If \(K^{3/2}\) is \(50\%\) bigger than \(K^{5/4},\) what is the value of \(K?\)

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If \(K^{3/2}\) is \(50\%\) bigger than \(K^{5/4},\) what is the value of \(K?\)

(A) \(\dfrac{\sqrt3}{\sqrt2}\)

(B) \(\dfrac32\)

(C) \(\dfrac94\)

(D) \(\dfrac{27}8\)

(E) \(\dfrac{81}{16}\)

[spoiler]OA=E[/spoiler]

Source: Magoosh
Source: — Problem Solving |

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M7MBA wrote:
Thu Jul 02, 2020 12:52 am
If \(K^{3/2}\) is \(50\%\) bigger than \(K^{5/4},\) what is the value of \(K?\)

(A) \(\dfrac{\sqrt3}{\sqrt2}\)

(B) \(\dfrac32\)

(C) \(\dfrac94\)

(D) \(\dfrac{27}8\)

(E) \(\dfrac{81}{16}\)

[spoiler]OA=E[/spoiler]

Source: Magoosh
Given that \(K^{3/2}\) is \(50\%\) bigger than \(K^{5/4},\) we have

\(K^{3/2} = 1.50 *K^{5/4}\)

\(=> \dfrac{K^{3/2} }{K^{5/4}} = 1.50 \)

\(=> K^{3/2 – 5/4} = 1.50\)

\(=> K^{1/4} = 3/2\)

\(K = (3/2)^4\)

\(K = 81/16\)

Correct answer: E

Hope this helps!

-Jay
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M7MBA wrote:
Thu Jul 02, 2020 12:52 am
If \(K^{3/2}\) is \(50\%\) bigger than \(K^{5/4},\) what is the value of \(K?\)

(A) \(\dfrac{\sqrt3}{\sqrt2}\)

(B) \(\dfrac32\)

(C) \(\dfrac94\)

(D) \(\dfrac{27}8\)

(E) \(\dfrac{81}{16}\)

[spoiler]OA=E[/spoiler]

Solution:

We can create the equation:

K^(3/2) = 1.5K^(5/4)

Multiplying both sides by 2, we have:

2K^(3/2) = 3K^(5/4)

Raising both sides to the 4th power, we have:

16K^6 = 81K^5

K^6/K^5 = 81/16

K = 81/16

Answer: E

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