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niteshwaghray
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Posts: 59
29 May 2017, 10:16
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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In a certain sequence of 10 positive numbers, each number after the first is square root of the previous number. If the first number is 9, how many numbers are less than 1?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
29 May 2017, 12:39
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niteshwaghray
Here is a little trick: any positive integer root from a number more than 1 will be more than 1. So, it does not matter how many times we take the square root from 9, we'll still won't get a number less than 1. Meaning: \(\sqrt[1000000]{9}>1\).
Answer: A.
General Discussion
29 May 2017, 23:08
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So in this sequence we are going to start from 9 and take square roots.. so square root of 9 which is 3, .. then square root of 3 which is approx. 1.732, .. then sq root of 1.732 .. and so on...
Its tricky but interesting - as long as a number is > 1, its square root will also be > 1... so in this list no matter how many times we take the square root - ALL the numbers that we end up with will always be > 1 (though the numbers will keep on decreasing)
Hence all numbers are > 1. None of the numbers is < 1.
Thus option A
Its tricky but interesting - as long as a number is > 1, its square root will also be > 1... so in this list no matter how many times we take the square root - ALL the numbers that we end up with will always be > 1 (though the numbers will keep on decreasing)
Hence all numbers are > 1. None of the numbers is < 1.
Thus option A
