weird question from private tutor)

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wiut: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Fri Apr 02, 2010 11:48 am; Thanked: 2 times

by wiut » Mon Jan 10, 2011 9:23 am

if x and y are integers and x<y, what is the value of x+y?

1. x^y=4
2. x absolute = y absolute

it seems weird as obviously the only y that there is not Y which is less than X that satisfies either/both 1 and 2.....

what is your advise?
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Source: Beat The GMAT — Data Sufficiency | Mon Jan 10, 2011 9:23 am

clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Mon Jan 10, 2011 9:37 am

my answer is D
(1) the only pair of x,y such that y>x and x^y=4, is x=-2, y=2, -2+2=0 suff
(2)|x|=|y|
if x>0 and y>0 then x=y but according to the problem y>x not possible
if x<0 and y>0 then -x=y and we can find the sum x+(-x)=0
if x>0 and y<0 then x>y not possible
if x<0 and y<0 then -x=-y x=y again not possible
so st2 seems suff

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Jan 10, 2011 9:41 am

wiut wrote:if x and y are integers and x<y, what is the value of x+y?

1. x^y=4
2. x absolute = y absolute

Statement 1: (x^y) = 4
Possible ways to break 4 in (integer)^(integer) forms are:

1. (-2)^(2) ---> -2 < 2
2. (2)^(2) ---> 2 = 2
3. (4)^(1) ---> 4 > 1

Note that only the first one follows the condition, x < y. Hence a single set of values for x and y. Thus (x + y) can be determined, which is in fact (-2 + 2) = 0.

Sufficient

Statement 2: |x| = |y|
Only possible way to have |x| = |y|, while x is also less than y is, x = -y and y > 0. Which is equivalent to (x + y) = 0

Sufficient

The correct answer is D.

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Mon Jan 10, 2011 9:43 am

(-2)^2 =4

so, 1 is sufficient .

now

|x| = |y|

and x < y

so, 2 is sufficient would mean x+y =0

The solution is D

Last edited by arora007 on Mon Jan 10, 2011 9:51 am, edited 1 time in total.

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wiut: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Fri Apr 02, 2010 11:48 am; Thanked: 2 times

by wiut » Mon Jan 10, 2011 9:46 am

clock60 wrote:my answer is D
(1) the only pair of x,y such that y>x and x^y=4, is x=-2, y=2, -2+2=0 suff
(2)|x|=|y|
if x>0 and y>0 then x=y but according to the problem y>x not possible
if x<0 and y>0 then -x=y and we can find the sum x+(-x)=0
if x>0 and y<0 then x>y not possible
if x<0 and y<0 then -x=-y x=y again not possible
so st2 seems suff

But, x absolute could equal y absolute with any integer (one is obviously negative), say 3 and -3, 4 and -4. From (2) we can not infer the answer.

Now X^Y=4 (I hope you understood that ^ means degree, sorry I don't know how to put symbols in messages) it is not clear , because we know that x<y. So, it can not be 2. If we put both statements together, it seems like the answer is 2 and -2, but again, then the condition 1 will not hold, because if we put x into degree of -2, the answer is not 4.

What do you think, maybe the question itself is not correctly stipulated and there is no solution for this problem?

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Jan 10, 2011 9:50 am

But, x absolute could equal y absolute with any integer (one is obviously negative), say 3 and -3, 4 and -4. From (2) we can not infer the answer. (Look at your numbers! In each case they add up to zero.)

Now X^Y=4 (I hope you understood that ^ means degree, sorry I don't know how to put symbols in messages) it is not clear , because we know that x<y. So, it can not be 2. If we put both statements together, it seems like the answer is 2 and -2, but again, then the condition 1 will not hold, because if we put x into degree of -2, the answer is not 4.

(As I explained, there is one and only possible set of values for x and y, which is : x = -2 and y = 2)

What do you think, maybe the question itself is not correctly stipulated and there is no solution for this problem?

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wiut: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Fri Apr 02, 2010 11:48 am; Thanked: 2 times

by wiut » Mon Jan 10, 2011 9:54 am

Anurag@Gurome wrote:
But, x absolute could equal y absolute with any integer (one is obviously negative), say 3 and -3, 4 and -4. From (2) we can not infer the answer. (Look at your numbers! In each case they add up to zero.)

Now X^Y=4 (I hope you understood that ^ means degree, sorry I don't know how to put symbols in messages) it is not clear , because we know that x<y. So, it can not be 2. If we put both statements together, it seems like the answer is 2 and -2, but again, then the condition 1 will not hold, because if we put x into degree of -2, the answer is not 4.

(As I explained, there is one and only possible set of values for x and y, which is : x = -2 and y = 2)

What do you think, maybe the question itself is not correctly stipulated and there is no solution for this problem?

Thanks a lot. Now this question seems to obvious..sad part is that questions solved by others always look simple post-factum ((( Hope that soon I could master these concepts myself - I should say recall them from my school's years...thanks, great work by Experts and others! Great site!

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sushantgupta: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Sun Mar 27, 2011 9:54 am; Thanked: 2 times

by sushantgupta » Mon Jul 04, 2011 10:53 am

both alone are sufficient

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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members; Location: Ahmedabad; Thanked: 31 times

by ronnie1985 » Thu Mar 29, 2012 8:17 am

(D) is the answr

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Ganesh hatwar: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Sun Jun 24, 2012 11:23 pm

by Ganesh hatwar » Wed Sep 12, 2012 10:07 pm

narrowly missed .. Did not work on option B

Good question

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eski: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Wed Sep 05, 2012 11:36 amFollowed by: 4 members; Thanked: 2 times

by eski » Tue Sep 18, 2012 11:15 am

gr8 pvt tutor , nice throw , u caught me this time ...

what is the best way to do this?

table format?

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rajeshsinghgmat: Master | Next Rank: 500 Posts; Posts: 171; Joined: Tue Jan 08, 2013 7:24 am; Thanked: 1 times

by rajeshsinghgmat » Wed Mar 27, 2013 11:32 pm

B

|-2|=|2|

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Java_85: Master | Next Rank: 500 Posts; Posts: 234; Joined: Tue Jul 16, 2013 9:00 am; Location: West Virginia; Thanked: 9 times

by Java_85 » Mon Oct 21, 2013 8:37 am

D, 2^2 , -2^2 and 4^1

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jaspreetsra: Master | Next Rank: 500 Posts; Posts: 164; Joined: Sat Sep 20, 2014 10:26 pm; Thanked: 1 times

by jaspreetsra » Mon Dec 29, 2014 2:18 am

If x and y are integers and x < y, what is the value of x + y?

1. x^y = 4
2. |x| = |y|

Question Stem: y>x and x and y integers.

1)x^y = 4
=> (-2)^2 = 4 (bcz. y>x)
So, x+y = -2+2 = 0 Sufficient
2)|x| = |y|
if x = -2, then y = 2 (bcz. y>x)
if x = -3, then y = 3 (bcz. y>x)
So, x+y = -2+2 = 0
or -3+3 = 0 Sufficient
IMO: D

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