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24 Sep 2012, 04:53
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
90% (00:55) correct
10%
(01:19)
wrong
based on 5641
sessions
History
Date
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Result
Not Attempted Yet
The sum \(\frac{7}{8}+\frac{1}{9}\) is between
(A) \(\frac{1}{2}\) and \(\frac{3}{4}\)
(B) \(\frac{3}{4}\) and 1
(C) 1 and \(1\frac{1}{4}\)
(D) \(1\frac{1}{4}\) and \(1\frac{1}{2}\)
(E) \(1\frac{1}{2}\) and 2
(A) \(\frac{1}{2}\) and \(\frac{3}{4}\)
(B) \(\frac{3}{4}\) and 1
(C) 1 and \(1\frac{1}{4}\)
(D) \(1\frac{1}{4}\) and \(1\frac{1}{2}\)
(E) \(1\frac{1}{2}\) and 2
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24 Sep 2012, 04:53
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SOLUTION:
The sum \(\frac{7}{8}+\frac{1}{9}\) is between
(A) \(\frac{1}{2}\) and \(\frac{3}{4}\)
(B) \(\frac{3}{4}\) and 1
(C) 1 and \(1\frac{1}{4}\)
(D) \(1\frac{1}{4}\) and \(1\frac{1}{2}\)
(E) \(1\frac{1}{2}\) and 2
________________________________
\(\frac{7}{8}+\frac{1}{9}=\frac{63}{72}+\frac{8}{72}=\frac{71}{72}\).
Now, \(\frac{71}{72}\) is less but very close to 1, so the answer is B.
Answer: B.
Else, one can notice that since \(\frac{1}{9}\) is less than \(\frac{1}{8}\), thus \(\frac{7}{8}+\frac{1}{9}\) is less than 1. Therefore we can eliminate answer choices C, D, and E. Next, since \(\frac{7}{8}>\frac{3}{4}=\frac{6}{8}\), then answer choice A is also out. So, only answer choice B remains.
Answer: B.
The sum \(\frac{7}{8}+\frac{1}{9}\) is between
(A) \(\frac{1}{2}\) and \(\frac{3}{4}\)
(B) \(\frac{3}{4}\) and 1
(C) 1 and \(1\frac{1}{4}\)
(D) \(1\frac{1}{4}\) and \(1\frac{1}{2}\)
(E) \(1\frac{1}{2}\) and 2
________________________________
\(\frac{7}{8}+\frac{1}{9}=\frac{63}{72}+\frac{8}{72}=\frac{71}{72}\).
Now, \(\frac{71}{72}\) is less but very close to 1, so the answer is B.
Answer: B.
Else, one can notice that since \(\frac{1}{9}\) is less than \(\frac{1}{8}\), thus \(\frac{7}{8}+\frac{1}{9}\) is less than 1. Therefore we can eliminate answer choices C, D, and E. Next, since \(\frac{7}{8}>\frac{3}{4}=\frac{6}{8}\), then answer choice A is also out. So, only answer choice B remains.
Answer: B.
24 Sep 2012, 10:45
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Bunuel
\(\frac{7}{8}+\frac{1}{9}<\frac{7}{8}+\frac{1}{8}=1\) and \(\frac{7}{8}+\frac{1}{9}>\frac{6}{8}=\frac{3}{4}\).
Hence, Answer C.
