A manufacturing company's monthly total production cost C, in dollars, is related to the n… — GRE General Test

A manufacturing company's monthly total production cost

C

, in dollars, is related to the number of units produced,

x

, by the linear equation:

\frac{3(C - 1,200)}{5} = 2x + 180

If the total production cost in a given month was between

\$4,000

and

\$5,000

, inclusive, which of the following could be the number of units produced in that month? Select all such numbers.

The numbers of units that could be produced are 800, 950, and 1,000.

Solving the linear equation for

x

gives

x = \frac{3C - 4,500}{10}

. Substituting the boundary cost values

C = 4,000

and

C = 5,000

yields

x = 750

and

x = 1,050

, respectively. Therefore, any valid number of units must lie in the range

[750, 1,050]

. The values 800, 950, and 1,000 fall strictly within this interval.

Express x in terms of C by isolating the variable x in the linear equation.

3(C - 1,200) = 5(2x + 180) \implies 3C - 3,600 = 10x + 900 \implies 10x = 3C - 4,500 \implies x = \frac{3C - 4,500}{10}

Isolating

x

allows direct evaluation of the unit limits based on the cost boundary values.

Calculate the minimum value of x corresponding to the minimum cost C = 4,000.

x_{\min} = \frac{3(4,000) - 4,500}{10} = \frac{12,000 - 4,500}{10} = \frac{7,500}{10} = 750

Determines the lower bound of units produced.

Calculate the maximum value of x corresponding to the maximum cost C = 5,000.

x_{\max} = \frac{3(5,000) - 4,500}{10} = \frac{15,000 - 4,500}{10} = \frac{10,500}{10} = 1,050

Determines the upper bound of units produced.

Filter the given choices to find all values within the interval [750, 1,050].

The valid values are 800, 950, and 1,000.

Only values between 750 and 1,050 inclusive satisfy the cost constraint.

Linear Equations in One Variable

Soru