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by sanju09 » Sat Sep 04, 2010 2:14 am
How often does the sum of positive integers x and y contain their difference?

(1) x - y, with x > y, is a factor of x + y.

(2) x is of the order of 81 thousand and y is of the order of 54 thousand.


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The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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Source: — Data Sufficiency |

by neerajkumar1_1 » Sat Sep 04, 2010 2:55 am
sanju09 wrote:How often does the sum of positive integers x and y contain their difference?

(1) x - y, with x > y, is a factor of x + y.

(2) x is of the order of 81 thousand and y is of the order of 54 thousand.


Source: [email protected]
From what i understood from the question stem
we are looking for
x+y=k(x-y)

the problem will occur when x<y
then u will get a -ve number and that cannot be present in the sum of 2 +ve numbers...

statement 1 gives us exactly that... hence 1 is sufficient...

statement 2... the numbers could be anything so the difference will not always be present in the sum...

hence IMO A
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by sanju09 » Sat Sep 04, 2010 3:06 am
neerajkumar1_1 wrote:
sanju09 wrote:How often does the sum of positive integers x and y contain their difference?

(1) x - y, with x > y, is a factor of x + y.

(2) x is of the order of 81 thousand and y is of the order of 54 thousand.


Source: [email protected]
From what i understood from the question stem
we are looking for
x+y=k(x-y)

the problem will occur when x<y
then u will get a -ve number and that cannot be present in the sum of 2 +ve numbers...

statement 1 gives us exactly that... hence 1 is sufficient...

statement 2... the numbers could be anything so the difference will not always be present in the sum...

hence IMO A
May I help you?

Q. How often (many number of times) 12 contains 4?

A. 12/4 = 3 times.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion