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

Source:Grockit

IMO D. OA?

selango 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

Source:Grockit

--------------------Avg---------------------
-8 X Y

On the number line if X is closer to -8 then X is definitely less than the average of y and -8
SUFF

(2) x = 4y

is 4y > (y-8)/2
is 8y > y-8
is 7y > -8
is y> -8/7

Hence when y> -8/7, the answer is YES and when y < -8/7 answer is NO
INSUFF

pick A

IMO A[/spoiler]

kvcpk wrote:

selango 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

Source:Grockit

--------------------Avg---------------------
-8 X Y

On the number line if X is closer to -8 then X is definitely less than the average of y and -8
SUFF

(2) x = 4y

is 4y > (y-8)/2
is 8y > y-8
is 7y > -8
is y> -8/7

Hence when y> -8/7, the answer is YES and when y < -8/7 answer is NO
INSUFF

pick A

y is a negative number greater than -8 implies that y=-1 ... - 7...

case1: avg of -1 and -8 = -4.5... x is closer to - 8 means that x should be <-4.6... hence x is less...

case 2: avg of -7 and - 8 = -7.5... x is closer to -8 means that x should be <-4.6... in this case x can be > - 7.5 and < - 4.6...

A cannot give the answer... wat is the OA... please lemme know if my understanding is incorrect...

OA?? IMO it's A.

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 there!

We know -8 < y < 0, and we must focus on the question: x > (y+(-8))/2 ?

(1) SUFFICIENT:

Draw a real line with "points" -8 (left), y (right) and their middle-point (say) M.

The question is, therefore, x > M ?

From sttm (1) we know that x is less than M, because it is (-8 or less) or (between -8 and M, M excluded).


(2) INSUFFICIENT:

Please note that the question is equivalent to y < 2x+8 or y < 8(y+1) or (y+1)/y < 1/8 ? (Remember that y<0)

> Take y negative but (below and) nearer 0, say y = -1. We get (y+1)/y = 0/-1 = 0 answering positively.
> Take y negative but (above and) nearer -8, say y = -7. We get (y+1)/y = -6/-7 = 6/7 answering negatively.

Regards,
Fabio.

1. Statement #1 - "On the number line, x is closer to -8 than it is to y".
y can be between -1 and -7 and the average would always be middle point of the two (e.g. if y=-2 then average = -5, y=-6, average=-7 etc.)
given that x is closer to -8 than y then this means that it is more negative than the average of -8 and y. hence it is smaller than the average (which is the middle point).
Hence sufficient data to say that answer is NO

2. Statement #2 - x = 4y
this means that whatever y is, x is 4 times as negative (e.g. if y=-6, x = -24)
this means that it is smaller than y foresure.
now average of -8 and y = (-8+y)/2 ==> -4+y/2 ==> between -4 and -8 somewhere depending on y
if y == -1 then x = -4 and average = -4.5 but if y = -6 then x = -24 and average = -7
so insufficient

I am not able to get the solution here. Can anybody help me, where I am wrong??

How A is sufficient?
Our problem can be rephrased as Is x>(y-8)/2?

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

So I took x = -7(which is closer to -8) and y = - 2
Now, this will give me -
-7 > (-2-8)/5 -> -7 > -2 -> False.

then put x= -5(which is closer to -8) and y = -3

-5 > (-3-8)/2 -> -5 > -5.5 -> True

How can A be sufficient. Can anybody help me?

HRIMJAM,

TAKE A LOOK AT STATEMENT 1 AGAIN. I HAD THE SAME ANSWER AS YOU BUT AFTER A FEW MINUTES I REALIZED MY REASONING WAS WRONG.

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

THE NUMBERS YOU SELECTED DO NOT SATISFY THE STATEMENT.

"then put x= -5(which is closer to -8) and y = -3

-5 > (-3-8)/2 -> -5 > -5.5 -> True "

HERE X (-5) IS 3 AWAY FROM -8 AND X(-5) IS ONLY 2 AWAY FROM Y(-3). THE STATEMENT IS SAYING THAT THE DIFFERENCE BETWEEN X AND -8 IS LESS THAN THE DIFFERENCE X AND Y ...HERE (-5)X - 8 = 3 AND (-5) - 3 = 2 ... CLEARLYTHESE NUMBERS DO NOT SATISFY THE EQUATION AS X IS CLOSER TO Y THAN IT IS TO -8 ... WE NEED TO USE DIFFERENT #'S TO SATISFY THE EQUATION.

HOPE THIS HELPS[/quote]

+1 for A .

lets see if i can draw this one .

------------(-8)---x--(am)--y-------(0)--------

hope this answers the question

For B. ( insufficient , just plug positive and negative values for y . you ll know

DELETE

Guys...

its actually very simple...when you draw the exact number line....

option A will be easily solved...whether yes or no...
Option B : just try taking any two numbers..eg -4 and 2 as Y , in both cases ...X will be either less or greater than the mean....

Hope it helps someone..!

hrimahajan wrote:I am not able to get the solution here. Can anybody help me, where I am wrong??

How A is sufficient?
Our problem can be rephrased as Is x>(y-8)/2?

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

So I took x = -7(which is closer to -8) and y = - 2
Now, this will give me -
-7 > (-2-8)/5 -> -7 > -2 -> False.

then put x= -5(which is closer to -8) and y = -3

-5 > (-3-8)/2 -> -5 > -5.5 -> True

How can A be sufficient. Can anybody help me?

Hi hrimahajan,

There are two mistakes which I find.

1. The data which you have taken y = -3 and x = -5 is wrong. The mean of -8 and -3 is -5.5, so the acceptable range of x is -8 < x < -5.5 . So x = -5 value does not fit in here.

Another way of understanding the problem is that the mean is equidistant from both the numbers, say A & B. So if a third number(C) is closer to one of these numbers(A,B), so it can either be greater or lesser than the mean. It cannot be both at the same time.

2. Second mistake

hrimahajan wrote:
-7 > (-2-8)/5 -> -7 > -2 -> False.

Here you have divided (-2-8) by five, it should be by 2 to take the mean.


Hope this helps,

Best Regards,
JOHN

Digging a question from past.
"x is closer to -8 than y" - Doesnt this mean that x can lie on either side of -8 while still being closer to 08 than y. ?