Is |x – 3| > |y – 3|?

Is |x – 3| > |y – 3|?

To make true the statement we have to alternatives:
x>y (both x and y positives)
x<y (both x and y negatives)

(1) x > y. If both are positive the answer will be YES but if they are negatives the answer will be NO
(2) xу is not equal to 0. This statement is irrelevant.

(1)+(2) We can not determine. Not sufficient.

E

Picking smart numbers is the best approach for this problem:
(1)
N: x=6, y=-6
Y: x=6, y=5
NS

(2)
Same numbers as (1)
NS

(1)+(2)
NS

Hello Experts,
Since both sides of the eqn are positive I assumed we can approach the problem as below:
(x-3)^2 - (y-3)^2 >0 which leads to (x-3-y-3)(x-3-y+3)>0 and further (x-y-6)(x-y)>0 ie, we get 2 solutions : x>y or (x-y)>6.

Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.

Thanks & Regards,
Nab

Hi Nab,
you have done two mistakes in the highlighted portion..

1) \((x-3)^2 - (y-3)^2 >0.............. (x-3+y-3)(x-3(y-3))>0...... (x+y-6)(x-y)>0.....\) and NOT (x-y-6)(x-y)>0

2) It is NOT x>y or (x-y)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6..
so two cases-
a) BOTH (x+y-6) and (x-y) are +ive or BOTh are -ive

Hi Chetan,
Oops yes I'm sorry that was a typo, i did get (x+y-6)(x-y)>0.
From 2) Do you mean that it is not an OR condition but an AND condition? I didn't understand that quite well, can you please explain again. Especially the part BUT x>y and (x+y)>6 or x<y and (x+y)<6. I didn't understand how the equality signs changed.

Thanks for your reply.

Regards,
Nab

Hi,

the equation is-

\((x+y-6)(x-y)>0\)...

The Left Hand Side can be >0 under two cases..

1) when both (x+y-6) and (x-y) are greater than 0... since Positive * Positive = Positive..
so x-y>0 or x>y................and x+y-6>0 or x+y>0.........
Example x = 5, y=2.. x>y and x+y>6 ... so (5+2-6)(5-2)>0.....1*3>0...YES

2) when both (x+y-6) and (x-y) are lesser than 0... since Negative * Negative = Positive..
so x-y<0 or x<y................and x+y-6<0 or x+y<6.........
x= 2 and y =3.....x<y and x+y<6 ... so (3+2-6)(2-3)>0.....(-1)(-1)>0..........1>0......YES

Yes i realize where I went wrong!
Thanks a tonne for the explanation.

Thanks & Regards,
Nab

(1) x > y.
Let's say x=4 and y=3, then |x – 3| > |y – 3|
but it x=-3 and y=-4, then the aboveineqzuality will not hold true.
Insufficient
(2) xу is not equal to 0
This statement means either x or y is not equal to 0. It can be +ve or -ve. not sufficient.

Combining both statements doesn't give a unique answer. Hence E is the answer

DS: Is |x – 3| > |y – 3|?

Statement 1 : x>y
Lets say x =6 , y = 4 |x-3| = 3>|y-3| = 1
Lets say x = 1, y = -4 |x-3|= 2<|y-3| = 7

NOT SUFFICIENT

Statement 2 : xy is not equal to 0 . So, neither x nor y is not equal to 0 .
Lets say x =6 , y = 4 |x-3| = 3>|y-3| = 1
Lets say x = 1, y = -4 |x-3|= 2<|y-3| = 7
NOT SUFFICIENT

Combined : same examples
NOT SUFFICIENT

Answer E

Is |x – 3| > |y – 3|?

(1) x > y.
(2) xу is not equal to 0

2. XY not equal to 0 we don't know the value of X and Y so Insufficient

2. If x =10 and Y = -100 then |Y-3| is greater if x=10 and Y= 5 then |X-3| is greater so insuffiecient

together the same example holds good, not sufficient hence Answer should be E

For DS AbsValue questions I try to read the question and spend about 10 seconds to see if I can understand what it is asking before I even look at the statements.

Is |x – 3| > |y – 3|? When would this be true and when would it be not true?

Well, if they are both positive and x>y, then yes.

If they are both negative and x>y, then no.

Read statement (1) and (2), neither account for the option of a Negative Y. Example: x=1 y=-900, satisfies both statements, answer to question is no.

Yes/No= E

Hello

You are correct that answer is E. However, I would like to point out the highlighted part in your analysis. (I have highlighted)

|x-3| > |y-3| is NOT the same as |x|>|y|.

Eg., if x=-1 and y=5, then |x-3| > |y-3| BUT |x| < |y|

Actually |x-3| > |y-3| means that the distance of 'x' and '3' is more than the distance of 'y' and '3'.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2):
x = 4, y = 3: Yes
x = 3, y = 2: No

Since we have two answer, "yes" and "no", both conditions together are not sufficient.

Therefore, the answer is E.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

I got the solution explained above. My doubt is shouldn't be we checking the other 2 scenarios also which are
-(x-3)>(y-3) and
-(x-3) > -(y-3)

my thinking is as there are two modulus involved we should check all the 4 possible scenario
x-3 > y-3
x-3 > -(y-3)
-(x-3)>(y-3)
-(x-3) > -(y-3)

Can anyone explain to me why |x-3| > |y-3| is only true if x>y AND 6>x+y? Why is it not sufficient to just have one of the conditions met?

Is |x – 3| > |y – 3|?

(1) x > y.
x = -2, y = 3 NO
x = 5, y = 4 YES
Insufficient
(2) xу is not equal to 0.
This just tells us neither of x and y are zero.
Insufficient.

C: We don't know anything new by combining

Answer is E.