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15 Nov 2021, 07:15
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (02:26) correct
30%
(02:35)
wrong
based on 191
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The sum of a certain number and its reciprocal is equal to 2.9. What is the absolute value of the difference of this number and its reciprocal ?
A. 2.2
B. 2.1
C. 1.2
D. 0.3
E. 0.4
A. 2.2
B. 2.1
C. 1.2
D. 0.3
E. 0.4
General Discussion
PyjamaScientist
Admitted - Which School Forum Moderator
Joined: 25 Oct 2020
Last visit: 04 Apr 2026
Posts: 1,125
Given Kudos: 633
Schools: Ross '25 (M$)
GMAT 1: 740 Q49 V42 (Online)
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15 Nov 2021, 08:17
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Maybe my method is not the best. And I am myself waiting for the likes of Chetan2u/IanStewart to post shorter answers, but this is how I went about it-
Given \(a \) + \(\frac{1}{a}\) = \(2.9 \) ---------- (1)
We need to find the value of \(a\) - \(\frac{1}{a}\)
We can re-write it as-> \(\frac{(a^2 - 1)}{a}\) => \(\frac{(a+1)(a-1)}{a}\) ---------- (2)
So, we need values of (a+1) and (a-1). We can get those values through manipulation of (1),
\(a \) + \(\frac{1}{a}\) = \(2.9\)
Add and subtract \(2a \) on L.H.S,
\(a \) + \(\frac{1}{a}\) = \(\frac{(a^2 + 1)}{a}\) = \(\frac{(a^2 + 1 + 2a - 2a)}{a}\) \(= 2.9\)
From above we get,
\((a+1)^2 = 4.9a\) & \((a-1)^2 = 0.9a\)
or
\((a+1) = \sqrt{4.9a}\) & \((a-1) = \sqrt{0.9a} \)
Substitute in (2), we get,
\(\frac{(\sqrt{4.9a}*\sqrt{0.9a})}{a}\) = \(\sqrt{\frac{(49*9)}{100}}\) = \(\frac{7*3}{10}\) = \(2.1\).
Given \(a \) + \(\frac{1}{a}\) = \(2.9 \) ---------- (1)
We need to find the value of \(a\) - \(\frac{1}{a}\)
We can re-write it as-> \(\frac{(a^2 - 1)}{a}\) => \(\frac{(a+1)(a-1)}{a}\) ---------- (2)
So, we need values of (a+1) and (a-1). We can get those values through manipulation of (1),
\(a \) + \(\frac{1}{a}\) = \(2.9\)
Add and subtract \(2a \) on L.H.S,
\(a \) + \(\frac{1}{a}\) = \(\frac{(a^2 + 1)}{a}\) = \(\frac{(a^2 + 1 + 2a - 2a)}{a}\) \(= 2.9\)
From above we get,
\((a+1)^2 = 4.9a\) & \((a-1)^2 = 0.9a\)
or
\((a+1) = \sqrt{4.9a}\) & \((a-1) = \sqrt{0.9a} \)
Substitute in (2), we get,
\(\frac{(\sqrt{4.9a}*\sqrt{0.9a})}{a}\) = \(\sqrt{\frac{(49*9)}{100}}\) = \(\frac{7*3}{10}\) = \(2.1\).
