Mean and Product

breaking it down...

((a^6)^2+(a^7)^2)/2=K
A6*A7=L

It seems like K will always be greater than A based on the above. Any other take?

dtweah wrote:The Number K is the arithmetic mean of a^12 and b^14 and L is the product of (a^6 ) and ( b^7), where a and b are positive integers greater than 1. Which of the following CANNOT be False?

A. L>K
B. L> or equal to K
C. K>L
D. K> or equal to L
E. K=L

The arithmetic mean of a given set of n integers is greater than or equal to the geometric mean of that set of n integers.

(a^12+b^12)/2 is greater than or equal to (a^12 x b14)^.5= (a^6)(b^7)= geometric mean.

D cannot be false.

dtweah wrote:The Number K is the arithmetic mean of a^12 and b^14 and L is the product of (a^6 ) and ( b^7), where a and b are positive integers greater than 1. Which of the following CANNOT be False?

A. L>K
B. L> or equal to K
C. K>L
D. K> or equal to L
E. K=L

First, let's change "CANNOT be False" to "MUST be True". So, we're looking for the choice that's always correct.

Seemed very complicated conceptually, so I jumped in and picked some easy numbers.

Nowhere does it say that a and b are distinct, so I just picked a=b=2.

K = (2^12 + 2^14)/2 = 2^11 + 2^13

L = 2^6*2^7 = 2^13

Well, 2^13 + 2^11 is certainly more than 2^13, so we have K > L.

If it's possible that K>L, then (A), (B) and (E) are right out as "MUST BE TRUE".

Next, let's apply some logic (often underused on the GMAT). "K is greater than L" is a subset of "K is greater than or equal to L".

So, if it MUST BE TRUE that K>L, then it also MUST BE TRUE that "K>=L". Since they can't both be the right answer, we can eliminate (C) and choose (D).

In fact, merely by applying that logic we could have eliminated (A), (C) and (E) without doing any math at all. ((A) and (E) are subsets of (B); (C) and (E) are subsets of (D).)

((a^6)^2+(a^7)^2)/2=K
A6*A7=L

First, let's change "CANNOT be False" to "MUST be True". So, we're looking for the choice that's always correct

good starting point(s).

(a^6 - b^7)^2 = a^12 + b^14 - 2a^6b^7 = 2(K-L)

since the left side of the equation must be >=0, it follows that K-L >= 0, i.e., K>=L

.

dtweah wrote:The Number K is the arithmetic mean of a^12 and b^14 and L is the product of (a^6 ) and ( b^7), where a and b are positive integers greater than 1. Which of the following CANNOT be False?

A. L>K
B. L> or equal to K
C. K>L
D. K> or equal to L
E. K=L

Folks, this is simple if you know that AM >=GM>=HM

Since AM>=GM

we have (a^12 + b^12)/2 >= (a^12 x b^12)^1/2

i.e. (a^12 + b^12)/2 >= (a^6 x b^6)

i.e K >=L. (the equality holds if a =b)

Hence D cannot be false