x and y are positive integers, is x=y ?

When you use the critical points to open the modulus, you get

|x-a| = |y-a| ----> \(\pm\)(x-a) = \(\pm\)(y-a) ----> you get the following 4 cases

1) x-a=y-a ---> x=y
2) x-a=-(y-a)=-y+a---> x+y=2a
3) -x+a = y-a --->x+y=2a
4) -x+a=-y+a ---> x=y

Thus , you get a yes for x=y based on cases 1 and 4 while you get a "no" based on cases 2 and 3. Thus this statement is not sufficient.

(1) |x - a| = |y - a|
--> Distance of point 'x' from point 'a' is equal to distance of point 'y' from point 'a' on the number line
2 Cases are possible (as shown)


Insufficient

(2) a < 0
Doesn't talk about x or y

Insufficient

Combining (1) & (2)
Since a is negative and both (x, y) are positive; only case 2 is possible (as shown)


Sufficient

IMO Option C

Pls Hit Kudos if you like the solution

Once you get that none of the statements are sufficient on their own, on combining the statements you get

a<0 and 2 possible cases of x=y or x+y=2a

You can now either follow the number line approach that Bunuel has mentions above or refer below for an alternate method.

Let's evaluate x+y=2a in light of a<0 ----> x+y < 0 and this is NOT ALLOWED as the question stem tells us that both x and y are positive integers. Thus you can eliminate this case , leaving you with x=y ONLY.

Hence you can now say "yes" to the question asked with certainty.

Hope this helps.

Or since a is negative and both x and y are positive, then |x - a| = x - a and |y - a| = y - a, so from (1) x - a = y - a, which leads to x = y.

hello
Can any one solve my problem
squaring both side in Statement (1)
i got (x-y)(x+y-2a)=0
either x=y or x+y=2a

so statement (1) should be sufficient.........
anyone Plz explain my fault................

Thnx

given x & Y are positive. is x =y?

a) |x-a| =|y-a|
distance of x & y from a point will be same only if a is the mid point of x & y or x & y are same point.
more than 1 answer possible.
Not sufficient (cancel DE)

b) a<0 no info. on x & y. Not sufficient (cancel B)

together 1 & 2. a<0 & distance of x&y from a is same. with this, including given condition as x&y>0. only 1 case is possible, when a<0, x=y. then only distance of a to x & y will be same.

sufficient (answer C)

Posted from my mobile device

Why can you just say that so easily? Is this something one needs to learn by heart?

reto, it is given that x and y are both positive and a<0.

|x-a| = distance of x from a negative number 'a' and |y-a| = distance of y from a negative number 'a'.

Thus , the only case possible for 2 positive numbers to be equidistant from another negative number is for x = y. Or in other words, x and y are on the right side of 0 while a is on the left. In this particular case, distance of both x and y from a can be equal ONLY IF x = y.

VeritasPrepKarishma
i understood the suggested solutions.
PLease help me in figuring out what i did wrong or missed.

in normal case we solve the modulus by considering critical point, accordingly we take +ve and -ve
in statement A cosidering both +ve and -ve we get x=y.

please advise.

Even with my heavy math background, formulas with multiple absolute values always have confused the hell out of me. The wording here about comparing distances seems to be key in a lot of these problems. Definitely worth remembering when faced with absolute values. =)

Thanks a lot, New lesson learned.

The problem with your solution is the understanding that comes after you see that the 2 possible cases are either x=y or x+y=2a. The question asks you whether x=y?

In DS questions, you must have a definite "yes" or a definite "no" to consider this statement sufficient.

If you would have got x=y as the ONLy possible case after solving for |x-a|=|y-a|, then yes, you could have called this statement as sufficient. But in reality, even x+y=2a is ALSO possible. What if x=0.5a and y=1.5a.

x \(\neq\) y but x+y=2a . In this case you will get a "no" for the question asked.

But if you have x=1=y, then you get "yes" for the question asked.

Thus, you get both "yes" and "no" for the same statement and as such this statement becomes NOT SUFFICIENT.

Hope this helps.

Hi,
you have done everything correct except answering it..


so in one case x=y.. answer to teh Q "x = y?" will be YES..
x+y=2a.. x can be a/2 and y=3a/2..answer to the Q "x = y?" will be NO..

so different answers... statement 1 cannot be sufficient unless it gives the same answer irrespective of changes in values..

Chetan & Engr2012

thanks for replying
But please also let me know how i deduce that now both statement are necessary from the same view of solution which i used for statement(1).
Thanks again

Bunuel wrote:

x and y are positive integers, is x = y?

(1) |x - a| = |y - a|. This statement basically tells us that the distance between x and a is the same as the distance between y and a. Now, it's certainly possible x to be equal y, for example consider x = y = a = 3 but it's also possible x not to be equal y, for example, consider x = 1, y = 3 and a = 2. Not sufficient.

(2) a < 0. Not sufficient.

a is negative and both x and y are positive, so both x and y are in red territory. Now, can you place x and y there so that the distance between them and a to be equal and x and y not to be equal?[/quote]

Or since a is negative and both x and y are positive, then |x - a| = x - a and |y - a| = y - a, so from (1) x - a = y - a, which leads to x = y.[/quote]


Dear Bunuel,
Thanks again for the great explanation - using the number line (distance) method.
Please allow me to ask two small questions:
1). In the explanation above, in red, "Or since a is negative and both x and y are positive, then |x - a| = x - a and |y - a| = y - a, so from (1) x - a = y - a, which leads to x = y"; as a is negative, shouldn't |x - a| = x+a and similarly |y - a| = y+a..
I am thinking this as lets say x is 5 and a is -3 (negative), then |x-a| = |5-(-3)| = |5+3| = 5+3.
Do let me know if I am overlooking something.

2). Using a different (longer) method, we reach a point from Statement (1), where we get 4 cases:
a) x-a = y-a ---> x=y
b) x-a = -(y-a)=-y+a---> x+y=2a
c) -x+a = y-a --->x+y=2a
d) -x+a = -y+a ---> x=y
Since we get a YES for x=y based on cases 1 and 4 while NO based on cases 2 and 3, this statement is not sufficient.
My question is, how can we fit the information from Statement (2) that a<0 in these 4 equations to get an answer (C).

Thanks a lot for your great explanations & help, always!

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