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08 Apr 2005, 10:17
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A list contains 11 consecutive integers. Whats is the greatest integer on the list?
1) If X is the smallest integer on the list, then (x+72)^(1/3)=4
2) If X is the smallest integer on the list, and Y is the greatest integer on the list, then 16 x^(-2)=y^(-2)
1) If X is the smallest integer on the list, then (x+72)^(1/3)=4
2) If X is the smallest integer on the list, and Y is the greatest integer on the list, then 16 x^(-2)=y^(-2)
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08 Apr 2005, 11:51
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I hope your X and x are referring to the same integers and the variation isn't intentional. In GMAT I'd consider them different.
(1) Specifies x+72 = 64 =>x = smallest integer in the list = -8.
This gives the largest integer as +2.
(2) Here y = x+10
Therefore, 16x^(-2) = (x+10)^(-2)
16 x^2 + 1600 + 320x = x^2
15x^2 + 320x + 1600 = 0
3x^2 + 64x + 340 = 0
3x^2 + 34x + 30x + 340 = 0
x(3x + 34) + 10(3x+34) = 0
x = -10, x=34/3.
Since x is an integer, 34/3 is inadmissible, and so we get a unique value of x here too.
Hence solution possible by either statement independently.
(D).
(1) Specifies x+72 = 64 =>x = smallest integer in the list = -8.
This gives the largest integer as +2.
(2) Here y = x+10
Therefore, 16x^(-2) = (x+10)^(-2)
16 x^2 + 1600 + 320x = x^2
15x^2 + 320x + 1600 = 0
3x^2 + 64x + 340 = 0
3x^2 + 34x + 30x + 340 = 0
x(3x + 34) + 10(3x+34) = 0
x = -10, x=34/3.
Since x is an integer, 34/3 is inadmissible, and so we get a unique value of x here too.
Hence solution possible by either statement independently.
(D).
09 Apr 2005, 05:26
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D)...
1) (x+72)^(1/3)=4 => sqrt(x+72)=4 => x+72=64 => x=-8 => suff
2) 16 x^(-2)=y^(-2) => 16/(x^2)=1/(y^2) => 16y^2=x^2 => 4y=x and y=x+10 => solve for x and y => suff
1) (x+72)^(1/3)=4 => sqrt(x+72)=4 => x+72=64 => x=-8 => suff
2) 16 x^(-2)=y^(-2) => 16/(x^2)=1/(y^2) => 16y^2=x^2 => 4y=x and y=x+10 => solve for x and y => suff
