equations / squares problem

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topspin360: Master | Next Rank: 500 Posts; Posts: 103; Joined: Sat Jun 02, 2012 9:46 pm; Thanked: 1 times

by topspin360 » Sun Nov 04, 2012 12:19 pm

can someone please explain why the answer is E Is it b/c y is indeterminable when r = 5? is "indeterminable" an ambiguous answer?

(y+3)(y-1) - (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25
(2) r = 5

thanks

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Source: Beat The GMAT — Data Sufficiency | Sun Nov 04, 2012 12:19 pm

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Sun Nov 04, 2012 5:20 pm

topspin360 wrote:can someone please explain why the answer is E Is it b/c y is indeterminable when r = 5? is "indeterminable" an ambiguous answer?

(y+3)(y-1) - (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25
(2) r = 5

thanks

Yes,

factoring out (y-1) will give y+3-y+2=r and further r=5 only when (y-1)=!0 (not equal to 0)
in case of y-1=0 and y=1 we may have r=any real value
st(1) r^2=25 implies r=|5| Insuff
st(2) r=5 Insuff
combined also Insuff

answer is e)

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Nov 04, 2012 9:15 pm

topspin360 wrote:can someone please explain why the answer is E Is it b/c y is indeterminable when r = 5? is "indeterminable" an ambiguous answer?

(y+3)(y-1) - (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25
(2) r = 5

thanks

(y+3)(y-1) - (y-2)(y-1) = r(y-1)

(y-1)( (y+3) - (y-2) ) = r(y-1)

(y-1)(5) = r(y-1)

5(y-1) - r(y-1) = 0

(5-r)(y-1) = 0.

r=5 is the only value that satisfies both statements.
If r=5, y can be any value.
INSUFFICIENT.

The correct answer is E.

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