"Give them the gift of words"

GRE Vocabulary 2 – Word Parts

It does work sometimes to use the parts of the word to guess its meaning if you don’t know.

In this video, we’ll discuss how to learn and use parts of words to help guess word meaning.

GRE Vocabulary 1 – Learning Vocabulary

So, having a strong vocabulary is really important for success in the GRE’s verbal reasoning section. Reading will help you build vocabulary over time but you’re also gonna want to study words that you don’t know.

In this video, we’ll give you a few ideas about how to study vocabulary effectively.

Prime & Composite: Dividing a Number

Problems involving prime and composite numbers are often not too hard. It’s good to know prime numbers at least till 50, or even better, 100.

As a recap, the prime numbers till 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

Q. Amy is given a number x. She picks two numbers a and b and then generates a sequence of numbers x, x +a, x + a + b, x + 2a + b, x + 2a + 2b, i.e. she alternately adds a to get a new number and then b to get another number. Which two numbers should she choose for a and so that she is guaranteed to generate a prime number, regardless of the choice of x?

a.) 2, 3
b) 3, 6
c) 2, 4
d) 7, 14

So we don’t want to get stuck with composite numbers alone. This happens if aand bhave a common factor. For example if a = 3 and b = 6, then picking x = 3 leaves us with composite numbers alone. So option a is the correct answer.

Q. There are k students in a class. What could be the value of k if every attempt to sort them perfectly into teams of greater than 2, but less than k fails?

a) 6
b) 17
c) 18
d) 301

k is clearly a prime. So the answer is 17.

This was a really easy question. A slightly hard question would be something like this:

Q. Which of the following could be a value of k if k has an odd number of factors?

a) 72
b) 36
c) 45
d) 60

You can factorise each number above. But a useful trick is to observe that factors come in pairs. So if a is a factor of n, so is n/a. So factors will be counted in pairs, unless we have a repeated factor, i.e. a factor a such that a=n/a. Or we need a2=n i.e. n should be a perfect square. So the answer is b.

This is a useful fact. In the previous question, the answer choices were small. But imagine if the same question were rephrased as follows.

Q. Which of the following could be a value of k if k has an odd number of factors?

a) 100000000
b) 2210269
c) 373176
d) 1264845

The answer is a because 100000000 is a perfect square.

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Fun Math Problems With Permutations & Combinations

Problems related to permutations and combinations are often more fun to solve. There are generally no equations, no diagrams, no graphs. You just have to translate the words into the situation at hand. Words like ‘at least’ and ‘at most’ make a huge difference, as you know. For example,

Q: 10 students in a class of 60 students study French, while 47 students study German.What is the minimum/maximum number of students who study Japanese if at the most/ at least 15 students study more than one language and everybody studies at least one language?

As you can see, there is a variety of answers possible, based on the choice of words. But this was a fairly straightforward question that you can solve by plugging in answer values, or actually deriving the answer. It was also easy to translate the words into relations.

A slightly harder question would require some extraneous knowledge, maybe knowledge of geometry or algebra.

Q: What is the number of possible regular geometric figures with integer sides and an area of 16 units? Assume that the number of sides is less than 7.

A regular geometric figure means an equilateral triangle, a square, a regular pentagon, a regular hexagon and so on. We need not consider heptagons as the number of sides is less than 7. For a regular geometric figure, the area is completely determined by the number of sides and the length of a side. So we just need to find out if an area of 16 units corresponds to a side of integer length for the 4 regular geometric figures.

As it turns out, only the square can have both, an integral side and an integral length.

This is highly unlikely to be asked in the GRE, mainly because you need to separately calculate an expression for the area of a regular pentagon using some trigonometry. The takeaway is that counting problems can be deceptively simple but there are a lot of factors involved.

Permutation problems seem easier, mainly because it is clear that there is some sequence involved, some sort of ordering. But let’s try a hard permutation example:

Q: Consider the numbers from 1 to 10. How many ways is it possible to build a sequence of 5 numbers such that all the numbers are in ascending order?

This is actually a combination problem masquerading as a permutation problem. Given a set of distinct numbers, there is only one way to arrange all the numbers in ascending order. So what the question is really asking is the number of ways to select 5 numbers from a set of 10 numbers.

(Another variation: ‘How many orderings of 5 numbers are possible such that the second number is the largest number while the other numbers are in ascending order?’ Here, too, for a given set of numbers, this type of ordering is uniquely determined. So it’s just another combination problem.)

At a higher level, permutation and combination problems are deceptively simple. They are often conceptually harder than algebra and especially, geometry problems. Also, they are inherently open-ended, as they can involve concepts from algebra, geometry or other areas. So it’s really not enough to just know the factorial formulas, it helps if you have an intuition of what exactly is required.

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Probability Tip Worth Remembering

There is one type of probability questions which are basically permutation-combination questions. So here I’ll talk about other types of probability questions that are not directly related to counting.

Q. A and B are two independent events. The probability of both occurring is 0.08. What is the probability that neither A nor B occurs given that A is twice as likely as B?

This is a straightforward question involving quadratic equations.
b2=0.04, b =0.2, a = 0.4.

So the answer is 1 – (a + b – ab) = 1 – (0.4 + 0.2 – 0.08) = 0.48

This was a really easy question.