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# If m > 0 and n > 0, is (m+x)/(n+x) > m/n?

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If m > 0 and n > 0, is (m+x)/(n+x) > m/n? [#permalink]

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26 Jul 2011, 21:58
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If m > 0 and n > 0, is (m+x)/(n+x) > m/n?

(1) m < n.
(2) x > 0.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-0-and-n-0-is-m-x-n-x-m-n-93967.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jul 2012, 05:21, edited 5 times in total.
Edited the question.

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26 Jul 2011, 22:18
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Revision:

Please see ~fluke's solution below. Thanks.
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Last edited by GyanOne on 02 Aug 2011, 00:40, edited 4 times in total.

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27 Jul 2011, 06:47
Don't we need to know if X is positive or negative before multiplying both sides by (N + X)?
What if N is negative so that N + X < 0? This would call for us to change the sign of the inequality when we take the 1st step you did above in restating the question.

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27 Jul 2011, 08:36
I would go for C

(1) Gives us that any side could be greated depending on sign of X
(2) Gives us condition on X being +ve.

Hence combine together and we get C as answer.

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27 Jul 2011, 08:37

Just check with the numbers. eg 4/3 < 3/2.

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27 Jul 2011, 08:42
GyanOne is correct. The OA is A. I just don't understand why we don't need to know that X is positive.

Revision:

GyanOne is incorrect. The OA is C. We do need to know the sign of X.

Last edited by Yalephd on 27 Jul 2011, 09:48, edited 2 times in total.

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27 Jul 2011, 08:51
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This is why answer IS NOT B:

II) x>0
Case 1: let m=2, n=3 and x=1
(2+1)/(3+1) = 3/4 and 3/4 > 2/3

Case 2: let m=3, n=2 and x=1
(3+1)/(2+1) = 4/3 and 4/3 < 3/2

Since we can find find values to make the statement true and false, II is not sufficient.

From I) we know that m<n, also, not sufficient on it's own because X can be either positive or negative.

However, taking I and II together we can answer the question, so the answer is C.

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27 Jul 2011, 09:22
Yalephd wrote:
GyanOne is correct. The OA is A. I just don't understand why we don't need to know that X is positive.

I second this. GyanOne's algebra seems correct. And apparently OA is A. However what about cases 1. x=1, m=1, n=2; and 2. x=-1, m= 1, n=2. First case resolves to 2/3, which is greater than 1/2. Second case resolves to 0, which is less than 1/2. Insufficient.

So what am I missing?

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27 Jul 2011, 09:42
elementbrdr, you're right. We definitely need to know the sign of X. I have to go back and check the OA. Maybe I made a mistake. It looks like it has to be C.

M < N tells us that this is a proper fraction. If we add the same number to the numerator and the denominator, the fraction increases. If we subtract the same number from the numerator and the denominator, the fraction decreases. Hence, we need to know whether M < N and whether X < 0.

Last edited by Yalephd on 27 Jul 2011, 09:45, edited 1 time in total.

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27 Jul 2011, 09:57
Yalephd wrote:
OA is C

Well that makes sense based on plugging numbers. I still don't see what is wrong with GyanOne's algebra though...

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27 Jul 2011, 10:00
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$$\frac{m+x}{n+x} > \frac{m}{n}$$

$$\frac{m+x}{n+x}-\frac{m}{n} > 0$$

$$\frac{mn+nx-mn-mx}{n+x}> 0$$

$$\frac{x(n-m)}{n+x}> 0$$

1. $$m<n$$

Thus, n-m>0

Now, if x>0; the fraction will be greater than 0.
If x<0 but |x|<n; the fraction will be less than 0.

2. $$x>0$$

If x>0;
m<n; the fraction will be greater than 0
m>n; the fraction will be less than 0

Together;
The fraction is greater than 0.

Ans: "C"
********************************************************************

Since, we don't know the sign of x; we can't cross multiply as x can be negative and |x|>n
***********************************************************************
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28 Jul 2011, 07:56
After plug-in various numbers for m,n,x .I will go with c.

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31 Jul 2011, 10:08
fluke wrote:
$$\frac{m+x}{n+x} > \frac{m}{n}$$

$$\frac{m+x}{n+x}-\frac{m}{n} > 0$$

$$\frac{mn+nx-mn-mx}{n+x}> 0$$

$$\frac{x(n-m)}{n+x}> 0$$

1. $$m<n$$

Thus, n-m>0

Now, if x>0; the fraction will be greater than 0.
If x<0 but |x|<n; the fraction will be less than 0.

2. $$x>0$$

If x>0;
m<n; the fraction will be greater than 0
m>n; the fraction will be less than 0

Together;
The fraction is greater than 0.

Ans: "C"
********************************************************************

Since, we don't know the sign of x; we can't cross multiply as x can be negative and |x|>n
***********************************************************************

Shouldnt the divisor should be n(n+x) or am i missing something with the N?

how did you get rid of it after combining n+x and n?

thanks.
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02 Aug 2011, 00:39
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144144,

As n>0, we can exclude n from the divisor.
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02 Aug 2011, 01:09
tnx.
i did it at around 2am so i was "floating"... +1
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02 Aug 2011, 05:50
How do we go from this
$$\frac{m+x}{n+x}-\frac{m}{n} > 0$$

to this
$$\frac{mn+nx-mn-mx}{n+x}> 0$$

I don't see how we dropped the N from the denominator.

"As n>0, we can exclude n from the divisor. " Why is that?

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02 Aug 2011, 07:30
Yalephd wrote:
How do we go from this
$$\frac{m+x}{n+x}-\frac{m}{n} > 0$$

to this
$$\frac{mn+nx-mn-mx}{n+x}> 0$$

I don't see how we dropped the N from the denominator.

"As n>0, we can exclude n from the divisor. " Why is that?

because its positive, you are allowed to multiple both sides of the equation times N. when u multiple the right side it stays 0, when you multiple the left side, N cancels the denominator.

If we had no information about whether it is +ve or -ve, we couldnt do it bc this might change the > sign (If N was -ve)

hope this is clear.
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02 Aug 2011, 07:35
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If the product of two numbers is positive, they must either both be positive or both be negative.

Since n is > 0 (given in the question), (nx-mx)/(m+x) must also be > 0. Therefore we check only for the condition where (nx-mx)/(m+x) is > 0 as it will also make the entire expression > 0.
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02 Aug 2011, 07:44
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GyanOne wrote:
If the product of two numbers is positive, they must either both be positive or both be negative.

Since n is > 0 (given in the question), (nx-mx)/(m+x) must also be > 0. Therefore we check only for the condition where (nx-mx)/(m+x) is > 0 as it will also make the entire expression > 0.

GyanOne - I can say from a user of this amazing forum for few month now, we are very happy to have you here with us. Your answers are clear and very helpful. Thanks. +1
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02 Aug 2011, 07:54
Got it. Thanks, guys. +1

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Re: M versus N   [#permalink] 02 Aug 2011, 07:54

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