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|X| and |Y|

Expert replies
Source: — Problem Solving |

Re: |X| and |Y|

by logitech » Wed Oct 29, 2008 10:24 pm
canuckclint wrote:If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) -4x – 12y = 0

(2) |x| – |y| = 16
(1) -4x – 12y = 0

-4x = 12y

x=-3y This tells us X and Y have different signs. ( -,+)

we can also say |x| = 3|y|

Using this and |x| + |y| = 32,

We can find that |x| = 24 an |y| = 8

we don't know which one is NEGATIVE and which one is positive but since we know that they have different signs...xy= - 192

Hence statement is SUFFICIENT

(2) |x| – |y| = 16

Using |x| + |y| = 32, we can find that |x| = 24 an |y| = 8

But we have no information given for their signs so xy can be either - 192 or +192

So this tricky statement is INSUFFICIENT

IMO the answer should be A
LGTCH
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by gmat009 » Thu Oct 30, 2008 12:48 pm
I agree with logitech solution
IMO answer is A
Can you plz. confirm OA
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Re: |X| and |Y|

by abcdefg » Sat Jul 18, 2009 1:33 pm
OA is A .

Just out of curiosity, why is it that we can make x=-3y into |x| = 3|y| ?

logitech wrote:
canuckclint wrote:If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) -4x – 12y = 0

(2) |x| – |y| = 16
(1) -4x – 12y = 0

-4x = 12y

x=-3y This tells us X and Y have different signs. ( -,+)

we can also say |x| = 3|y|

Using this and |x| + |y| = 32,

We can find that |x| = 24 an |y| = 8

we don't know which one is NEGATIVE and which one is positive but since we know that they have different signs...xy= - 192

Hence statement is SUFFICIENT

(2) |x| – |y| = 16

Using |x| + |y| = 32, we can find that |x| = 24 an |y| = 8

But we have no information given for their signs so xy can be either - 192 or +192

So this tricky statement is INSUFFICIENT

IMO the answer should be A
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by rah_pandey » Sat Jul 18, 2009 10:23 pm
This is a property of nos
if ax=by
than
|ax|=|by|

if you ask why did we convert to mod than answer is simple, since the original equation is in modular form so it makes sense to convert the second also into modular form and solve.
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