Danger of multiplying/dividing by variables (w/inequalities)

[kjberkemeyer]

In class we were given an example of the danger of multiplying/dividing by variables in dealing with inequalities.

This was the Data Sufficiency example from class

Is ab>xy?
(1) a/x>y/b
(2) xb>0

Could you please explain how to go about answering this question?

Thanks for your help,

kjb

[guest]

Is ab>xy?
(1) a/x>y/b
(2) xb>0

(2) looks a bit easier to deal with, so I'd start there. xb>0 means that x,b are either both positive, or both negative. By itself however, it doesn't tell you whether ab>xy and so you can rule out B and D and move onto (1). It's tempting to manipulate (1) algebraically into ax>yb and declare it sufficient and select A. However, you don't know which way the inequality sign should be pointing since you don't know the sign of x and b (the variables you have to multiply by to get to the from ax < yb). So (1) by itself is insufficient and you can rule out A. And like many difficult DS problems it comes down to C and E.

To see whether (1) and (2) are sufficient together we need to consider the two cases that (2) gives us: either x, b are both positive or both negative:

Both positive, then the sign in (1) doesn't change direction when you multiply by the variables and you get ax<yb.

Both negative and the sign changes direction TWICE and you still end up at ax<yb.

So (1) and (2) together are sufficient and the answer is (C).

[RonPurewal]

Is ab>xy?
(1) a/x>y/b
(2) xb>0

(2) looks a bit easier to deal with, so I'd start there. xb>0 means that x,b are either both positive, or both negative. By itself however, it doesn't tell you whether ab>xy and so you can rule out B and D and move onto (1). It's tempting to manipulate (1) algebraically into ax>yb and declare it sufficient and select A. However, you don't know which way the inequality sign should be pointing since you don't know the sign of x and b (the variables you have to multiply by to get to the from ax < yb). So (1) by itself is insufficient and you can rule out A. And like many difficult DS problems it comes down to C and E.

To see whether (1) and (2) are sufficient together we need to consider the two cases that (2) gives us: either x, b are both positive or both negative:

Both positive, then the sign in (1) doesn't change direction when you multiply by the variables and you get ax<yb.

Both negative and the sign changes direction TWICE and you still end up at ax<yb.

So (1) and (2) together are sufficient and the answer is (C).

well played.

[rajkapoor]

2nd Statement easier
xb > 0 / non-conclusive
but tells us x and b are of same signs.

1st statement
a/x > y/b
a/x - y/b > 0
(ab-xy)/xb > 0

doesnt help much either

combined , we know xb > 0 , therefore , ab - xy > 0 => ab > xy

hence C

[RonPurewal]

2nd Statement easier
xb > 0 / non-conclusive
but tells us x and b are of same signs.

1st statement
a/x > y/b
a/x - y/b > 0
(ab-xy)/xb > 0

doesnt help much either

combined , we know xb > 0 , therefore , ab - xy > 0 => ab > xy

hence C

that works too.