Is the sum of 7x and 7y divisible by 14?

(1) x = y
(2) x and y are both even integers

(7x + 7y )/14 = 7(x + y)/14 = (x + y)/2

Effectively question becomes is x + y a multiple of 2 .

1 ) x =y --> 2y/2 or 2x/2 = both are multiple of 2 .hence sufficient

2) x,y are both even integer --> both will be divisible by 2
hence sufficient

D ans
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Question stem:
\(\frac{7x+7y}{14}\)
\(=> \frac{7(x+y)}{14}\)
\(=> \frac{x+y}{2}\)
So basically we need to find whether x+y is even

S1: \(x = y\)
This will work if x and y are integers. And will not work otherwise.
eg:
if \(x=3, y=3\)
\(=> x+y = 3+3 = 6\) Even
But if
\(x=0.75, y=0.75\)
\(=> x+y = 0.75+0.75 = 1.50\) Not even
Hence Insufficient.

S2: \(x\) and \(y\) are both even integers
This will always work. \(x+y\) will always yield an even integer
\(even+even = even\)
Hence Sufficient.

Answer B

The question asks whether 7x+7y=14p , p \(\in\) non negative integer ---> x+y=2p = even integer?

Per statement 1, x=y . "yes" to "x+y=2p = even integer" if x=y=1 but "no" for "x+y=2p = even integer" if x=y=1/3. Not sufficient.

Per statement 2, x and y both are even integer ---> clearly sufficient.

B is the correct answer.
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Hi All,

This DS question, like many DS questions on Test Day, is meant to test the 'thoroughness' of your thinking. It's important to think about what you KNOW about the variables and what you DON'T know...

We're asked if the sum of 7X and 7Y is divisibly by 14. This is a YES/NO question. It's important to remember that we were NOT told anything about X and Y (maybe they're integers, but maybe they're not). TESTing VALUES will prove to be quite useful here.

1) X = Y

This Fact tells us that the two variables are equal to one another, but we don't know what they actually are.

IF...
X = Y = 1/2
Then 7(1/2) + 7(1/2) = 7 and the answer to the question is NO.

IF...
X = Y = 1
Then 7(1) + 7(1) = 14 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) X and Y are both even integers

This Fact gives us a very specific restriction - X and Y are BOTH EVEN integers. With this restriction, ANY possible values that you choose for X and Y will lead to a total that is divisible by 14.

For example...
IF...
X = Y = 2
Then 7(2) + 7(2) = 28 and the answer to the question is YES.

IF...
X = 0
Y = 2
Then 7(0) + 7(2) = 14 and the answer to the question is YES.

IF...
X = -2
Y = -4
Then 7(-2) + 7(-4) = -42 and the answer to the question is YES.
Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi Rich,

I'm also a member of your website so I follow your posts quite frequently. One question I had was why we can't use the scenario for statement 2 with x=0 and y=0. There's nothing in statement two which says that we can't set both integers equal to 0. In that case 7(0) + 7(0) equals 0 which is not divisible by 14. Therefore, my thought is that B would not be sufficient and hence the correct choice would be (E).

I would love to hear your feedback when you have a moment.

--BH

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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Hi quietreader,

You CAN use X=0 and Y=0 for Fact 1, Fact 2 or both. 0 IS divisible by 14, so the answer to the question would be YES.

As a broader Number Property worth knowing, 0 is divisible by EVERY number (barring 0/0, which is a concept that you won't see on the GMAT).

GMAT assassins aren't born, they're made,
Rich
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Hi Rich,

Why Can't we try with 2 and -2 ?

Thanks And Regards
NSS
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Hi NandishSS,

For Fact 2, you CAN use 2 and -2. When you TEST those two values, the answer to the question is still YES.

GMAT assassins aren't born, they're made,
Rich
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7(x+y) mod 14 = 0
x+y mod 2 = 0

1) x=y
x=y=1 ; ans = yes
x=y=1/2 ; ans = no

2) x and y are even so 2 can is a common divisor of both, hence can be taken out of the brackets making the entity outside brackets as 14. Hence, sufficient.
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ha! tricky one..you got me here...
never assume that you deal with integers on DS...never!!!

Hi mvictor,

When working through DS questions, before jumping to the two Facts, I always write down whatever information I've been given and the specific question that I'm attempting to answer. In addition, if I have not been given ANY information about the variables in the prompt, then I will physically write down that information.... "X, Y can be ANYTHING." In that way, when I'm dealing with each of the Facts and doing whatever work is required, I have a constant reminder (on the PAD) that I need to consider more than just the obvious possibilities (re: positive integers).

GMAT assassins aren't born, they're made,
Rich
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thanks..i sometimes forget about this one..will definitely write on my scratch paper just the way you do.

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