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Re: if x and y are integers and 2<x<y, does y =16 [#permalink]

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09 Mar 2013, 12:09

I know that each statements are insufficient by itself. But i think Even taken together they are insufficient. We know that GCF*LCM = X*Y So X*Y = 96 Hence , when y = 24 , x =4 Similarly when y = 16 , x = 6 So i think both statements are insufficient taken together.

If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2. (2) The LCM of X and Y is 48.

I know that each statements are insufficient by itself. But i think Even taken together they are insufficient. We know that GCF*LCM = X*Y So X*Y = 96 Hence , when y = 24 , x =4 Similarly when y = 16 , x = 6 So i think both statements are insufficient taken together.

Archit

Notice that the greatest factor of 24 and 4 is 4, not 2 as given in the second statement.

Re: if x and y are integers and 2<x<y, does y =16 [#permalink]

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19 Oct 2013, 19:07

Archit143 wrote:

I know that each statements are insufficient by itself. But i think Even taken together they are insufficient. We know that GCF*LCM = X*Y So X*Y = 96 Hence , when y = 24 , x =4 Similarly when y = 16 , x = 6 So i think both statements are insufficient taken together.

Archit

If GCF is 2 and LCM is 48, than the Pair of X and Y can be: X=2,Y=48 and X=6,Y=16 and X=16, Y=6 and X=48, Y=2

I know that each statements are insufficient by itself. But i think Even taken together they are insufficient. We know that GCF*LCM = X*Y So X*Y = 96 Hence , when y = 24 , x =4 Similarly when y = 16 , x = 6 So i think both statements are insufficient taken together.

Archit

If GCF is 2 and LCM is 48, than the Pair of X and Y can be: X=2,Y=48 and X=6,Y=16 and X=16, Y=6 and X=48, Y=2

Statement #1: The GCF of x and y is 2 This leave open a wide array of possibilities. All we know is that x and y are two even numbers, both bigger than 2, with no common factors other than two: they could be x = 4, y = 6 x = 6, y = 8 x = 6, y = 10 x = 6, y = 16 So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.

Statement #2: The LCM of x and y is 48 Without any other information, we could have x = 3, y = 16 x = 4, y = 48 So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.

Combined: this is where it gets interesting. The GCF of x and y is 2 The LCM of x and y is 48 This is a tricky combination. First, let's list all the factors of 48 --- in order to have a LCM of 48 with another number, each number must be a factor of 48. factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Those are the possible candidates for x & y. We can eliminate 1 & 2, because x > 2, and we can eliminate 3, because that cannot have a GCF of 2 with anything else. Possibilities for x & y = {4, 6, 8, 12, 16, 24, 48} If y = 48, then every other number in the set is factor of 48, so the GCF would be the smaller number --- e.g. the GCF of 6 and 48 is 6. Therefore, we can't use 48. If y = 24, then the first four numbers are factors of 24, so they don't work, and the GCF of 16 & 24 is 8. Therefore, we can't use 24. Possibilities for x & y = {4, 6, 8, 12, 16} Suppose y = 16 x = 4, y = 16 ===> GCF = 4, doesn't work x = 6, y = 16 ===> GCF = 2 --- this is one possible pair!! x = 8, y = 16 ===> GCF = 8, doesn't work x = 12, y = 16 ===> GCF = 4, doesn't work Suppose y = 8 x = 4, y = 8 ===> GCF = 4, doesn't work x = 6, y = 8 ===> GCF = 2, but LCM = 24, doesn't work Suppose y = 6 x = 4, y = 6 ===> GCF = 2, but LCM = 12, doesn't work So, after all that, the only pair that satisfies both statements is x = 6, y = 16, so it turns out, y does in fact equal 16. We are able to give a definitive answer to the prompt question, so the combined statements are sufficient.

Statement #1: The GCF of x and y is 2 This leave open a wide array of possibilities. All we know is that x and y are two even numbers, both bigger than 2, with no common factors other than two: they could be x = 4, y = 6 x = 6, y = 8 x = 6, y = 10 x = 6, y = 16 So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.

Statement #2: The LCM of x and y is 48 Without any other information, we could have x = 3, y = 16 x = 4, y = 48 So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient.

Combined: this is where it gets interesting. The GCF of x and y is 2 The LCM of x and y is 48 This is a tricky combination. First, let's list all the factors of 48 --- in order to have a LCM of 48 with another number, each number must be a factor of 48. factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Those are the possible candidates for x & y. We can eliminate 1 & 2, because x > 2, and we can eliminate 3, because that cannot have a GCF of 2 with anything else. Possibilities for x & y = {4, 6, 8, 12, 16, 24, 48} If y = 48, then every other number in the set is factor of 48, so the GCF would be the smaller number --- e.g. the GCF of 6 and 48 is 6. Therefore, we can't use 48. If y = 24, then the first four numbers are factors of 24, so they don't work, and the GCF of 16 & 24 is 8. Therefore, we can't use 24. Possibilities for x & y = {4, 6, 8, 12, 16} Suppose y = 16 x = 4, y = 16 ===> GCF = 4, doesn't work x = 6, y = 16 ===> GCF = 2 --- this is one possible pair!! x = 8, y = 16 ===> GCF = 8, doesn't work x = 12, y = 16 ===> GCF = 4, doesn't work Suppose y = 8 x = 4, y = 8 ===> GCF = 4, doesn't work x = 6, y = 8 ===> GCF = 2, but LCM = 24, doesn't work Suppose y = 6 x = 4, y = 6 ===> GCF = 2, but LCM = 12, doesn't work So, after all that, the only pair that satisfies both statements is x = 6, y = 16, so it turns out, y does in fact equal 16. We are able to give a definitive answer to the prompt question, so the combined statements are sufficient.

Answer = (C)

Does all this make sense? Mike

No need to test that many cases, let's see

x,y are positive integers and 2<x<y. Is y=16?

First statement GCF (x,y) is 2

Well we could have:

x=4, y=6 answer is NO or x=6, y=16 answer is YES

Hence insufficient

Second Statement LCM (x,y) is 48 48 = 2^4 * 3

Now we can test cases here too: Either x=3 and y=2^4 when answer is YES OR x=7 y = 48 answer is NO

Both together Since GCF =2 and LCM = 48 and since 2<x<y we can only have x=6, y=16

Hence C is the correct answer Please ask if anything remains unclear Cheers J

If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2. (2) The LCM of X and Y is 48.

Given: 2 < X < Y Question: Does Y = 16?

(1) The GCF of X and Y is 2. X = 2a; Y = 2b (a and b are co prime integers) Y may or may not be 16 e.g. X = 6, Y = 16 OR X = 6, Y = 8 etc

(2) The LCM of X and Y is 48. \(48 = 2^4 * 3\) One of X and Y must have 2^4 = 16 as a factor and one must have 3 as a factor. Again, Y may or may not be 16 e.g. X = 6, Y = 16 OR X = 1, Y = 48 etc

Using both together, X = 2a, Y = 2b. Since a and b need to be co-prime and both X and Y need to be greater than 2, one of a and b must be 3 and the other must be 8 (since X and Y already have a 2 to make 16). X = 6, Y = 16 (since X is less than Y). Y must be 16.

If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]

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10 Jan 2016, 08:24

If x and y are integers and \(2 < x < y\), does \(y = 16\) ? (1) The GCF of X and Y is 2.

since we have many solutions such as \((6, 8)\), \((6,16)\), etc.

we cannot say whether \(y=16\) or not.

Statement 1 is insufficient. (2) The LCM of X and Y is 48.

since we have many solutions such as \((16, 24)\), \((3,16)\), etc.

we cannot say whether \(y=16\) or not.

Statement 2 is insufficient.

Combining 1 and 2,

we know that \(LCM * GCF\)= product of the integers \(x * y\)

so \(xy=96=2^5*3\)

since GCF is 2, we know that both are even numbers indicating a single 2 in both x and y.

Thereby we get values of \((x,y)\) as \((6, 16)\), \((8, 12)\) and \((4, 24).\)

out of these 3 pairs only \((6,16)\) has \(LCM\) of \(48\).

So \(y=16\) and thus \(C\) is the correct answer.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2. (2) The LCM of X and Y is 48.

In the original condition, there are 2 variables(x,y) and 1 equation(2<x<y), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1equation, which is likely to make D the answer. For 1), (x,y)=(4,6) -> no, (x,y)=(6,16) -> yes, which is not sufficient. For 2), (x,y)=(3,48) -> no, (x,y)=(6,16) -> yes, which is not sufficient. When 1) & 2), (x,y)=(6,16) -> yes, which is sufficient. Therefore, the answer is C.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x and y are integers and 2 < x < y, does y = 16 ? [#permalink]

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12 Oct 2016, 11:40

Archit143 wrote:

If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2. (2) The LCM of X and Y is 48.

FROM 1 GCF is 2 ( common prime factor with the lowest power )

(x,y) could be (4, 28) , (6,10) (14,16) ... insuff

from 2

LCM = (2^4 *3) ( factors unique to each integer * common prime factor to the largest power) ... insuff

both

since 2 is the GCF then one one of the 2 numbers has 2^1 and the other 2^4 and we need to know where the 3 factor belongs ( it only belongs to one of the two numbers since GCF is 2) . from stem 2<x<y , thus the smallest of x and y has to be bigger than 2^1 by the 3 factor , thus (x,y) = (6,16)

gmatclubot

Re: If x and y are integers and 2 < x < y, does y = 16 ?
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12 Oct 2016, 11:40

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