Hello
I have been able to work this problem out but do not know the fastest way to do this can anyone help pls
thanks[/img]
M<n easier way to solve the porblem
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- Maciek
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Hi venmic!
Are you sure that C is correct?
problem:
if x > y, is zx > yx ?
(1) z > 0
(2) y < 0
Let as look at zx > yx
We can deduct yx on both sides of an inequality zx - yx > 0,
we can further reshape it x(z - y) > 0
(1) z > 0
at this point we can plugin numbers
z=5 x=8, y=7
z(x - y) = 8(5-7) = -16
thus x(z - y) < 0
let us try other numbers
z=5, x=4, y=2
z(x - y) = 4(5 - 2) = 12
thus x(z - y) > 0
It is insufficient
(2) y < 0
now let us try to plugin numbers
z=5 x=-2, y=-3
z(x - y) = -2(5+3) = -16
thus x(z - y) < 0
let us try other numbers
z=3, x=1, y=-2
z(x - y) = 1(3+2) = 5
thus x(z - y) > 0
it is insufficient
According to me correct answer is E
hope it helps!
Best
Are you sure that C is correct?
problem:
if x > y, is zx > yx ?
(1) z > 0
(2) y < 0
Let as look at zx > yx
We can deduct yx on both sides of an inequality zx - yx > 0,
we can further reshape it x(z - y) > 0
(1) z > 0
at this point we can plugin numbers
z=5 x=8, y=7
z(x - y) = 8(5-7) = -16
thus x(z - y) < 0
let us try other numbers
z=5, x=4, y=2
z(x - y) = 4(5 - 2) = 12
thus x(z - y) > 0
It is insufficient
(2) y < 0
now let us try to plugin numbers
z=5 x=-2, y=-3
z(x - y) = -2(5+3) = -16
thus x(z - y) < 0
let us try other numbers
z=3, x=1, y=-2
z(x - y) = 1(3+2) = 5
thus x(z - y) > 0
it is insufficient
According to me correct answer is E
hope it helps!
Best
"There is no greater wealth in a nation than that of being made up of learned citizens." Pope John Paul II
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- Maciek
- Master | Next Rank: 500 Posts
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- Joined: Sun Jul 18, 2010 5:26 am
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we have to include the possibility that x can be positive or negative
if you multiply (or divide) both sides by a negative number "<" becomes ">"
I hope it is better explanation
if you multiply (or divide) both sides by a negative number "<" becomes ">"
I hope it is better explanation
"There is no greater wealth in a nation than that of being made up of learned citizens." Pope John Paul II
if you have any questions, send me a private message!
should you find this post useful, please click on "thanks" button
if you have any questions, send me a private message!
should you find this post useful, please click on "thanks" button
ah, you're right.Maciek wrote:we have to include the possibility that x can be positive or negative
if you multiply (or divide) both sides by a negative number "<" becomes ">"
I hope it is better explanation
if x > y, is zx > yx ?
(1) z > 0
(2) y < 0
(1)
First thing I did with (1) is suppose z=1 which tells me nothing about the equation. If z = 1 then zx>yx = x>yx
Three possibilities, x +ve, y +ve .. x +ve, y -ve .. x -ve, y -ve.
So is +ve greater than +ve x +ve, yea it can be. ( 1/2 > 1/2 x 1/4) but ( 4 < 4 x 3) so insuf
(2)
yx could be positive or negative and zx could be positive or negative independent of each other so insuf
(together)
Let's put z = 1 so it drops out, then I could have x is positive or negative while y is negative.
+ve > +ve x -ve
-ve < -ve x -ve
insuff
(1) z > 0
(2) y < 0
(1)
First thing I did with (1) is suppose z=1 which tells me nothing about the equation. If z = 1 then zx>yx = x>yx
Three possibilities, x +ve, y +ve .. x +ve, y -ve .. x -ve, y -ve.
So is +ve greater than +ve x +ve, yea it can be. ( 1/2 > 1/2 x 1/4) but ( 4 < 4 x 3) so insuf
(2)
yx could be positive or negative and zx could be positive or negative independent of each other so insuf
(together)
Let's put z = 1 so it drops out, then I could have x is positive or negative while y is negative.
+ve > +ve x -ve
-ve < -ve x -ve
insuff