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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

2

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

1

This post received KUDOS

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

2

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Expert's post

1

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.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...