\(x=y+2\) ---->if \(x\)-even then \(y\) must be even too

\(x+y = y+y+2 = 2y+2 = 2(y+1) = 2(even+1) = 2*odd\)

\(\frac{x+y}{2} = odd\)

A) 140/2 = 70 even
B) 146/2 = 73 odd
C) 148/2= 74 even
D) 152/2 = 76 even
E) 156/2 = 78 even

Answer B

Fabulous..!! chetan2u
+1 to you.

Here is what i did-->
I took the long route.
x-> Even
So y->Even

Option 1->
x+y=140
2y+2= 140
2y=138
y=> odd => Rejected

Option 2->
x+y=146
2y=144
y=72
Hence x=74
Acceptable.
Hence B
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Hi..

You pick up any two consecutive even integer, one will be multiple of 4 & other not..
Say you pick y=2, x=2+2=4..... Only x is div by 4
Pick y as 4, x=4+2=6..... Only y is MULTIPLE of 4
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We see that x + y must be the sum of 2 consecutive even integers, or y + (y + 2) = 2y + 2. We also see that both x and y are even integers.

If we subtract 2 from each answer choice we have:

138, 144, 146, 150, and 154

When we divide those answers by 2, the only one that provides an even quotient is 144, since 144/2 = 72.

Alternate Solution:

We know that x = y + 2, and x is even, so we know that y is also even. By substitution, we see that x + y = (y + 2) + y = 2y + 2.

Let’s check each answer choice. We are looking for an answer choice that yields an even integer value for y:

Choice A: Does 2y + 2 = 140 yield an even integer value for y? No. In this case, y = 69.

Choice B: Does 2y + 2 = 146 yield an even integer value for y? Yes! In this case, y = 72.

Answer: B
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