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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|
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