Is |p - q| > |p - r|?
(1) |q| > |r|
(2) p < 0

|p - q| > |p - r| -----(1)
statement 1) |q| > |r|

there can be two cases here

1:
(i)P > 0 and when q and r both are greater than 0

as p is common in both we test by simple scenarios

p = 100 , q = 50 and r = 20 (q > r)
p-q = 50 and p-r = 80
(1) is not satisfied

p = 100 q = 150 and r = 130

p-q = -50
|p-q| = 50
similarly
|q-r| = 30
(1) is satisfied
we are getting contradictory answers, insuff

statement 2) p < 0
case 1 : p <0 , q>0 and r>0
lets say p = -100, q = 50 and r = 30

|p-q| = 150
|q-r| = 130

(1) is satisfied

p = -100 , q = -50 , r = -20
|p-q| = 50
|q-r| = 80
(1) is not satisfied

insuff (rule out A,B,D)

lets combine st1 and st2

lets say p = -100, q = 50 and r = 30 (q>r)

|p-q| = 150
|q-r| = 130
(1) is satisfied
p = -100 , q = -20 and r = -50 (-q > -r)

|p-q| = 80
|p-r| = 50
(1) is satisfied
ans (C)

Let me know if I have done something wrong

From 1


Above table is prepared for the first condition..clearly unless we know the sign of P nothing can be concluded....not sufficient

From 2

if p<0 and |q|>|r|...from the image |p - q| > |p - r| if |q|<|r| we will have |p - q| < |p - r|...not sufficient

combining
|p - q| > |p - r| is true...sufficient
Ans option C

1:
(i)P > 0 and when q and r both are greater than 0

as p is common in both we test by simple scenarios

p = 100 , q = 50 and r = 20 (q > r)
p-q = 50 and p-r = 80
(1) is not satisfied

p = 100 q = 150 and r = 130

p-q = -50
|p-q| = 50
similarly
|q-r| = 30
(1) is satisfied
we are getting contradictory answers, insuff

statement 2) p < 0
case 1 : p <0 , q>0 and r>0
lets say p = -100, q = 50 and r = 30

|p-q| = 150
|q-r| = 130

(1) is satisfied

p = -100 , q = -50 , r = -20
|p-q| = 50
|q-r| = 80
(1) is not satisfied

insuff (rule out A,B,D)

lets combine st1 and st2

lets say p = -100, q = 50 and r = 30 (q>r)

|p-q| = 150
|q-r| = 130
(1) is satisfied
p = -100 , q = -20 and r = -50 (-q > -r)

|p-q| = 80
|p-r| = 50


wht if |-6| > |-5| in q < r but |q| > |r|
-100 + 6 = -94 and |-94| = 94 will be less then |-100 -(-5)| = 95

hence (E)

to |p - q| > |p - r|
squaring (p-q)² > (p-r)² or p²+q²-2pq > p²+r²-2rp
q²-2pq>r²-2rp
1)│q│>│r│ so q²>r²
2) p is -ve
so is q²+2pq>r²+2rp
still the sign of q and r is not known
so IMO E.