If x < p < q < y, is |q-x| < |q-y| ?
(1) |p-x| < |p-y|
(2) |q-x| < |p-y|
Absolute Value
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hi i`ll try
if x<p<q<y,does |q-x|<|q-y|?
with given range, |q-x|=(q-x), and |q-y|=-(q-y)=y-q. so we need to estimate, does
q-x<y-q, or 2q<x+y, q<(x+y)/2
(1)|p-x|<|p-y|
|p-x|=(p-x), and |p-y|=-(p-y)=y-p and as a result
p-x<y-p, 2p<x+y.or p<(x+y)/2,
we are give that p<q, so to me two possible cases
p<q<(x+y)/2,the answer is yes, or
p<(x+y)/2<q the answer is no
so to me 1 st insuff
(2)|q-x|<|p-y|
|q-x|=(q-x), and |p-y|=-(p-y)=y-p
q-x<y-p, and q+p<x+y we are also given that p<q let us add undelined parts
q+p+p<q+x+y, cancel q and left with 2p<x+y the same as in st1
even both are insuff
so my answer is E
if x<p<q<y,does |q-x|<|q-y|?
with given range, |q-x|=(q-x), and |q-y|=-(q-y)=y-q. so we need to estimate, does
q-x<y-q, or 2q<x+y, q<(x+y)/2
(1)|p-x|<|p-y|
|p-x|=(p-x), and |p-y|=-(p-y)=y-p and as a result
p-x<y-p, 2p<x+y.or p<(x+y)/2,
we are give that p<q, so to me two possible cases
p<q<(x+y)/2,the answer is yes, or
p<(x+y)/2<q the answer is no
so to me 1 st insuff
(2)|q-x|<|p-y|
|q-x|=(q-x), and |p-y|=-(p-y)=y-p
q-x<y-p, and q+p<x+y we are also given that p<q let us add undelined parts
q+p+p<q+x+y, cancel q and left with 2p<x+y the same as in st1
even both are insuff
so my answer is E