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If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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 18 Mar 2012, 04:40
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If xy < 4, is x < 2 ? (1) y > 1 (2) y > x
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Last edited by Bunuel on 18 Mar 2012, 06:12, edited 1 time in total.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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18 Mar 2012, 06:25
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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18 Mar 2012, 21:38
Bunuel wrote: If xy < 4, is x < 2 ?
Notice that in order xy<4 to hold true at least one of the multiples must be less than 2 (if both x and y are more than or equal to 2 then xy>=2).
(1) y > 1. If for example \(y=1.5>1\) then \(x\) can be 1, so less than 2 or 2 so not less than 2. Not sufficient.
(2) y > x. According to above: \(y>x\geq{2}\) is not possible since in this case \(xy>2\), so \(x\) must be less than 2. Sufficient.
Answer: B.
Hope it's clear. Yes Bunuel. It's clear. Thanks! 
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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17 Jun 2013, 05:03
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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18 Jun 2013, 00:40
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Smita04 wrote: If xy < 4, is x < 2 ?
(1) y > 1
(2) y > x From F.S 1, when y = 4, x = 0.5, xy<4 and x<2.Hence a YES for the question stem. However, for y = 3/2 and x = 2, xy<4 and x=2, hence a NO for the question stem.Insufficient. From F. S 2, we know that y>x. Had y been equal to x, we would have x*x<4 --> |x|<2 --> -2<x<2 However, as y>x, to maintain the above given inequality, the value of x will have to reduce even further. Thus, x <2. Sufficient. B.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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24 Jul 2013, 12:54
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Bunuel wrote: If xy < 4, is x < 2 ?
Notice that in order xy<4 to hold true at least one of the multiples must be less than 2 (if both x and y are more than or equal to 2 then xy>2).
(1) y > 1. If for example \(y=1.5>1\) then \(x\) can be 1, so less than 2 or 2 so not less than 2. Not sufficient.
(2) y > x. According to above: \(y>x\geq{2}\) is not possible since in this case \(xy>2\), so \(x\) must be less than 2. Sufficient.
Answer: B.
Hope it's clear. Hi Bunuel, I guess the xy>2 should be replaced by xy>4 in above solution
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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24 Jul 2013, 12:57
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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Have a somewhat different solution for this one. Given that XY <4 ...... Is X<2??? (1). Y > 1 => Y = GT 1 (Where GT = Greater than) XY < 4 => X < (4/Y) (We can divide by a number >1 both sides) => X < (4/GT1) = > X < LT4 = > X < 4 Not sufficient as X can be 3 , 2 ,1 and so on so forth. (2). Y>X => X GTX < 4 Substituting values X=2 , 2*3 < 4 Not true X=1 , 1*2 < 4 true Hence (B) !!
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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27 Jul 2013, 19:08
Smita04 wrote: If xy < 4, is x < 2 ?
(1) y > 1
(2) y > x Alternate solution (rather alternate interpretation) for this problem - xy < 4 or x < 4/y (1) y > 1: Interpret 4/y as 4/(1 and increasing). This implies x can be <2 or >2 - So not sufficient (2) y > x: Interpret xy < 4 as x\(2\)<4 and since y is greater than x, it implies that x < 2 - So sufficient Ans.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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27 Jul 2013, 21:13
- A is insufficient.
- Option B. what if x or y is negative integer. I don't see you mentioned it in your answer.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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27 Jul 2013, 21:52
nguyenduong wrote: - A is insufficient.
- Option B. what if x or y is negative integer. I don't see you mentioned it in your answer. nguyenduong, Interpretation should be based on given (true) statements. Given here xy<4 and for pt. (2) y>x. Take any signs, y has to be greater than x or greater part of the multiplication with x resulting in a product smaller than 4. Consider if xy was 4 (xy=4) and y=x then both x & y would have been + /- 2. Now when the product reduces (product < 4), x has to be always less than the equal contribution i.e. 2. Hope I could explain.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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28 Jul 2013, 04:48
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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06 Aug 2014, 07:50
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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15 Jul 2015, 11:39
Hi Folks, does the method below looks ok?
1) xy <4
-4y<-4
y(x-4) <0
y<0 and x < 4
not suff
2) xy<4
xy^2 <4y
4x < 4y
x(y^2 -4) <0
x <0 suff.
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If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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15 Jul 2015, 11:53
getbetter wrote: Hi Folks, does the method below looks ok?
1) xy <4
-4y<-4
y(x-4) <0
y<0 and x < 4
not suff
2) xy<4
xy^2 <4y
4x < 4y
x(y^2 -4) <0
x <0 suff. 1) Where are you getting the above text in red from? How do you get y (x-4) < 0 ? Even if I assume that what you have written in correct, where are you getting y<0 from? 2) Text in blue is only possible if y>0. You need to reverse the inequality if y <0. Statement 2 just mentions that y>x and it does not tell us that x>0 to make y>0 at the same time. Also, how are you getting x(y^2-4)<0 from 4x<4y? Based on your solution, you were lucky to get to the correct answer and it does not look correct. Maybe, if you could write out all the steps clearly, we can take a look again.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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15 Jul 2015, 15:49
Sure let me explain step by step:
1) xy <4 ———(1)
y > 1 -> 1<y -> -y<-1
multiply both sides by 4 : -4y<-4 ————(2)
(1)+(2)
xy - 4y < 0
y(x-4) <0
y<0 and x < 4
not suff.
2) xy<4
multiply both sides by ‘y’: xy^2 < 4y ————(1)
y > x
Multiply both sides by 4: 4x < 4y —————(2)
(1)-(2):
x(y^2 -4) <0
x <0 suff.
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If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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15 Jul 2015, 16:09
getbetter wrote: Sure let me explain step by step:
1) xy <4 ———(1)
y > 1 -> 1<y -> -y<-1
multiply both sides by 4 : -4y<-4 ————(2)
(1)+(2)
xy - 4y < 0
y(x-4) <0
y<0 and x < 4
not suff.
2) xy<4
multiply both sides by ‘y’: xy^2 < 4y ————(1)
y > x
Multiply both sides by 4: 4x < 4y —————(2)
(1)-(2):
x(y^2 -4) <0
x <0 suff. Ok, this looks better now. Text in green: correct! Text in red: incorrect 1) xy <4 ———(1)
y > 1 -> 1<y -> -y<-1
multiply both sides by 4 : -4y<-4 ————(2)
(1)+(2)
xy - 4y < 0
y(x-4) <0 ------ (3) y<0 and x < 4
not suff.From 3, you know that y (x-4)<0 ---> this means 2 cases: either y<0 and x>4 (not possible as given y>1) or y>0 and x<4 . Even if we have x<4, x can be 3 or 1.5 , thus x <2 may or may not be true. Thus this statement is not sufficient. 2) xy<4 multiply both sides by ‘y’: xy^2 < 4y ————(1)
y > x
Multiply both sides by 4: 4x < 4y —————(2)
(1)-(2):
x(y^2 -4) <0
x <0 suff.How do you know whether y < or > 0 when you multiply xy<4 by y? if y is >0 then yes what you did is correct, but if y < 0 , then xy *y > 4y. The thing to remember here is that DO NOT multiply an inequality by a variable when you do not know the sign of that variable.Alternately, what you can do is : xy < 4 and y >x , ---> as x<y ----> \(x^2 < 4\) (as x<y and xy < 4 ----> \(x^2< xy <y^2\)) ----> -2<x<2 and thus this statement is sufficient to say x<2. Thus B is the correct answer. Hope this answers your question.
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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18 Jul 2015, 09:52
given: xy<4 ;is x<2? from(1) y>1=> y can be any +ve number(not specifically Integer);say y= 1.5 than x>2 but if y=3 than x<2 Hence Not sufficientfrom(2) y>x; Take the worst case when y=x than both will have value 4; but as y grows more than 2 value of x has to go below 2 in order for Y to be greater than x. Since already xy<4 hence x has to be less than 2 Hence Sufficient
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x [#permalink]
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Re: If xy < 4, is x < 2 ? (1) y > 1 (2) y > x
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