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I'm studying using Jeff Sackman's "Total GMAT Math" and I feel like one of the DS question answers is wrong in the answer guide. Here is the question:

If m and n are negative, is m/n less than 1? (1) mn<1 (2) m-n>n

I say B. The guide says E. Can anyone explain?

Hi, welcome to the Gmat Club. Below is the solution for your question:

Given \(m<0\) and \(n<0\). Q: is \(\frac{m}{n}<1\)? --> is \(m>n\)? (when multiplying by negative \(n\), switch the sign).

(1) \(mn<1\). As both are negative --> \(0<mn<1\). Clearly insufficient to conclude which one is greater.

(2) \(m-n>n\) --> \(m>2n\). If \(m=-3\) and \(n=-2\) answer is NO but if \(m=-3\) and \(n=-4\) answer is YES. Not sufficient.

(1)+(2) \(m=-0.2\) and \(n=-1\), both conditions satisfied --> \(m>n\) BUT \(m=-1.1\) and \(n=-0.6\), again both conditions satisfied --> \(m<n\). Two different answers. Not sufficient.

Re: Can anyone answer this DS question? [#permalink]

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26 May 2010, 10:19

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Bunuel- Wow. Thanks for the quick response to that. I'm still new to DS questions. I'm only about a week and a half deep into my GMAT studying, so I have a long way to go. It makes more sense to me now. I feel like I'm making a lot of careless mistakes, especially when it comes to DS questions. I guess I need to just keep practicing. Anyway, seriously thanks.

Re: Can anyone answer this DS question? [#permalink]

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30 Oct 2010, 20:17

Bunuel wrote:

MateoLibre wrote:

I'm studying using Jeff Sackman's "Total GMAT Math" and I feel like one of the DS question answers is wrong in the answer guide. Here is the question:

If m and n are negative, is m/n less than 1? (1) mn<1 (2) m-n>n

I say B. The guide says E. Can anyone explain?

Hi, welcome to the Gmat Club. Below is the solution for your question:

Given \(m<0\) and \(n<0\). Q: is \(\frac{m}{n}<1\)? --> is \(m>n\)? (when multiplying by negative \(n\), switch the sign).

(1) \(mn<1\). As both are negative --> \(0<mn<1\). Clearly insufficient to conclude which one is greater.

(2) \(m-n>n\) --> \(m>2n\). If \(m=-3\) and \(n=-2\) answer is NO but if \(m=-3\) and \(n=-4\) answer is YES. Not sufficient.

(1)+(2) \(m=-0.2\) and \(n=-1\), both conditions satisfied --> \(m>n\) BUT \(m=-1.1\) and \(n=-0.6\), again both conditions satisfied --> \(m<n\). Two different answers. Not sufficient.

Answer: E.

Hope it helps.

I solved it, but it took me many minutes. I still wondering how you know the right steps to solve it so fast
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Re: Can anyone answer this DS question? [#permalink]

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05 Dec 2010, 04:00

Hi bunuel,

I could not understand statement 2 ; m>2n you have taken m=-3 and n =-2 this makes -3>-4 which will be 3<4 true and in second case you have taken m=-3 and n=-4 this will make -3>-8 3<8this is also true ..i could not understand Pleas explain

I could not understand statement 2 ; m>2n you have taken m=-3 and n =-2 this makes -3>-4 which will be 3<4 true and in second case you have taken m=-3 and n=-4 this will make -3>-8 3<8this is also true ..i could not understand Pleas explain

Given \(m<0\) and \(n<0\). Question: is \(m>n\)?

(2) \(m-n>n\) --> \(m>2n\). If \(m=-3\) and \(n=-2\), (these values satisfy both the stem and statement: both are negative and \(-3>2*(-2)\)), then as \(m=-3<-2=n\) the answer is NO; If \(m=-3\) and \(n=-4\), (these values satisfy both the stem and statement: both are negative and \(-3>2*(-4)\)), then as \(m=-3>-4=n\) the answer is YES. Not sufficient.

Re: Can anyone answer this DS question? [#permalink]

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05 Dec 2010, 18:24

I have a random question on this problem.

For the second statement, you're given that m - n > n, so why can't you just divide the whole thing out by n, in which case you get an expression with m/n? I can intuitively sense it's wrong but I can't place my finger on why.

For the second statement, you're given that m - n > n, so why can't you just divide the whole thing out by n, in which case you get an expression with m/n? I can intuitively sense it's wrong but I can't place my finger on why.

It's not wrong at all, in fact it's a valid algebraic way to deal with statement (2).

Statement (2): \(m-n>n\) --> \(m>2n\) --> divide both parts by \(n\) (note that as given that \(n\) is negative then we should switch sign): \(\frac{m}{n}<2\) --> but this is not sufficient to say whether \(\frac{m}{n}<1\).
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Re: Can anyone answer this DS question? [#permalink]

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05 Dec 2010, 23:37

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Friends,

In GMAT, to avoid the confusion with the negative #s, i use my own and simple technique that is as below.

PLEASE NOTE THAT: DEALING WITH THE INEQUALITIES THAT CONTAIN +VE #s IS EASIER THAN DEALING WITH THOSE CONTAIN -VE#S.

Given: m and n are -ve #s I take: x and y as +ve #s.

now i write m = -x and n = -y

FROM NOW ON I AM GONNA DEAL WITH THE +VE #S ONLY. I CAN NOT BE TRAPPED BY THE TESTMAKER.

qtn: is m/n less than 1? == > is -x/-y < 1 ==> is x/y<1 ==> is x < y (NOTE: as x and y both are +ve, i can safely cross multiply).

(1) mn<1

==> (-x)(-y)<1 ==> xy<1 : our qtn is "is x<y", why is y a greatest person than x or vice-versa. As their product is < 1 ==> means, they can always inerchange their values with out effecting thier product. So x could b < y or vice-versa. NOT SUFF.

(2) m-n>n

==> -x+y>-y ==> x-y<y ==> x< 2y ==> x/y < 2 (Again note that x and y both are +ve and hence i can safely cross multiply)

hence x/y is < 2, aren't there any other #s between 2 and 1, how can i say that x/y is < 1 knwoing that x/y is < 2....NOT SUFF.

1 & 2 together , xy<1 and x/y<2...Can't even be combined...NO LUCK...NOT SUFF.

Re: Can anyone answer this DS question? [#permalink]

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06 Dec 2010, 06:52

Ah, thanks Bunuel. The sign conversion was what tripped me up. Forgot to read the part that said they were negative numbers; should pay more attention.

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06 Dec 2010, 21:48

Bunuel, I was wondering about the -.2 and -1 and the -1.1 and -.6. Did you use these particular figures for a purpose or were they just easy numbers you knew configured into a number less than 1. Allowing you to make M as big as possible then M as small as possible?

Bunuel, I was wondering about the -.2 and -1 and the -1.1 and -.6. Did you use these particular figures for a purpose or were they just easy numbers you knew configured into a number less than 1. Allowing you to make M as big as possible then M as small as possible?

Question asks is \(\frac{m}{n}<1\)? or is \(m>n\)?

From (1) we have: \(0<mn<1\), which is useless (whether alone or combined with 2) to answer which variable is greater.

From (2) we have: \(\frac{m}{n}<2\), so \(\frac{m}{n}\) is less than 2 but we can not say whether it's less than 1, so again this statement is useless to answer the question.

So even not testing the numbers we could say that the answer is E. But to show this, to demonstrate that the answer is E I just picked 2 sets of numbers, first set m>n and the second m<n, also with little trial and error it's not hard to get the numbers to satisfy stem and the statements (remember on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another).
_________________

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14 Feb 2013, 14:17

rite2deepti wrote:

Hi bunuel,

I could not understand statement 2 ; m>2n you have taken m=-3 and n =-2 this makes -3>-4 which will be 3<4 true and in second case you have taken m=-3 and n=-4 this will make -3>-8 3<8this is also true ..i could not understand Pleas explain

rite2deepti,

I thought the same, but you have to look at the question itself m>n?

Plugging m = -3 and n = -2 this yields -3 > -4 which is correct, however, the question is -3>-2 and this is incorrect, thats why its a NO.

For the second case m = -3 and n = -4 this yields -3 > -8 which is correct, and YES as an answer -3>-4. Having this, we can conclude its INSUFFICIENT, NO AND YES.
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