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Given (0.08/270)^(-1/2) = 15^n, then n = ?

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Bunuel
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Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink] New post   17 Jun 2022, 01:30
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Given \((\frac{0.08}{270})^{-\frac{1}{2}} =15^n\), then n = ?

(A) \(\frac{1}{2}\)

(B) 1

(C)\(\frac{ 4}{3}\)

(D) \(\frac{3}{2}\)

(E) 2



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D
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oyejangir
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Re: Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink] New post   17 Jun 2022, 01:46
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Answer = D
(8/27000)^-1/2 = 15^n
(27000/8)^1/2 = 15^n
(3*10/2)^3/2 = 15^n
(3*5)^3/2 = 15^n
(15)^3/2 = 15^n
15^3/2 = 15^n
=> n = 3/2
Which is option D

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Re: Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink] New post   17 Jun 2022, 02:39
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(270/0.08)^0.5 = (270*25/2)^0.5 = (135X25)^0.5 = 15*(15)^0.5 = 15^1.5

The answer is thus (D).
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Re: Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink] New post   17 Jun 2022, 05:13
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(0.08/270)^-1/2
= (27000/8)^1/2
= (3375)^1/2

Options have 15 to the power 1,2,3 and 4

15^1 = 15
15^2 = 225
15^3 = 225 × 15 = 3375 (2250+1125)

Thus,
(3375)^1/2 = 15^3/2

So answer = Option D

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Re: Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink] New post   18 Jun 2022, 00:16
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Bunuel wrote:
Given \((\frac{0.08}{270})^{-\frac{1}{2}} =15^n\), then n = ?


\((\frac{0.08}{270})^{-\frac{1}{2}} = (\frac{270*50}{4})^{\frac{1}{2}} = 15*15^{\frac{1}{2}} = 15^\frac{3}{2}\)

Hence, Option D is correct.
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Re: Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink] New post   23 Feb 2024, 12:47
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Re: Given (0.08/270)^(-1/2) = 15^n, then n = ? [#permalink]
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