is |y-a|<|y-a|<|y-b|

I think there's an error in the question. Is there an OA?

typo..

The question should be:

If a<y<z<b, is |y-a|<|y-b|
(1) |z-a|<|z-b|
(2) |y-a|<|z-b|

I'd go with A:

a<y<z<b

Q: is |y-a| < |y-b|?

rephrasing, since y-a >0 and y-b<0:

is y-a < -(y-b)?, i.e., is y < (a+b)/2

Statement 1:

|z-a| < |z-b|, i.e, z-a < -(z-b)), and then z < (a+b)/2

Since z>y, it follows that y < (a+b)/2 SUFFICIENT


Statement 2:

|y-a|<|z-b| , i.e., y-a <-(z-b) ==> y< a+b-z

Not sufficient

What's the OA??

(A)

my answer will be D.
plz post OA.

I have solved this one thru a number line.I want to know if the answer is D or not? I will post the solution if it is correct.

go for D as well

I go with A.

|y-a|<|y-b| can be re-written as (y-a)^2<(y-b)^2

expanding and simplifying gives you y<(a+b)/2

stmt I actually provides this information:

|z-a|<|z-b| re-written as (z-a)^2<(z-b)^2

expanding and simplifying gives you z<(a+b)/2

we already know that z is bigger than y, so if (a+b)/2 is bigger than z then it is also bigger than y.

what is OA?

given in the question

a<y<z<b
hence the distance b/w y and b (|y-b|) will always be greater than b/w z and b(|z-b|).
so we can arrange them in the order....
|y-b| > |z-b|.....and .....|z-a| > |y-a|...(equ..a,b)


now

1) |z-a|<|z-b|......hence |z-b| > |y-a|....so...|y-b| > |y-a|......suff.

2) |y-a|<|z-b|......hence from equations a and b.....|y-a|<|y-b|....suff.

D...plz post OA.