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10 Dec 2021, 03:15
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Difficulty:
65%
(hard)
Question Stats:
41% (02:41) correct
59%
(01:45)
wrong
based on 17
sessions
History
Date
Time
Result
Not Attempted Yet
A line 10 cm long is drawn. A square is constructed with this line segment as diagonal. Then a second square is drawn with a side of the first square as diagonal. This process is repeated ‘n’ times, until the length of a side of the nth square is less than 1/100. What is the least value of ‘n’?
(A) 26
(B) 22
(C) 20
(D) 11
(E) 10
Are You Up For the Challenge: 700 Level Questions
(A) 26
(B) 22
(C) 20
(D) 11
(E) 10
Are You Up For the Challenge: 700 Level Questions
22 Dec 2021, 22:58
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Bunuel
The question means:
Side of the bigger square becomes the diagonal of smaller square each time.
Thus, if the side of bigger square is a, the diagonal of smaller square becomes a, making the sides as \(\frac{a}{\sqrt{2}}\).
Similarly, the next smaller square will have a side equal to \(\frac{1}{\sqrt{2}}(\frac{a}{\sqrt{2}})=\frac{a}{2}\).
So, \(\frac{10}{(\sqrt{2})^n}<\frac{1}{100}\)
\(\frac{1}{(\sqrt{2})^n}<\frac{1}{1000}\)
\(\sqrt{2}^n>1000..............2^\frac{n}{2}>1000\)
As \(2^9=512.......2^{10}=1024\), the least value of \(\frac{n}{2}\) is 10. Thus, n is 20.
C
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