Data Sufficiency

If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8?

1. On the number line, x is closer to -8 than it is to y
2. x = 4y

IMO E

Given that, 0 > y > -8
Asked: Is x > (y - 8) /2 => 2x > y -8 => 2x - y > -8 =>IS y - 2x < 8??

stmnt 2: x = 4y
case I: if y = -7 => x = -28, therefore y - 2x = -7 + 56 = 48 ans NO
case II: if y = -1/4 => x = -1 therefore y - 2x = -1/4 + 2 = 7 / 4 ans YES
conclusion insuff

cross off B D

stmnt 1: On the number line, x is closer to -8 than it is to y
case I: if x = -7 and y = -1 then y - 2x = -1 + 14 = 13 ans NO
case II: if x = -6 and y = -5 then y - 2x = -5 +12 = 7 ans YES
conclusion insuff

cross off A

Together, same thing
if y = -1/4 => x = -1; so y - 2x = -1/4 + 2 = 7/4 ans YES
if y = -3/2 => x = -6; so y - 2x = -3/2 + 12 = 10.5 and NO
conclusion insuff

cross off C

Answer E


hope this helps...

please post OA and OE as well.

Hi,

stmnt 1: On the number line, x is closer to -8 than it is to y
case I: if x = -7 and y = -1 then y - 2x = -1 + 14 = 13 ans NO
case II: if x = -6 and y = -5 then y - 2x = -5 +12 = 7 ans YES
conclusion insuff

This doesn't work because in this case, x is closer to y(distance=1) than it is to -8(distance=2).

Yes indeed. Thanks for pointing it out. I was thinking about the distance between -8 and x and -8 and y.
Let me re-do it.

abhishekswamy wrote:If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8?

1. On the number line, x is closer to -8 than it is to y
2. x = 4y

Hi,
It is easy to visualize if you can draw a number line.
From(1):

case-1: x<=-8, x is always less than y and x is at most -8.
So, it will be less than the average as well.
case-2: -8<x<y
x is closer to -8 than y means it is to the left of mid point of y and -8.
So, x is less than average of y and -8
case-3: x>y
No x satisfies this condition.
So, x is always less than the average of y and -8
Sufficient

From(2):
y=-1, x= -4. x>(y-8)/2
y=-2, x=-8. x<(y-8)/2
Not sufficient

Hence, A

OA: A

OK. I think the answer must be A. here's why -
given constraint, 0> y > -8

stmnt 1: On the number line, x is closer to -8 than it is to y
this means x < -4
cause if x = -4 then y would be +ve which is not allowed.
so is y - 2x < 8? the answer will always be NO

suff.

I would love to see the algebraic solution for this problem. can someone please post it?

Thanks,

sumgb wrote: I would love to see the algebraic solution for this problem. can someone please post it?

Hi,
-8<y<0
As you have mentioned : IS y - 2x < 8??
From(1):
|x+8| < |x-y|
1)For x<-8,
|x+8| < |x-y| => -x-8 < y-x
So, y> -8(always true)
2)-8<=x<y
|x+8| < |x-y| => x+8 < y-x
So, y-2x > 8
3) x>y
|x+8| < |x-y| => x+8 < x-y
=>y < -8(never happens. so x>y is not feasible)

It is almost same.

Thanks Frankenstein. I understand the algebra you have posted. but I thought the algebraic solution must have 4 cases. I understand that you have taken cases based on values of x.

here's what I did on my scratch paper. Can you please point out what is wrong here and why? Thanks in advance-
|x+8| < |x-y|

case 1:
x + 8 < x - y
8 < -y
y > -8 (always true as you said)

case 2:
x + 8 < -(x-y)
x + 8 < -x + y
2x - y < -8
y - 2x > 8

case 3:
-(x+8) < x - y
-x -8 < x - y
-8 < 2x - y
8 > y - 2x

case 4:
-(x+8) < -(x-y)
-x - 8 < -x +y
-8 < y
y > -8 (same as case 1)

I am confused between case 2 and 3 here. Is this approach wrong?

I understand the solution using number line. but I expected same result using this approach as well. but I got exactly opposite results in case 2 and 3.

Please let me know what's wrong here?

Much appreciated..

case 3:
-(x+8) < x - y
-x -8 < x - y
-8 < 2x - y
8 > y - 2x

Hi,
Well, considering all 4 cases is always going to be time-consuming. When you know the ranges in which the value changes, it is better to consider only those as that would generally reduce your cases.
Coming to case-3:
LHS : |x+8| = -(x+8)
So, you are assuming x+8<0. So, x<-8
RHS: |x-y| = x-y
So, you are assuming x>y
We know that y>-8. So, x>y>-8
So, x<-8 and x>-8 contradict each other making this case invalid.

Understood. one last que

x<-8 and x>y contradict each another making this case invalid.

do you mean x <-8 and x >-8 (rather than x >y) contradict here?

Thanks a bunch.

Great. Thanks a lot. Algebra works!!!

sumgb wrote:Understood. one last que

x<-8 and x>y contradict each another making this case invalid.

do you mean x <-8 and x >-8 (rather than x >y) contradict here?

Thanks a bunch.

x<-8 and x>y contradict each other because by saying x>y, we are implicitly saying x>-8(since x> y and y>-8 implies x>-8)

yeah, A should be the answer.. Easy to do this question if you draw the number line and do the question with actual numbers.

.