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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, 13:09

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, 20:07

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, 09:24

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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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, 12:40

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)

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