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If y is an integer and y =lxl + x, is y = 0?

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If y is an integer and y =lxl + x, is y = 0? [#permalink] New post 03 Jun 2007, 19:42
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A
B
C
D
E

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Q19:
If y is an integer and y =lxl + x, is y = 0?
(1) x < 0
(2) y < 1

Q34:
What is the median number of employees assigned per project for the projects at Company Z?
(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project.
(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project.
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 [#permalink] New post 03 Jun 2007, 22:11
Q19: (D) for me :)

y =lxl + x and y integer

is y = 0?

From(1)
As x < 0, we have
y =lxl + x
<=> y = -x + x = 0

SUFF.

From(2)
As y < 1 and as y is an integer, y is forced to be equal to 0. Y cannot be negative, the minimum value of y is 0 reached when x < 0.

o If x > 0, then |x| + x > 0 and as y < 1 : there is no solution.
o If x =< 0, then |x| + x = 0 and y = 0.

SUFF.
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 [#permalink] New post 04 Jun 2007, 04:09
for me the first prob. is A and second is E
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 [#permalink] New post 04 Jun 2007, 07:17
C for Q34

Let there be 100 projects. Line them up, from smallest to largest and number them p1 to p100. (where p1 is the project with fewest employees, p100 with the most)

1) The largest 25% have >=4 people. so p75 through p100 have four or more employees. This says nothing about p50 and p51, which form the median. not suff.

2) The smallest 35% have less than 2 people, so p1 through p35 have 1 or 2 employees. This again says nothing about p50 and p51. not suff.

1 and 2) We know the projects are sorted by size, so if every project before p35 has less than 3, and every project after p75 has more than 3, then that means p36 - p74 must have 3 employees. This includes p50 and p51. So the median = 3 employees.

Answer C
  [#permalink] 04 Jun 2007, 07:17
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