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Inequality DS and veritas teacher answer

Problem Solving — algebra and arithmetic
Expert replies
by diegocuenca » Wed Jul 06, 2011 6:47 am
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).
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Source: — Quantitative Reasoning |

by Brent@GMATPrepNow » Wed Jul 06, 2011 9:56 am
diegocuenca wrote: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.
For statement 1, you have essentially multiplied both sides of the inequality by y and by 3 to get 3x > 3y. From here, if we divide both sides by 3, we get x > y
It seems reasonable, since it would be okay to do this if we had an equation rather than an inequality.

Another way to look at it is to say that if we multiply both sides by 3y, we get x > y

The problem with this step (or your step) is that, when we multiply both sides by 3y (or by y as you did in your solution), we need to know whether we are multiplying both sides be either a positive number or a negative number. Remember that, if we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality.

e.g., If we take the inequality -2x < 6 and divide both sides by -2, we get x > 3
However, if we take the inequality 2x < 6 and divide both sides by 2, we get x < 3

So, when we take x/(3y) > 1/3 and multiply both sides by 3y, we need to know whether y is positive or negative.

If y is positive (which means 3y is positive ), we get x > y once we multiply both sides by 3y
If y is negative(which means 3y is negative), we get x < y once we multiply both sides by 3y

Since we cannot answer the target question (Is x>y?) with certainty, statement 1 is not sufficient.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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