If a<y<z<b, is |y-a|<|y-a|<|y-b|
(1) |z-a|<|z-b|
(2) |y-a|<|z-b|
My answer is (E). Can someone help to explain? thanks.
inequation
This topic has expert replies
I'd go with A:
a<y<z<b
Q: is |y-a| < |y-b|?
rephrasing, since y-a >0 and y-b<0:
is y-a < -(y-b)?, i.e., is y < (a+b)/2
Statement 1:
|z-a| < |z-b|, i.e, z-a < -(z-b)), and then z < (a+b)/2
Since z>y, it follows that y < (a+b)/2 SUFFICIENT
Statement 2:
|y-a|<|z-b| , i.e., y-a <-(z-b) ==> y< a+b-z
Not sufficient
What's the OA??
a<y<z<b
Q: is |y-a| < |y-b|?
rephrasing, since y-a >0 and y-b<0:
is y-a < -(y-b)?, i.e., is y < (a+b)/2
Statement 1:
|z-a| < |z-b|, i.e, z-a < -(z-b)), and then z < (a+b)/2
Since z>y, it follows that y < (a+b)/2 SUFFICIENT
Statement 2:
|y-a|<|z-b| , i.e., y-a <-(z-b) ==> y< a+b-z
Not sufficient
What's the OA??
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(A)
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GMATPowerPrep Test1= 740
GMATPowerPrep Test2= 760
Kaplan Diagnostic Test= 700
Kaplan Test1=600
Kalplan Test2=670
Kalplan Test3=570
I go with A.
|y-a|<|y-b| can be re-written as (y-a)^2<(y-b)^2
expanding and simplifying gives you y<(a+b)/2
stmt I actually provides this information:
|z-a|<|z-b| re-written as (z-a)^2<(z-b)^2
expanding and simplifying gives you z<(a+b)/2
we already know that z is bigger than y, so if (a+b)/2 is bigger than z then it is also bigger than y.
what is OA?
|y-a|<|y-b| can be re-written as (y-a)^2<(y-b)^2
expanding and simplifying gives you y<(a+b)/2
stmt I actually provides this information:
|z-a|<|z-b| re-written as (z-a)^2<(z-b)^2
expanding and simplifying gives you z<(a+b)/2
we already know that z is bigger than y, so if (a+b)/2 is bigger than z then it is also bigger than y.
what is OA?
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given in the question
a<y<z<b
hence the distance b/w y and b (|y-b|) will always be greater than b/w z and b(|z-b|).
so we can arrange them in the order....
|y-b| > |z-b|.....and .....|z-a| > |y-a|...(equ..a,b)
now
1) |z-a|<|z-b|......hence |z-b| > |y-a|....so...|y-b| > |y-a|......suff.
2) |y-a|<|z-b|......hence from equations a and b.....|y-a|<|y-b|....suff.
D...plz post OA.
a<y<z<b
hence the distance b/w y and b (|y-b|) will always be greater than b/w z and b(|z-b|).
so we can arrange them in the order....
|y-b| > |z-b|.....and .....|z-a| > |y-a|...(equ..a,b)
now
1) |z-a|<|z-b|......hence |z-b| > |y-a|....so...|y-b| > |y-a|......suff.
2) |y-a|<|z-b|......hence from equations a and b.....|y-a|<|y-b|....suff.
D...plz post OA.