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If x - y > 10, is x - y > x + y ? (1) x = 8 (2) y = -20
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vrajesh
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If x - y > 10, is x - y > x + y ? (1) x = 8 (2) y = -20 [#permalink] New post  21 Nov 2010, 15:44
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If x - y > 10, is x - y > x + y ?

(1) x = 8
(2) y = -20
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Bunuel
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Re: If x - y > 10, is x - y > x + y ? (1) x = 8 (2) y = -20 [#permalink] New post  21 Nov 2010, 15:54
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vrajesh wrote:
if x - y > 10, is x - y > x + y
(1) x = 8
(2) y = -20


Can you please explain each statement. Not just the answer.


Given: \(x-y>10\). Question: is \(x-y>x+y\)? --> is \(2y<0\)? --> is \(y<0\)?

(1) \(x=8\) --> \(8-y>10\) --> \(y<-2\), hence \(y<0\). Sufficient.
(2) \(y=-20\), hence \(y<0\). Sufficient.

Answer: D.
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Re: If x - y > 10, is x - y > x + y ? (1) x = 8 (2) y = -20 [#permalink] New post  18 Jul 2011, 03:40
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siddhans wrote:
if x-y > 10 , is x -y > x+ y?

1)x=8

2) y =- 20


To prove: x-y>x+y
or x-y-x-y>0
or -2y>0
or y<0 ?

Stmt1: x=8
x-y > 10
8-y>10
-2-y>0
2+y<0
y<-2
Hence y < 0. Sufficient.

Stmt2: y=-20
Hence y <0. Sufficient.

OA D.
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Re: If x - y > 10, is x - y > x + y ? (1) x = 8 (2) y = -20 [#permalink] New post  07 Oct 2019, 02:11
Bunuel wrote:
If x - y > 10, is x - y > x + y?

Given: \(x-y>10\). Question: is \(x-y>x+y\) --> is \(2y<0\) --> is \(y<0\).

(1) x = 8 --> \(8-y>10\) --> \(y<-2\), hence \(y<0\). Sufficient.
(2) y = -20, hence \(y<0\). Sufficient.

Answer: D.


So we can do the arithmetic, but we can't divide?
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Re: If x - y > 10, is x - y > x + y ? (1) x = 8 (2) y = -20 [#permalink] New post  07 Oct 2019, 02:17
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dcummins wrote:
Bunuel wrote:
If x - y > 10, is x - y > x + y?

Given: \(x-y>10\). Question: is \(x-y>x+y\) --> is \(2y<0\) --> is \(y<0\).

(1) x = 8 --> \(8-y>10\) --> \(y<-2\), hence \(y<0\). Sufficient.
(2) y = -20, hence \(y<0\). Sufficient.

Answer: D.


So we can do the arithmetic, but we can't divide?


Yes, we can add/subtract a number to both sides of an inequality but we cannot multiply/divide unless we know the sign.
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