# A car manufacturer produced a car at a cost of d dollars and sold it

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. If the dealer sold the car to a consumer for \$15\$ percent more than the dealer paid for it, what did the car cost the consumer, in dollars?

1. \$1.05d\$
2. \$1.23d\$
3. \$1.30d\$
4. \$1.35d\$
5. \$1.38d\$

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 18 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of Percents can be kind of tricky, but never fear, PrepScholar has got your back!

## Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on your paper.

We see the word “percent” mentioned a couple of times, so let’s keep in mind what we’ve learned so far about Percents as we approach this question.

## What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

1. The production cost of a car is \$d\$ dollars
2. The dealer bought the car at a price that was \$20\$ percent higher than the production cost
3. The consumer bought the car at a price that was \$15\$ percent higher than the price the dealer bought it for
4. We want to know how much the consumer paid for the car

## Develop a Plan

The car goes through two consecutive percentage increases in its price. Let’s first recall the equation with percentage increase.

\$\$\New \Price = (1+{\Percentage \Change}/100)·\Old \Price\$\$

Alright, we know that we can solve this question if we add in the first percentage change and then do the second percentage change. We do remember that we cannot just add the two percentages together. For example, an increase of \$25\$ percent followed by an increase of \$20\$ percent will not give us an increase of \$45\$ percent. Instead, we’ll use our percent change equation twice to solve for the final price of the car.

## Solve the Question

The problem gives us the car’s initial production cost, \$d\$. The dealer’s price is \$20\$ percent higher than the production cost:

 \$\Dealer \Cost\$ \$=\$ \$(1+{\Percentage \Change}/100)·\Production \Cost\$ \$ \$ \$ \$ \$\Dealer \Cost\$ \$=\$ \$(1+20/100)·d\$ \$ \$ \$ \$ \$\Dealer \Cost\$ \$ \$ \$(1+0.2)·d\$ \$ \$ \$ \$ \$\Dealer \Cost\$ \$=\$ \$1.2d\$

Excellent! Next, we know that the consumer cost is \$15\$ percent higher than the dealer cost. Let’s use our percent change equation again here.

 \$\Consumer \Cost\$ \$=\$ \$(1+{\Percentage \Change}/100)·\Dealer \Cost\$ \$ \$ \$ \$ \$\Consumer \Cost\$ \$=\$ \$(1+15/100)·1.2d\$ \$ \$ \$ \$ \$\Consumer \Cost\$ \$ \$ \$(1+0.15)·1.2d\$ \$ \$ \$ \$ \$\Consumer \Cost\$ \$=\$ \$1.15·1.2d\$ \$ \$ \$ \$ \$\Consumer \Cost\$ \$=\$ \$1.38d\$

This gives us the final answer, showing that the correct answer is E \$1.38d\$.

## What Did We Learn

Whenever we have consecutive percent changes in a question, we can’t just add the percents together. Instead, we need to apply our percent change equation more than once.

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