GMAT Data Sufficiency Discussions

vikshen · September 8, 2009, 4:05am

Question:
If w, x and y are different positive integers, and wxy = 24, is xy divisible by 6?

(1) wy is not divisible by 6.

(2) x is even.

rohitforever · September 8, 2009, 6:14am

Question:
If w, x and y are different positive integers, and wxy = 24, is xy divisible by 6?

(1) wy is not divisible by 6.

(2) x is even.

1=> possible values of x=3,6,12

when x=3 {y,z}={(1.

,(8,1),(2,4),(4,2)}
now or x,y,z= 3,1,8 xy= 3---not divisible by 6
but for x,y,z= 3,2,4 xy= 6--- divisible by 6
so NOT SUFF

2=> possible values of x=2,4,8,12
now for x,y,z=2,4,3 xy =8 not divisible by 6
now for x,y,z=2,3,4 xy=6 divisible by 6
so NOT SUFF

1&2=> possible values of x =6,12
in any cas enow xy is divisible by 6
so SUFF

ans is D IMO

deepakraam · September 8, 2009, 6:33am

Statement 1:
============
w= 8, y=1,x=3
w= 2, y=4, x=3..Nt suff

Statement 2:
============
x=4,y=1,w=6
x=4,y=6,w=1..Nt suff..

Combining both we get possible values..

So I wud go with option C

pawank_iit · September 8, 2009, 7:06am

Question:
If w, x and y are different positive integers, and wxy = 24, is xy divisible by 6?

(1) wy is not divisible by 6.

(2) x is even.

I am getting C as the answer.

aric.parker · September 8, 2009, 7:51pm

Question:
If w, x and y are different positive integers, and wxy = 24, is xy divisible by 6?

(1) wy is not divisible by 6.

(2) x is even.

IMO - E;

the options, neither independently nor together are sufficient.
Explanation - Look at Statement 2 first, "x is even". Now if you factorize 24 you get only "3" as odd factor. It means only the placement of "3" will get us the answer.

Now as per Statement 1, for wy not to be divisible by 6... the most troublesome options would be

W X Y
3 8 1
1 8 3

Neither independently nor together can the two statements reduce the above two cases from 2 to one so that we are certain of the answer. Only way is -if it is known that whether 'w' or 'y' is even. Hence, E.

aric.parker · September 8, 2009, 8:04pm

Got stumped in the following DS questions. : Need some detail explanation:

2> S is a set of integers such that
i) if a is in S, then a is in S, and
ii) if each of a and b is in S, then ab is in S.
Is 4 in S?

(1) 1 is in S.
(2) 2 is in S.

IMO = B

Statement - 1
1 in S => -1 in S which means 1 X -1 = -1 is also in S (which is of no help). NOT SUFF
Statement 2
2 in S => -2 is also in S, which means 2 X -2 = -4 is also in S.
SUFFICIENT

aric.parker · September 8, 2009, 8:23pm

Is X greater than Y?
1. 6X=5Y
2. X^2 x
NOT SUFF

Sq X Sq X - Sq Y (x-y) (x+y) -y x -y. That is, depend upon the sign of y.
NOT SUFF

Together also they are not sufficient..

vikshen · September 9, 2009, 7:28am

Question:
If w, x and y are different positive integers, and wxy = 24, is xy divisible by 6?

(1) wy is not divisible by 6.

(2) x is even.

@aric.parker is right. The OE is E.

Consider statement (1): wy is not divisible by 6. Lets look at the following ex:
w=1,x=6, y=4 : Is xy divisible by 6 ?- Yes
w=3, x=8 , y=1: Is xy divisible by 6 ? - No

So by using statement(1), we cannot figure out whether xy is divisible by 6 or not.

Considering statement (2): x is even
w=1,x=6, y=4 : Is XY divisible by 6 ?- Yes
w=3, x=8 , y=1: Is XY divisible by 6 ? - No
So by using just statement B, we cannot figure out whether xy is divisible by 6 or not.

Now looking at statement 1 and 2 together: wy is not divisible by 6 and x is even
w=1,x=6, y=4 : Is xy divisible by 6 ?- Yes
w=3, x=8 , y=1: Is xy divisible by 6 ? - No

So we have an example which proves that even 1 and 2 combined are not enough to answer the question. So the answer is E.

deepakraam · September 10, 2009, 10:28am

I am stumped with the following... help please.

If z = xn - 19, is z divisible by 9?
1) x = 10; n is a positive integer
2) z + 981 is a multiple of 9

When I did this, I got that the answer was B. However, the official question answer say something different. I can easily see how z + 981 is a multiple of 9 tells you z is divisible by 9 since both 981 and 9 are divisible by 9, but I am apparently not understanding Statement (1). I will avoid going into further detail about Statement (1) so as to not spoil it for others yet.

Statement 1:
============
Z = 10n-19
n = 1 ; z = -9 divisible by 9
n = 2 , z = 1 not divisible by 9

Nt suff

Statement 2:
============
z + 981 is divisible by 9
since 981 is divisible by 9 ,(Z = 9n or Z = 0).Nt suff

Combining both we can ans the q.So my take on this q is option C

aric.parker · September 10, 2009, 7:18pm

I am stumped with the following... help please.

If z = xn - 19, is z divisible by 9?
1) x = 10; n is a positive integer
2) z + 981 is a multiple of 9

When I did this, I got that the answer was B. However, the official question answer say something different. I can easily see how z + 981 is a multiple of 9 tells you z is divisible by 9 since both 981 and 9 are divisible by 9, but I am apparently not understanding Statement (1). I will avoid going into further detail about Statement (1) so as to not spoil it for others yet.

Hey Swati...

I feel the answer is "D"

Please visit your "Answer" link... it says ---

The Question is --- If z = x^n - 19, is z divisible by 9? (That is, 'x' raise to the power 'n')

So with z = x^n - 19, try n=1, 2, 3.....

You get 10 - 19 = -9 ; 100 - 19 = 81 ; 1000 - 19 = 981 ; 9981; 99981.... and all are divisible by 9 (there sum of digits being div by 9)

Hence, Statement A is also SUFFICIENT

Statement 2 you\ve already proved..So Answer is
"D"

aric.parker · September 10, 2009, 7:24pm

Statement 2:
============
z + 981 is divisible by 9
since 981 is divisible by 9 ,(Z = 9n or Z = 0).Nt suff

Combining both we can ans the q.So my take on this q is option C

Hi Deepak,

try this way...

z + 981 = 9k (as it is divisible by 9)
=> z = 9k - 981 => z = 9 (k - 109) => z = 9 p , where p = (k - 109)

Hence, Statement 2 is also sufficient.

deepakraam · September 11, 2009, 7:10am

Statement 2:
============
z + 981 is divisible by 9
since 981 is divisible by 9 ,(Z = 9n or Z = 0).Nt suff

Combining both we can ans the q.So my take on this q is option C

Hi Deepak,

try this way...

z + 981 = 9k (as it is divisible by 9)
=> z = 9k - 981 => z = 9 (k - 109) => z = 9 p , where p = (k - 109)

Hence, Statement 2 is also sufficient.

Hi Aric,
I agree that z should be a multiple of 9 ONLY WHEN IT IS NOT EQUAL TO ZERO.But the q dint mention that z is not equal to zero.Even when z=0 statement will hold true.So it is alone not suff.Do you agree?

Statement 2:
==========
Z+981 is divisible by 9.

sdt83 · September 11, 2009, 10:28am

1. Does b=c?
(a) c-b = b -c
(b) b/c = c/b

2. x & y are non zero. what is the value of y/x?
(a) x =6
(b) y^2 = x^2
3. Does the integer K have a factor P such that 1(a) K> 4!
(b) 13! +2

deepakraam · September 11, 2009, 10:43am

1. Does b=c?
(a) c-b = b -c
(b) b/c = c/b

2. x & y are non zero. what is the value of y/x?
(a) x =6
(b) y^2 = x^2
3. Does the integer K have a factor P such that 1(a) K> 4!
(b) 13! +2

1. Does b=c?
(a) c-b = b -c.Suff
(b) b/c = c/b Nt suff.

option A

2. x & y are non zero. what is the value of y/x?
(a) x =6 . Nt suff
(b) y^2 = x^2.Nt suff

combining we get +/- 1.

Option E

nemo1 · September 11, 2009, 10:47am

1. Does b=c?
(a) c-b = b -c
(b) b/c = c/b

2. x & y are non zero. what is the value of y/x?
(a) x =6
(b) y^2 = x^2
3. Does the integer K have a factor P such that 1(a) K> 4!
(b) 13! +2

1) From the first condition, we get that b=c by rearranging the terms. The 2nd condition is not sufficient as we do not have info on the respective signs. Therefore: (A)

2) The first condition is definitely insufficient. From the 2nd condition we are clueless about the signs. i.e. - if we do assume x=6, we are not sure whether y = +6 or -6. Therefore, y/x cannot be got conclusively even after using both the options. Hence: (E)

sdt83 · September 11, 2009, 10:51am

Hey Deepak.. these are some of d qs from GMATPrep test.. i think Q1 & Q2 have the same concepts.. While d OA for Q1 is given as D , OA for Q2 is E..I was bit confused... thats why i have posted these Qs together..

deepakraam · September 11, 2009, 12:28pm

sdt83 Says
Hey Deepak.. these are some of d qs from GMATPrep test.. i think Q1 & Q2 have the same concepts.. While d OA for Q1 is given as D , OA for Q2 is E..I was bit confused... thats why i have posted these Qs together..

For the first q, I feel D can't certainly be the ans bcos statement 2 gives no clue about the question.So statement 2 is useless which also drives out option D.If b and c are given as positive nos then I cud go with option D

deepakraam · September 11, 2009, 12:33pm

3. Does the integer K have a factor P such that 1(a) K> 4!
(b) 13! +2

For the above q

statement 1 : Nt Suff ( K can be composite or prime)

statement 2 : possible values

K = 13!+2
K = 13!+3
.
.
.
.
K = 13!+13

For all the above values it is possible to get one factor which will be less than K.Suff

So my take for the above q is option B.
OA please

subash_r · September 11, 2009, 12:51pm

1. Does b=c?
(a) c-b = b -c
(b) b/c = c/b

2. x & y are non zero. what is the value of y/x?
(a) x =6
(b) y^2 = x^2
3. Does the integer K have a factor P such that 1(a) K> 4!
(b) 13! +2
1.
Statement A:
=>2b = 2c
=> b = c

Statement B:
b^2 = c^2

So statement is not sufficient as 'b' and 'c' can be of opposite signs as well.

2.
Statement A:
Don't know the value of 'y' here. So not sufficient.

Statement B:
Alone is not sufficient.

Statement A and B together:
We don't know the value of 'y' again.

subash_r · September 11, 2009, 12:58pm

statement 2 : possible values

K = 13!+2
K = 13!+3
.
.
.
.
K = 13!+13

For all the above values it is possible to get one factor which will be less than K.Suff

So my take for the above q is option B.
OA please

Deepak..can you please explain the Bold and italised statement.....
how can it be said that we will have one factor which will be less than K ??