Question 249564: I am having problems with this question. I'm not sure what equation to use or how to go about solving it. Could you please help me figure out how to work out the question. Thank-you.
When the Vancouver choir performed at BC place, the 330 singers were arranged in rows of equal lenth. The choir next performed in Coquitlam where the stage was narrower. They arranged the choir with four fewer singers in each row, but they filled up one extra row. However, when they counted the singers there were eleven less than had sung at BC Place.
How many rows of singers were there when the choir sang at BC Place?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = number of singers in each row.
let y = number of rows.
at BC we have x*y = 330
at Coquitlam we have (x-4)*(y+1) = 319
we need to solve both these equations simultaneously to get our answer.
use the first equation to solve for one of the variable in terms of the other.
you get y = 330/x
substitute for y in the second equation to get:
(x-4) * (330/x + 1) = 319
simplify by performing indicated operations to get:
330 + x - (4*330)/x - 4 = 319
multiply both sides of the equation by x to get:
330*x + x^2 - (4*330) - 4*x = 319*x
subtract 319*x from both sides of the equation to get:
330*x + x^2 - (4*330) - 4*x - 319*x = 0
simplify by performing indicated operations to get:
330*x + x^2 - 1320 - 4*x - 319*x = 0
simplify by combine like terms to get:
x^2 + 7*x - 1320 = 0
factor to get:
(x+40) * (x-33) = 0
solve for x to get:
x = -40 and x = 33
since x can't be negative, x must equal 33.
x = 33 means there were 33 singers in each row at BC.
this means y = 10 which means there were 10 rows at BC.
at Coquitlam they had 4 less singers in each row and 1 more row.
This makes 29 singers and 11 rows at Coquitlam.
29 * 11 = 319 singers total at Coquitlam.
Everything checks out so your answer is:
there were 10 rows of singers at BC.
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