Data Sufficiency

Your right... I must be misunderstanding some fundamental rule of math. I thought you could go from (x/y) > 1 --> x >y (by multiplying both sides by y).

Therefore, -2 > -2.5, but your way is correct....

Why am I not allowed to convert (x/y) > 1 --> x > y?

shaden wrote:Your right... I must be misunderstanding some fundamental rule of math. I thought you could go from (x/y) > 1 --> x >y (by multiplying both sides by y).

Therefore, -2 > -2.5, but your way is correct....

Why am I not allowed to convert (x/y) > 1 --> x > y?

Because when u multiply on a negative number, u should flip the sign ( <,>)

Fab wrote:Hey guys, just one question:

What happen if X=-0.5 and Y=-1

Condition1: 2X - 2Y =1 -----> -1 - 2(-1) = 1 OK
Condition 2: x/y>1 -------> -0.5 > -1 OK

Both could be negative either, so why not E ??

THANKS.

You cannot take X=-0.5 and Y=-1 because (-0,5)/(-1)=0,5 which is not more that 1 according to (2)

If I can't solve an inequality with math, I try to plug in numbers such as -2,-½,0,½,2 in a table.
(1)2x - 2y = 1
(2)x/y > 1

Statement 1: 2x - 2y = 1
On the left side of the table we put the the x values and across the top we put useful test cases.
We can manipulate the equation so we can solve for y.
(2x - 1)/2 = y

x_______y_________ Are both x and y positive?
-2______-5/2_________no
-½_____-1___________no
0_______½ __________no
½_______0___________no
2_______3/2__________yes

From the table above we can see that there are times when x and y are not positive and there is a time when it is. Therefore it is not decisive.
We conclude Statement (1) is not sufficient.

Statement 2: x/y > 1

x_______y_________ Are both x and y positive?
-2______-1___________no
-½_____-1/4_________ no
0_______X ___________X
½______1/4__________no
2_______ 1__________yes

From the table above we can see that there are times when x and y are not positive and there is a time when it is. Therefore it is not decisive.
We conclude Statement (2) is not sufficient.

To combine the statements we substitute in the y value in statement 1 in statement 2 like this:
From Statement1:
y= (2x - 1)/2
Plug in y value into Statement2:
x/[(2x-1)/2] > 1=
x/[x-½] > 1
We construct another table using the values in table 1 to see if they meet the condition in statement 2.

x_______y_______x/[x-½] > 1?___ Are both x and y positive?
-2______-5/2_________4/5-no____________no
-½_____-1___________1/2-no___________no
0_______½ ___________X___________no
½_______0____________X___________no
2_______3/2_________4/3-yes__________yes

The only set of x and y's in our table that meet criteria 1 and 2 are x = 2 and y = 3/2. Both of them are positive.
We conclude that both statements together are sufficient.

II wrote:Please illustrate your logic behind the answer.

Thanks.

1) x = y + 1/2 OKAY WE KNOW WHO IS THE DADDY! X>Y

2) X/Y > 1 HERE IS THE TRAP!

(Y+1/2) / Y > 1

OR 1 + Y/2 > 1 hmmm...so, Y/2>0 ---> Y>0

Remember X > Y

So these guys are both POSITIVE


Thing to remember:

X/Y > 1

X/Y - 1 > 0 Learn this!

(X-Y)/Y > 0

Y < 0 and X-Y < 0

OR

Y > 0 and X-Y> 0

Even I intially chose E as the choice. But we need to be careful. If we are substituting values, Please see that we are not changing the choice. For example,

x/y > 1

Please dont change the choice to x > y.

keep it as it is.

Everyone by now must be clear why A is not sufficient.

x - y = 1/2

(i) 3/2 - 1 = 1/2 both +ve (3/2, 1)
(ii) -3/2 - (-2) = 1/2 both -ve (-3/2, -2)

So unsufficient.

x/y > 1

(3/2)/1 > 1 for sure.
(-3/2)/(-2) < 1 for sure. --> so when we take two negative values to satisfy the first condition x-y=1/2. They can never satisfy the second condition x/y > 1.
The trap is changing the given condition to x > y and substituting the negative values in it.