I need help with this one math problem.

Question

A mechanic bought several high performance distributors for a total of $2000. When his unit cost decreased by $100, he was able to buy twice as many for the same total price. How many distributors did he buy originally?

A. 10

B. 5

C. 40

D. 20

Answer 1

Let's try to express this word problem as an algebraic equation.

n = number of distributors originally purchased

x = original cost per distributor

n*x = 2000 (the mechanic bought several distributors for 2000)

2n*(x-100) = 2000 (when the unit price dropped by 100, the mechanic bought twice as many distributors)

We can see that:

2000 = 2000

n*x = 2n*(x-100)

x = 2*(x-100)

x/2 = x - 100

-x/2 = -100

x = 200

So the original unit price was $200/distributor.

We know that n*x = 2000, so n*(200) = 2000, so n = 10.

Hence, the mechanic originally purchased 10 distributors at $200/distributor for a total of $2000.