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Sub 505 (Easy)|   Inequalities|                        
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Bunuel
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Bunuel
If u > 0 and v > 0, which is greater, u^v or v^u?

(1) u = 1
(2) v > 2

Statement 1
when u = 1, u raised to any power is always 1. Also v^u is always v.

If v>1 then v^u > u^v

If v[b]<1, then v^u < u^v [/b]

Clearly not sufficient.

Statement 2

v > 2

say u = 0.5

(3)^0.5 = 1.732
(0.5)^3 = 0.125

v^u > u^v


v=4; u=-1

v^u = 4^(-1) = (1/4)
u^v = (-1)^4 =1

v^u < u^v

Not sufficient.

Combining these two.
v > 2; u = 1
when u = 1, u raised to any power is always 1. Also v^u is always v.

since v is always > 2, and u is always 1. hence

v^u is always greater than u^v
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Bunuel Thanks for your reply.

In statement 2:

u=1 and v=3 Then v^u is greater
u=4 and v=3 Then v^u is greater

Please tell a situation when it will not hold true.
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Bunuel Thanks for your reply.

In statement 2:

u=1 and v=3 Then v^u is greater
u=4 and v=3 Then v^u is greater

Please tell a situation when it will not hold true.

You can consider the case when u = v. In this case v^u is NOT greater than u^v.
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If u > 0 and v > 0, which is greater, u^v or v^u?

(1) u = 1

Let v = 10........then 1^10 < 10^1

Let v= 1...........then 1^1 = 1^1

Insufficient

(2) v > 2

same example as above:
Let v = 10........then 1^10 < 10^1

Let v= 10...........then 10^10 = 10^10

Insufficient

Combine 1 & 2

V > 2 & u= 1..........So first example applies

Let v = 10........then 1^10 < 10^1

Sufficient

Answer: C
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Bunuel
If u > 0 and v > 0, which is greater, \(u^v\) or \(v^u\)?

(1) u = 1
(2) v > 2

\(u^v\) ?? \(v^u\)

Statement 1 : u=1

\(1^v\) ?? \(v^1\)

Since we don't know anything about v (except it is positive) statement 1 is insufficient.

Statement 1 : v>2. Let's say 3

\(u^3\) ?? \(3^u\)

Since we don't know anything about u (except it is positive) statement 2 is insufficient.

Combined :

\(1^3\) < \(3^1\)

Hence Sufficient.

Note: Even if you take value of v as 2.5, the relation will still hold true.
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If u > 0 and v > 0, which is greater, u^v or v^u?

(1) u = 1

Since we are not told if u and v are different numbers there are two possibilites:
a) u=v= 1 in which case u^v = v^u
b) They are different numbers in which case v^u will always be greater since u^v = 1 (1 raised to anything is 1)

(2) v > 2

It does not give us any information about u. For example u could be 3 and v could be 4 and vice versa

On combining we know u^v will always be 1 . Therefore v^u is greater. SUFFICIENT

Ans = C

You wrote "...b) They are different numbers in which case v^u will always be greater since u^v = 1 (1 raised to anything is 1)"

I think this is not correct, since v could be less than 1, in which case \(v^{u} < u^{v}, i.e. (1/2)^1 < 1^{1/2}\)
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Bunuel
If u > 0 and v > 0, which is greater, \(u^v\) or \(v^u\)?

(1) u = 1
(2) v > 2


Answer: Option C

Video solution by GMATinsight

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Bunuel
If u > 0 and v > 0, which is greater, \(u^v\) or \(v^u\)?

(1) u = 1
(2) v > 2
Solution:

Question Stem Analysis:

We need to determine which of u^v and v^u is greater, given that both u and v are positive.

Statement One Alone:

Since u = 1, u^v = 1^v = 1 (1 raised to any power is 1) and v^u = v^1 = v (any number raised to the power of 1 is itself). However, we can’t determine which one is greater. For example, if v = 0.5, then u^v > v^u since 1 > 0.5. However, if v = 2, then v^u > u^v since 2 > 1. Statement one alone is not sufficient.

Statement Two Alone:

Knowing only that v > 2 is not sufficient to determine whether u^v > v^u or v^u > u^v. For example, if u = 1, and v = 3, then u^v = 1^3 = 1 and v^u = 3^1 = 3. However, if u = 3 and v = 3, then both u^v and v^u are equal to 27. Statement two alone is not sufficient.

Statements One and Two Together:

From statement one, we see that u^v = 1 and v^u = v. From statement two, we see that v > 2, so v^u > 2, and hence it’s greater than u^v. Both statements together are sufficient.

Answer: C
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Hi vipsgmat,

The doubt that's been running through this thread (the one anurag16 raised when they argued for B) is: why isn't Statement (2), v > 2, sufficient on its own? It feels sufficient because every quick case you try seems to make v^u the bigger one. Let me clear up exactly where that breaks.

What Statement (2) actually gives you: only that v > 2. It says nothing about u. So to test sufficiency, you have to push on u - try values of u that might flip the result, not just the convenient ones.

Here's the minimal pair of cases, both obeying v > 2:

- u = 1, v = 3 - u^v = 13 = 1, v^u = 31 = 3. Here v^u is greater.
- u = 3, v = 3 - u^v = 33 = 27, v^u = 33 = 27. Here they're equal - v^u is not greater.

Same statement, two different answers to "which is greater?" - not sufficient. That equal case (u = v) is exactly the situation Bunuel pointed anurag16 toward, and it's the one most people skip.

Why the trap is so easy to fall into: when you only test cases where u is small or clearly different from v, you keep landing on "v^u wins" and feel done. The DS rule to lock in: one set of valid numbers isn't enough - you have to actively hunt for a second set that gives a different answer. Here, letting u equal v does it.

That's why you need Statement (1) too: u = 1 forces u^v = 1, and then v > 2 guarantees v^u = v > 1. Only together do they pin the answer down - hence C.

Answer: C
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