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28 Oct 2019, 09:07
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C
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If x and y are integers and xy = 6, what is the value of x?
(1) x > y + 1
(2) x = 6y
(1) x > y + 1
(2) x = 6y
Chetan's questions
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28 Oct 2019, 09:55
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Will go with C
1. x> 6/x + 1. Simplify !
---> [(x-3)(x+2)]/x > 0 so the range is x is between -2 and 0 i.e. x = -1 or x = more than 3 i.e. x= 6 . Not sufficient
2. x^2 = 36 ----> x = 6 or -6 . Insufficient
Combining 1 and 2 ---> unique x is 6.
1. x> 6/x + 1. Simplify !
---> [(x-3)(x+2)]/x > 0 so the range is x is between -2 and 0 i.e. x = -1 or x = more than 3 i.e. x= 6 . Not sufficient
2. x^2 = 36 ----> x = 6 or -6 . Insufficient
Combining 1 and 2 ---> unique x is 6.
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28 Oct 2019, 14:59
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chetan2u
If x and y are integers and xy = 6, what is the value of x?
(1) x > y + 1
(2) x = 6y
(1) x > y + 1
(2) x = 6y
Chetan's questions
Since x and y are integers, they can take the values of (2,3),(6,1),(-2,-3),(-6,-1) for xy = 6.
Statement 1, x > y + 1. x can be either -1 or 6. NOT SUFFICIENT.
Statement 2, x = 6y. x can be 6 or -6. NOT SUFFICIENT.
Combining both statement, 6 is the single value that meets the requirements. SUFFICIENT.
ANSWER IS C.
