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25 Aug 2023, 01:30
Kudos
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C
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Difficulty:
15%
(low)
Question Stats:
80% (01:20) correct
20%
(01:30)
wrong
based on 92
sessions
History
Date
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If \(x=10^{-1}\) what is the value of \((3+x)(3-x)(\frac{1}{x})\) ?
A. 1/90
B. 89/10
C. 89.9
D. 90
E. 99.9
A. 1/90
B. 89/10
C. 89.9
D. 90
E. 99.9
25 Aug 2023, 18:57
Kudos
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Bunuel
\((3+x)(3-x)(\frac{1}{x})=\frac{9-x^2}{x}=\frac{9-(0.1)^2}{0.1}=(9-0.01)*10=8.99*10=89.9\)
Prk1410
Joined: 01 Aug 2023
Last visit: 20 Jun 2024
Posts: 9
Location: India
Concentration: Marketing, Strategy
Schools: LBS MiM "26 (II) INSEAD MiM "26 (D) Duke Fuqua (S) McDonough Kellogg MiM "26 Booth (Chicago)
GMAT 1: 710 Q50 V34
GPA: 3.88
Schools: LBS MiM "26 (II) INSEAD MiM "26 (D) Duke Fuqua (S) McDonough Kellogg MiM "26 Booth (Chicago)
GMAT 1: 710 Q50 V34
Posts: 9
26 Aug 2023, 06:37
Kudos
Bookmarks
We know that \((a-b) (a+b) = a^2 - b^2\)
Hence \((3+x)(3-x)(\frac{1}{x})\) = \({9-x^2}(\frac{1}{x})\) = (9-0.01) x 10 = 89.9
Hence \((3+x)(3-x)(\frac{1}{x})\) = \({9-x^2}(\frac{1}{x})\) = (9-0.01) x 10 = 89.9
