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31 Oct 2022, 01:30
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D
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Difficulty:
35%
(medium)
Question Stats:
70% (01:50) correct
30%
(02:08)
wrong
based on 54
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\(A = \{ \sqrt{n}, \ n, \ n^2 \}\)
If n is a nonnegative number, is the median and the average (arithmetic mean) of list A equal ?
(1) \(\sqrt{n}= n\)
(2) \(\frac{\sqrt{n}}{3} +\frac{ n}{3} + \frac{n^2}{3} - n = 0\)
If n is a nonnegative number, is the median and the average (arithmetic mean) of list A equal ?
(1) \(\sqrt{n}= n\)
(2) \(\frac{\sqrt{n}}{3} +\frac{ n}{3} + \frac{n^2}{3} - n = 0\)
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31 Oct 2022, 05:10
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Bunuel
(1) \(\sqrt{n}= n\)
Either n = 0 or n=1.
In both cases, all three terms are same, so median is equal to mean, which is equal to n.
(2) \(\frac{\sqrt{n}}{3} +\frac{ n}{3} + \frac{n^2}{3} - n = 0\)
\(\frac{\sqrt{n}+n+n^2}{3}=n\)
Thus mean is n, which is also the median.
Sufficient
D
31 Oct 2022, 14:12
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Bunuel
Given : n is a nonnegative number
Question : Is the median and the average (arithmetic mean) of list A
Statement 1
Squaring both sides
\(n = n^2\)
\(n - n^2 = 0\)
\(n(1-n) = 0\)
Therefore n = 0 or n = 1
For both of these values, the median is equal to the arithmetic mean.
Hence eliminate B, C and E
Statement 2
\(\frac{n+\sqrt{n}+n^2}{3} = n\)
\(\frac{n+\sqrt{n}+n^2}{3}\) is the arithmetic mean
However, let's see if we can conclude n is the median.
We know that n >= 0, so n can lie in between 0 and 1 or n > 1
- Between 0 and 1, \(n^2 < n < \sqrt{n}\)
- When n > 1, \(\sqrt{n} < n < n^2\)
So we can conclude that whatever the value may be, n will be the middle term.
Hence n is the median.
Also, when n = 0, or n = 1; we already know from A that mean = median.
Thus statement 2 represents mean = median
Option D
