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Hard looking algebra problem

by minkathebest » Tue Nov 13, 2012 9:35 pm
If x and y are positive integers, is (2+x) / (3+y) greater than (2+y)/(3+x)?


1) x+y = 3
2) x > y

This is from question pack 1 of gmatprep. There is an algebra way of doing this but I was wondering if the plug-in approach or any other approaches are useful here?

Answer is B
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by FLUID » Tue Nov 13, 2012 11:34 pm
minkathebest wrote:If x and y are positive integers, is (2+x) / (3+y) greater than (2+y)/(3+x)?


1) x+y = 3
2) x > y

This is from question pack 1 of gmatprep. There is an algebra way of doing this but I was wondering if the plug-in approach or any other approaches are useful here?

Answer is B

Explanation:

Statement 1: x+y = 3

This statement does not help to determine (2+x) / (3+y) greater than (2+y)/(3+x)

Statement 2: x > y
=> Number which has X in the numerator and Y in denominator is greater than
Number which has Y in the numerator and X in denominator since X > Y

So answer is (b)
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by anjalimanas » Sat Apr 13, 2013 2:01 am
1) x+y = 3
2) x > y

From statement (1) ,the expressions become
5 - y / y + 3
Y + 2 / 6 - y

Check for boundary conditions
Y < 5 and y > 6
For y <5 say y = 3 2/5 and 5/1 , No
For y > 6 say y = 7, -1/10 and -9, Yes
Insufficient

From statement (2), x = 3 and y = 2 , boundary condition when difference between x and y is 1, for differences greater than 1 , it is always No
Substituting values, The expressions become
5 / 5 = 1
4 / 6 = 2 / 3
Hence No
Sufficient
Hence (B)
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by GMATGuruNY » Sat Apr 13, 2013 3:18 am
minkathebest wrote:If x and y are positive integers, is (2+x) / (3+y) greater than (2+y)/(3+x)?


1) x+y = 3
2) x > y

Answer is B
Since x and y are positive, we can rephrase the question stem by cross-multiplying:
(2+x)/(3+y) > (2+y)/(3+x)

(2+x)(3+x) > (2+y)(3+y)

6 + 5x + x² > 6 + 5y + y²

5x + x² > 5y + y².

Notice that the two sides of the inequality are almost a perfect match: the only difference is the variable.
Thus, the sum on the lefthand side will be greater than the sum on the righthand side if x>y.

Question rephrased: Is x>y?

Statement 1: x+y =3
If x=1 and y=2, then x<y.
If x=2 and y=1, then x>y.
INSUFFICIENT.

Statement 2: x>y
SUFFICIENT.

The correct answer is B.
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by Mission2012 » Mon Apr 15, 2013 12:06 pm
GMATGuruNY wrote:
minkathebest wrote:If x and y are positive integers, is (2+x) / (3+y) greater than (2+y)/(3+x)?


1) x+y = 3
2) x > y

Answer is B
Since x and y are positive, we can rephrase the question stem by cross-multiplying:
(2+x)/(3+y) > (2+y)/(3+x)

(2+x)(3+x) > (2+y)(3+y)

6 + 5x + x² > 6 + 5y + y²

5x + x² > 5y + y².

Notice that the two sides of the inequality are almost a perfect match: the only difference is the variable.
Thus, the sum on the lefthand side will be greater than the sum on the righthand side if x>y.

Question rephrased: Is x>y?

Statement 1: x+y =3
If x=1 and y=2, then x<y.
If x=2 and y=1, then x>y.
INSUFFICIENT.

Statement 2: x>y
SUFFICIENT.

The correct answer is B.
Hi Mitch,

After step -

5x + x² > 5y + y².

I actually went ahead
5x + x²+ 5y - y² > 0
(x-y)(x+y+5) > 0

hence we can say x-y > 0 or x+y+5 > 0

therefore x > y or b solves the problem.

I am not sure how to tackle (1).. could you help
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by GMATGuruNY » Mon Apr 15, 2013 2:16 pm
Mission2012 wrote:
minkathebest wrote:If x and y are positive integers, is (2+x) / (3+y) greater than (2+y)/(3+x)?


1) x+y = 3
2) x > y

Answer is B
Hi Mitch,

After step -

5x + x² > 5y + y².

I actually went ahead
5x + x²- 5y - y² > 0
(x-y)(x+y+5) > 0

hence we can say x-y > 0 or x+y+5 > 0

therefore x > y or b solves the problem.

I am not sure how to tackle (1).. could you help
The conclusion in red is not quite correct.
For it to be true that (x-y)(x+y+5) > 0, there are two cases:

Case 1: x-y < 0 AND x+y+5 < 0
Here, both factors are NEGATIVE.
Since the question stem indicates that x and y are both positive, it is not possible that x+y+5 < 0.
Thus, this case is not viable.

Case 2: x-y > 0 AND x+y+5 > 0
Here, both factors are POSITIVE.
x-y > 0 will be true if x>y.
ANY positive values for x and y will satisfy the constraint that x+y+5 > 0.
Thus, Case 2 will be true if x>y.

Question rephrased: Is x>y?

From here, we can proceed as I did in my solution above.
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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].
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