Pricing

PP1 (Shorter) Quant Section 2 (Hard) Q13

Solving the Problem We can solve this using two methods: Method 1 (Algebra) <blockquote> A car manufacturer produced a car at a cost of d dollars and sold it to a dealer at a price 20 percent higher than the production cost. </blockquote> If the dealer bought the car at $20\%$ higher than the production cost, then the dealer would have paid $d + 20\%d = d + 0.2d = 1.2d$. <blockquote> If the dealer sold the car to a consumer for 15 percent more than the dealer paid for it,  </blockquote> If the customer bought the car at $15\%$ more than the dealer's cost, then the customer would have paid $dealer's\,cost + 15\%(dealer's\,cost) = dealer's\,cost + 0.15(dealer's\,cost) = 1.15(dealer's\,cost)$. <blockquote> what did the car cost the consumer, in dollars? </blockquote> Since the answer choices are given in terms of the production cost $d$, we have to substitute the $dealer's\,cost$ in terms of the production cost $d$ into the equation above: $1.15(dealer's\,cost) = 1.15(1.2d) = 1.38d$ Method 2 (Choosing Numbers) <blockquote> A car manufacturer produced a car at a cost of d dollars and sold it to a dealer at a price 20 percent higher than the production cost. </blockquote> Let's say that the car costs $\$100$ to produce. In this case, the dealer would have paid $\$100 + 20\%\,of\,\$100 = \$100 + \$20 = \$120$. <blockquote> If the dealer sold the car to a consumer for 15 percent more than the dealer paid for it,  </blockquote> If the customer bought the care at $15\%$ more than the dealer's cost, then the customer would have paid $dealer's\,cost + 15\%(dealer's\,cost) = dealer's\,cost + 0.15(dealer's\,cost) = \$120 + \$18 = \$138$. <blockquote> what did the car cost the consumer, in dollars? </blockquote> We can see that only when we substitute our production cost $d$ of $\$100$ into the expression $1.38d = 1.38(100)$, we get our desired value of $\$138$. So, the answer here is E.