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diegocuenca
- Senior | Next Rank: 100 Posts
- Posts: 32
- Joined: Thu Feb 24, 2011 1:17 pm
I don't quite understand this problem.
Is x > y?
1. x/(3y) > 1/3, cross multiply which would give you a 3x > 3y. I put sufficient, however it says insufficient because x and y could be both positive or negative. I don't understand why that would throw it off, it clearly says 3x is bigger than 3y. All I can think is that I should have gone further with the manipulation? So divide both sides by 3 which yields that x > y?
2. -x + p < -y + p
No problems here, multiply everything by -1 so x - p > y - p so here I didn't go further, should I have added + p to both sides to cancel it. I feel this step is not necessary or am I wrong.
I put D, the answer is B.
Remember! The trick you're almost always being tested on in any inequality problem is the fact that if you divide of multiply by a negative number, you have to flip the direction of the inequality sign. I would avoid cross-multiplying inequalities in general because it tends to obscure whether or not you are multiplying by negatives (never, once, or twice) and reserve cross-multiplying for instances where you've got a proportion (i.e. where there's an equal sign in between). Here, if rather than cross-multiplying, you multiply through on both sides by 3y, that 3y may be positive or negative, so you'll have two different scenarios: x > y if y is positive; x < y if y is negative. That's why it's not sufficient. Repeat, this is nearly ALWAYS relevant in inequalities problems (especially on DS), so watch out for it!
Awesome, that's a nice little trick. I got that on the second one but not the first because the cross mult. messed me up. I will avoid that from now on. On the second one is it necessary once you get x - p > y - p to add + p to cancel out the p on both sides? Or is x - p > y - p to come up with a conclusion?
Hmm, I guess *technically* it's necessary to add p to both sides, but if you can see it from the x-p > y-p, you don't need to. If you're seeing it from that, it's probably because you're sort of subconsciously adding p to both sides in your head (or thinking about how p will drag both x and y down by an equal amount, so if x-p still comes out on top, it must have started off as the greater quantity, or something).
Is x > y?
1. x/(3y) > 1/3, cross multiply which would give you a 3x > 3y. I put sufficient, however it says insufficient because x and y could be both positive or negative. I don't understand why that would throw it off, it clearly says 3x is bigger than 3y. All I can think is that I should have gone further with the manipulation? So divide both sides by 3 which yields that x > y?
2. -x + p < -y + p
No problems here, multiply everything by -1 so x - p > y - p so here I didn't go further, should I have added + p to both sides to cancel it. I feel this step is not necessary or am I wrong.
I put D, the answer is B.
Remember! The trick you're almost always being tested on in any inequality problem is the fact that if you divide of multiply by a negative number, you have to flip the direction of the inequality sign. I would avoid cross-multiplying inequalities in general because it tends to obscure whether or not you are multiplying by negatives (never, once, or twice) and reserve cross-multiplying for instances where you've got a proportion (i.e. where there's an equal sign in between). Here, if rather than cross-multiplying, you multiply through on both sides by 3y, that 3y may be positive or negative, so you'll have two different scenarios: x > y if y is positive; x < y if y is negative. That's why it's not sufficient. Repeat, this is nearly ALWAYS relevant in inequalities problems (especially on DS), so watch out for it!
Awesome, that's a nice little trick. I got that on the second one but not the first because the cross mult. messed me up. I will avoid that from now on. On the second one is it necessary once you get x - p > y - p to add + p to cancel out the p on both sides? Or is x - p > y - p to come up with a conclusion?
Hmm, I guess *technically* it's necessary to add p to both sides, but if you can see it from the x-p > y-p, you don't need to. If you're seeing it from that, it's probably because you're sort of subconsciously adding p to both sides in your head (or thinking about how p will drag both x and y down by an equal amount, so if x-p still comes out on top, it must have started off as the greater quantity, or something).
