What is the decreasing order from left to right of the following numbe

Question

What is the decreasing order from left to right of the following numbers?

\sqrt {3}, \sqrt {6},\sqrt {12},\sqrt {45}

(A)  \sqrt {3}, \sqrt {6}, \sqrt {12}, \sqrt {45}

(B)  \sqrt {6}, \sqrt {12}, \sqrt {45}, \sqrt {3}

(C)  \sqrt {6}, \sqrt {45}, \sqrt {3}, \sqrt {12}

(D)  \sqrt {6}, \sqrt {3}, \sqrt {45}, \sqrt {12}

(E)  \sqrt {3}, \sqrt {45}, \sqrt {6} , \sqrt {12}

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chetan2u · Accepted Answer

What is the decreasing order from left to right of the following numbers?

\(\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}\)

Let us get all the terms without roots, so raise each to the power of LCM(4,5,10,15) = 60
\(\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}\)=\(3^{15},6^{12},12^{6},45^4\)
(I) let us check between \(3^{15},6^{12}\)
\(3^{15}=3^{12}*3^3\) and \(6^{12}=3^{12}*2^{12}=3^{12}*(2^4)^3=3^{12}*16^3\).....\(3^{15}<6^{12}...\sqrt[4]{3}<\sqrt[5]{6}.\)..
Choices A and E can be eliminated
Similarly we can test other pairs..

Another way to test \(3^{15},6^{12},12^{6},45^4\) is to try to get them in same power
\(3^{15}=(3^5)^3=243^3,6^{12}=(6^4)^3=(1296)^3,12^{6}=(12^2)^3=(144)^3,45^4\)...
So, descending order \( \sqrt[5]{6},\sqrt[4]{3},\sqrt[10]{12},\), and we have to check where \(\sqrt[15]{45}\) fits in...
\(6^{12}=(6^3)^4=216^4,45^4\)......6^{12} >45^4
\(12^{6}=(12^3)^2=1728^2,45^4=(45^2)^2=(2025)^2.....45^4>12^6\)
\(3^{15}=3^8*3^7=3^8*3*(3^3)^2=3^8*3*27^2,45^4=3^8*5^4=3^8*25^2......3^{15}>45^4\)

(D) \(\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}\)

ScottTargetTestPrep · Answer

If we raise each root to the 60th power, we have:

3^15, 6^12, 12^6, 45^4

3^15, 3^12 x 2^12, 2^12 x 3^6, 3^8 x 5^4

Now let’s divide each of them by 3^6 (which is their GCF), and we have:

3^9, 3^6 x 2^12, 2^12, 3^2 x 5^4

Let’s divide each them by 3^2 (since three of the four has 3^2 as a factor):

3^7, 3^4 x 2^12, 2^12/3^2, 5^4

Now let’s approximate each one:

3^7 = 3^4 x 3^2 = 81 x 27 > 80 x 25 = 2000

3^4 x 2^12 = 81 x 4 x 2^10 > 80 x 4 x 1000 = 320,000 (notice that 2^10 = 1,024)

2^12/3^2 < 2^12/2^3 = 2^9 = 512

5^4 = (5^2)^2 = 25^2 = 625

So the order of the 4 quantities, from the largest to the smallest, is:
3^4 x 2^12, 3^7, 5^4, 2^12/3^2

Going back to the original quantities, we have:

^5√6, ^4√3, ^15√45, ^10√12

Answer: D

sthahvi · Answer

but is it alright to treat 1/n roots as n roots ?

chetan2u · Answer

Hi

All the terms are positive and more than 1.
Property of all the numbers greater than 1 is that as you increase the power, the number becomes larger. Also, larger the number, the larger it will become when raised to same power as a smaller number than that. 
So if  \sqrt {x}   and  \sqrt {y}  are raised to SAME power, the value of whichever of the terms was initially more will remain more. 
Using this property, we raise them to a bigger power to simply the terms.

A_Nishith · Answer

Need some Fast Technique to get answer.

bumpbot · Answer

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What is the decreasing order from left to right of the following numbe

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