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If u(u+v)

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Source: — Data Sufficiency |
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by jrakhe » Wed May 16, 2012 1:55 pm
This question can be solved by putting numbers:-

1) u + v > 0 and u>0 (given in the question)

A) If we assume u=5 and v= -2 then u + v =3
So 1/3 < 1/5 -2 => 1/3 < -9/2 (which is not true)

B) If we assume u=5 and v= 2 then u + v =7
So 1/7 < 1/5 +2 (which is true)

So here we have one true and one false answer so first statement is not sufficient.

2) v >0 and u>0 (given in the question)

So here we can use the second example and can see that 1/(u+v) < 1/u + v.
So statement 2 is sufficient to answer this question.

Hope this helps :)
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by eagleeye » Wed May 16, 2012 2:31 pm
Hi Massi:
we are given that, u>0, u(u+v) >0 ; and asked if 1/(u+v) < 1/u + v

First observe that since u>0; 2) v>0 implies that u+v > 0 (if you add two positive numbers, you get a positive number). This is the same condition as in 1). Hence 1 is redundant. Therefore, if 2) can give us the answer, the correct answer will be B, otherwise it will be E.

Now u(u+v)>0 , since this is positive, we can multiply both sides with u(u+v) without changing the inequality.

Then we get
u(u+v) * 1/(u+v) < u(u+v) (1/u + v )
which becomes after cancellation :

u < u+v + u(u+v)/v
subtracting u from both sides

0 < v + u(u+v)/v

now we know that for both u and v, u>0, and v>0. hence the right hand side ( v + u(u+v)/v ) is greater than zero. Hence, the inequality is true. and the answer is a definite YES. Since we have a definite answer, the condition in 2) was sufficient, hence the correct answer should be B.

Let me know if this helps :)
Last edited by eagleeye on Thu May 17, 2012 2:51 am, edited 1 time in total.
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by massi2884 » Thu May 17, 2012 2:38 am
HI eagleeye,
thanks for you reply. However, if we can multiply both sides without changing the inequality why did you change from < to > ? Could you please clarify? Thanks.

Thanks.
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by eagleeye » Thu May 17, 2012 2:55 am
My bad! It was a typo. Good catch massi. I've edited the original derivation. Fortunately, in this case, the final answer is still B.

Cheers.
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by GMATGuruNY » Thu May 17, 2012 8:11 am
massi2884 wrote:If u(u+v) different from 0 and u >0, is 1/(u+v) < 1/u + v?
1) u+v >0
2) v>0

OA B Please explain. Thanks
Source: GMATPrep Pack 1

Statement 1: u+v > 0
Plugging u=1 and v=1 into 1/(u+v) < 1/u + v, we get:
1/(1+1) < 1/1 + 1
1/2 < 2.
YES.

Plugging u=2 and v=-1 into 1/(u+v) < 1/u + v, we get:
1/(2-1) < 1/2 - 1
1 < -1/2.
NO.

Since in the first case the answer is YES, and in the second case the answer is NO, INSUFFICIENT.

Statement 2: v>0
Since all of the values are positive, we can rephrase the questions stem:
1/(u+v) < (1+uv)/u [Putting the right side over a common denominator]
u < (u+v)(1+uv) [Cross multiplying]
u < u + u²v + v + uv².
0 < u²v + v + uv².
Since all of the values on the right side are positive, SUFFICIENT.

The correct answer is B.

Two take-aways:
1. When we're plugging values into a Yes/No DS question, we need to determine what types of values could change the answer from YES to NO or vice versa. Since statement 2 requires that v>0, the sign of v seems to be an issue here. Thus, we should consider NEGATIVE values when we plug into statement 1.

2. When an inequality problem is restricted to positive values, algebra becomes much easier to implement, since we don't have to worry about the direction of the inequality.
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by George7 » Mon Nov 05, 2012 1:56 pm
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Last edited by George7 on Mon Nov 05, 2012 2:36 pm, edited 1 time in total.
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by Whitney Garner » Mon Nov 05, 2012 2:17 pm
George7 wrote:Image

This is the OA, how can they assume that if v<0 then -v/u(u+v) > 0 ?
Thanks
Notice that we know that we know from the stem that U is positive, and from Statement (1) that (u+v) is positive. That means that in the fraction -v/[u(u+v)], the denominator is a (+)(+) = (+). Okay, then if V is a positive number -(v) = -(+) = -, and a (-)/(+) means the entire fraction would be negative <0. However, if we know that V<0, then -(-) = +, and a (+)/(+) means the entire fraction would be positive >0.

Hope this helps!
:)
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by FLUID » Tue Nov 06, 2012 12:03 am
If u(u+v) different from 0 and u >0, is 1/(u+v) < 1/u + v?
1) u+v >0
2) v>0



From 1, u+v > 0 but we will still don't know if v > 0 but as per the question U > 0.
we can't decide until we know the sign of V.

Form 2, V > 0 => V is positive so obviously 1/(u+v) < 1/u + v

Option (B)
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by Nina1987 » Wed Dec 16, 2015 9:37 am
GMATGuruNY wrote: Statement 1: u+v > 0[/b]
Plugging u=1 and v=1 into 1/(u+v) < 1/u + v, we get:
1/(1+1) < 1/1 + 1
1/2 < 2.
YES.

Plugging u=2 and v=-1 into 1/(u+v) < 1/u + v, we get:
1/(2-1) < 1/2 - 1
1 < -1/2.
NO.

Since in the first case the answer is YES, and in the second case the answer is NO, INSUFFICIENT.
I really like the approach to St1. However, number picking doesnt come naturally to me. I have to hold the following three statements in mind:
u(u+v)# 0
u >0
u+v > 0
And then think of numbers that will give two different answers to 1/(u+v) < 1/u + v?
This entire process flips me out.

Where as algebra is far more natural. but algebra is not always obvious and could be really time consuming especially on options where the statement is insufficient. That is probably also why it takes me a lot of time solve DS problems where answer is E.

Any suggestions to improve number picking skills? Any drills would you guys suggest?

Thanks
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by [email protected] » Wed Dec 16, 2015 9:57 am
Hi Nina1987,

TESTing VALUES is a skill that's quite useful to have in the Quant section (and on certain IR questions), but it's one that many Test Takers have not ever learned to use before. As such, it takes time to practice and hone that skill.

1) How long have you been studying?
2) What resources have you been using?
3) How have you been scoring on your CATs (including the Quant and Verbal Scaled Scores)?

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