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Variable Problems

Expert replies
by hyang4 » Sun Feb 15, 2009 2:08 pm
Hey beatthegmat community, I am new here and I am having problems figuring out these type of questions.

Do you guys have any good strategies in tackling these types of problems?

Much thanks in advance.



1. Is 6 a factor of n + 3?

(1) n is even and divisible by 3
(2) n is divisible by 6

OA:
D

-------------------------

2. If x and y are positive integers, is xy evenly divisible by 4?

(1) y + 2 is divisible by 4.
(2) x - 2 is divisible by 4.

OA:
C
Join the discussion
Source: — Data Sufficiency |

by cramya » Sun Feb 15, 2009 2:32 pm
1)

Q: Is 6 * some integer k = n+3

(1) n is even and divisible by 3

For 6 to be a factor of n+3 2 and 3 have both go to be its factors

n is even therfore n+3 is odd since even + odd is odd . 2 can never be a factor of an odd number. Therefor 6 is not a factor of n+3

SUFF

(2) n is divisible by 6

If n is divisible by 6 then n+6 is divisible by 6 . Therefore n+3 is not.
Therefor 6 is not a factor of n+3

SUFF

Choose D
Join the discussion

by cramya » Sun Feb 15, 2009 2:40 pm
q 2)

Stmt I

INSUFF No info on x

Stmt II

INSUFF No info on y

Together

y+2 is divisible by 4
x-2 is divisible by 4


Y+2 = 4K => Y = 4K-2 (where k is some integer)
x-2 = 4k => x = 4k+2
xy = (4k+2) (4k-2) = 16k^2 - 4 = 4(4k^2-1) which is always divisible by 4

Choose C
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by cramya » Sun Feb 15, 2009 2:43 pm
Q 2)

Another variation

Stmt I and Stmt II together

y+2 = 4k (where k is some integer) => y = 4k-2
x-2 = 4z(where z is some integer) => x = 4z+2

xy
= (4k-2) (4z+2)
= 16kz - 8z + 8k - 4 which is divisible by 4 since each of the terms involved 16kz,8z,8k and 4 are divisible by 4

SUFF
Join the discussion