# 3 Ways of Looking at “Profit” Questions on the GMAT As someone who is about to shell out hundreds of dollars in MBA application fees, you know that money makes the GMAT-world go round. Profit is an essential concept for any aspiring MBA admissions applicant. The GMAT tests this concept in both Problem Solving and Data Sufficiency questions in three main ways. Let’s examine the need-to-know formulas with three GMAT practice questions.

1. A firm increases its revenues by 10% between 2008 and 2009. The firm’s costs increase by 8% during this same time. What is the firm’s percent increase in profits over this period, if profits are defined as revenues minus costs?

(1) The firm’s initial profit is \$200,000.

(2) The firm’s initial revenues are 1.5 times its initial costs.

In this question from Grockit, we can start with our most basic Profit formula:

Profit = Revenue – Cost

Using Statement (1), we can say that 200,000 = R – C.
(1.1)r – (1.08)c = 200,000(1 + x), where x equals the amount of the increase. We still do not know R and C so we can’t find x. Insufficient.

Using Statement (2), 1.5c – c = p and (1.1)(1.5)c – (1.08)c = (1 + x)P. Here we can simplify.

.5c = p

.57c = (1 + x)p
Without continuing to solve, we can see that we can solve for x using substitution. .57c = (1 + x)(.5c), and dividing both sides by c will cancel out that variable and allow us to isolate x. Statement 2 is sufficient. Now to a more challenging question!

2. A store purchased 20 coats that each cost an equal amount and then sold each of the 20 coats at an equal price. What was the stores gross profit on the 20 coats?

(1) If the selling price per coat had been twice as much, the store’s gross profit on the 20 coats would have been \$2400.

(2) If the selling price per coat had been \$2 more, the store’s gross profit on the 20 coats would have been \$440.

Gross Profit = Selling price – Cost

For the value Data Sufficiency question, we need to know the price of each coat and the selling price of each coat. From the given information, we can use our known formula to set us the equation: P = 20 (s – c). So either we’ll need a value for s and a value for c, or we’ll need the value of (s – c).

Statement (1) tells us that \$2400 = (20(2s – c)) or 2400 = 40s – 20c. We can divide both sides by 20 and simplify it to: 120 = 2s – c. We still don’t know s and c. Insufficient.

Statement (2) tells us that 440 = 20(s + 2 – c). Let’s simplify: 440 = 20s + 40 – 20c. 400 = 20s – 20c. 400 = 20 (s – c). 20 = s – c. Sufficient. Even though we didn’t solve for s and c separately, we were able to find the value of (s – c). Sometimes DS will surprise you!

3. If the cost price of 20 articles is equal to the selling price of 25 articles, what is the % profit or loss made by the merchant?

A. 25% loss
B. 25% profit
C. 20% loss
D. 20% profit
E. 5% profit

Profit/Loss % = (Sales Price – Cost Price) / Cost Price x 100

The question asks about % profit or loss. It tells us that 20c = 25s, or 4c = 5s. So the ratio of the sales price to the cost price is 4/5.

Let’s simplify our Profit/Loss % formula by dividing each term by the cost price:

Profit/Loss % = (S/C – C/C) x 100

P/L% = (S/C – 1) x 100
We know that S/C = 4/5 for this problem. So we can plug in and solve:

P/L% = (4/5 – 1) x 100

P/L% = (-1/5) x 100

P/L% = -20%. The answer is a 20% loss.

# Breaking the GMAT Word Problem Barrier!

“Expect problems and eat them for breakfast.”

– Alfred A. Montapert Word problems on the GMAT get an unfair reputation for being especially challenging. Once you’re able to effortlessly translate the key phrases into algebra, you’ll be able to handle any word problem with aplomb!

Here’s my favorite resources from around the web all in one convenient Learnboard to help you overcome your fear of word problems: GMAT – Word Problems. And, no worries if Word Problems are the bane of your existence, we’ll start with the very basics!

# Compound Interest on the GMAT “Compound interest” occurs when interest earned is added to the principal, which then earns interest. If you’re an investor, compound interest is a very good thing! Usually interest appears on the GMAT in the form of a word problem. Let’s solve one together!

An amount is deposited into an account accruing interest annually at a fixed percentage rate. It is valued at \$900 in the third year (after interest has compounded twice), and \$1080 in the fourth year (after interest has compounded three times). What is the original amount?

A) 540
B) 600
C) 625
D) 750
E) 800

To calculate the amount earned under compound interest under a certain period of time, you need to know three pieces of information: 1) the initial principle “P”, 2) the rate of interest “R”, and 3) the number of times it compound per year. You will always raise the rate to a power equal to the number of times it compounds.

For example, the amount in an account started with an initial investment of \$200 and earning compound interest at a rate of 8% compounded three times a year would be found by the equation:

1 + (8/100)^3 * 200 = Final Amount

Knowing this, we can set up these formulas with the given information from this problem:

(1+(R/100)^2)*P = 900
(1+(R/100)^3)*P = 1080

We can put R/100 so that the R will be the percent and not a decimal. Then we can divide the 2nd equation by the 1st to get:

(1+ R/100)= 1080/900
1 + R/100 = 1.2

Now we can plug in 1.2 for (1 + R/100) into our original equation to find the “original amount”:

1.2^2*P = 900
1.44 * P = 900
P = 900/1.44
P = 625

# GRE Quant Question of the Day: Rates! Try this “rates” question on your own!

A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?

(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph

First, let’s consider Day 1 and Day 2’s hours.

If x = hours on Day 1, then x + 2 = hours on Day 2. The question said he walked 18 hours total, so we can set up a simple equation:

x + (x + 2) = 18
2x + 2 = 18
2x = 16
x = 8

Therefore he walked 8 hours on Day 1 and 10 hours on Day 2.

We are told he went 1mph FASTER on Day 2. So if Day 1’s mph is y, then Day 2’s mph is y + 1.

Let’s look at the D = R x T formula.

D1 = R1 x T1

D2 = R2 x T2

If we plug in what we know:

D1 = (y) x 8 hrs

D2 = (y + 1) x 10 hrs

We know that D1 + D2 must equal 64, so we can sum the two equations and set them equal to 64.

(y) x 8hrs + (y + 1) x 10hrs = 64

Simplifying…

64 = 8y + 10y + 10

64 = 18y + 10

54 = 18y

3 = y