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Question
the tax on the sale of a car in michigan is 6%. at a dealership in ann arbor, a car purchase also involves $120 in miscellaneous charges.
- there are several quantities in this situation: the original car price, sales tax, miscellaneous charges, and total price. write an equation to describe the relationship between all the quantities when:
a. the original car price is $9,500.
b. the original car price is $14,699.
c. the total price is $22,480.
d. the original price is $p$.
- how would each equation you wrote change if the tax on car sales is $r%$ and miscellaneous charges are $m$ dollars?
Step1: Recall the relationship formula
The total price $T$ of the car is the sum of the original car - price $P$, the sales - tax (which is a percentage of the original price), and the miscellaneous charges. The sales tax rate is $6\%=0.06$ and the miscellaneous charges are $120$. So the formula is $T = P+0.06P + 120=1.06P + 120$.
Step2: Solve part a
When $P = 9500$, substitute $P$ into the formula:
$T=1.06\times9500 + 120=1.06\times9500+120 = 10070+120=10190$.
Step3: Solve part b
When $P = 14699$, substitute $P$ into the formula:
$T=1.06\times14699+120=15580.94 + 120=15700.94$.
Step4: Solve part c
We know that $T = 22480$, and the formula is $T = 1.06P+120$. So we solve for $P$:
$22480=1.06P + 120$.
First, subtract 120 from both sides: $22480 - 120=1.06P$, so $22360 = 1.06P$.
Then, divide both sides by 1.06: $P=\frac{22360}{1.06}=21094.34$.
Step5: Solve part d
When the original price is $p$, the total - price formula is $T = 1.06p+120$.
Step6: Solve part 2
If the tax rate is $r\%= \frac{r}{100}$ and the miscellaneous charges are $m$ dollars, the new formula for the total price $T$ in terms of the original price $P$ is $T=P+\frac{r}{100}P + m=(1 + \frac{r}{100})P + m$.
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a. $T = 1.06\times9500+120 = 10190$
b. $T = 1.06\times14699+120 = 15700.94$
c. $P=\frac{22480 - 120}{1.06}\approx21094.34$
d. $T = 1.06p+120$
- $T=(1+\frac{r}{100})P + m$