If a < y < z < b, is |y - a| < |y - b| ?
(1) |z - a| < |z - b|
(2) |y - a| < |z - b|
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Hi All,
Well this question would get a bit complicated & lengthy if we try solve it by picking up numbers & trying out various combinations like
all 4 nos +ve
all 4 -ve
a -ve, y -ve, z +ve
z =0
y = 0 etc
However a faster & better approach would be just visualise the 4 nos as 4 points on a number scale which encompasses +ve nos ,0 (origin), -ve nos
now since a < y < z < b hence whatever points you choose on the number
scale to represent these nos (-ve or +ve points or origin) , the absolute difference between the points will always remain same ,this can be visulaised as the distance between the points on the number scale. It will not matter if you shift this number system left of origin to make all values -ve, right of origin to make all +ve, or coincide pts with origin, the distance between the points will always remain the same THIS is nothing but the ABSOLUTE DIFFERENCE between the points.
Now coming up to the solution
stmt 1 says |z-a| < |z-b|
now visualise this as distance between pts on the number scale,
We can say that distance between pts z & a is < distance between pts z & b
then distance between pts y & a i.e. |y-a| will be lesser than z & a as y is towards the left of z on the nos scale i.e closer to "a"
i.e. |y-a| < |z-a|
also the distance between y & b |y-a| i.e. will be greater that z & b again as y is towards the left of z, hence to travel to b from y we need to travel through z i.e travel more
hence we can say |y-b| > |z-b|
hence we can say that |y-a| < |y-b|
SO SUFF
stmt 2: says that
|y - a| < |z - b|
i.e distance between y & a is less that distance between z & b
hence we can say that distance between y & b will be greater than z & b i.e. |y-b| > |z-b| ( same reason to travel to b from y we need
to reach z first and then y)
hence sufficient
so the ans should be "D" each alone is sufficient
Pls correct me if I'm wrong.
Thanks & Sorry for the lenghty explanation.
Well this question would get a bit complicated & lengthy if we try solve it by picking up numbers & trying out various combinations like
all 4 nos +ve
all 4 -ve
a -ve, y -ve, z +ve
z =0
y = 0 etc
However a faster & better approach would be just visualise the 4 nos as 4 points on a number scale which encompasses +ve nos ,0 (origin), -ve nos
now since a < y < z < b hence whatever points you choose on the number
scale to represent these nos (-ve or +ve points or origin) , the absolute difference between the points will always remain same ,this can be visulaised as the distance between the points on the number scale. It will not matter if you shift this number system left of origin to make all values -ve, right of origin to make all +ve, or coincide pts with origin, the distance between the points will always remain the same THIS is nothing but the ABSOLUTE DIFFERENCE between the points.
Now coming up to the solution
stmt 1 says |z-a| < |z-b|
now visualise this as distance between pts on the number scale,
We can say that distance between pts z & a is < distance between pts z & b
then distance between pts y & a i.e. |y-a| will be lesser than z & a as y is towards the left of z on the nos scale i.e closer to "a"
i.e. |y-a| < |z-a|
also the distance between y & b |y-a| i.e. will be greater that z & b again as y is towards the left of z, hence to travel to b from y we need to travel through z i.e travel more
hence we can say |y-b| > |z-b|
hence we can say that |y-a| < |y-b|
SO SUFF
stmt 2: says that
|y - a| < |z - b|
i.e distance between y & a is less that distance between z & b
hence we can say that distance between y & b will be greater than z & b i.e. |y-b| > |z-b| ( same reason to travel to b from y we need
to reach z first and then y)
hence sufficient
so the ans should be "D" each alone is sufficient
Pls correct me if I'm wrong.
Thanks & Sorry for the lenghty explanation.