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

Price of each car be P
After price increment P+2500
total price of each car after tax (P+2500)*8%

Total price before increment P*8%
15*(P+2500)*8%=18*P*8%

which P =12500
New price 12500+2500=15000
ST 8%
Total price 15000*8% =16200


Answer D

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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Hii how did you come up with the 1/5?

H = 6/5 * L, so H = L + 1/5 * L. Essentially, H = 6/5 * L is the same as H = 1.2 * L, implying the new price is 20% (or 1/5) higher than the old price.