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180. Is |x – 3| > |y – 3|? (1) x > y. (2) xу is not equal to 0.

Is |x – 3| > |y – 3|?

You need no algebra for this one. Question basically asks whether point x on the number line is further from 3 than point y (as |x-3| is the distance between points x and 3 on the number line and |y-3| is the distance between y and 3).

(1) x > y. Totally irrelevant.

(2) xу is not equal to 0. Also irrelevant, it just means that neither x nor y equals to zero.

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.

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.

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
_________________

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.

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
_________________

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.

(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
_________________

I welcome critical analysis of my post!! That will help me reach 700+

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