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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by
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14 Jun 2022, 02:00
14 Jun 2022, 02:00
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Question Stats:
83% (01:56) correct
16% (01:10) wrong based on 6 sessions
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If \(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) and \(s = 1 + \frac{1}{3}r\), then \(s\) exceeds \(r\) by
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)
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Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by
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14 Jun 2022, 05:24
14 Jun 2022, 05:24
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Carcass wrote:
If \(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) and \(s = 1 + \frac{1}{3}r\), then \(s\) exceeds \(r\) by
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)
Given: \(s = 1 + \frac{1}{3}r\)
Replace \(r\) with \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) to get: \(s = 1 + \frac{1}{3}(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27})\)
Expand to get: \(s= 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}\)
Since we're told \(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\), we can see that \(s\) exceeds \(r\) by \(\frac{1}{81}\)
Answer: E
Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by
[#permalink]
14 Jun 2022, 11:31
14 Jun 2022, 11:31
1
If we multiply 1/3 by r, we get one more term which is 1/81
Answer is (E)
Answer is (E)
