Data sufficiency question - Pls help

[priyasaibaba]

Can anyone pls answer this data sufficiency question?

Is x an odd integer?
(1) x + 3 is an even integer.
(2) x/3 is an odd integer.

My solution:

for stmt 1 to be true, 'x' can take values 1,3,5,7...or negative values like -1,-3,-5 etc. (I'm considering 0 as even here)
so stmt 1 is sufficient

for stmt 2 to be true, 'x' can take values 3,9,15,21..or negative values -3,-9,-15,-21 etc.
so whenever x/3 is odd, x is always odd.
stmt2 is also sufficient

So answers d - Both statements are alone sufficient

This is a question of GMAT set 5. The given answer is C.

Can anyone tell me what mistake I'm doing?

Thanks,
Priya

[gmatmachoman]

Can anyone pls answer this data sufficiency question?

Is x an odd integer?
(1) x + 3 is an even integer.
(2) x/3 is an odd integer.

My solution:

for stmt 1 to be true, 'x' can take values 1,3,5,7...or negative values like -1,-3,-5 etc. (I'm considering 0 as even here)
so stmt 1 is sufficient

for stmt 2 to be true, 'x' can take values 3,9,15,21..or negative values -3,-9,-15,-21 etc.
so whenever x/3 is odd, x is always odd.
stmt2 is also sufficient

So answers d - Both statements are alone sufficient

This is a question of GMAT set 5. The given answer is C.

Can anyone tell me what mistake I'm doing?

Thanks,
Priya

@Priya
If u take zero to be even ,try plugging that in st 2. It will become even number rite??

But according to st 2......x/3 is an odd integer.

Let us focus on st 2.
x/3 is an odd integer

Now plug in X =1...
It wont give a Odd integer at all( its a fraction). So we cannot use X=1 or -1.


Coming to st 1, we can use X =1 or -1 to make x + 3 is an even integer ...Agreed?

Now try to combine both the statements 1 & 2.

apart from {-1,0,1} we can precisely say whether X is odd or not.

[godspeed]

Hi

Have you considered values like 30,36,42 .. for the statement 2 ? When you consider these statement 2 will not be sufficient

[gmatmachoman]

Hi

Have you considered values like 30,36,42 .. for the statement 2 ? When you consider these statement 2 will not be sufficient

Multiples of X follows a pattern :

3,9,15,21,27,33,39.......( with a difference of 6).

the values 30,36,42 is NOT at all a issue.

Issue was whether u can use 0 or NOT. I said Drop 0 while taking values of X as it is giving Contradictory results.
{-1,0,1) need not be considered.

Values of X are {3,9,15...27....39...45...}( Series with a difference of 6 between the elements) Same holds true for negative integers.

So we need to use both the sts!

[fibbonnaci]

@gmatmachoman,
as u say in statement 2, we cannot use x= 1 , -1 etc since it does not satisfy the criteria.
But our concern is not what numbers satisfy but to identify if the numbers that satisfy the criteria are odd or not.

According to statement 2: all the numbers that satisfy the criteria are odd because of the simple reason, odd * odd => odd

x = 3* odd, therefore X is odd.
This statement in itself is sufficient. we dont even have to consider zero, coz 0/3 -> 0 (which is even and our statement 2 is different)

and from statement 1:
odd+ odd => even
x+3 -> even
therefore this statement in itself is sufficient.

considering zero, you do not disprove the sufficiency. IMO i would go for D and not C

[gmatmachoman]

@gmatmachoman,
as u say in statement 2, we cannot use x= 1 , -1 etc since it does not satisfy the criteria.
But our concern is not what numbers satisfy but to identify if the numbers that satisfy the criteria are odd or not.

According to statement 2: all the numbers that satisfy the criteria are odd because of the simple reason, odd * odd => odd

x = 3* odd, therefore X is odd.
This statement in itself is sufficient. we dont even have to consider zero, coz 0/3 -> 0 (which is even and our statement 2 is different)

and from statement 1:
odd+ odd => even
x+3 -> even
therefore this statement in itself is sufficient.

considering zero, you do not disprove the sufficiency. IMO i would go for D and not C

Fibo, gimme a moment , I shall revert back to ur concerns!

[rockeyb]

Is x an odd integer?
(1) x + 3 is an even integer.
(2) x/3 is an odd integer.


Now lets rephrase statement (1)

x + 3 = even .

Now you can get an even number result only in following two cases :

(1) odd + odd = even

(2) even + even = even

Now since 3 here is odd "x" has to be odd .

Sufficient.


Now lets see statement (2)

x/3 = odd integer .

Now this statement says "x" is a multiple of 3 more importantly its a odd multiple why ?

Because you will not get odd integer if you divide an even multiple of 3.

lets see if x is an even multiple of 3 then x will have factors (3, 2 .....) that is x will have at least a 3 and 2 as factors .

Now if you divide this even multiple of 3 with 3 you will cancel out one factor 3 and the result will have at least one factor 2 . Hence the result will be even.

Hence if x/3 has to be odd x has to be odd multiple of 3 .

in other words x is odd .

Sufficient .

The answer should be D .

[gmatmachoman]

Fibo & Rockeyb

to simplify our discussion , I request you to drop down the probable values of X.


Can X = 1 or -1?? , if so plz explain!

[gmatmachoman]

Is x an odd integer?
(1) x + 3 is an even integer.
(2) x/3 is an odd integer.


Now lets rephrase statement (1)

x + 3 = even .

Now you can get an even number result only in following two cases :

(1) odd + odd = even

(2) even + even = even

Now since 3 here is odd "x" has to be odd .

Sufficient.

Govi's explanation:My Issue : Agreed X has to be ODD. If we take X=1, we get X+3 =4. Now try plugging the value of X=1 in St2.It wont give a Odd integer. So you can't use X={-1,0,1}

I am pretty sure in DS, we are supposed to get same Values if we select Option D.
.

[harshavardhanc]

Fibo & Rockeyb

to simplify our discussion , I request you to drop down the probable values of X.


Can X = 1 or -1?? , if so plz explain!

Macho bhai,

yes, x can be 1 or -1. both these values will be present in the set X = { x ; x+3 is even }.

however, they won't be present in the set given by statement 2, say Y, where Y = { y ; y/3 is an odd integer}.
This doesn't affect the answer choice ( D).


Look at the question in this way :

There are two sets of integers given by statement 1 and statement 2.

If a number is included in either one of them, you need to tell whether the number is Odd?

The number may be present in both of them. It will be a coincidence with which we are not concerned.


In our case, you will find that both these sets contain only odd values. Hence, each statement is sufficient on its own to tell that the number is odd.

tried to make it clearer. let me know if you are now satisfied.

[rockeyb]

Is x an odd integer?
(1) x + 3 is an even integer.
(2) x/3 is an odd integer.


Now lets rephrase statement (1)

x + 3 = even .

Now you can get an even number result only in following two cases :

(1) odd + odd = even

(2) even + even = even

Now since 3 here is odd "x" has to be odd .

Sufficient.

Govi's explanation:My Issue : Agreed X has to be ODD. If we take X=1, we get X+3 =4. Now try plugging the value of X=1 in St2.It wont give a Odd integer. So you can't use X={-1,0,1}

I am pretty sure in DS, we are supposed to get same Values if we select Option D.
.

Govi aka ,

I will try to make a simple explanation to this .

This is a fact that in GMAT DS statements are always true .

That is both the statements will always give you the same value . If you have not got same values then we have made a mistake .


Now having said that lets try and see statement 1 : X has to be odd .

Agreed we are clear till this point every one agrees to this .

Now lets go to statement 2 :

It not only says that x has to be odd but also says that x has to be a multiple of 3 .

(2) x/3 = odd integer
x = 3(odd integer)

Bingo !!!! and now we know why statement 2 is not true when we put in {-1,0,1} in place of x , simple in order for statement 2 to be true x has to be a multiple of 3 and thus it satisfies statement 1 also since 3(odd ) x (odd integer) will give you odd integer.


Another point to note here is if you take values {-1,0,1} for x and put it in statement 2 .

In order for statement to be true the result has to be an INTEGER first let alone odd or even , we have to make the equation as it is the basic requirement of the statement .

So indirectly statement 2 says :

x is an odd integer its a multiple of 3 AND x not equal to (-1,0,1).

I hope I have explained well .

[gmatmachoman]

Guys,(@harsha,Fibo,Rockey)

Can u plz tell me for st 1 & st 2 what are the range of values of X??

St 1: {-----------}

St2:{--------}

If u post the values, I think our discussion can come to end..

[rockeyb]

Guys,(@harsha,Fibo,Rockey)

Can u plz tell me for st 1 & st 2 what are the range of values of X??

St 1: {-----------}

St2:{--------}

If u post the values, I think our discussion can come to end..

Any integer that is a multiple of 3 will be set of values for X that will satisfy statement 1 and 2 .

[gmatmachoman]

Guys,(@harsha,Fibo,Rockey)

Can u plz tell me for st 1 & st 2 what are the range of values of X??

St 1: {-----------}

St2:{--------}

If u post the values, I think our discussion can come to end..

Any integer that is a multiple of 3 will be set of values for X that will satisfy statement 1 and 2 .

hehehehehhhehehhehe

If so is the case, what abt values of 1,-1 & 0 in st 1??? Now plz dont start with old versions of our discusion bro...Plz!

I know u r also tired of this drama pranabji!

[harshavardhanc]

Guys,(@harsha,Fibo,Rockey)

Can u plz tell me for st 1 & st 2 what are the range of values of X??

St 1: {-----------}

St2:{--------}

If u post the values, I think our discussion can come to end..

Any integer that is a multiple of 3 will be set of values for X that will satisfy statement 1 and 2 .

just a little correction!

Both, St1-set and St2-set will contain multiples of 3, but the sets will not be the same. St2-set is a subset of St1-set.



macho bhai,

the sets need not be the same, HENCE each statement alone is sufficient to answer the question. Moreover, 0 is neither odd nor even, that's why out of discussion.