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A perfect square is an integer whose square root is an integ
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23 May 2020, 02:40
23 May 2020, 02:40
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Question Stats:
69% (01:30) correct
30% (01:28) wrong based on 113 sessions
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A perfect square is an integer whose square root is an integer.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The average (arithmetic mean) of the first 100 positive perfect squares |
The median of the first 100 positive perfect squares |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: A perfect square is an integer whose square root is an integ
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15 Nov 2021, 19:55
15 Nov 2021, 19:55
3
Sum of square of integers starting from 1 to n = \(1^2 + 2^2 + 3^2 + ..... + n^2 = \frac{n(n+1)(2n+1)}{6}\)
Average of sum of square of integers starting from 1 to n = \(\frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}\)
Average of sum of the first 100 positive perfect squares = \(\frac{(100+1)(2(100)+1)}{6} = \frac{101*201}{6} = \)3383.5
Median of the first 100 positive perfect squares = (50th term + 51st term)/2 = \(\frac{50^2 + 51^2}{2} = \)2550.5
Hence, Answer is A
Average of sum of square of integers starting from 1 to n = \(\frac{n(n+1)(2n+1)}{6n} = \frac{(n+1)(2n+1)}{6}\)
Average of sum of the first 100 positive perfect squares = \(\frac{(100+1)(2(100)+1)}{6} = \frac{101*201}{6} = \)3383.5
Median of the first 100 positive perfect squares = (50th term + 51st term)/2 = \(\frac{50^2 + 51^2}{2} = \)2550.5
Hence, Answer is A
General Discussion
Re: A perfect square is an integer whose square root is an integ
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23 May 2020, 18:09
23 May 2020, 18:09
4
Not sure if I'm using the right language here, but: Think about how quickly the numbers blow up as you go up with each square. The larger the numbers are in a distribution, the greater the average of those numbers will be because the numbers will skew right (and the more likely the average is going to be greater than the median.)
Alternatively, if you want to write this out to see it blow up just for the first 20 perfect squares (the same thing will happen but on a larger scale from 1 to 100):
First 10 perfect squares: 1, 4, 9, 16, 25 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400
Average: 143, Median: 110.5
^ Unless you've memorized these, you won't have time to calculate all of these on the exam. So you'll have to stick to having that number sense (unless someone has another way to think of this)
Alternatively, if you want to write this out to see it blow up just for the first 20 perfect squares (the same thing will happen but on a larger scale from 1 to 100):
First 10 perfect squares: 1, 4, 9, 16, 25 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400
Average: 143, Median: 110.5
^ Unless you've memorized these, you won't have time to calculate all of these on the exam. So you'll have to stick to having that number sense (unless someone has another way to think of this)
