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If u(u+v)neq{0} and u >0, is 1/(u+v) < 1/u + v?

Expert replies
by Asma77 » Wed Feb 11, 2015 5:55 pm
If u(u+v)neq{0} and u >0, is 1/(u+v) < 1/u + v?


(1) u+v >0
(2) v>0

I answered the above equestion as follow please correct me if I am wrong

1/(u+v)<1/u+v

1/(u+v) -1/u -v <0

1/(u+v) - (1-uv/u) <0

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

u-u-uv-uv/u<0

-2uv/u <0

-2v<0 or does v positive?

can I devided both sides by -2 and it will be does v>o ?
Join the discussion
Source: — Data Sufficiency |

by GMATGuruNY » Thu Feb 12, 2015 4:48 am
If u(u+v) ≠ 0 and u > 0, is 1/(u+v) < 1/u + v?
1) u+v >0
2) v>0
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, but in the second case the answer is NO, INSUFFICIENT.

Statement 2: v>0
Since all of the values are positive, we can rephrase the question stem.
Is 1/(u+v) < 1/u + v?

Put the right side over a common denominator:
1/(u+v) < (1+uv)/u

Cross-multiply:
u < (u+v)(1+uv)

Distribute on the right-hand side:
u < u + u²v + v + uv².

Subtract u from each side:
0 < u²v + v + uv².

Question stem, rephrased:
Is u²v + v + uv² > 0?
Since all of the values on the left-hand side are positive, the answer is YES.
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.
Asma77 wrote:
I answered the above equestion as follow please correct me if I am wrong

1/(u+v)<1/u+v

1/(u+v) -1/u -v <0

1/(u+v) - (1-uv/u) <0

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

u-u-uv-uv/u<0


-2uv/u <0

-2v<0 or does v positive?

can I devided both sides by -2 and it will be does v>o ?
The steps in red contain algebraic errors and yield an incorrect rephrase of the question stem.
I suggest that you check your work.
Last edited by GMATGuruNY on Sun Sep 18, 2016 3:03 am, edited 2 times in total.
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by Asma77 » Thu Feb 12, 2015 6:01 am
GMATGuruNY wrote:
massi2884 wrote:If u(u+v) ≠ 0 and u > 0, is 1/(u+v) < 1/u + v?
1) u+v >0
2) v>0
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.
Is 1/(u+v) < 1/u + v?

Put the right side over a common denominator:
1/(u+v) < (1+uv)/u

Cross-multiply:
u < (u+v)(1+uv)

Distribute on the right-hand side:
u < u + u²v + v + uv².

Subtract u from each side:
0 < u²v + v + uv².

Question stem, rephrased:
Is u²v + v + uv² > 0?
Since all of the values on the left-hand side are positive, the answer is YES.
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.
Asma77 wrote:
I answered the above equestion as follow please correct me if I am wrong

1/(u+v)<1/u+v

1/(u+v) -1/u -v <0

1/(u+v) - (1-uv/u) <0

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

u-u-uv-uv/u<0


-2uv/u <0

-2v<0 or does v positive?

can I devided both sides by -2 and it will be does v>o ?
The steps in red contain algebraic errors and yield an incorrect rephrase of the question stem.
I suggest that you check your work.
OK, I see my mistake here so it will be u-(u+v)+uv(u+v)/u(u+v)<o

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

-v+uv/u<0

v(-1+u)/u<0

is that correct
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by GMATGuruNY » Thu Feb 12, 2015 7:53 am
Asma77 wrote:
OK, I see my mistake here so it will be u-(u+v)+uv(u+v)/u(u+v)<o

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

-v+uv/u<0

v(-1+u)/u<0

is that correct
Unfortunately, there are still algebraic errors.
When the lefthand side is put over a common denominator, you are neglecting to enclose the ENTIRE NUMERATOR in brackets.
As a result, the steps that follow are incorrect.

Putting all three terms over a common denominator seems quite messy, making the problem harder rather than easier.
I recommend that you opt for an alternate approach.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
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