# Ratios and Proportions on the GMAT

A ratio expresses the relationship between two or more things. A proportion is a relationship that is formed by setting two ratios equal. Learn how to solve proportion problems using equivalent ratios on the GMAT….like a rockstar on this Learnboard!

Once you’ve reviewed the board, try this Data Sufficiency problem on your own:

For each month, the number of accounts, a, that a certain salesman has contracted that month is directly proportional to his efficiency score, e, which is directly proportional to his commission rate, c. What is a if c = 3.0?

(1) Whenever c = 4.0, e = 0.3

(2) Whenever c = 6.0, a = 80

Explanation:

It will be helpful to first note that because a is directly proportional to e, which is in turn directly proportional to c, a is then directly proportional to c. To say that a is directly proportional to c is just to say that there is a constant k such that ck = a, or, perhaps more simply, that there is a fixed ratio between a and c. A statement, or set of statements, will be sufficient if and only if it determines that ratio.

Statement (1): From this, the proportional relationship between e and c can be determined. However, a is directly proportional to e, and nothing is said about that relationship; therefore, the value of a when c = 3.0 cannot be found; NOT sufficient.

Statement (2): This gives you the ratio you want. You don’t need to actually calculate the value of a if c = 3.0. You just need to know that it’s possible. Don’t believe me? Because a is directly proportional to c: a/c = 80/6.0. Since the question asks for the value of a when c = 3.0, divide the numerator and denominator each be 2. a = 40. Or, if you’re determined to cross multiply: substitute the given value for c: a/3.0 = 80/6.0. By cross multiplication, 6a = 240. Therefore, a = 40; SUFFICIENT. The credited response is B.

# Learnist: Basic Algebra on the ACT and SAT

Expert Katie Cantrell over at Learnist offers a great look at the fundamental algebra concepts on the ACT and SAT exams. Refresh how to isolate variables, solve basic equations, and apply the order of operations on Test Day!

This video in particular covers several ways to conceive of variables, how to solve linear equations with fractions, and how to check your work to ensure that you have found the correct answer.

# 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