SOLUTION: A party rental company has chairs and tables for rent. The total cost to rent 2 chairs and 5 tables is $33. The total cost to rent 8 chairs and 3 tables is $30. What is the cost to
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Question 1172900: A party rental company has chairs and tables for rent. The total cost to rent 2 chairs and 5 tables is $33. The total cost to rent 8 chairs and 3 tables is $30. What is the cost to rent each chair and each table?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
The first given rental is for 2 chairs and 5 tables, for a cost of $33. The second is for 8 chairs and 3 tables, for a cost of $30.
Observe that the second rental has 4 times as many chairs as the first.
So imagine a third rental that is 4 times the first; it has 8 chairs and 20 tables, for a total of $132.
Now the second and third rentals have the same numbers of chairs; the second had 3 tables and the third has 20 tables. The difference in the number of tables is 17; and the difference in cost is $132-$30 = $102. That means the cost of each table is $102/17 = $6.
So the first rental of 2 chairs and 5 tables, for a cost of $33, includes 5($6) = $30 for the tables. So the two chairs cost $3, and then each chair costs $1.50.
ANSWERS: $6 for each table; $1.50 for each chair.
A solution using formal algebra uses exactly the same calculations as the informal solution above.
2c+5t = 33 (1)
8c+3t = 30 (2)
8c+20t = 132 (3) [(1), multiplied by 4]
17t = 102 [the difference between (2) and (3)
t = 6
2c+5(6) = 33
2c = 3
c = 1.5
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