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tricky DS

by sanjib » Wed Jun 24, 2009 6:20 am
If x and y are positive integers and x is a multiple of y, is y = 2?
(1) y ≠ 1
(2) x + 2 is a multiple of y.
the OA is c
Need help please
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Source: — Data Sufficiency |
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Re: tricky DS

by ssmiles08 » Wed Jun 24, 2009 6:34 am
sanjib wrote:If x and y are positive integers and x is a multiple of y, is y = 2?
(1) y ≠ 1
(2) x + 2 is a multiple of y.
the OA is c
Need help please
1) y can be 2 and x can be 4
y can be 3 and x can be 9
INSUFF

2) y = 1, and x = 3
y = 1 and x = 5
INSUFF

together:

if x increases by a factor of 2 and still is a multiple of y, then y = 2.

Also another way to look at it is: x/y = integer, (x+2)/y = integer.

(C)
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by david4431 » Wed Jun 24, 2009 9:55 am
Answer: C.

S1 is clearly not sufficient.

S2: (x+2)/y. Break the faction into two separate fractions. x/y and 2/y. x/y is an integer. In order for 2/y to be an integer, y must be 1 or 2. Insufficient.

Combined statements: y = 2 since y cannot equal 1.
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by rah_pandey » Wed Jun 24, 2009 10:46 pm
By the question
x=k*y ----(A)
now y<>1 is not sufficient

by second condition
x+2=k'*y

Use (A)
2=(k'-k)*y

this means 2 is a multiple of y so y can be 1 or 2 itself not sufficient

using both conditions together we get

y=2

Hence C
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by Domnu » Thu Jun 25, 2009 1:23 pm
Clearly, 1 and 2 alone aren't sufficient. Now, let

x = n1 * y

Then,

x + 2 = n2 * y,

for n1 and n2 integers. Now, we have

(n1 * y) + 2 = n2 * y ---> y * (n2 - n1) = 2 ---> n2 - n1 = 1

since y and n2 - n1 are integers and y isn't 1. So, y = 2.
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