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# IF X and Y are positive integers, is (4^X) (1/3)^Y < 1?

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IF X and Y are positive integers, is (4^X) (1/3)^Y < 1? [#permalink]

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15 Jun 2005, 15:48
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IF X and Y are positive integers, is (4^X) (1/3)^Y < 1?

(A) Y= 2X

(B) Y = 4
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15 Jun 2005, 17:08
A = insufficient

0 and positive numbers are fine. But the equation is greater then 1 if x and y are negative.

B = insufficient

X could = a HUGE number or a very small number

A and B = Sufficient

If Y = 4 then x must equal 2. Only one solution.
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15 Jun 2005, 18:00
TooLegitToQuit wrote:
IF X and Y are positive integers, is (4^X) (1/3)^Y < 1?

(A) Y= 2X

(B) Y = 4

A. 1)(4/9)^X < 1 if X >1 suff

2) is insuff since X can vary
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15 Jun 2005, 18:45
I will go with A
Statement 1: (2/3)^2x < 1 for any positive integer
Statement 2: Depending on x's value it could be either.
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15 Jun 2005, 19:49
The OA is A.

I was confused because I read first statement as X = 2Y... doh~
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15 Jun 2005, 19:53
A) If x = 1, y = 2, then 4(1/9) < 1
If x = 2, y = 4, then 16(1/81) < 1

A is sufficient

B) is not sufficient since no information about x is given.

Ans: A
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21 Jun 2005, 18:43
tkirk32 wrote:
A = insufficient

0 and positive numbers are fine. But the equation is greater then 1 if x and y are negative.
.

The question stem specifically states that both X and Y are positive integers. So you cannot use negative numbers as examples.

A is thus sufficient because you'll always get fractions with values < 1.

Test using Y= 1,2,3,4...etc to see it.
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21 Jun 2005, 18:47
sparky wrote:
TooLegitToQuit wrote:
IF X and Y are positive integers, is (4^X) (1/3)^Y < 1?

(A) Y= 2X

(B) Y = 4

A. 1)(4/9)^X < 1 if X >1 suff

2) is insuff since X can vary

Sparky that was good way to approach this problem.....

Yes it is A
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21 Jun 2005, 20:31
S1

4^X * (1/3)^2X = 2^2X / 3^2X = (2/3)^2X which is < 1 when X is a positive integer.

So Strike out B,C,E

S2

4^X * 1/3^4 = 4^X/81.This is < 1 if 1 <= X <= 3 and > 1 if X > 3

Strike out D

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22 Jun 2005, 02:31
got C
st1 Y=2X does not give any value for x and y
insufficient

St2 y=4 DOES Not tell much except 1/3^4 = 1/81
that a may be answer depending on x
hence insufficient

TOgether yes we know Y= 4 AND Y/2 = x hence x=2
HENCE we can plug back the x and y values to answer the question
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22 Jun 2005, 03:03
TooLegitToQuit wrote:
The OA is A.

I was confused because I read first statement as X = 2Y... doh~

it seems I am wrong can anyone explain me why the first statement is sufficient thanks
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22 Jun 2005, 11:36
ywilfred's explanation covers what I did but here's just detailed breakdown:

-from the stem, we know X and Y are positive so we only need to test pos numbers

(A) Y= 2X

-from this, we know Y is twice X. I used (x=1,y=2) and(x=2,y=4). You can select any number so long as it fits (A). (x=5,y=10) etc. But small numbers are easier to work with.

Quote:
A) If x = 1, y = 2, then 4(1/9) < 1
If x = 2, y = 4, then 16(1/81) < 1

A is sufficient

Now the only possible choices are A and D.

(B) Y=4

-this says nothing about X so it's insufficient

The logic you are following (two unknowns require two equations) doesn't apply here because we're not asked to find a specific value for either X or Y. We only need to know if

(4^X) (1/3)^Y < 1

And from (A), we know that the left side of this inequality will always be less than 1.
22 Jun 2005, 11:36
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