to me this is hard one, not convinced with my answer but E
if x < p < q < y, is |q - x| < |q - y| ?
after little tranforming the problem asks
does 2q<x+y
(1) says 2p<x+y but it is not enough to answer
(2) says q+p<x+y again different outcomes are possible
both to me insuff
but want verification

I tend to E
question stem If x < p < q < y, is |q - x| < |q - y| ? Since modes are positive at all times we can square both parts
st(1) (p-x)^2 < (p-y)^2 OR like [a^2-b^2=(a+b)(a-b)] ---> [(p-x)-(p-y)]*[(p-x)+(p-y)] <0 , (y-x)(2p-x-y) <0
Either (y-x)<0 AND (2p-x-y)>0 OR Vice Versa. We know that y<x is Not True, as x<y. Hence (y-x)>0 AND (2p-x-y)<0

Our question stem could be simplified as st(1)
|q - x| < |q - y| ? OR (q - x)^2 < (q - y)^2 --> [(q-x)-(q-y)]*[(q-x)+(q-y)], (y-x)(2q-x-y)<0 ?
from statement (1) we know (y-x)>0, BUT we don't know about (2q-x-y) which can be +ve, -ve OR even 0, Not Sufficient;

st(2) the same method as one applied for st(1) --> [(q-x)-(p-y)]*[(q-x)+(p-y)] <0, (q-x-p+y)(q-x+p-y)<0
Two ways are (q-x-p+y)<0 AND (q-x+p-y)>o OR Vice Versa. We don't know the conditions, both sides can be +ve OR -ve, Not Sufficient;

Combined st(1&2): y-x>0 AND (y-x+q-p)(q-x+p-y)<0 We can only deduce that q-p>0 always, BUT this is true when q and p are negative OR positive too... Not Sufficient

IOM E

maihuna wrote:If x < p < q < y, is |q - x| < |q - y| ?
(1) |p - x| < |p - y|
(2) |q - x| < |p - y|

IMO - E

maihuna wrote:If x < p < q < y, is |q - x| < |q - y| ?
(1) |p - x| < |p - y|
(2) |q - x| < |p - y|

If you draw them on number lines, there are two possible representation based on the facts given :

---x---p---(x+y)/2---q----y---

OR

---x---p---q-----(x+y)/2-------------y----

Q: |q-x| < |q-y| => is distance of x from q is less than distance of y from q ?
or, can we for sure say which side of (x+y)/2 does q lie ?

Statement 1:
|p-x| < |p-y|
Means : distance of x from p is less than the distance of p from y - Both the scenarios drawn above fulfill it , so - Insufficient

Statement 2:
|q - x| < |p - y|
Means : distance of q from x is less than the distance of p from y - Both satisfy - So insufficient

Combining 1 and 2 :
We have already seen that both of the drawing satisfy the two statements, and we are still left with two possibilities - So still insufficient E