Hi,
Some important observations that could be helpful if we internalize this.
\(\frac{ODD}{EVEN}\) = never an integer and remainder is always odd.
\(\frac{3}{2}\) = never an integer value and reminder is 1,
\(\frac{191}{60}\) = never an integer remainder is 11.
\(\frac{ODD}{ODD}\) = either Odd Integer or never integer, then remainder can be either Odd or even
\(\frac{3}{1}\)= Odd integer remainder 0,
\(\frac{3}{5}\)= not an integer , remainder is 3,
\(\frac{5}{3}\)= not an integer and reminder is 2.
\(\frac{EVEN}{EVEN}\)= either Integer ( could be even or could be odd ) or could not be an integer, then remainder is always even.
\(\frac{6}{2}\)= 3 which is odd integer and remainder is 0 which is even .
\(\frac{12}{6}\)= 6 which is even integer and remainder is 0 which is even.
\(\frac{14}{4}\)= not an integer value and remainder is 2 which is even
\(\frac{EVEN}{ODD}\)= even integer or not an integer,then remainder is even or odd.
\(\frac{12}{3}\)= 4 even integer and remainder is 0
\(\frac{12}{5}\)= not an integer & reminder is 2 which is even
\(\frac{12}{7}\)= not an integer & remainder is 5 which is Odd
Now back to question,
we are given that w,x,y,z are integers and \(\frac {w}{x}\)and \(\frac{y}{z}\) are integers is \(\frac {w}{x}\)+ \(\frac{y}{z}\) = odd?
So lets say \(\frac{w}{x}\)= a , & \(\frac{y}{z}\) = b
then is a+b = odd integer . { note a, b are integers, and integer+integer= integer}
if we rearrange the terms \(\frac{wz+xy}{zx}\) = odd integer
we know that \(\frac{wz+xy}{zx}\)= integer,then we can write wz+xy= zx( integer). let this integer be denoted by k.
wz+xy= zx(k) -------(i)
Stm1 :wx+yz is odd.
adding wx+yz on both sides of equation(i)
then, wx+yz+wz+xy= k(zx)+wx+yz
so odd+ wz+xy= k(zx)+odd
or odd-odd+wz+xy= k(zx)
or even +wz+xy= k(zx)
so if wz+xy = odd then yes, if wz+xy = even then no.
So stmt 1 Insufficient .
Stmt 2: wz+xy = odd.
substituting in equation (i) we have
odd= k(zx) ,this implies k(zx) is odd { Note neither K, nor z or x is even}
So stmt 2 Sufficient .
Though , i did not use the observations above, but felt useful sharing as many Friends out here were using this approach.
Probus
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Probus
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