First, break the absolute for all the inequations using the given order: x < p < q < y

lq - xl < lq - yl => q- x< y- q => 2*q < y +x

1. lp - xl < lp - yl => p-x < y- p => 2*p < y + x
As p<q then 2p<2q, then y+x> 2*p does not ensure y+x always larger than 2*q
=> insuff.

2. lq - xl < lp - yl => q-x < y- p => q+p < y +x
p<q then p + q < q+q or p+q < 2q.
As p+q < 2q, again, insuff.

1&2. y+x>p+q>2p
however, 2q is larger than both p+q and 2p, so the above inequation can not secure that y+x > 2q.
Insuff.

Pick E.

hi limestone, i admire your solution, honestly said my own initial solving was too time-consuming, when i squared both parts, and made others borrowing operations,
your way is smart, and thank you for showing. only one question concerning st 2 and further....
(2) |q-x|<|p-y|,
q-x<-(p-y), q-x<y-p, and finally
q+p<x+y, as in your solution. and we are given that p<q, now we can simply add two inequalities, and recieve
q+p+p<x+y+q, from here cancel q, and left with 2p<x+y,the same as in st 1, so both st insufficient as they are equal
what do you say?
i did not get at all you solution concerning both st

q+p<x+y, as in your solution. and we are given that p<q, now we can simply add two inequalities, and recieve
q+p+p<x+y+q, from here cancel q, and left with 2p<x+y,the same as in st 1, so both st insufficient as they are equal

In my opinion, adding two different numbers to both side of an inequation is risky.

As you said: q+p<x+y AND p<q then obviously, q+p+p<x+y+q or 2p<x+y. This is correct.

However, you can tranform q+p<x+y into 2p<x+y by using the relationship p<q. However, you cannot do the reverse : tranform 2p<x+y back into q+p<x+y (p<q then you CANNOT say x+y>2p then x+y>p+q)

To get it cleared, take this example:
Is x+y> 6? Given information: q+p = 6, q>p

(1) x+y>2p
(2) x+y> q+p

(2) obviously gives out a YES answer as q+p = 6
(1), if solved as you mentioned above, then (1) is the same to (2). Then answer to (1) must be YES too.

However, let p be 2, then x+y>4, this cannot answer the question "Is x+y>6?" => cannot define

Thus (2) is different from (1) in that: there's a gap between 2p and q+p, and x+y can be a value in that gap. Such x+y will ensure (1) but will not ensure (2).

In my example: x+y can be 5, and this x+y satisfy (1) but do not satisfy (2), then (1) is not (2).

hi limestone
this is nice remark to solve other problems, but in this one, it seems as you overcomplicate topic.Any way thank you for sharing.

If x < p < q < y, is lq - xl < lq - yl ?

(1) lp - xl < lp - yl
(2) lq - xl < lp - yl

lq - xl < lq - yl => q- x< y- q => 2*q < y +x

2. lq - xl < lp - yl => q-x < y- p => q+p < y +x

If y+x is greater than q+p then then it y+x is always greater than 2*q (as q is greater than p) unless they are fractions.

Please clarify.

Sorry, guys.
i assumed the other way.