Gprep-Inequality..

[eravi11]

If a<y<z<b, Is |y-a|<|y-b| ?

1)|z-a|<|z-b|
2)|y-a|<|z-b|

SPOILER: D

[Fiver]

Given a<y<z<b
Is |y-a|<|y-b|?
Here y-a>0 as given y>a and y-b<0 as given b>y.
Hence the question is: Is y-a > b-y. Or is 2y>b+a ?

1)|z-a|<|z-b|
Here z-a>0 as given z>a and z-b<0 as given z<b
Hence this becomes: z-a>b-z. Or 2z>b+a.
Now since given that z>y and that 2z>b+a we cannot be sure whether 2y>b+a. Hence Insufficient.

2)|y-a|<|z-b|
Here y-a>0 as given a<y and z-b<0 as given z<b.
Hence this becomes: y-a>b-z Or z+y>b+a. Again Insufficient.

Together the 1st statement does not give us any more information than the 2nd one. Infact the 1st statement can be derived from the 2nd statement as when we know from st2 that z+y>b+a and given that z>y it is natural that 2z>b+a.
My answer is E. What's the OA?

[abhasjha]

IMO - D

Because we know that a < y < z < b, we know
abs(y - a) = y - a
abs(y - b) = b - y (since y - b is negative)

We can rephrase the question:
Is y - a < b - y?
or
Is 2y < a + b?

Statement (1) can be rephrased: z - a < b - z, so 2z < a + b. We also know that since y < z, then 2y < 2z. So 2y < 2z < a + b. (1) is sufficient.

Statement (2) can be rephrased: y - a < b - z, so y + z < a + b. Since y < z, we can add y to both sides: 2y < y + z. So 2y < y + z < a + b, so (2) is sufficient as well.

[clock60]

Originally Posted by Fiver

Given a<y<z<b
Is |y-a|<|y-b|?
Here y-a>0 as given y>a and y-b<0 as given b>y.
Hence the question is: Is y-a > b-y. Or is 2y>b+a ?

1)|z-a|<|z-b|
Here z-a>0 as given z>a and z-b<0 as given z<b
Hence this becomes: z-a>b-z. Or 2z>b+a.
Now since given that z>y and that 2z>b+a we cannot be sure whether 2y>b+a. Hence Insufficient.

2)|y-a|<|z-b|
Here y-a>0 as given a<y and z-b<0 as given z<b.
Hence this becomes: y-a>b-z Or z+y>b+a. Again Insufficient.

Together the 1st statement does not give us any more information than the 2nd one. Infact the 1st statement can be derived from the 2nd statement as when we know from st2 that z+y>b+a and given that z>y it is natural that 2z>b+a.
My answer is E. What's the OA?

hi fiver
it seems to me that you accidentally change the sign of expression
we need to know is 2y<a+b but not is 2y>a+b
the same typo is in st 1
2z<a+b but not 2z>a+b...

[clock60]

agree with D my reasoning
a<y<z<b
is |y-a|<|y-b|
|y-a|=y-a, if y>a, according to the task y>a
|y-b|=-(y-b),if y<b so
is (y-a)<-(y-b) or y-a<-y+b,
is 2y<a+b
(1) |z-a|=z-a if z>a
|z-b|=-(z-b), if z<b, so
z-a=-z+b,
2z<a+b,
y<z -multiply by 2, 2y<2z and add with 2z<a+b
2z+2y<a+b+2z, cancel 2z so we left with 2y<a+b-sufficient
(2)|y-a|=(y-a), if y>a
|z-b|=-(z-b) is z<b so st 2
y-a<-z+b, and y+z<a+b
according problem y<z- add with y+z<a+b
y+z+y<a+b+z, cancel z from both parts we left with 2y<a+b sufficient

[Fiver]

Originally Posted by clock60

hi fiver
it seems to me that you accidentally change the sign of expression
we need to know is 2y<a+b but not is 2y>a+b
the same typo is in st 1
2z<a+b but not 2z>a+b...

You are right champ!
Last night, I spent over half an hour in vain trying to figure this out.
Thanks for bringing this to my notice.

[nverma]

GUYS

I REQUEST you to go through the following attachment. I have tried to solve this question graphically. I think i hv solved it in much less time than algebraically.

PLEASE COMMENT.

[missweizhang]

agree with D, my approach is similar to nverma's:

number line: -inf ----------a---------y--------z------------b-------------- +inf

Is |y-a|<|y-b| ?

1)|z-a|<|z-b|
from line we know |z-a|>|y-a|, and |z-b|<|y-b|,
putting together we have |y-a|<|z-a|<|z-b|<|y-b|
SUF

2)|y-a|<|z-b|
from line we know |z-b|<|y-b|
putting together we have |y-a|<|z-b|<|y-b|
SUF

choice D.

What does IMO stand for?

[atishree]

Originally Posted by missweizhang

What does IMO stand for?

IMO stands for In My opinion

[touche]

Originally Posted by abhasjha

IMO - D

Because we know that a < y < z < b, we know
abs(y - a) = y - a
abs(y - b) = b - y (since y - b is negative)

We can rephrase the question:
Is y - a < b - y?
or
Is 2y < a + b?

Statement (1) can be rephrased: z - a < b - z, so 2z < a + b. We also know that since y < z, then 2y < 2z. So 2y < 2z < a + b. (1) is sufficient.

Statement (2) can be rephrased: y - a < b - z, so y + z < a + b. Since y < z, we can add y to both sides: 2y < y + z. So 2y < y + z < a + b, so (2) is sufficient as well.

Good one!