If x is an even number, is y also an even number?

Question

(1) a/(x + y) = 2.1, where a is a positive integer.

(2) a(x + y) = 210, where a is a positive integer.

chetan2u · Answer

hi..

(1) a/(x + y) = 2.1, where a is a positive integer.
so  a=(x+y)*2.1.. 
 since a is positive integer x+y has to be a multiple of 10..
 so if x is EVEN, y has to be EVEN as ODD + even cannot be 10  
sufficient

(2) a(x + y) = 210, where a is a positive integer
 a=10, x=2, y=19...NO
a=19,x=2,y=8...Yes 
insuff

A

mohshu · Answer

stat2 : not suff
depends on the value of a and y...

stat 1: a = 2.1 * (x+y)
given a is integer,,, we shud be removing the decimal from 2.1,, hence multiply by 10 atleast...
if x is even hen y has to be even,,, and x cannot be odd

hence suff..

ans A

GTExl · Answer

1) a/(x + y) = 2.1

a/(x + y)= 21/10

10a=21(x+y) that means 21(x+y) = even number

21=odd so (x+y) should be even to make 21*(x+y) an even number

as x is even so y should also be even

sufficient

2) a(x + y) = 210

a(x + y) =even number
 
3 possible cases

i) a is even, (x+y) is odd (x+y will be odd if y is odd)

ii) a is odd, (x+y) is even (x+y will be even if y is even)

iii) a is even ,(x+y) is even (x+y will be even if y is even)

so whether y is even or odd we can get an even number

not sufficient.

Answer is A

gps5441 · Answer

1) a/x+y=2.1=21/10 ie x+y=10 now x is even ie 2, 4, 6, 8 only hence y can only be 8, 6, 4, 2-even sufficient
2) a(x+y)=210=2x105 or 2x5x21 or 2x5x7x3 
a can be 2 and x+y can be (105)
a can be 10 and x+y can be (21)
ie multiple combinations of odd even is possible-not sufficient

leanhdung · Answer

From (1), we have: 10a = 21(x+y)

Let a = 6, x = 2, y = 60/21 - 2 = 6/7 --> y is not even number.

How can we deduce that (1) is sufficient ? Does the stimulus assume that  y is integer  ?

If x is an even number, is y also an even number?

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