Is X>0 and Y>0, is (X-a)/(Y-a) < X/Y?
1. X>Y
2. a>0
Fractions
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- sivaelectric
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- Geva@EconomistGMAT
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With this type of yes no question, you generally want to try plugging in and elimination - show that the statements allow examples of both a yes and a no answer to the question stem, and are thus insufficient.sivaelectric wrote:Is X>0 and Y>0, is (X-a)/(Y-a) < X/Y?
1. X>Y
2. a>0
Let's go with an initial plug in of x=3, y=2 (x>Y to satisfy stat. (1). X/Y = 3/2, and from here it depends on the value of a.
IF a=1, then (3-1)/(2-1) = 2/1 = 2, and the answer is "no": 2 is not smaller than 3/2.
BUT
IF a=3, then (3-3) / 2-3) = 0/-1 = 0, and the answer is "yes": 0 is smaller than 3/2.
(BTW, the question has a bug, as it allows y-a to equal zero, thereby dividing by 0).
These values of a=1, a=3 satisfy both statements, so even the combination of statements still allows by a yes and a no answer, meaning that the combination is also insufficient and the answer is E.
- sivaelectric
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Hey Geva,
Thanks for the explanation .
BTW can you please provide your opinion on this also https://www.beatthegmat.com/percentages- ... tml#370301
Thanks for the explanation .
BTW can you please provide your opinion on this also https://www.beatthegmat.com/percentages- ... tml#370301
If I am wrong correct me , If my post helped let me know by clicking the Thanks button .
Chitra Sivasankar Arunagiri
Chitra Sivasankar Arunagiri
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Given that both X & Y are positive, is (X-a)/(Y-a) < X/Y?sivaelectric wrote:Is X>0 and Y>0, is (X-a)/(Y-a) < X/Y?
1. X>Y
2. a>0
Since it is given that Y>0, multiply by Y on both sides
Is Y(X-a)/(Y-a) < X
St1]X>Y
So if Y(X-a)/(Y-a) < Y, then automatically the answer to our question is YES.
Is (X-a)/(Y-a) < 1?
Here if X = a, then the answer to our question is YES, while If X>a, when a>0 then the answer to our question is NO.
St2] a>0
Original question=> (X-a)/(Y-a) < X/Y?
Again the former of the above conditions will yeild YES as an answer, while if X>Y>a, then the answer to the question would be a NO.
Combining both again both conditions are possible.
Agree with E.