if x and y are integers and x > 0, is y > 0?

[Radar]

if x and y are integers and x > 0, is y > 0?

1) 7x - 2y >0
2) -y < x







Ans: E

[GMAT700]

1) 7x-2y>0

7(8)-2(3) = positive
7(8)-2(-3) = positive

y can be neg or pos. NOT sufficient.

2) -y<x

x = 8 , y can equal 3
x = 8, -3<8, so y can be positive

x = 8, y can equal -3
x=8, -(-3)<8, 6<8, y can be negative too. Not sufficient.

c) combined: Look at the breakdown from statement (1) with numbers that also satisfy statement (2)'s criteria. y can be neg OR positive.

Answer is Choice E.

[Sudhan]

BDACE Grid,

From B,
-y<x

Since y is negative, we can swap the signs to make it positive.Hence,
x>y.

X >0 but y can be either -ve integer or +ve integer. Hence Insuff.

From A,
7x-2y>0
--> 7x>2y. X >0, y can be -ve or +ve. Hence Insuff.

From A and B, we can say y can be either +ve or _ve.

Hence E.

[RonPurewal]

one very important note to the post above:

From B,
-y<x

Since y is negative, we can swap the signs to make it positive.Hence,
x>y.

actually, you don't know that the quantity (-y) is negative: if y itself is a negative number, then (-y) is positive, and if y itself is positive, then (-y) is negative.
you must take care NOT to assume that (-y) is necessarily a negative quantity; that only has to be the case if y is positive.

[claudia.minoiu]

Can someone solve this graphically? I did and got C. Thank you.

[tim]

show us your graph, and we'll help you figure out where you went wrong..

[Jazmet]

But what if you take both the equations simultaneously as following

7x - 2y > 0 - Stat 1
7x + 7y > 0 - Stat 2 (multiplied by 7)

-9y > 0
9y < 0

Hence, y < 0

According to this its C. Can you solve like this?

(How I arrived to 7x+ 7y > 0 from Stat 2)
-y < x ---- y > -x ---- y + x > 0 --- x + y > 0 --- 7x + 7y > 0

[RonPurewal]

you can't subtract inequalities if they have the same sign.

you can only subtract inequalities if they have opposite signs. however, that's rather confusing, so, if you get that sort of situation, it's best to multiply one of the inequalities by -1 and then add them.

e.g., if x > 10 and y < 6
then you can write x > 10 and -y > -6
then add to get x - y > 4.
(you can also just do this by common sense: if something costs more than $10, and you have a coupon worth less than $6, then the item will clearly still cost more than $4 after the coupon is applied.)

if you have x > 10 and y > 6, then you can say for sure that x + y > 16, but you can't say anything at all about x - y.
in fact, if x > 10 and y > 6, then x - y can have absolutely any value whatsoever.
same thing is true for the two inequalities that you have here.

--

you can also prove that the answer is (e) just by finding cases that go both ways.
e.g.
x = 1, y = 2 --> satisfies both statements; yes, y is greater than 0
x = 1, y = 0 --> satisfies both statements; no, y is not greater than 0
done. (e). not much pain there.

[Jazmet]

Thank you Ron!

Its crystal clear now.

[RonPurewal]

Thank you Ron!

Its crystal clear now.

sweet.

[Virender KumarM746]

Hi Ron,
Followed the chain. Isn't the answer A ? No idea if it makes sense or not but this is what I did.

(1) 7x - 2 y > 0
7x > 2y
x / y > 2 / 7
If x > 0 is already given, then y must be greater than zero (positive). Hence y > 0.

Can you please clarify what's wrong here or some basic mathematical rule is not followed here.

[HarmeetS612]

Picking values always come handy in questions like these.

St1: 7x - 2y>0 - - -> 7x > 2y

x y St1(Satisfied) Q(y>0)
1 1 Yes Yes
1 -1 Yes No

We are getting two different answers. Hence Insufficient

St2: -y<x - - -> x>-y

X Y St2(Satisfied) Q(y>0)
1 1 Yes Yes
2 -1 Yes No


We are getting two different answers. Hence Insufficient

Combining both the statements.
We can use the same values again.

Hence answer is E.

[RonPurewal]

Hi Ron,
Followed the chain. Isn't the answer A ? No idea if it makes sense or not but this is what I did.

(1) 7x - 2 y > 0
7x > 2y
x / y > 2 / 7
If x > 0 is already given, then y must be greater than zero (positive). Hence y > 0.

nope. you can't divide by y, because the sign of y is unknown. (in doing this step-- in which you leave the ">" sign untouched-- you're assuming, without justification, that y is positive.)

[RonPurewal]

also, if x is positive, you should be able to tell from a quick glance at "7x > 2y" that y can be positive, zero, or negative.

why try to transform the inequality into a less straightforward form? hmm.

[Virender KumarM746]

Thanks Ron. this makes sense. Its clear.