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Re: DS - Is 2x a multiple of y [#permalink]
10 Aug 2011, 13:21

allabout wrote:

This one isn't clear to me...Like to explain why it is E and not A

IF x and y are positive integers, is 2x a multiple of y? a. 2x + 2 is a multiple of y b. y is a multiple of x

From a:

If x = 2 and y = 3, (2x+2) is a multiple of y but not 2x. If x = 2 and y = 2, (2x+2) and 2x are multiple of y.

So not sufficient.

From b: y is a multiple of x means x is equal to or smaller than y.

If y = x = 3, 2x is a multiple of y. If y = 3 and x = 1, 2x is not a multiple of y. not sufficient.

From a and b: (2x + 2) is a multiple of y and y is a multiple of x.

If x = 2 and y = 2, (2x+2) is multiple of y and y is also a multiple of x. So 2x is also a multiple of y. If y = 6 and x = 2, (2x+2) is multiple of y and y is also a multiple of x but 2x is not a multiple of y.

Re: DS - Is 2x a multiple of y [#permalink]
10 Aug 2011, 15:27

statement 1. just means that y is even, but it does not tell us anything about the direct relationship between x and y. Simplified is really saying 2x + 2 =yk , k being an unknown multiplication of y

as the earlier post did, plug numbers.

2(2) + 2 = yk : when x is 2, y can be 2 multiplied 3 times, and so 2x will be a multiple of y BUT we could also assume y is 3 multiplied 2 times, in which case 2x ie. 4 is not a multiple of 3. this gives a yes/no answer so its insufficient.

2(x) + 2 = y (k times) 2(2) + 2 = 2 (3 times) ...........4 is a multiple of 2 BUT 2(2) + 2 = 3 (2 times)...........4 is not a multiple of 3

Re: DS - Is 2x a multiple of y [#permalink]
13 Aug 2011, 15:10

1

This post received KUDOS

x, y > 0, integers

"Is \frac{2x}{y} an integer?" "Is 2x a multiple of (divisible by) y?" "Is y a factor of 2x?"

Statement 1) This statements says that \frac{2x + 2}{y} = \frac{2(x+1)}{y} is an integer. In other words, 2(x+1) is a multiple of y.

In order for both 2x and 2(x+1) to be divisible by y, y must be either 1 (factor of 2 and both x and (x+1)) or 2 (factor of 2). Thus if y equals 1 or 2, then \frac{2x}{y} is an integer. If y has any other value under this constraint, \frac{2x}{y} is not an integer.

You can also break this statement into two expressions: \frac{2x + 2}{y} = \frac{2x}{y} + \frac{2}{y}. If \frac{2x}{y} + \frac{2}{y}=integer, then in order for \frac{2x}{y} to be an integer, \frac{2}{y} must also be an integer. Thus if y equals 1 or 2, then \frac{2x}{y} is an integer. If y has any other value, \frac{2x}{y} is not an integer.

Insufficient.

Statement 2) This statement says that \frac{y}{x} is an integer. So y = x*k, where k is some integer.

In order for \frac{2x}{y} = \frac{2*x}{k*x} to be an integer, \frac{2}{k} must be an integer, ie k must equal either 1 or 2. Thus if y = 2x or y = x, 2x will be a multiple of y. With any other value of k, 2x will not be a multiple of y.

Insufficient.

Combined) From statement 1, \frac{2x}{y} is an integer when y={1,2}, and not an integer when y has another value. From statement 2: \frac{2x}{y} is an integer when y=2x or y=x, and not an integer when y is a different multiple of x than 2.

So under the constraints of both statements, \frac{2x}{y} is an integer when (x,y) = (1,1), (2,2), or (1,2).

It only takes one example that follows the two constraints but is not one of these 3 possibilities to prove the combined statements insufficient. A couple examples are (x,y) = (1,4), or (2,6).

Re: Is 2x a multiple of y? [#permalink]
31 May 2012, 00:24

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kashishh wrote:

Is 2x a multiple of y?

(1) 2x+2 is a multiple of y (2) y is a multiple of x

stmnt1:

If y= 1 x=2 then 2x+2 = 6 and multiple of y. 2x =4 and multiple of y but if y=3 and x=2 then 2x+2=6 and multiple of y but 2x=4 and not a multiple of y. Hence insuff

stmnt 2:

y = 2 and x= 1 then 2x is multiple of y y= 3 and x= 1 then 2x is not multiple of y. Hence insuff

taking together x=1 and y =2 we have y is multiple of x, 2x+2 = 4 multiple of y and 2x= 2 is multiple of y (2) . x=1 and y =4 we have y is multiple of x, 2x+2 = 4 multiple of y but 2x= 2 is not multiple of y (4)

Re: Is 2x a multiple of y? [#permalink]
15 Jul 2014, 05:54

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Re: Is 2x a multiple of y? [#permalink]
15 Jul 2014, 22:29

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Expert's post

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This post was BOOKMARKED

kashishh wrote:

Is 2x a multiple of y?

(1) 2x+2 is a multiple of y (2) y is a multiple of x

Use some logic and some number plugging:

Question: Is 2x a multiple of y?

(1) 2x+2 is a multiple of y

(2x + 2) is a multiple of y. 2x and (2x+2) are two consecutive even numbers and hence share only 2 factors: 1 and 2 So 2x will also be a multiple of y only if y is 1 or 2. Not sufficient.

(2) y is a multiple of x Here are the multiples: x....y....2x If y is x or 2x, 2x will be a multiple of y, else it will not be. Not sufficient.

Using both together: If y = 2 and x = 2, all conditions are met and 2x IS a multiple of y. If y = 6 and x = 2, all conditions are met but 2x IS NOT a multiple of y. Not sufficient.

Re: Is 2x a multiple of y? [#permalink]
17 Jul 2014, 02:26

Statment 1 tells that y is either 1 or 2. Statment 2 shows that y is 2 as its a multiple of x and 1 cant be a multiple of any number, thats what understand, tell me where am going wrong!

Re: Is 2x a multiple of y? [#permalink]
17 Jul 2014, 05:03

Expert's post

sunaimshadmani wrote:

What if x=1 and y=2..it satisfies all the conditions??

sunaimshadmani wrote:

Statment 1 tells that y is either 1 or 2. Statment 2 shows that y is 2 as its a multiple of x and 1 cant be a multiple of any number, thats what understand, tell me where am going wrong!

From (1), from 2x+2 is a multiple of y, we don't have that y is either 1 or 2. It can take many other values.

Is 2x a multiple of y?

(1) 2x+2 is a multiple of y. If x=1 and y=1, then the answer is YES but if x=1 and y=4, then the answer is NO. Not sufficient.

(2) y is a multiple of x. If x=1 and y=1, then then the answer is YES but if x=1 and y=4, then the answer is NO. Not sufficient.

(1)+(2) The same here: if x=1 and y=1, then then the answer is YES but if x=1 and y=4, then the answer is NO. Not sufficient.