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Expert replies
Source: — Data Sufficiency |

by singhpreet1 » Fri Jun 11, 2010 8:23 pm
f2001290 wrote:Please explain this ..
hi..

4^x(1/3)^y<1

St.1 : y=2x: this only tells me what i can use to substitute y or x in the equation, therefore not sufficient.
St.2: x=4: this tells me that x=4, but no value for y is available to solve the equation.

Now if i use both the equations, i know the value of x and can substitute for y and solve the equation, i.e.

4^4(1/3)^8<1, this is sufficient to answer the problem, remember we do not need to provide the answer, only see if it possible to arrive at a conclusion wit the data given.

FREE HUGS,
Preet
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by selango » Mon Jun 14, 2010 10:25 pm
Rephrasing the equation,we get

4^x.3^(-y)<1

-->4^x/3^y<1

From stmt 1 ,y=2x

-->4^x/3^2x=4^x/9^x

x can be any number,so insufficient.

From stmt 2,y=4

4^x/3^4<1

Again x can be any number,so insufficient.

Combining 1 and 2,we know x=2,sufficient.


Hence C
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by kvcpk » Tue Jun 15, 2010 3:24 am
I believe that the answer should be A.

Because, (4/9)^x will be <1 for all possible positive integer values of X.

x range is 1,2,3,4,.......

contrasts OA......
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by selango » Tue Jun 15, 2010 5:49 am
I am also having the same doubt.

If we rephrase the equation,


-->4^x/3^y<1

4^x<3^y

from stmt 1,y=2x

4^x<3^2x

4^x<9^x

Since x is +ve,whatever value of x,4^x will be less than 9^x.

Hence A is sufficient.
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by GMATGuruNY » Tue Jun 15, 2010 6:34 am
With DS problems, I always ask myself the following three questions:

What do I want?
What do I have?
What do I need?


The goal is not to solve but only to determine whether the statement gives you sufficient information to solve. In other words: Is the statement giving me what I need?

In this case:

What do I want? To know whether (4^x)[(1/3)^y] < 1
What do I have? Before I look at the two statements, what information have I been given? In this case, that x and y have to be positive integers.
What do I need? In this case, more information about x and y.

Since the answer to the question "Is (4^x)[(1/3)^y] < 1?" can be yes or no, I need to do the following:

Try different kinds of numbers in order to see whether the answer to the question stays consistently YES or NO or could go either way.

As I try out different kinds of numbers:

If the answer to the question stays consistently YES, the statement is SUFFICIENT.
If the answer to the question stays consistently NO, the statement is SUFFICIENT.
If the answer to the question can be YES or NO, the statement is INSUFFICIENT.


Important: I have no stake in whether the answer is yes or no, as long as it stays consistent; if the answer is always YES or always NO, the statement is SUFFICIENT.

Statement 1:

If x = 1, y = 2. Is (4^x)[(1/3)^y] < 1? Yes, because (4^1)[(1/3)^2] = 4/9, and 4/9 < 1.
If x = 2, y = 4. Is (4^x)[(1/3)^y] < 1? Yes, because (4^2)[(1/3)^4] = 16/81, and 16/81 < 1.
If x = 10, y = 20. Is (4^x)[(1/3)^y] < 1? Yes, because (4^10)[(1/3)^20] = a messy fraction, so (4^10)[(1/3)^20] < 1.

Since the answer stays consistently YES, the statement is SUFFICIENT.

Statement 2:
Tells us only that y = 4, but nothing about x.

If x = 2 and y = 4, is (4^x)[(1/3)^y] < 1? Yes, because (4^2)[(1/3)^4] = 16/81, and 16/81 < 1.
If x = 4 and y = 4, is (4^x)[(1/3)^y] < 1? No, because (4^4)[(1/3)^4] = 256/81, and 256/81 > 1.

Since the answer can be YES or NO, the statement is INSUFFICIENT.

The correct answer is A.
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by GMATGuruNY » Tue Jun 15, 2010 6:45 am
selango wrote:Rephrasing the equation,we get

4^x.3^(-y)<1

-->4^x/3^y<1

From stmt 1 ,y=2x

-->4^x/3^2x=4^x/9^x

x can be any number,so insufficient.

From stmt 2,y=4

4^x/3^4<1

Again x can be any number,so insufficient.

Combining 1 and 2,we know x=2,sufficient.


Hence C
Perfect logic until you get here:

-->4^x/3^2x=4^x/9^x

x can be any number,so insufficient.


4^x/9^x = (4/9)^x

When you raise a positive fraction to a power, it becomes SMALLER.

So no matter what positive integer we use for x, (4/9)^x < 1, and statement 1 is SUFFICIENT.
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I unlock the best way for YOU to solve problems.

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by mj78ind » Tue Jun 15, 2010 6:50 am
Another way to look at this .......

per stmt 1, y = 2x hence (4^x)*(1/3)^y can be written as 4^x*(1/3)^x*(1/3)^x which translates into (4/3)^x*(1/3)^x

which is (1.333)^x*(1/3)^x which is (1.333/3)^x which will always be less than 1 for integer x>0.

Hope this makes sense.
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by Haaress » Tue Jun 15, 2010 3:59 pm
Stmt 1 would have been suff by itself ; however, however, falls apart if both x and y are zeros . The trick in this question is to understand that both x and y can not be equal to zero. If both x and y are zeroes then the equation equals to 1.

Thats why we need stmt 2 because stmt 2 states that at least one is not zero.

Thus C.
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by mj78ind » Tue Jun 15, 2010 5:46 pm
Haaress wrote:Stmt 1 would have been suff by itself ; however, however, falls apart if both x and y are zeros . The trick in this question is to understand that both x and y can not be equal to zero. If both x and y are zeroes then the equation equals to 1.

Thats why we need stmt 2 because stmt 2 states that at least one is not zero.

Thus C.
Interesting I thought the stem states x,y are positive integers hence they can not be 0, although I did not think about that possibility when making then answer.

Thanks
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by kvcpk » Tue Jun 15, 2010 11:21 pm
Haaress wrote:Stmt 1 would have been suff by itself ; however, however, falls apart if both x and y are zeros . The trick in this question is to understand that both x and y can not be equal to zero. If both x and y are zeroes then the equation equals to 1.
Thus C.
List of positive integers doesnt include 0
0 is neither positive nor negative. So answer should be A.
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by Haaress » Wed Jun 16, 2010 8:19 am
Thanks kvcpk and mj78ind.

I must have overlooked the positive powers and that should exclude zero.
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by kvcpk » Wed Jun 16, 2010 8:37 am
Haaress wrote:Thanks kvcpk and mj78ind.

I must have overlooked the positive powers and that should exclude zero.
That Happens!! Dont worry abou it. ;)
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