A dealer once sold identical cars at a piece of x dollars each (includ

Let L = the lower price and H = the higher price.

18 cars purchased at the old price would cost the same as purchasing three fewer cars at the new price.
Thus:
18L = 15H
6L = 5H
\(H = \frac{6}{5}L\)

Implication:
The higher price is 1/5 greater than the lower price.
Thus, the $2,500 increase must be equal to 1/5 of the lower price:
\(2500 = \frac{1}{5}L\)
\(L = 12,500\)

Since the higher price is $2500 greater than the lower price, we get:
H = 12,500 + 2500 = 15,000
H + 8% tax = 15,000 + 1200 = 16,200

The number of cars that can be bought is inversely proportional to the pre-tax price of each car

In other words, if the pre-tax price goes up, since the tax rate is constant, the number of cars we can buy will go down proportionally


Under the old price X, we could buy 18 cars

Now we can buy 15 cars because of the $2,500 pre tax price increase.

This represents a Fractional Decrease of ———> (15 - 18) / 18 = -(3/18) = -(1/6)

Rule: when 2 quantities are inversely proportional to each other, a Decrease in one quantity by: -(n) / (d + n)

Results in an INCREASE in the other quantity by: +(n / d)


We have a decrease in the number of cars we can buy = -(1/6)

This corresponds to an Increase in the pre-tax price of the car by = + (1/5)

And this +(1/5) Proportional increase in the pre-tax price of the car corresponds to a value increase of +$2,500

Let the original pre-tax price of the vehicle be = P

Then:

(P) + (1/5)(P) = (P) + $2,500

Or

(1/5)(P) = $2,500

P = $12,500 = Original pre-tax price of the car

The new pre-tax price is therefore:

$12,500 + $2,500 = $15,000


We also need to add in the 8% sales tax

1% ———> is $150

*8

8% ———> is $1,200


$15,000 + ($1,200 tax) =


16,200

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