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Question

If xy < 5, is x < 1 ? (1) |y| > 5 (2) x/y > 0

Bunuel · Answer

If xy < 5, is x < 1 ?

(1) |y| > 5 --> if y=10 and x=0 then the answer is YES but if y=-10 and x=2 then the answer is NO. Not sufficient.

(2) x/y > 0 --> x and y have the same sign. Still insufficient: if y=-2 and x=-1 then the answer is YES but if y=2 and x=2 then the answer is NO.

(1)+(2) From (2) x and y have the same sign. Now, if from (1) y>5 then 0<x<1 (in order xy<5 to hold true) and if from (1) y<-5 then -1<x<0 (again in order xy<5 to hold true). So in both cases x<1. Sufficient.

Answer: C.

prathyushaR · Answer

(1)-> if y is +ve (y=6,7 etc.. check) x < 1 , if y is -ve x > -1 ( y= -6, -7 etc..) here it can be both less than one or greater than one (2) -> both x and y are same sign : we still dont know (take values of y as 1 or ten , x can be both >1 or <1) combined : if y is +ve we know x<1 , if y is -ve x>-1 but now, x is also negative that gives us 0>x>-1 so here too x<1 , thus we can say that x<1 for sure hence, C

chetan2u · Answer

xy<5 .....Is x<1? (1) |y| > 5 Two cases a) y>5......xy<5..... Add both the equations.... y+5>xy+5......xy-y<0.....y(x-1)<0 We know y>0, so x-1<0 or x<1....YES b) y<-5......xy<5..... Add both the equations.... 5-5>xy+y......xy+y<0.....y(x+1)<0 We know y<0, so x+1>0 or x>-1.... x could be 10, 100, or 0....... Both YES and NO possible. Insuff (2) \frac{x}{y} > 0 This means x and y have same sign. (x,y) = (4,1)=(-4,-1)...So Yes and No both possible Combined.. Both have same sign y>5... x has to be positive, so 0

Lipun · Answer

Hi chetan2u , Need some clarification regarding the range of x in ST1. If we are obtaining x < 1, x > -1, doesn't it imply that -1 < x < 1 ? Or, is it because |y| > 5 gives us an OR result set (either less than -5 OR greater than 5) thus, the range for x will also be OR and not AND? If so, what will be the range of x in ST1? Thanks in advance! Lipun

chetan2u · Answer

Range of x is dependent on the value of y..
So we cannot take any specific range. 
If y>5, it is different and if y<-5, it is different. 
As the values of y is totally different in two cases, we cannot join the ranges of x.

If xy < 5, is x < 1 ? (1) | y | > 5 (2) x / y > 0

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