(√27+√243)/√54=?

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(√27+√243)/√54=?

by Max@Math Revolution » Wed Jun 01, 2016 5:48 am
(√27+√243)/√54=?
A. 2√2
B. 2√3
C. 3√2
D. 3√3
E. √2

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by 800_or_bust » Wed Jun 01, 2016 8:48 am
Max@Math Revolution wrote:(√27+√243)/√54=?
A. 2√2
B. 2√3
C. 3√2
D. 3√3
E. √2

*An answer will be posted in 2 days.
√27+√243 = √9√3 + √81√3 = 3√3 + 9√3 = 12√3

√54 = √27√2 = 3√3√2 = 3√6

12√3 / 3√6 = (12√3 x √6) / 3 x 6 = 12√18 / 18 = (2√18/3) = 6√2/3 = 2√2

Answer choice A
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by Brent@GMATPrepNow » Wed Jun 01, 2016 11:11 am
Max@Math Revolution wrote:(√27 + √243)/√54=?
A. 2√2
B. 2√3
C. 3√2
D. 3√3
E. √2
We'll use the following rule: √(xy) = (√x)(√y)

√27 = √[(9)(3)] = (√9)(√3) = 3√3
√243 = √[(81)(3)] = (√81)(√3) = 9√3
√54 = √[(9)(6)] = (√9)(√6) = 3√6

So, (√27 + √243)/√54 = (3√3 + 9√3)/(3√6)
= (12√3)/(3√6)
= 4/√2
Check answer choices.... not there.

Take 4/√2 and multiply top and bottom by √2.
We get: (4/√2)(√2/√2) = (4√2)(2) = [spoiler]2/√2 = A[/spoiler]

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by Max@Math Revolution » Sun Jun 05, 2016 4:49 am
(√27+√243)/√54=(3√3+9√3)/3√6=12√3/3√6=4/√2=2√2. Hence, the correct answer is A.

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by Matt@VeritasPrep » Tue Jun 07, 2016 10:56 pm
Or we could just say that

(√27 + √243)/√54 =

√27/√54 + √243/√54 =

√(1/2) + √(9/2) =

(√1 + √9)/√2

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by nasahtahir » Fri Dec 02, 2016 7:45 am
Brent would you kindly explain if we take 4/√2 and turn it into 2^2/√2= 2^2/ 2^1/2 we have a common base do we not? We are allowed to subtract the powers at will? Yes? Or do I have it all wrong?
appreciate your help.

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by Brent@GMATPrepNow » Fri Dec 02, 2016 8:15 am
nasahtahir wrote:Brent would you kindly explain if we take 4/√2 and turn it into 2^2/√2= 2^2/ 2^1/2 we have a common base do we not? We are allowed to subtract the powers at will? Yes? Or do I have it all wrong?
appreciate your help.

HT
Sure, we can think of 4/√2 as being equal to (2^2)/[2^(1/2)]
Then, because we have the same base, we can subtract the exponents (2 - 1/2) to get 2^(3/2).
From here, 2^(3/2) = (2^3)^(1/2) = 8^(1/2) = √8 = 2√2
So, your idea works. It just requires us to know how to deal with fractional exponents.


Alternatively, what I did was create a fraction that's equivalent to 4/√2
So, I took 4/√2 and multiplied numerator and denominator by √2 to get:4√2/(√2)(√2)
This simplifies to 4√2/2, which equals 2√2

Does that help?

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by Jeff@TargetTestPrep » Sat Dec 10, 2016 5:27 am
Max@Math Revolution wrote:(√27+√243)/√54=?
A. 2√2
B. 2√3
C. 3√2
D. 3√3
E. √2
We are given (√27+√243)/√54. Let's simplify each term.

√27 = √9 x √3 = 3√3

√243 = √81 x √3 = 9√3

√54 = √9 x √6 = 3√6

Thus:

(√27+√243)/√54 = (3√3 + 9√3)/(3√6) = (12√3)/(3√6) = 4/√2

Finally, we can rationalize 4/√2 by multiplying the numerator and denominator by √2 and we obtain:

(4/√2)(√2/√2) = (4√2)/2 = 2√2

Answer: A

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