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aishwaryasingh0311
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Post subject: Algebra question (men and women in choir) Posted: Wed Jul 11, 2012 12:20 pm |
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If there are x men and y women in a choir, and there are z more men than there are women in that choir, what is z?
(1) x2 – 2xy + y2 – 9 = 0
(2) x2 + 2xy + y2 – 225 = 0
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.
Now the ans is a)statement one is sufficient
Can you please explain me how is the second equation not sufficient?
In the explanation it is written that when we solve the second equation it could be x+y=15 or -15 But it is very obvious that the no. of men and women in a choir cannot be-(minus) So according to that we have to chose a positive no. only, which is 15 and which qualifies the second option.
Please help...
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campb441
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Post subject: Re: Algebra question (men and women in choir) Posted: Wed Jul 11, 2012 5:24 pm |
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First you should make the equation when comparing men and women in the choir.
x = men y = women and there are z more men than women Therefore:
x = y + z or you can rewrite this as x - z = y
1) Sufficient
x^2 - 2xy + y^2 - 9 = 0 x^2 - 2xy + y^2 = 9 additive inverse (x - y)(x - y) = 9 (x - y)^2 = 9 this creates a perfect square x - y = 3 square root each side x -(x - z) = 3 substitute for y from the equation at the top z = 3 solve for z
2) NOT sufficient
Start by repeating the same steps but you'll see after you substitute for y you won't cancel out your x variables like you did above. This is all because of the + 2xy in this problem instead of the - 2xy like the problem above.
x^2 + 2xy + y^2 - 225 = 0 x^2 + 2xy + y^2 = 225 additive inverse (x + y)(x + y) = 225 (x + y)^2 = 225 perfect square (x + y) = 15 square root each side x + (x - z) = 15 substitute for y 2x - z = 15 ... not solvable
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jnelson0612
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Post subject: Re: Algebra question (men and women in choir) Posted: Sun Jul 15, 2012 10:46 pm |
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Posts: 2412
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Wow, very nice! Thanks campb!
_________________ Jamie Nelson ManhattanGMAT Instructor
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thulsy
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Post subject: Re: Algebra question (men and women in choir) Posted: Mon Jul 16, 2012 8:20 pm |
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I have a question regarding this problem: Why can't z be negative? i.e. why can't we say "there are -3 more men than there are women in that choir"? I thought that was equivalent to "there are 3 more women than there are men in that choir". Just as we can say that "the population increase is -2.5%" or "the population increases by -2.5%". I mean, here the figure can be negative, meaning the population actually decreases. So here I suppose that a negative integer (-3) associated with "more" is supposed to mean the corresponding positive integer (3) associated with "less". I'm not a native speaker of English, and it seems like I'm missing here. Please kindly explain. Thanks.
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CurtisVincentBorns
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Post subject: Re: Algebra question (men and women in choir) Posted: Thu Jul 19, 2012 9:31 pm |
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I solved the problem pretty similar to campb but it seems more simplistic to me. Please correct me if I'm wrong...
The question states that "z more men than there are women in that choir"
So if:
x=men y=women and there are z more men than woman.
The equation be written y+z=x which then can be rewritten as z=x-y
So then the rephrasing of the question should be "what is x-y?"
1) sufficient x^2 - 2xy + y^2 - 9 = 0 x^2 - 2xy + y^2 = 9 additive inverse (x - y)(x - y) = 9 (x - y)^2 = 9 this creates a perfect square x - y = 3 square root each side
***can we not just stop here? we now have x-y = 3 = z
2) NOT sufficient
x^2 + 2xy + y^2 - 225 = 0 x^2 + 2xy + y^2 = 225 additive inverse (x + y)(x + y) = 225 (x + y)^2 = 225 perfect square (x + y) = 15 square root each side
***this gives us x+y, which is not what we were looking for
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tim
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Post subject: Re: Algebra question (men and women in choir) Posted: Fri Jul 20, 2012 5:03 pm |
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Posts: 4462 Location: Southwest Airlines, seat 21C
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thulsy, i agree with you. the problem should stipulate that z is a positive integer..
_________________ Tim Sanders Manhattan GMAT Instructor
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