Is xy > 0?
(1) x - y > -2
(2) x - 2y < -6
Is the following approach correct?
First I multiplied the statement (2) with -1:
(2) -x + 2y > 6
Now I added statement (1) and statement (2):
(1) x - y > -2
(2) -x + 2y > 6
results in y > 4
Then I use this expression in combination with statement (1) (I add it):
(1) x - y > -2
y > 4
results in x > 2
Therefore, x > 2 and y > 4 and hence xy = positive
Correct answer: C
I am not sure whether I can do these calculations like this. Could somebody clarify? Or also give a better strategy to solve this question?
Thx a lot
Is xy > 0?
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- cans
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Your method is correct.
1) x > y-2. x=1,y=-2. xy <0 also x=4,y=3,xy>0 insufficient
2) x < 2y-6. y=3, x=-1, xy<0 and y=4,x=1, xy>0 insufficient
1&2) y-2 < x < 2y-6.
thus y>4. and x>y-2 >2
xy>0
Sufficient
IMO C
1) x > y-2. x=1,y=-2. xy <0 also x=4,y=3,xy>0 insufficient
2) x < 2y-6. y=3, x=-1, xy<0 and y=4,x=1, xy>0 insufficient
1&2) y-2 < x < 2y-6.
thus y>4. and x>y-2 >2
xy>0
Sufficient
IMO C
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Ok thanks! But I am not able to follow your approach! As far as I can see, you just plug in some different number in statement (1) and (2). But then I don't understand, how you combine these two statements? Could you tell me plz? Thx a lotcans wrote:Your method is correct.
1) x > y-2. x=1,y=-2. xy <0 also x=4,y=3,xy>0 insufficient
2) x < 2y-6. y=3, x=-1, xy<0 and y=4,x=1, xy>0 insufficient
1&2) y-2 < x < 2y-6.
thus y>4. and x>y-2 >2
xy>0
Sufficient
IMO C
- knight247
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@Fractal
When you combine both statements u can take an algebraic approach.
x - y > -2
and
-6> x - 2y
Add both
x-y-6 > x-2y-2
y>4
Lets assume y=5 which satisfies the above condition.
Put y=5 in statement 1 and 2
x-5>-2
x>3
x-2(5)<-6
x<4
Combining we get 3<x<4 Meaning x is +ve and we already know that y is +ve. So their product is +ve. Hence C
When you combine both statements u can take an algebraic approach.
x - y > -2
and
-6> x - 2y
Add both
x-y-6 > x-2y-2
y>4
Lets assume y=5 which satisfies the above condition.
Put y=5 in statement 1 and 2
x-5>-2
x>3
x-2(5)<-6
x<4
Combining we get 3<x<4 Meaning x is +ve and we already know that y is +ve. So their product is +ve. Hence C
- cans
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1 plugged in numbers in 1 and 2, to show they are not sufficient. ant for C),Fractal wrote:Ok thanks! But I am not able to follow your approach! As far as I can see, you just plug in some different number in statement (1) and (2). But then I don't understand, how you combine these two statements? Could you tell me plz? Thx a lotcans wrote:Your method is correct.
1) x > y-2. x=1,y=-2. xy <0 also x=4,y=3,xy>0 insufficient
2) x < 2y-6. y=3, x=-1, xy<0 and y=4,x=1, xy>0 insufficient
1&2) y-2 < x < 2y-6.
thus y>4. and x>y-2 >2
xy>0
Sufficient
IMO C
if x>a and x<b we can say that b>x>a or b>a...
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please note the following rules:
you CAN add inequalities with the same direction,
and you can subtract opposite inequalities, but not the other way
around. For example, since we know that:
5 > 4 and
3 > 1
we can add them and know that:
8 > 5
but we can't subtract them and get:
2 > 3 (WRONG!)
However, we can write it as:
5 > 4 and
1 < 3
and then we can subtract them:
4 > 1
but we can't add them:
6 > 7 (WRONG!)
you CAN add inequalities with the same direction,
and you can subtract opposite inequalities, but not the other way
around. For example, since we know that:
5 > 4 and
3 > 1
we can add them and know that:
8 > 5
but we can't subtract them and get:
2 > 3 (WRONG!)
However, we can write it as:
5 > 4 and
1 < 3
and then we can subtract them:
4 > 1
but we can't add them:
6 > 7 (WRONG!)