You are choosing between two plans at a discount warehouse. Plan A offers an annual membership fee of $120 and you pay 80% of the manufacturer's recommended list price. Plan B offers an annual membership fee of $40 and you pay 90% of the manufacturer's recommended list price. How many dollars of merchandise would you have to purchase in a year to pay the same amount under both plans? What will be the cost for each plan?

Respuesta :

Let $A be the cost of $q of merchandise under plan A and similarly $B under plan B.
A=120+0.80q, B=40+0.90q
When A=B, 120+0.80q=40+0.90q, 80=0.10q, q=80/0.10=$800.
So you would need to buy $800 of merchandise.
A=B=120+640=40+720=$760 for each plan.

An equation is formed when two equal expressions. The number of merchandise that you would have to purchase in a year to pay the same amount under both plans is 800.

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

Given that you are choosing between two plans at a discount warehouse.

The cost for x number of merchandise under plan A, which offers an annual membership fee of $120 and you pay 80% of the manufacturer's. Therefore, the cost will be,

Cost = 120 + 0.8x

The cost for x number of merchandise under plan B, which offers an annual membership fee of $40 and you pay 90% of the manufacturer's recommended list price. Therefore, the cost will be,

Cost = 40 + 0.9x

Now, in order to find the number of merchandise that you would have to purchase in a year to pay the same amount under both plans, you need to equate the two equations together. Therefore,

Cost from Plan A = Cost from Plan B

120 + 0.8x = 40 + 0.9x

120 - 40 = 0.9x - 0.8x

80 = 0.1x

x = 80 / 0.1

x = 800

Hence, the number of merchandise that you would have to purchase in a year to pay the same amount under both plans is 800.

Further, the cost for each plan for 800 merchandise can be written as,

PlanA,

Cost = 120 + 0.8(800)

        = $760

Plan B,

Cost = 40 + 0.9(800)

        = $760

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