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by n@resh » Fri Sep 02, 2011 8:35 am
knight247 wrote:Is y an integer?
(1) y³ is an integer
(2) 3y is an integer

OA is C
Statement 1: say Y^3 = -8 ..then y = -2 (integer) but if Y^3 = 2... then Y isn't integer!
Not sufficient

Statement 2: say 3Y = 3...then y = 1 ( integer) but if 3Y = 1...then Y isn't integer!
Not sufficient

combine both: 1 and 2,
we need a value such that, 3Y should be an integer and also Y3 must be an integer.

so only multiples of 3, 3Y becomes an integer which in turn should make Y^3 as an integer!..i.e. 3y = + or - ( 3,6,9... ) which gives us Y = + or - ( 1,2,3... ) which ensures Y^3 as an integer!

Hence Ans: C!
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by jarmitage23 » Fri Sep 02, 2011 8:40 am
The first statement is not sufficient on it's own, because if y is 3 then y^3 is 27, but we can also conclude that it isn't an integer because if y^3 equals 30 than we know y isn't an integer because it would fall between 3 and 4. Statement 2 is not sufficient because y could be an integer (3*2=6); also y could be 1/3 because 3*(1/3) equals 1.

If the two statements are combined they are not sufficient, y can be either an integer or fraction. If y equals 2 then both 3y and y^3 are integers, but if y equals 4/3 then 3y is 4 and y^3 is 64/9 which is not an integer. If y^3 is an integer, lets say 10 then y would be the 1/3 root of ten which is some non-integer and 3y would be a non-integer. Therefore the answer is E.

Don't know if this is correct, but thought I would give it a shot.
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by knight247 » Fri Sep 02, 2011 9:03 am
jarmitage23 wrote:The first statement is not sufficient on it's own, because if y is 3 then y^3 is 27, but we can also conclude that it isn't an integer because if y^3 equals 30 than we know y isn't an integer because it would fall between 3 and 4. Statement 2 is not sufficient because y could be an integer (3*2=6); also y could be 1/3 because 3*(1/3) equals 1.

If the two statements are combined they are not sufficient, y can be either an integer or fraction. If y equals 2 then both 3y and y^3 are integers, but if y equals 4/3 then 3y is 4 and y^3 is 64/9 which is not an integer. If y^3 is an integer, lets say 10 then y would be the 1/3 root of ten which is some non-integer and 3y would be a non-integer. Therefore the answer is E.

Don't know if this is correct, but thought I would give it a shot.
After combining both how can we consider y=4/3 when Statement 1 specifically says that y³ is an integer? If y=4/3 then y³ would not be an integer.
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by prateek_guy2004 » Fri Sep 02, 2011 12:05 pm
Statement 1 and 2 both sufficient



IMOC
Don't look for the incorrect things that you have done rather look for remedies....

https://www.beatthegmat.com/motivation-t90253.html
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by jarmitage23 » Fri Sep 02, 2011 4:42 pm
Thanks for the explanation, I definitely over thought the question.
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by samyukta » Fri Sep 02, 2011 4:50 pm
jarmitage23 wrote:Thanks for the explanation, I definitely over thought the question.
As per CR language , u assumed something to prove which is already true & proved to be true.
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