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

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Math Expert
Joined: 02 Sep 2009
Posts: 64243

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29 Apr 2019, 04:58
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Difficulty:

15% (low)

Question Stats:

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

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

A. 1/45
B. 1/40
C. 2/9
D. 5
E. 45

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Senior Manager
Joined: 22 Feb 2018
Posts: 404
Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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29 Apr 2019, 05:13
OA:E

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

$$(\frac{\frac{1}{45}+\frac{1}{5}}{10})^{-1}=(\frac{\frac{1+9}{45}}{10})^{-1}=(\frac{\frac{10}{45}}{10})^{-1}=(\frac{10}{45*10})^{-1}=(\frac{1}{45})^{-1}=45$$
Manager
Joined: 31 Jul 2017
Posts: 195
Location: Tajikistan
Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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29 Apr 2019, 05:31
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Intern
Joined: 03 Apr 2017
Posts: 14
Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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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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Joined: 31 Jul 2017
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Location: Tajikistan
Re: (45^(-1) + 5^(-1)/10)^(-1) =  [#permalink]

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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.
Re: (45^(-1) + 5^(-1)/10)^(-1) =   [#permalink] 29 Apr 2019, 07:09