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05 Jan 2015, 06:45
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B
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Difficulty:
25%
(medium)
Question Stats:
74% (01:33) correct
26%
(01:51)
wrong
based on 215
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If n is an integer and 101n^2 is less than or equal to 8100, what is the greatest possible value of n?
A. 7
B. 8
C. 9
D. 10
E. 11
A. 7
B. 8
C. 9
D. 10
E. 11
05 Jan 2015, 17:04
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Bunuel
101 * n^2 <=8100
n^2 <=8100/101 which will be less than 81 since 8100/100 = 81 which is the square of 9
Next closest value of n where n^2<=81 is 8
Ans B
06 Jan 2015, 00:55
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Answer = B. 8
\(101n^2 < = 8100\)
\(100n^2 + n^2 < = 8100\)
For n = 9, LHS would be greater than RHS
8100+81 > 8100
So, n = 8
6400 + 64 < 8100
\(101n^2 < = 8100\)
\(100n^2 + n^2 < = 8100\)
For n = 9, LHS would be greater than RHS
8100+81 > 8100
So, n = 8
6400 + 64 < 8100
