SOLUTION: A rectangular auditorium seats 1505 people. The number of seats in each row exceeds the number of rows by 8. Find the number of seats in each row.
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Question 818108: A rectangular auditorium seats 1505 people. The number of seats in each row exceeds the number of rows by 8. Find the number of seats in each row. Found 4 solutions by mananth, amalm06, ikleyn, greenestamps:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! seats in each row =x
number of rows = y
x=y+8
x-y=8....................(1)
xy=1505
x= 1505/y..................(2)
1505/y -y = 8
1505-y^2=8y
y^2+8y-1505=0
y^2+43y-35y-1505=0
y(y+43)-35(y+43)=0
(y+43)(y-35)=0
y=-43 OR 35
There are 35 seats in each row and 43 rows
You can put this solution on YOUR website! .
There is much more elegant solution which allows to avoid boring calculations and solve the problem MENTALLY.
Let "m" be an unknown number which is 4 more than the number of rows: m = rows +4.
Then "m" is 4 less then the number of seats in a row: m = seats - 4.
The product seats*rows is 1505, or (m+4)*(m-4) = 1505.
Hence, = 1505 + 16 = 1521, and m = = 39.
Then the number of rows is m-4 = 39-4 = 35 and the number of seats in each row is m+4 = 39+4 = 43.
I definitely concur with tutor ikleyn that this kind of problem is much better solved using a little logical reasoning, instead of formal mathematics. Indeed, when you try to solve it using the traditional algebraic approach, you end up having to factor a quadratic by finding two numbers whose product is 1505 and whose difference is 8. But THAT IS WHAT THE ORIGINAL PROBLEM asks you to do.
So the formal algebra is a waste of time....
And the "trick" tutor ikleyn uses is a very useful one. If you know that
then you know that
and so the answers are 39-4=35 and 39+4=43.
It did not occur to me to use that trick on this problem. I just started playing with numbers to find a pair with a product of 1505 and a difference of 8. There weren't many possibilities....
I of course first saw
Then I had to find a factorization of 301; when I found it, I had my answer.
