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(45^(-1) + 5^(-1)/10)^(-1) =

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(45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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New post 29 Apr 2019, 04:58
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Question Stats:

81% (01:01) correct 19% (01:38) wrong based on 26 sessions

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Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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New post 29 Apr 2019, 05:13
OA:E

\((\frac{45^{-1} + 5^{-1}}{10})^{-1}\)

\((\frac{\frac{1}{45}+\frac{1}{5}}{10})^{-1}=(\frac{\frac{1+9}{45}}{10})^{-1}=(\frac{\frac{10}{45}}{10})^{-1}=(\frac{10}{45*10})^{-1}=(\frac{1}{45})^{-1}=45\)
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Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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New post 29 Apr 2019, 05:31
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Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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New post 29 Apr 2019, 06:56
Is there a reason you cannot distribute the exponent to all values within the parenthesis and then solve?

Once you distribute the exponent, you get (45^1 + 5^1 / 10^-1) = 50(10) = 500

Can someone correct my thinking here?
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Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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New post 29 Apr 2019, 07:09
peterpark wrote:
Is there a reason you cannot distribute the exponent to all values within the parenthesis and then solve?

Once you distribute the exponent, you get (45^1 + 5^1 / 10^-1) = 50(10) = 500

Can someone correct my thinking here?



Hi peterpark,
Let me try to clear your doubt. According to PEMDAS rule, you need to first take care inside the parenthesis and only then apply exponent outside the parenthesis. Let me know if you need further assistance.
GMAT Club Bot
Re: (45^(-1) + 5^(-1)/10)^(-1) =   [#permalink] 29 Apr 2019, 07:09

(45^(-1) + 5^(-1)/10)^(-1) =

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