If a, b, x, and y are positive integers, is a^(-x) b(-y)?

Question

If a, b, x, and y are positive integers, is a^{(-x)} > b^{(-y)} ? (1) a < b (2) x < y GmatPrepDS.jpg I chose E because if a=1/4, b=1/2, x=1, y=2, wouldn't a^(-x)=b^(-y)?

mau5 · Answer

The question stem mentions specifically that a,b,x and y are positive INTEGERS. Your choice of a,b doesn't subscribe to that.

From F.S 1, we know that b>a . Thus, for a=2,b=3 and x=y=1, we have  b^y>a^x and thus a YES for the question stem.Again, for a=2,b=3 and x=10,y=1, we have  b^y<a^x , and a NO.Insufficient.

Similarly for F.S 2, Insufficient.

Taking both together, we know that b>a and y>x.Thus:
1. b^y>a^y 
2. a^y>a^x

Thus,  b^y>a^x.  Sufficient.

C.

unceldolan · Answer

Question: a^-x > b^-y? --> 1/a^x > 1/b^y?

We need to find a definit YES or NO.

(1) a < b If a=2 and b = 8 and x = 1 and y = 1 then YES. But if a = 2 and b = 8 and x = 100 and y = 1 then NO. IS
(2) x < y If a = 2 and b = 8 and x = 1 and y = 2 then YES. But if a = 100 and b = 1 then NO. IS

TOGETHER a < b and x < y ==> plug in numbers Suff. Hence C.

stonecold · Answer

Here is my approach 
W need to prove that x1/a^x >1/b^x
now we can flip the inequality if they are of same sign while doing the reciprocal..
hence we need to prove => a^x<b^y
statement 1 => no clue of x and y => not sufficient
statement 2 => no clue of a and b => not sufficient 
combing them we can say that the bae an exponent of rhs are always greater hence rhs would be greater 
thus C

coolnuts · Answer

We can redefine the question as (1/a)^x > (1/b)^y

S1 a < b We know nothing about x and y here. So if a=3 and b = 4 and x = 1 and y = 1 then YES. But if a = 3 and b = 4 and x = 10 and y = 1 then NO. This statement is insufficient.

S2 x < y We know nothing about a and b here.So if a = 3 and b = 4 and x = 1 and y = 2 then YES. But if a = 10 and b = 1 and x=1 and y=2 then NO. This statement is insufficient.

Taking S1 and S2 together is sufficient to get the answer.

Hence C.

BrentGMATPrepNow · Answer

Target question : Is a^(-x) > b^(-y)? This is a great candidate for rephrasing the target question. Aside: At the bottom of this post, you can find a video with tips on rephrasing the target question First recognize the following: a^(-x) = 1/(a^x) and b^(-y) = 1/(b^y) So, we can ask Is 1/(a^x) > 1/(b^y)? Also, since a and b are POSITIVE, we can be certain that (a^x) is POSITIVE and (b^y) is POSITIVE So, we can safely take the inequality 1/(a^x) > 1/(b^y) and multiply both sides by (a^x) to get: 1 > (a^x)/(b^y) Next, we can multiply both sides by (b^y) to get: (b^y) > (a^x) So, we can now ask... REPHRASED target question : Is (b^y) > (a^x)? Statement 1 : a < b No information about x or y. So, statement 1 is NOT SUFFICIENT Statement 2 : x < y No information about a or b. So, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined The key here is that all 4 variables are positive . If b is greater than a AND y is greater than x, we can be certain that (b^y) > (a^x) Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT Answer: C RELATED VIDEO https://www.youtube.com/watch?v=MyG1K3ee69w

ScottTargetTestPrep · Answer

Rephrasing the question we have: Is 1/a^x > 1/b^y ? Is b^y > a^x ? Statement One Alone: a < b Since we do now know anything about x and y, statement one alone is not sufficient to answer the question. Statement Two Alone: x < y Since we do now know anything about a and b, statement two alone is not sufficient to answer the question. Statements One and Two Together: Since b is greater than a, and y is greater than x, we know that b^y is greater than a^x. Answer: C

MathRevolution · Answer

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 4 variables: Let the original condition in a DS question contain 4 variables. Now, 4 variables would generally require 4 equations for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 4 equations to match the numbers of variables and equations in the original condition, the logical answer is E.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find whether \(a^{-x} > b^{-y}\) .

=> \(\frac{1}{ a^{x}} > \frac{1 }{ b^{y}} OR \\
\\
=> b^{y} > a^{x}\)

=> Given that a, b, x, and y are positive integers.

Second and the third step of Variable Approach: From the original condition, we have 4 variables (a, b, x, and y).To match the number of variables with the number of equations, we need 4 equations. Since conditions (1) and (2) will provide 2 equations, E would most likely be the answer.

Let’s take look at both condition together.

Condition(1) tells us that a < b.

Condition(2) tells us that x < y .

=> when b > a then \(b^{y} > a^{y}\)

=> When y > x, then \(a^{y} > a^{x} \)

=> Therefore, \(b^{y} > a^{x}\) - YES

Since the answer is a unique YES, both conditions combined together are sufficient by CMT 1.

Both conditions combined together are sufficient.

So, C is the correct answer.

Answer: C

Basshead · Answer

Is  \frac{1}{a^x} > \frac{1}{b^y} ?

Is  a^x < b^y ?

(1) We don't know the value of x or y; INSUFFICIENT.

(2) We don't know the value of a or b; INSUFFICIENT.

(1&2) We have  a < b ;  x < y

Therefore  a^x < b^y  (b is larger and raised to a higher exponent). SUFFICIENT.

Answer is C.

v12345 · Answer

a, b, x, and y are positive integers
 a^{(-x)} > b^{(-y)} 
=  a^{x} < b^{y}

(1) a < b . Insufficient
let a=3,b=4
if x=3,y=1 then false
if x=1,y=2 then true

(2) x < y . Insufficient
let x=1,y=2
if a=3,b=1 then false
if a=1,b=2 then true

Combining both, we get . Sufficient
a < b and x < y
 a^{x} < b^{y}

Hence, OA is (C).

bumpbot · Answer

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If a, b, x, and y are positive integers, is a^(-x) > b(-y)?

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