How Many Seats?Date: 01/15/2003 at 11:25:35 From: Shay Subject: Word problem How do you increase without using powers? There are 40 rows in an auditorium. Each has two more seats than the previous row. There are 1960 seats in all. How many seats are in the first row of the auditorium? Date: 01/18/2003 at 12:16:49 From: Doctor Edwin Subject: Re: Word problem Hi, Shay. We need a formula to that gives us the number of seats, based on the number of rows and the number of seats in the starting row. If we call the number of seats in the first row X, then we can write, seats = (some formula involving X) 1960 = (some formula involving X) and then solve for X and we're done. The formula has to be something like: 1960 = X + (X + 2) + (X + 4) + (X + 6) ... (X + 78) We can simplify that to: 1960 = 40X + (2 + 4 + 6 + 8 + ... + 78) Now you can just use a calculator to add all the even numbers from 2 to 78, solve the equation for X, and you'll have your answer. But there's a nice shortcut for adding up a series of numbers that increases evenly. Let's call N the number of numbers: sum = N * (the average of all the numbers) How do we get the average of all the numbers? We just add them all up and... oops. That wasn't much of a shortcut, was it? Fortunately, we have another trick. The average of a series of numbers that increases evenly is the same as the average of the first and last numbers (I'll explain why below). That leaves us with: (first number + last number) sum = N * ---------------------------- 2 If we substitute in the information from your problem, we get (X + (number of seats in last row)) 1960 = 40 * ----------------------------------- 2 Now, I said that the average of a series of numbers is the same as the average of the first and last numbers if they increase evenly. Why is that true? Well, think of it this way. If you have a sequential group of numbers, the average is going to be in the middle. Let's say that you take a number that's 5 higher than the average away from the group. The average would go down, right? What if you took away one that was 5 higher and one that was 5 lower? The average would stay the same. So you could take away matched pairs of numbers from the set until you were left with just the first and last ones, without changing the average at all. If there is an odd number of numbers in the group, you'll be left with one right smack in the middle. Taking that one out won't change the average either. So how do you add up a bunch of integers? Well, let's start simply: sum = (all the integers added up) Now, if I divide something by a number and then multiply it by the same number, I haven't changed it, right? so I can do this: (all the integers added up) sum = (how many integers) * --------------------------- (how many integers) But "all the integers added up divided by the number of integers" is just a way of saying, "the average integer." So I can rewrite my equation: sum = (how many integers) * (the average integer) (seats in first row + seats in last row) seats = rows * ---------------------------------------- 2 (seats in first row + seats in last row) seats = rows * ---------------------------------------- 2 We know that there are 40 rows and 1960 seats: (seats in first row + seats in last row) 1960 = 40 * ---------------------------------------- 2 Let's call the number of seats in the first row F. How many seats are there in the last row, 39 rows later? F + something, right? So we have: (F + (F + something)) 1960 = 40 * --------------------- 2 Can you take it from here? Write back if you need more help. - Doctor Edwin, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/