFFICIENT
Hence, D

The only doubt I have is - usually in GMAT DS q's, when the answer is D, both 1 and 2 lead to the same answer, but here they are not. I think I am going wrong somewhere.??

[madboy]

#2:


For Div B let no of customers = x and no of service errors = y. The question is (y/x)=?
From a we get -> (86 + y)/(4850 + x) = 1.5/100 => not sufficient for determining y/x
From b we get x=9359 =>not sufficient
Combining a and b, we can determine y and hence y/x
Hence, C

[madboy]

# 3:


Let the equations of K and L be
y=m1.x +c1
y=m2.x +c2
The question is m1.m2<0 ?

From a, (-c1/m1)(-c2/m2)>0 or (c1.c2)/(m1.m2)>0. To satisfy this c1c2 and m1m2 should be of the same sign , but they can be +ve or -ve . Hence NOT SUFFICIENT

From b, c1.c2<0 => there is no way we can tell sign of m1m2 from this and the point of intersection (4,3), at least I can't find any. Hence NOT SUFFICIENT

Combining a and b, if (c1.c2)/(m1.m2)>0 and c1.c2<0 then m1.m2 <0. Hence sufficient
IMO, answer is C

[clock60]

Originally Posted by madboy

# 1:

From a
X=12U
Y=(3/2)(U-1) [By substituting X=12U in X=8Y+12 and solving for Y]
Since Y is an integer (U-1)/2 has to be an integer (U should be odd in this case). Since the greatest common divisor of U and (U-1)/2 is 1, the GCD of X and Y has to be 3. Hence SUFFICIENT

From b
X=12(2Z+1) [By substituting Y=12Z in X=8Y+12 and solving for X]
Y=12Z
Again, the GCD of z and 2z+1 is 1, so the GCD of X and Y has to be 12.Hence SUFFICIENT
Hence, D

The only doubt I have is - usually in GMAT DS q's, when the answer is D, both 1 and 2 lead to the same answer, but here they are not. I think I am going wrong somewhere.??

what if x=60 y=6 then GCF=6
but if x=36 y=3 GCF=3
so first is insuff
(2)y=12 x=108 GCF=12
y=24 x=204 204=12*17 GCF=12
y=36 x=300 300=12*25 GCF=12
y=48 x=396 396=12*33 GCF=12
so B

[givinggmat]

B B C are the answers..!

waiting for the explinations...!!

[madboy]

How can answer for #2 be B? No info on service errors in div B..

[clock60]

i also think that it is a mistake in oa for #2
the answer is C

[ss87]

i think that answer for #1 is B

1. If X and Y are positive integers such that X = 8Y + 12, what is the greatest common divisor of X and Y?
a. X = 12U, where U is an integer
b. Y = 12Z, where Z is an integer

Statement A says X=12U --> X=12U=8Y +12. Therefore, since X and U are integers, 8Y + 12 is also divisible by 12. So 8Y can be any multiple of 12 .
X is also a multiple of 12, since X=12U.
So statement A alone is not sufficient.

Y = 12Z
For statement B,
X = 8Y + 12 --> X= 96Z + 12 = 12(8Z+1).Now 8Z+1 is ALWAYS ODD!
WHY? because 8Z is always even.
If 8Z+1 is always odd, then 8Z+1 can NEVER be a multiple of 12.
So in X= 96Z + 12 = 12(8Z+1) , the only EVEN multiple is 12.
Since Y=12Z, 12 is always a multiple of Y.
even then however, GCD cannt be 12..cos there are exceptions such as when Z=3, etc when GCD is 36

[indianwarrior23]

1. If X and Y are positive integers such that X = 8Y + 12, what is the greatest common divisor of X and Y?
a. X = 12U, where U is an integer
b. Y = 12Z, where Z is an integer

a. X = 12U, where U is an integer
Solving for Y, Y=3(U-1)/2

Solving for U, U=1

X=12, Y=0. GCD of (12,0)=12

b.Y = 12Z, where Z is an integer
Solving for X, X=12(8Z+1)

Solving for Z, Z=0

(X,Y)=(12,0) - GCD is 12