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Bunuel
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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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chetan2u
Bunuel
If x = y + 2 and x is an even integer, which of the following could be the sum of x and y?

A. 140
B. 146
C. 148
D. 152
E. 156

Another way to look at it..
Both x and y are even but ONLY one will be divisible by 4 and other will be not, so their SUM will not be divisible by 4..
ONLY B is not div by 4
B

Hi chetan2u
I dont understand why only 1 will be divisible by 4 and other will not be. Can you please explain ?
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chetan2u
Bunuel
If x = y + 2 and x is an even integer, which of the following could be the sum of x and y?

A. 140
B. 146
C. 148
D. 152
E. 156

Another way to look at it..
Both x and y are even but ONLY one will be divisible by 4 and other will be not, so their SUM will not be divisible by 4..
ONLY B is not div by 4
B

Hi chetan2u
I dont understand why only 1 will be divisible by 4 and other will not be. Can you please explain ?

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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Bunuel
If x = y + 2 and x is an even integer, which of the following could be the sum of x and y?

A. 140
B. 146
C. 148
D. 152
E. 156

Given that x is even then y must be even.

x + y = y + 2 + y = 2y +2

2 (y+1) = even number

try \(\frac{140}{2} = 70 - 1 = 69\) not even not possible.

Try \(\frac{146}{2} = 73 - 1 = 72\) even and makes x = 74 possible.

\(\frac{148}{2} = 74 - 1 = 73\) not possible

\(\frac{152}{2} =76 - 1 = 75\) not possible.

\(\frac{156}{2} = 78 - 1 = 77\) not possible.

Only option B
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Bunuel
If x = y + 2 and x is an even integer, which of the following could be the sum of x and y?

A. 140
B. 146
C. 148
D. 152
E. 156

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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x is even, so x=2k
x=y + 2 --> y=2k - 2
this means:
x + y = 2k + 2k -2 = 4k -2

only 146 satisfies 4k - 2.
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Bunuel
If x = y + 2 and x is an even integer, which of the following could be the sum of x and y?

A. 140
B. 146
C. 148
D. 152
E. 156

x is an even integer, which is derived after adding 2 with y. So, y is an even integer too, and they are consecutive even integers.

Thus their sum is= y+y+2=2y+2

Now, trying the answer options:

A. 2y+2=140;2y=138; y=69 Odd (out)

B. 2y+2=146;2y=144; y=72 Even, x=74, Total 146.

The answer is B.
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