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Of the C condominiums in a certain building complex, 2 3 have at least
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23 Aug 2021, 05:13
23 Aug 2021, 05:13
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85% (01:33) correct
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Of the C condominiums in a certain building complex, 2/3 have at least two bedrooms. If, of those, 1/4 have at least two bathrooms, which of the following expressions represents the number of condominiums in the complex with at least two bedrooms that do not have at least two bathrooms?
(A) C/2
(B) C/3
(C) C/4
(D) C/6
(E) C/12
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
(A) C/2
(B) C/3
(C) C/4
(D) C/6
(E) C/12
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
Of the C condominiums in a certain building complex, 2 3 have at least
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24 Aug 2021, 04:02
24 Aug 2021, 04:02
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Expert Reply
Total condos = \(C\)
2-bed condos as a fraction of total = \(\frac{2}{3}C\)
2-bedroom condos with at least 2 bathrooms = \(\frac{ 1}{ 4 }\) of 2-bedroom condos
\(\implies\) 2-bedroom condos without at least two bathrooms = \(1 - \frac{ 1}{ 4 } = \frac{ 3 }{4 }\) of 2- bedroom condos.
2-bed condos without at least 2-bathrooms in total = 2-bed condos without at least 2 washrooms as a fraction of total 2-bedroom condos x 2-bedroom condos as a fraction of total condos
\( \implies \) 2-bed condos without 2-bathrooms in total = \(\frac{ 3 }{4 } \frac{ 2 }{ 3 } C = \frac{ C }{ 2 } \)
Thus the answer is Option A.
2-bed condos as a fraction of total = \(\frac{2}{3}C\)
2-bedroom condos with at least 2 bathrooms = \(\frac{ 1}{ 4 }\) of 2-bedroom condos
\(\implies\) 2-bedroom condos without at least two bathrooms = \(1 - \frac{ 1}{ 4 } = \frac{ 3 }{4 }\) of 2- bedroom condos.
2-bed condos without at least 2-bathrooms in total = 2-bed condos without at least 2 washrooms as a fraction of total 2-bedroom condos x 2-bedroom condos as a fraction of total condos
\( \implies \) 2-bed condos without 2-bathrooms in total = \(\frac{ 3 }{4 } \frac{ 2 }{ 3 } C = \frac{ C }{ 2 } \)
Thus the answer is Option A.
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Re: Of the C condominiums in a certain building complex, 2 3 have at least
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24 Aug 2021, 10:07
24 Aug 2021, 10:07
Carcass wrote:
Of the C condominiums in a certain building complex, 2/3 have at least two bedrooms. If, of those, 1/4 have at least two bathrooms, which of the following expressions represents the number of condominiums in the complex with at least two bedrooms that do not have at least two bathrooms?
(A) C/2
(B) C/3
(C) C/4
(D) C/6
(E) C/12
(A) C/2
(B) C/3
(C) C/4
(D) C/6
(E) C/12
Let \(C\) be 1200
So, Condominiums with atleast 2-BRs = \(\frac{2}{3}(1200) = 800\)
Now, out of these Condominiums, ones with atleast 2-Baths = \(\frac{1}{4}(800) = 200\)
And, out of these Condominiums, ones which do NOT have atleast 2-Baths = \(800 - 200 = 600\)
Plug in C as 1200 and check for option choices.
Hence, option A
