Could any of the math specialists please explain how to approach such a question?

stm1
aba = a
a(ab -1) = 0
a = 0, or ab = 1

stm2
Similar to stm1, b = 0, or ab = 1

Why cannot we derive ab=1 from both these statements and say answer is C?

Neeraj is correct for stm2
...looks like answer is D

Re: From GMAT Prep

If m*v < p*v < 0 , is it v > 0 ? 1) m < p 2) m < 0 mv<pv<0 implies mv-pv<0, i.e. v(m-p)<0 This is possible when either v<0 and (m-p)>0, i.e. m>p or, v>0 and (m-p)<0, i.e., m<p Stm 1: m<p, so v>0 suff Stm2: m<0...tells us nothing about p, hence in suff Answer is A

From the figure: pqr + qps =90 =>pqr+ (qpr+rps) =90 ....(a) and, rps + prs =90...(b) We need to find |prs-pqr| stm 1: qpr=30 from (a), pqr + 30 + rps =90, or rps = 60-pqr from (b), rps =90 -prs So, 60-pqr=90 -prs, hence we know the value of |prs-pqr|. Sufficient stm 2: pqr + (180-prs) = 150, again t...

Set up an arithmatic progression:
a=4 (4ft tall to begin with)
d = rate of growth every year

End of 1st year, height = a+d
End of 2nd year, height = a+2d
End of 4th year, height = a+4d
End of 6th year, height = a+6d

it is given that a+6d = (6/5)(a+4d)

we know a, solve for d to get the answer.

If z is the rent in 1997, then rent in 1998 is z+(xz/100) and the rent in 1999 is [z+(xz/100)] -(y/100)[z+(xz/100)] =[z+(xz/100)] - (yz/100) - (xyz/10000) = z+(z/100)(x-y-(xy/100)) The question requires us to compare z+(z/100)(x-y-(xy/100)) (1999)w/ z (1997) Stm1: x>y This gives us the diff values o...

Totally agree, very methodical approach.

Re: Stumped

muzali wrote:If x, y, and z are integers and xy+z is an odd integer, is x an even integer?

1. xy+xz is an even integer
2. y+xz is an odd integer

OA=A

Would appreciate a detailed solution.