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by sars72 » Sun Jan 31, 2010 8:04 pm
tallazndood wrote:Question:
If x is NOT equal to -y, is (x-y)/(x+y)>1?
1. x>0
2. y<0
i'll go about this by getting both yes & no answers to each statement, thereby proving that the statements are insufficient

1. x>0 --> x=2,y=2 --> (2-2)/(2+2) = 0 which is less than 1, giving NO
x=2, y=-1 --> (2--1)/(2-1) = 3 which is greater than 1, giving YES
therefore 1 alone is insufficient

2.y<0 --> x=2, y=-1 --> (2--1)/(2-1) = 3 which is greater than 1, giving YES
x=-1, y=-1 --> (1--1)/(-1-1) = 0 which is less than 1, giving NO
therefore 1 alone is insufficient

combinining 1 & 2
-> x>0 & y<0
x=2, y=-1 --> (2--1)/(2-1) = 3 which is greater than 1, giving YES
x=1, y=-5 --> (1--5)/(1-5) = -6/4 which is less than1, giving NO

Therefore, the answer is E
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by tallazndood » Sun Jan 31, 2010 8:34 pm
sars72, Thank you for the quick response. Plugging in numbers is definitely one way to go about this problem. I was hoping for someone to show me an algebraic approach?
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by ajith » Sun Jan 31, 2010 10:00 pm
tallazndood wrote:Question:
If x is NOT equal to -y, is (x-y)/(x+y)>1?
1. x>0
2. y<0

OA: E


1. x>0 in this case the expression can be greater than 1 only when y is -ve [e.g x=3 y= -1] since the answer gives no indication of the sign of y not sufficient

2. y<0 in this case also if it depends of |x| and sign of x (e.g. x = -2,y = -3 and x=2 and y=3) Insufficient

Combining 1&2 then also it depends on whether |x|>|y| ( eg x= 2 y = -1 and x= 2, y= -3) no indication of absolute value is given hence insufficient
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by sars72 » Sun Jan 31, 2010 10:14 pm
tallazndood wrote:sars72, Thank you for the quick response. Plugging in numbers is definitely one way to go about this problem. I was hoping for someone to show me an algebraic approach?
for these type of questions, you cannot solve algebraically. This is because you don't know whether (x+y) is positive or negeative and hence cannot cross multiply.
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by thephoenix » Sun Jan 31, 2010 11:59 pm
simplifying the ineq
[(x-y)/(x+y)]-1>0
-2y/(x+y) >0????

now x#y
and
s1) x>0
two cases
case1
if y<0,then for x<y the denominator will be -ve
if y<0--->-2y>0
+numerator/-ve denominator=a number<0
case 2
if y<0, then for lxl>lyl.....the deno.. is +ve and hence number>0
hence insuff

s2)
same can be done for s2
combining also we shown above tw ans are possible and hence not suff
and hence E
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