Question
The cost of making a table partly varies as the area and partly varies as square of the length. lf the cost of a table 2m by 3m is $50000 and the cost of a table 1.5m by 4m is $64000. Find the cost of a table 2m by 4m.
Answers
GPT 3.5
Let's assume that the cost of the table is directly proportional to the area and the square of the length. Therefore, we can write the equation:
Cost = k * (Area) * (Length^2)
where k is the constant of proportionality.
From the given information, we have two equations:
1) Cost1 = k * (Area1) * (Length1^2)
$50000 = k * (2m * 3m) * (2m^2)
2) Cost2 = k * (Area2) * (Length2^2)
$64000 = k * (1.5m * 4m) * (4m^2)
Let's divide equation 2 by equation 1 to eliminate k:
Cost2/Cost1 = (Area2 * Length2^2) / (Area1 * Length1^2)
($64000/$50000) = ((1.5m * 4m) * (4m^2)) / ((2m * 3m) * (2m^2))
Simplifying this expression:
1.28 = (6m^3) / (6m^3)
So, the value of 1.28 holds true for the ratio of the costs of two tables. Let's assume the cost of the table 2m by 4m is C:
C = 1.28 * Cost1
C = 1.28 * $50000
C = $64000
Therefore, the cost of a table 2m by 4m is $64000.
Cost = k * (Area) * (Length^2)
where k is the constant of proportionality.
From the given information, we have two equations:
1) Cost1 = k * (Area1) * (Length1^2)
$50000 = k * (2m * 3m) * (2m^2)
2) Cost2 = k * (Area2) * (Length2^2)
$64000 = k * (1.5m * 4m) * (4m^2)
Let's divide equation 2 by equation 1 to eliminate k:
Cost2/Cost1 = (Area2 * Length2^2) / (Area1 * Length1^2)
($64000/$50000) = ((1.5m * 4m) * (4m^2)) / ((2m * 3m) * (2m^2))
Simplifying this expression:
1.28 = (6m^3) / (6m^3)
So, the value of 1.28 holds true for the ratio of the costs of two tables. Let's assume the cost of the table 2m by 4m is C:
C = 1.28 * Cost1
C = 1.28 * $50000
C = $64000
Therefore, the cost of a table 2m by 4m is $64000.