Tag Archives: GRE quantitative

How to Rock Sequences on the GRE

On the GRE, there are two types of sequences to watch out for: arithmetic, and geometric. An arithmetic sequence occurs when there is a constant difference between terms. For example, in a sequence of 3, 5, 7, 9…, then the difference is +2. In a geometric sequence, there is a constant ratio and not a constant difference.

The common ratio is found by dividing the 1st term into the 2nd term. For example, in a sequence of 2, 4, 8, 16…, the ratio is 2, since each term is multiplied by 2 to get the next term.

The concept of sequences is fairly simple, but what to do when a question asks for an impossibly high term, such as the 149th term? There isn’t enough time to write the sequence out that far, so we’d use one of the following formulas:

For Arithmetic: an = a1 + (n – 1)d

For Geometric: an = a1 * r(n-1)

In these equations, an = nth term, a1 = first term in the sequence, d = difference, r = ratio, and n = the number of the term you want to find. For example, if we were asked to find the 33rd term in the geometric sequence above, we would plug in as follows:

an = 2 * 2(33 – 1)
an = 233

Let’s look at a practice question:

1. In the sequence of numbers, a1, a2, a3, a4, a5, each number after the first is 5 times the preceding number. If a4 – a1 is 93, what is the value of a1?
For this question, it is best to choose simple numbers to see the pattern. If a1 is 1, then we know that a2 = 5, a3 = 25 and a4 = 125, so we know that a4 will be 5*5*5 or 125 times the value a1,. No matter what we choose as a1, a4 will always be 125 times greater than a1. We need to find a value such that 125x – x = 93

124x = 93
x = 3/4

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Two Types of GRE “Averages”: Mean and Rates

The word “average” on the GRE can refer to two concepts: arithmetic mean, and the average speed (or average rate) formula. It’s important not to confuse the two on the Test Day, as they require different formulas to solve.

Mean is the mathematical average. This is defined as the sum of the terms divided by the number of terms. Mean = Sum / # of terms. For a list of consecutive integers or evenly spaced numbers, the mean is equal to the median, or the middle number. For example, the “average” of 3, 5, and 9 is 5.67.

Average Speed or Average Rate is often found in complex word problems. This type of question is one many students are less familiar with so you may not have seen it before. Let’s review two important equations to remember and look at how this concept appears on the GRE.

The first formula to memorize is: D = R x T. This stands for Distance = Rate x Time (referred to as the “DIRT” formula). It is perfectly acceptable to also think of it as Time = Distance / Rate or as Rate = Distance / Time as well. Usually the “Rate” is speed but it could be anything “per” anything. In a word problem, if you see the word “per” you know this is a question involving rates.

The second formula is: Average Rate = Total Distance / Total Time. This is its own special concept and you will notice that it is NOT a simple Average of the Speeds (which would be something like the Sum of the Speeds / the Number of Different Speeds or what we know as the Arithmetic Mean). Average Rate is a completely different concept, so do not let the common word “average” confuse you. Let’s look at a sample question from Grockit’s GRE question bank:

Question 1: The average (arithmetic mean) of four numbers is 30, after one of the numbers is removed, the average of the remaining three numbers is 10. What number was removed?

We know that the four original numbers sum to 30*4 = 120. The new equation becomes:

4*30 – x/3 = 10
120 – x/3 = 10
120 – x = 30 (add an x to each side and subtract a 30)
90 = x

Strategies and Formulas for Tough GRE Sets

For some advanced Data Analysis and Probability questions, it will help you achieve better scores to know the logic and formulas behind set theory. Set theory hinges on two concepts: union and intersection. The union of sets is all elements from all sets. The intersection of sets is only those elements common to all sets.

Let’s call our sets A, B, and C, and use a Venn diagram to express their relationship.

If n = intersection and u = union, then we can describe the relationship between the sets thusly:
P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)

To find the number of people in exactly one set: P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)

To find the number of people in exactly two sets: P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)

To find the number of people in exactly three sets: P(A n B n C)

To find the number of people in two or more sets: P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)

To find the number of people in at least one set: P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)

To find the union of all set: (A + B + C + X + Y + Z + O)

Number of people in exactly one set: (A + B + C)

Number of people in exactly two of the sets: (X + Y + Z)

Number of people in exactly three of the sets: O

Number of people in two or more sets: (X + Y + Z + O)

If you’re like me, and formulas like these sometimes seem complicated and intimidating, let’s look at how making a Venn diagram and applying it to a tough GRE question can provide a little relief!

In 1997, N people graduated from college. If 1/3 of them received a degree in the applied sciences, and, of those, 1/4 graduated from a school in one of six northeastern states, which of the following expressions represents the number of people who graduated from college in 1997 who did not both receive a degree in the applied sciences and graduate from a school in one of six northeastern states?

(A) 11N/12
(B) 7N/12
(C) 5N/12
(D) 6N/7
(E) N/7

The key to understanding this question lies in the last sentence:

…who did not both receive a degree in the applied sciences and graduate from a school in one of six northeastern states?

We have two categories to sum: the people who ONLY received a science degree but NOT from one of the 6 schools, and the people who ONLY went to the 6 schools but did NOT receive a science degree. I made up variables for these categories (x and y).

If N = 12, there are 4 applied science students, 1 of which is both. That means x = 3. If 4 students are applied science, then 12-4 = 8 are from one of the six states but NOT applied science. y = 8.

3 + 8 = 11

So we are looking for an answer choice that gives us 11 when N = 12; the answer is A.

You aren’t likely to see many questions at this difficulty level on the actual GRE, but if you continue to challenge yourself, the medium GRE sets questions will soon look easy!

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

The answer is (B).