SOLUTION: James went to the theatre to watch a concert with his family. The theatre had 4270 seats. There were 60% fewer $50-seats than $30-seats. 50% of the $30-seats and some $50-seats wer

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: James went to the theatre to watch a concert with his family. The theatre had 4270 seats. There were 60% fewer $50-seats than $30-seats. 50% of the $30-seats and some $50-seats wer      Log On


   



Question 1203093: James went to the theatre to watch a concert with his family. The theatre had 4270 seats. There were 60% fewer $50-seats than $30-seats. 50% of the $30-seats and some $50-seats were sold. A total of $47 500 was collected. How many seats were unsold?
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52887) About Me  (Show Source):
You can put this solution on YOUR website!
.
James went to the theatre to watch a concert with his family.
The theatre had 4270 seats.
There were 60% fewer $50-seats than $30-seats.
50% of the $30-seats and some $50-seats were sold.
A total of $47 500 was collected. How many seats were unsold?
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Let x be the number of the $30-seats in the theater.
Then the number of the $50-seats in the theater is (4270-x).


The problem says that

    4270-x = (1-0.6)x.    ("There were 60% fewer $50-seats than $30-seats")


From this equation, 

    4270-x = 0.4x

    4270 = x + 0.4x

    4270 = 1.4x

    x    = 4270/1.4 = 3050.


Thus, there were 3050 of the $30-seats in the theater and (4270-3050) = 1220 of the $50-seats.


Next, 50% of the $30-seats were sold, which is 3050/2 = 1525 seats,
providing 30*1525 = 45750 dollars of the total revenue.


Hence, the rest part of the revenue, or 47500-45750 = 1750 dollars, was the cost of the $50-seats sold.

So, the number of the $50-seats sold was 1750/50 = 35.


Thus, the number of the total sold seats was 1525 + 35 = 1560.


The number of unsold seats was the rest, or 4270 - 1560 = 2710.


ANSWER.  The number of the unsold seats was 2710.

Solved.



Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
this is what i get.

A = the number of 30 dollar seats available.
B = the number of 50 dollar seats available.
the total number of seats available is 4270.
the equation for that is A + B = 4270

there were 60% fewer 50 dollar seats available than 30 dollar seats.
the equation for that is B = A - .6 * A = .4 * A.
replace B with .4 * A in the equation of A + B = 4270 to get:
A + .4 * A = 4270
combine like terms to get:
1.4 * A = 4270
solve for A to get:
A = 4270 / 1.4 = 3050 thirty dollar seats available.
this means that there are 4270 - 3050 = 1220 fifty dollar seats available.

50% of the 30 dollar seats and some of the 50 dollar seats were sold for a total revenue of 47500.
50% times 3050 = 1525 thirty dollars seats that were sold.
let x equal the number of 50 dollar seats that were sold and the equation for the total revenue becomes:
1525 * 30 + 50 * x = 47500.
simplify to get 45750 + 50 * x = 47500
subtract 45750 from both sides of this equation to get 50 * x = 1750
solve for x to get x = 35.

1525 thirty dollar seats were sold and 35 fifty dollar seats were sold for a total revenue of 1525 * 30 + 35 * 50 = 47500.

the number of seats that were unsold was 5050 minus 1525 = 1525 thirty dollar seats and 1220 - 35 = 1185 fifty dollar seats for a total of 2710 seats that were unsold.

that shoud be your solution.