A salesperson is offered two different monthly salary plans. Under Plan A, the salary is $1,800 plus a 4% commission on…

Correct answer: C. 42,000

Let x be the amount of total sales in dollars. The monthly salary for each plan can be represented by a linear function. For Plan A, the salary S_A is the base salary plus 4% of sales: S_A = 1800 + 0.04x. For Plan B, the salary S_B is the base salary plus 6% of sales: S_B = 1200 + 0.06x. The question asks for the minimum sales for which the salary under Plan B is at least $240 more than the salary under Plan A. This can be written as an inequality: S_B ≥ S_A + 240 Substitute the expressions for S_A and S_B into the inequality: 1200 + 0.06x ≥ (1800 + 0.04x) + 240 First, simplify the right side of the inequality: 1200 + 0.06x ≥ 2040 + 0.04x Next, solve for x. Subtract 0.04x from both sides: 1200 + 0.02x ≥ 2040 Subtract 1200 from both sides: 0.02x ≥ 840 Finally, divide both sides by 0.02: x ≥ 840 / 0.02 x ≥ 42,000 The inequality shows that the sales must be greater than or equal to $42,000. Therefore, the minimum amount of sales is $42,000. Choice A is incorrect; it is the result of reversing the comparison and solving for when Plan A is at least $240 more than Plan B. Choice B is incorrect; it represents the break-even point where the two salaries are equal, without considering the '$240 more' condition. Choice D is incorrect; it's the result of a sign error when isolating the variable, adding 1200 to the right side instead of subtracting it.