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gobucks
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AREA88
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It's D.

1) is sufficient. X is prime and X>10 means that x odd. 3 is odd. Odd-Odd=even. If we have one number in multiplication even the product is even. => y(x-3) is even.

2) is sufficient. y<3 =>y=2.
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Good catch Natalya!

I disregarded A without even working it out. :oops:
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D is right!! I did'nt catch no.1 condition.

Thanks! ^^
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Answer is D.

since both statements can independently lead to answer.
if x> 10 and prime .. x will always be odd and in that case

x-3 will always be even and hence. we can come to answer

y < 3 and a prime number y = 2



y ( x-3) .. will always be even :D
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D

In both cases at least one of the term is even so the result will be even.
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Please explain how this could be D.


a) X > 10. Agree that A is sufficient

b) y< 3 i.e y=2 which is prime number. But we don't know the value of X.
X could be any number say 3

Y(x-3)=0. so expression gives 0. so B is insufficient.

So Answer is A
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baski6
Please explain how this could be D.


a) X > 10. Agree that A is sufficient

b) y< 3 i.e y=2 which is prime number. But we don't know the value of X.
X could be any number say 3

Y(x-3)=0. so expression gives 0. so B is insufficient.

So Answer is A

Whatever is the value of other term, anything multiplied my an even number is EVEN.
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gobucks
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If you plug in 11 for X, and 2 for Y you get:
I) x> 10 2(11-3) = 2(8) 16 which is not even.. How is A sufficient here?

II) y<3 2(11-3) = 2(8) 16 which again is not even..

What am I missing in understanding this problem...

Thanks for all your help....
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gobucks
If you plug in 11 for X, and 2 for Y you get:
I) x> 10 2(11-3) = 2(8) 16 which is not even.. How is A sufficient here?

II) y<3 2(11-3) = 2(8) 16 which again is not even..

What am I missing in understanding this problem...

Thanks for all your help....

16 is NOT even??? :roll: :roll: :horror :horror
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E,

S1)

If you use X = 11 and Y = 2, y(x-3) = Even
If you use x = 11 and Y = 1, y(x-3) = Odd

Insufficient

S2)

If you use X = 11 and Y = 2, y(x-3) = Even
If you use X = 11 and Y = 1, y(x-3) = Odd

Insufficient
Both 1 and 2 are prime and they are less then 3.

With both S1&S2:

If you use X = 11 and Y = 2, y(x-3) = Even
If you use X = 11 and Y = 1, y(x-3) = Odd

Insufficient
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ps_dahiya
gobucks
If you plug in 11 for X, and 2 for Y you get:
I) x> 10 2(11-3) = 2(8) 16 which is not even.. How is A sufficient here?

II) y<3 2(11-3) = 2(8) 16 which again is not even..

What am I missing in understanding this problem...

Thanks for all your help....
16 is NOT even??? :roll: :roll: :horror :horror



SORRY, Meant to say that it is NOT ODD?? So, I am still confusused because the question asks if it is ODD...?? :cry:
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gobucks
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The answer in the book is ..... D
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Whoa...I was way off, i see what the problem is now. Thanks.
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gobucks
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agsfaltex
Whoa...I was way off, i see what the problem is now. Thanks.


How do you get an ODD number to in order to get the answer D?? I substituted 11 for X, and 2 for Y
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gobucks
agsfaltex
Whoa...I was way off, i see what the problem is now. Thanks.

How do you get an ODD number to in order to get the answer D?? I substituted 11 for X, and 2 for Y

You have to answer if the term is odd or not.
Answer is it is NOT ODD. That means you can answer the question using both statements. Hence it is D.
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baski6 - zero is EVEN
so stmt 2 holds and the result is NO

So the answer is D



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