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Re: At a graduation, guests sit in rows such that there are p pe [#permalink]
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Please Sir, use the search button of the board. There is a high probability that a GRE question is already discussed on the board

https://gre.myprepclub.com/forum/search.php

Also, post a question but the consideration goes at the bottom NOt at the beginning. This way the question posed is more clear

here the post that shows you how to format a question in one step https://gre.myprepclub.com/forum/how-to-po ... 12752.html

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Re: At a graduation, guests sit in rows such that there are p pe [#permalink]
janani11 wrote:
# of guests in nth row = p+n-1
How many more people sit in row n than in the second row = p+n-1 - (p+1) = n-2


Thanks for the reply. I guess I just don't understand where the (n-1) is from. Could you elaborate a little bit, if you have time?
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Re: At a graduation, guests sit in rows such that there are p pe [#permalink]
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In row 3, there are p + 2; in row 12, there would be p + 11, we add one less than the row number to p to determine how many people sit in that row. So in row n there will be p + (n - 1) people.

The number of people who sit in row n is p + n - 1, and the number of people who sit in the second row is p + 1; therefore, the difference is:

p + n - 1 - (p + 1) = p + n - 1 - p - 1 = n - 2

Answer: E
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Re: At a graduation, guests sit in rows such that there are p pe [#permalink]
Thank u. Good question
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Re: At a graduation, guests sit in rows such that there are p pe [#permalink]
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# of guests in first row = p
# of guests in 2nd row = p+1
# of guests in 3rd row = p+2

# of guests in nth row = p+n-1

How many more people sit in row n than in the second row = p+n-1 - (p+1) = n-2

Answer - E
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At a graduation, guests sit in rows such that there are p pe [#permalink]
The pattern is

1st row = p
2nd row = p + 1
3rd row = p + 2
4th row = p + 3
..........
..........
nth row = p + (n-1)

Thus, number of people sitting in the nth row is \(p + (n-1)\)

How many more people sit in row n than in the second row?

It is given by

people in row n - people in row 2

\(p + (n-1) - (p + 1)\)

\(p + n - 1 - p - 1\)

\(n - 2\)

The answer is E
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Re: At a graduation, guests sit in rows such that there are p pe [#permalink]
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The number of people in each row increases by 1 compared to the previous row, so row n will have n more people than the second row
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