Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT!

Quarter Wit, Quarter WisdomYour first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:

The last digit of 12^12 + 13^13 – 14^14 × 15^15 =

(A) 0
(B) 1
(C) 5
(D) 8
(E) 9

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

(i) 100-29
100
-29
071

(ii) 29-100
100
-29
071
(But since the sign of 100 is negative, your answer is actually -71.)

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0
–  pq9
ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Linear Relations in GMAT Questions

Quarter Wit, Quarter WisdomWe have covered the concepts of direct, inverse and joint variation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.

We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?

(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)
60 = 24m + c ……(II)

(II) – (I)
30 = 18m
m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c
c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20
R = 48

48 (answer choice C) is our answer.

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Number Properties on the GMAT – Part VI

Quarter Wit, Quarter WisdomMost people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:

Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

Say N = 3
The numbers are 5, 6, 7 (any three consecutive numbers)
Their sum is 5 + 6 + 7 = 18
Their product is 5*6*7 = 210
Note that both the sum and the product are divisible by 3 (i.e. N).

Say N = 5
The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 2 + 3 + 4 + 5 + 6 = 20
Their product is 2*3*4*5*6 = 720
Again, note that both the sum and the product are divisible by 5 (i.e. N)

Say N = 4
The numbers are 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 3 + 4 + 5 + 6 = 18
Their product is 3*4*5*6 = 360
Now note that the sum is not divisible by 4, but the product is divisible by 4.

If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.
Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

(Multiple of N),
(Multiple of N) +1,
(Multiple of N) + 2,
… ,
(Multiple of N) + (N-2),
(Multiple of N) + (N-1)

In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:

(Multiple of 3) + 2 = 5
(Multiple of 3)       = 6
(Multiple of 3) + 1 = 7

In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:

(Multiple of 4) + 3 = 3
(Multiple of 4)        = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
etc.

What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will always add up in pairs to give the sum of N:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

So when you add up all the integers, you will get a multiple of N.

What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will add up to give integers of N but one will be leftover:

1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N

The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.

For example, let’s reconsider the previous example in which we had four consecutive integers:

(Multiple of 4)      = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
(Multiple of 4) + 3 = 3

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

Which of the following must be true?
1) The sum of N consecutive integers is always divisible by N.
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!

(A) 1, 4, 5
(B) 3, 4, 5
(C) 4 and 5
(D) 1, 2, 3, 4
(E) only 4

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

5) The product of K consecutive integers is divisible by K!

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts I, II, III, IV and V of this series.)

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Divisibility by Powers of 2

Quarter Wit, Quarter WisdomWe know the divisibility rules of 2, 4 and 8:

For 2 – If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2.

For 4 – If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4.

For 8 – If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8.

A similar rule applies to all powers of 2:

For 16 – If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16.

For 32 – If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32.

and so on…

Let’s figure out why:

The generic rule can be written like this: A number M is divisible by 2^n if the last n digits of M are divisible by 2^n.

Take, for example, a division by 8 (= 2^3), where M = 65748048 and n = 3.

Our digits of interest are the last three digits, 048.

48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.

A valid question here is, “What about the remaining five digits? Why do we ignore them?”

Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).

Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.

All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.

In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.

In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.

We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:

What is the remainder when 1990990900034 is divided by 32 ?

(A) 16
(B) 8
(C) 4
(D) 2
(E) 0

Breaking down our given number, 1990990900034 = 1990990900000 + 00034.

1990990900000 ends in five 0’s so it is divisible by 32. 34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. Our answer is D.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit Quarter Wisdom: What is Your Favorite Number?

Quarter Wit, Quarter WisdomFans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this  number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768
(B) 819
(C) 826
(D) 858
(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If  we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Attacking Gerunds on the GMAT!

Quarter Wit, Quarter WisdomA few weeks back, we talked about participles and how they are used on the GMAT. In that post, we had promised to discuss gerunds more in depth at another time. So today, as promised, we’ll be looking at gerunds. Before we do that, however, let’s examine Verbals.

A Verbal is a verb that acts as a different part of speech – not as a verb.

There are three types of verbals:

  • Infinitives – these take the form of “to + verb”
  • Gerunds – these are the “-ing” form of the verb
  • Participles – these can take the “-ing,” “-ed,” “-en” etc. forms

Gerunds end in “-ing” and act as nouns in the sentence. They can act as a subject, direct object, subject complement or object of a preposition. For example:

Running a marathon is very difficult. – Subject
I love swimming. – Direct object
The activity I enjoy the most is swimming. – Subject complement
She thanked me for helping her. – Object of a preposition

You don’t have to identify the part of speech the gerund represents in a sentence; you just need to identify whether a verb’s “-ing” form is being used as a gerund and evaluate whether it is being used correctly.

A sentence could also use a gerund phrase that begins with a gerund, such as, “Swimming in the morning is exhilarating.”

Let’s take a look at a couple of official questions now:

A recent study has found that within the past few years, many doctors had elected early retirement rather than face the threats of lawsuits and the rising costs of malpractice insurance.

(A) had elected early retirement rather than face
(B) had elected early retirement instead of facing
(C) have elected retiring early instead of facing
(D) have elected to retire early rather than facing
(E) have elected to retire early rather than face

Upon reading the original sentence, we see that there is a gerund phrase here – “rising costs of malpractice insurance” – which is parallel to the noun “threat of lawsuits.”

The two are logically parallel too, since there are two aspects that the doctors do not want to face: rising costs and the threat of lawsuits.

Note, however, that they are not logically parallel to “face.” Hence, the use of the form “facing” would not be correct, since it would put “facing” and “rising” in parallel. So answer choices B, C and D are incorrect.

Actually, “retire” and “face” are logically parallel so they should be grammatically parallel, too. Answer choice E has the two in parallel in infinitive form – to retire and (to is implied here) face are in parallel.

Obviously, there are other decision points to take note of here, mainly the question of “had elected” vs. “have elected.” The use of “had elected” will not be correct here, since we are not discussing two actions in the past occurring at different times. Therefore, the correct answer is E.

Take a look at one more:

In virtually all types of tissue in every animal species dioxin induces the production of enzymes that are the organism’s trying to metabolize, or render harmless, the chemical that is irritating it.

(A) trying to metabolize, or render harmless, the chemical that is irritating it
(B) trying that it metabolize, or render harmless, the chemical irritant
(C) attempt to try to metabolize, or render harmless, such a chemical irritant
(D) attempt to try and metabolize, or render harmless, the chemical irritating it
(E) attempt to metabolize, or render harmless, the chemical irritant

Notice the use of the gerund “trying” in answer choice A. “Organism’s” is in possessive form and acts as an adjective for the noun verbal “trying.” Usually, with possessives, a gerund does not work. We need to use a noun only. With this in mind, answer choices A and B will not work.

The other three options replace “trying” with “attempt” and hence correct this error, however options C and D use the redundant “attempt to try.” The use of “attempt” means “try,” so there is no need to use both. Option E corrects this problem, so it is our correct answer.

Unlike participles, which can be a bit confusing, gerunds are relatively easy to understand and use. Feeling more confident about them now?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Go From a 48 to 51 in GMAT Quant – Part VII

Quarter Wit, Quarter WisdomBoth a test-taker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a test-taker at the 51 level does skillfully:

  1. Uses holistic, big-picture methods to solve Quant questions.
  2. Handles questions he or she finds difficult in a timely manner.

We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IVV and VI of this series.)

Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level test-taker – in fact, the 51 scorers often have time left over after attempting all these questions.

Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too.

There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions.

Let’s take an official example:

Pam owns an inventory of unopened packages of corn and rice, which she has purchased for $17 and $13 per package, respectively. How many packages of corn does she have ?

Statement 1: She has $282 worth of packages.

Statement 2: She has twice as many packages of corn as of rice.

A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions.

After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force.

The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own $282 worth of packages.)

So, the test-taker will start working on finding the first solution (using the method discussed in this post). We are told:

Price of a packet of corn = $17
Price of a packet of rice = $13

Say Pam has “x” packets of corn and “y” packets of rice.

Statement 1: She has $282 worth of packages

Using Statement 1, we know that 17x + 13y = 282.

We are looking for the integer values of x and y.

If x = 0, y will be 21.something (not an integer)
If x = 1, y = 20.something
If x = 2, y = 19.something
If x = 3, y = 17.something

This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.

So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient.

Statement 2: She has twice as many packages of corn as of rice.

Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A.

I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using Prepositional Phrases on the GMAT

Quarter Wit, Quarter WisdomIn previous posts, we have already discussed participles as well as absolute phrases. Today, let’s take a look at another type of modifier – the prepositional phrase.

A prepositional phrase will begin with a preposition and end with a noun, pronoun, gerund, or clause – the “object” of the preposition. The object of the preposition might have one or more modifiers to describe it.

Here are some examples of prepositional phrases (with prepositions underlined):

  • along the ten mile highway…
  • with a cozy blanket…
  • without worrying…
  • about what he likes…

A prepositional phrase can function as an adjective or an adverb. As an adjective, it answers the question, “Which one?” while as an adverb it can answer the questions, “How?” “When?” or “Where?”.

For example:

  • The book under the table belongs to my mom. Here, the prepositional phrase acts as an adjective and tells us “which one” of the books belongs to my mom.
  • We tried the double cheeseburger at the new burger joint. Here, the prepositional phrase acts as an adverb and tells us “where” we tried the cheeseburger.

Like other modifiers, a prepositional modifier should be placed as close as possible to the thing it is modifying.

Let’s take a look at a couple of official GMAT questions to see how understanding prepositional phrases can help us on this exam:

The nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius in two letters to the historian Tacitus.

(A) The nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius in two letters to the historian Tacitus.
(B) To the historian Tacitus, the nephew of Pliny the Elder wrote two letters, being the only eyewitness accounts of the great eruption of Vesuvius.
(C) The only eyewitness account is in two letters by the nephew of Pliny the Elder writing to the historian Tacitus an account of the great eruption of Vesuvius.
(D) Writing the only eyewitness account, Pliny the Elder’s nephew accounted for the great eruption of Vesuvius in two letters to the historian Tacitus.
(E) In two letters to the historian Tacitus, the nephew of Pliny the Elder wrote the only eyewitness account of the great eruption of Vesuvius.

There are multiple prepositional phrases here:

  • of the great eruption of Vesuvius (answers “Which eruption?”)
  • in two letters (tells us “where” he wrote his account)
  • to the historian Tacitus (answers “Which letters?”)

Therefore, the phrase “to the historian Tacitus” should be close to what it is describing, “letters,” which makes answer choices B and C incorrect.

Also, “in two letters to the historian Tacitus” should modify the verb “wrote.” In options A and D, “in two letters to the historian Tacitus” seems to be modifying “eruption,” which is incorrect. (There are other errors in answer choices B, C and D as well, but we will stick to the topic at hand.)

Option E corrects the prepositional phrase errors by putting the modifier close to the verb “wrote,” so therefore, E is our answer.

Let’s try one more:

Defense attorneys have occasionally argued that their clients’ misconduct stemmed from a reaction to something ingested, but in attributing criminal or delinquent behavior to some food allergy, the perpetrators are in effect told that they are not responsible for their actions.

(A) in attributing criminal or delinquent behavior to some food allergy
(B) if criminal or delinquent behavior is attributed to an allergy to some food
(C) in attributing behavior that is criminal or delinquent to an allergy to some food
(D) if some food allergy is attributed as the cause of criminal or delinquent behavior
(E) in attributing a food allergy as the cause of criminal or delinquent behavior

This sentence has two clauses:

Clause 1: Defense attorneys have occasionally argued that their clients’ misconduct stemmed from a reaction to something ingested,

Clause 2: in attributing criminal or delinquent behavior to some food allergy, the perpetrators are in effect told that they are not responsible for their actions.

These two clauses are joined by the conjunction “but,” and the underlined part is a prepositional phrase in the second clause.

Answer choices A, C and E imply that the perpetrators are attributing their own behaviors to food allergies. That is not correct – their defense attorneys are attributing their behavior to food allergies, and hence, all three of these options have modifier errors.

This leaves us with B and D. Answer choice D uses the phrase “attributed as,” which is grammatically incorrect – the correct usage should be “X is attributed to Y,” rather than “X attributed as Y.” Therefore, option B is our answer.

As you can see, the proper placement of prepositional phrases is instrumental in creating a sentence with a clear, logical meaning.  Since that type of clear, logical meaning is a primary emphasis of correct Sentence Correction answers, you should be prepared to look for prepositional phrases (here we go…) *on the GMAT*.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Some GMAT Questions Using the “Like” vs. “As” Concept

Quarter Wit, Quarter WisdomToday we will look at some official GMAT questions testing the “like” vs. “as” concept we discussed last week.

(Review last week’s post – if you haven’t read it already – before you read this one for greater insight on this concept.)

Take a look at the following GMAT Sentence Correction question:

As with those of humans, the DNA of grape plants contains sites where certain unique sequences of nucleotides are repeated over and over.

(A) As with those of humans, the DNA of grape plants contains sites where
(B) As human DNA, the DNA of grape plants contain sites in which
(C) As it is with human DNA, the DNA of grape plants, containing sites in which
(D) Like human, the DNA of grape plants contain sites where
(E) Like human DNA, the DNA of grape plants contains sites in which

Should we use “as” or “like”? Well, what are we comparing? We’re comparing the DNA of humans to the DNA of grape plants. Answer choice E compares these two properly – “Like human DNA, the DNA of grape plants…” DNA is singular, so it uses the singular verb “contains”.

All other options are incorrect. Answer choice A uses “those of” for DNA, but DNA is singular, so this cannot be right. B uses “as” to compare the two nouns, which is also incorrect. C is a sentence fragment without a main verb. And D compares “human” to “DNA”, which is not the “apples-to-apples” comparison we need to make this sentence correct. Therefore, our answer must be E.

Let’s try another one:

Like Auden, the language of James Merrill is chatty, arch, and conversational — given to complex syntactic flights as well as to prosaic free-verse strolls.

(A) Like Auden, the language of James Merrill
(B) Like Auden, James Merrill’s language
(C) Like Auden’s, James Merrill’s language
(D) As with Auden, James Merrill’s language
(E) As is Auden’s the language of James Merrill

Here, we’re comparing Auden’s language to James Merrill’s language. Answer choice C correctly uses the possessive “Auden’s” to show that language is implied. “Like Auden’s language, James Merrill’s language …” contains both parallel structure and a correct comparison.

Answer choices A, B and D incorrectly compare “Auden” to “language,” rather than “Auden’s language” to “language,” so those options are out. The structure of answer choice E is not parallel – “Auden’s” vs. “the language of James Merrill”. Therefore, the answer must be C.

Let’s try something more difficult:

More than thirty years ago Dr. Barbara McClintock, the Nobel Prize winner, reported that genes can “jump,” as pearls moving mysteriously from one necklace to another.

(A) as pearls moving mysteriously from one necklace to another
(B) like pearls moving mysteriously from one necklace to another
(C) as pearls do that move mysteriously from one necklace to others
(D) like pearls do that move mysteriously from one necklace to others
(E) as do pearls that move mysteriously from one necklace to some other one

This is a tricky question – it’s perfect for us to re-iterate how important it is to focus on the meaning of the given sentence. Do not try to follow grammar rules blindly on the GMAT!

Is the comparison between “genes jumping” and “pearls moving”? Do pearls really move mysteriously from one necklace to another? No! This is a hypothetical situation, so we must use “like” – genes are like pearls. Answer choices B and D are the only ones that use “like,” so we can eliminate our other options. D uses a clause with “like,” which is incorrect. In answer choice B, “moving from …” is a modifier – “moving” doesn’t act as a verb here, so it doesn’t need a clause. Hence, answer choice B is correct.

Here’s another one:

According to a recent poll, owning and living in a freestanding house on its own land is still a goal of a majority of young adults, like that of earlier generations.

(A) like that of earlier generations
(B) as that for earlier generations
(C) just as earlier generations did
(D) as have earlier generations
(E) as it was of earlier generations

Note the parallel structure of the comparison in answer choice E – “Owning … a house… is still a goal of young adults, as it was of earlier generations.” It correctly uses “as” with a clause.

Answer choice A uses “that” but its antecedent is not very clear; there are other nouns between “goal” and “like,” and hence, confusion arises. None of the other answer choices give us a clear, parallel comparison, so our answer is E.

Alright, last one:

In Hungary, as in much of Eastern Europe, an overwhelming proportion of women work, many of which are in middle management and light industry.

(A) as in much of Eastern Europe, an overwhelming proportion of women work, many of which are in
(B) as with much of Eastern Europe, an overwhelming proportion of women works, many in
(C) as in much of Eastern Europe, an overwhelming proportion of women work, many of them in.
(D) like much of Eastern Europe, an overwhelming proportion of women works, and many are.
(E) like much of Eastern Europe, an overwhelming proportion of women work, many are in.

Another tricky question. The comparison here is between “what happens in Hungary” and “what happens in much of Eastern Europe,” not between “Hungary” and “much of Eastern Europe.” A different sentence structure would be required to compare “Hungary” to “much of Eastern Europe” such as “Hungary, like much of Eastern Europe, has an overwhelming …”

With prepositional phrases, as with clauses, “as” is used. So, we have two relevant options – A and C. Answer choice A uses “which” for “women,” and hence, is incorrect. Therefore, our answer is C.

Here are some takeaways to keep in mind:

  • You should be comparing “apples” to “apples”.
  • Parallel structure is important.
  • Use “as” with prepositional phrases.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using “Like” vs. “As” on the GMAT Verbal Section

Quarter Wit, Quarter WisdomIf you have seen the Veritas Prep curriculum, then you know we frequently highlight the strategy of “Think like the Testmaker” to answer GMAT questions. Recently, we had a student question the grammatical validity of this construct – this brought the “like” vs. “as” debate to mind, so we decided to tackle it this week.

When should you use “like” and when should you use “as” in a sentence?

Both words can be used in comparisons, however the structure of the sentence will be different in the two cases. This is because traditionally, “like” is a preposition and “as” is a conjunction – a preposition takes the form of an object while a conjunction takes the form of a clause. Therefore:

Using “like,” we compare nouns/pronouns (including gerunds). Usually, a single verb will be used.

Using “as,” we compare actual actions. There will be two verbs used when we compare using “as.”

So, this is how we are going to compare “like” and “as”:

  • He runs like a madman. – A single verb, “runs.”
  • He runs as a madman does. – Two verbs, “runs” and “does” (which is equivalent to “does run”).

In the same way, both of the following sentences are correct:

  • Think like the Testmaker.
  • Think as the Testmaker does.

But beware – “as” used with a noun or pronoun alone does not mean that this usage is incorrect. “As” can also be used to show a role or capacity. For example, in the sentence, “She works as a consultant,” the word “as” means that she works in the capacity of a consultant. There is no comparison here, but the sentence is still grammatically correct.

Also, we usually use “like” in the case of hypothetical comparisons. Take, for instance, the sentence, “She screams like a banshee.” Here, it would be odd to say, “She screams as a banshee does,” because we don’t really know how a banshee screams.

Let’s look at a few GMAT Sentence Correction questions now:

Like many self-taught artists, Perle Hessing did not begin to paint until she was well into middle age.

(A) Like
(B) As have
(C) Just as with
(D) Just like
(E) As did

In this sentence, the word “like” is correctly comparing “Perle Hessing” to “many self taught artists.” There is no clause after “like” and we are using a single verb. Hence, the use of “like” is correct and our answer is A.

Not too bad, right? Let’s try another question:

Based on recent box office receipts, the public’s appetite for documentary films, like nonfiction books, seems to be on the rise. 

(A) like nonfiction books 
(B) as nonfiction books 
(C) as its interest in nonfiction books 
(D) like their interest in nonfiction books 
(E) like its interest in nonfiction books

This sentence also has a comparison, and it is between “appetite” and “interest” and how they are both are on a rise. Answer choice E compares “appetite” to “interest” using “like” as a single verb. None of the answer choices have “as” with a clause so the answer must be E.

These were two simple examples of “like” vs. “as.” Now let’s look at a higher-level GMAT question:

During an ice age, the buildup of ice at the poles and the drop in water levels near the equator speed up the Earth’s rotation, like a spinning figure skater whose speed increases when her arms are drawn in

(A) like a spinning figure skater whose speed increases when her arms are drawn in 
(B) like the increased speed of a figure skater when her arms are drawn in 
(C) like a figure skater who increases speed while spinning with her arms drawn in 
(D) just as a spinning figure skater who increases speed by drawing in her arms 
(E) just as a spinning figure skater increases speed by drawing in her arms

There is a comparison here, but between which two things? Answer choice A seems to be comparing “Earth’s rotation” to “spinning figure skater,” but these two things are not comparable. Option E is the correct choice here – it compares “speed up Earth’s rotation” to “skater increases speed.” Therefore, our answer is E.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Are Official Answers Debatable on the GMAT?

Quarter Wit, Quarter WisdomLet’s begin with the bottom line: no, they are not. If you are thinking along the lines of, “This official answer cannot be correct! How can the answer be A? It must be C, or C is at least just as valid as A,” then you are wasting your time. The answer given is never debatable. What you should be thinking instead is, “The answer given is A, but  I thought it was C. I must find out where I made a mistake.”

The point is that since you are going to take GMAT, you must learn to think like the GMAT testmakers. The answers they give for these questions are the correct answers, so need to accept that – this way, the next step of figuring out the gap in your understanding will be far easier. Today, let’s take a look at an official question that is often debated:

The average hourly wage of television assemblers in Vernland has long been significantly lower than that in neighboring Borodia. Since Borodia dropped all tariffs on Vernlandian televisions three years ago, the number of televisions sold annually in Borodia has not changed. However, recent statistics show a drop in the number of television assemblers in Borodia. Therefore, updated trade statistics will probably indicate that the number of televisions Borodia imports annually from Vernland has increased.

Which of the following is an assumption on which the argument depends?

(A) The number of television assemblers in Vernland has increased by at least as much as the number of television assemblers in Borodia has decreased.
(B) Televisions assembled in Vernland have features that televisions assembled in Borodia do not have.
(C) The average number of hours it takes a Borodian television assembler to assemble a television has not decreased significantly during the past three years.
(D) The number of televisions assembled annually in Vernland has increased significantly during the past three years.
(E) The difference between the hourly wage of television assemblers in Vernland and the hourly wage of television assemblers in Borodia is likely to decrease in the next few years.

First, let’s look at the premises of the argument:

  • The hourly wage of assemblers in Vernland is much lower than that in Borodia.
  • 3 years ago, Borodia dropped all tariffs on TVs imported from Vernland.
  • The number of TVs sold annually in Borodia is same.
  • However, the number of assemblers in Borodia has decreased.

The conclusion is that the trade statistics will probably indicate that the number of televisions Borodia imports annually from Vernland has increased.

This conclusion might look logical, but it is full of assumptions.

Why does this conclusion seem so logical? Wages in Vernland are lower, so it would seem like TVs should be cheaper here. Borodia dropped all tariffs on imported TVs, which means there will be no artificial inflation of Vernland TV prices. Finally, the number of TVs sold in Borodia has not dropped, but number of assemblers in Borodia has dropped, which makes it look like fewer TVs are getting made in Borodia.

An onlooker might conclude that Borodia is importing more TVs from Vernland because they are cheaper, but here are some assumptions that come to mind:

  • The cost of a TV in Vernland is lower because assembler’s wage is lower. What if the raw material cost is higher in Vernland? Or other costs are higher? The cost to produce a Vernland TV could actually be higher than the cost to produce a Borodia TV.
  • Fewer TVs are getting made in Borodia, but that does not mean that Borodian assemblers have not become more productive. What if fewer assemblers are needed because they can actually complete the assembly process much faster? The number of TVs sold is the same, however, if each assembler is doing more work, fewer assemblers will be needed. In this case, the number of TVs made in Borodia might not have changed even though the number of producers dropped.

Coming to our question now: Which of the following is an assumption on which the argument depends?

We are looking for an assumption, i.e. a NECESSARY premise. We have already identified some assumptions, so let’s see if any of the answer choices gives us one of those:

(A) The number of television assemblers in Vernland has increased by at least as much as the number of television assemblers in Borodia has decreased.

This is the most popular incorrect answer choice. Test takers keep trying to justify why it makes perfect sense, but actually, it is not required for the conclusion to hold true.

The logic of test takers that pick this answer choice is often on the lines of, “If the number of workers from Borodia decreased, in order for Borodia to show an increased number of imports from Vernland, Vernland must have increased their number of workers by at least as much as the number of workers that left Borodia.”

Note that although this may sound logical, it is not necessary to the argument. There are lots of possible situations where this may not be the case:

Perhaps number of TVs being manufactured in Vernland is the same and, hence, the number of assemblers is the same, too. It is possible that out of the fixed number of TVs manufactured, fewer are getting locally bought and more are getting exported to Borodia. So, it is not necessarily true that number of TV assemblers in Vernland has increased.

(B) Televisions assembled in Vernland have features that televisions assembled in Borodia do not have.

This is also not required for the conclusion to hold – the TVs could actually be exactly the same, but the TVs assembled in Vernland could still be cheaper than the TVs assembled in Borodia due to a potentially lower cost of assembly in Vernland.

(C) The average number of hours it takes a Borodian television assembler to assemble a television has not decreased significantly during the past three years.

This is one of the assumptions we discussed above – we are assuming that the reduction in the number of assemblers must not be due to an increase in the productivity of the assemblers because if the assemblers have got more productive, then the number of TVs produced could be the same and, hence, the number of TVs imported would not have increased.

(D) The number of televisions assembled annually in Vernland has increased significantly during the past three years.

This is not required for the conclusion to hold. Perhaps the number of TVs being sold in Vernland has actually reduced while more are getting exported to Borodia, so the overall number of TVs being made is the same.

(E) The difference between the hourly wage of television assemblers in Vernland and the hourly wage of television assemblers in Borodia is likely to decrease in the next few years.

This is also not required for the conclusion to hold. What happens to the hourly wages of assemblers in Vernland and Borodia in the future doesn’t concern this argument – we are only concerned about what has been happening in the last 3 years.

Therefore, our answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Go From a 48 to 51 in GMAT Quant – Part VI

Quarter Wit, Quarter WisdomToday’s post the next part in our “How to Go From 48 to 51 in Quant” series. Again, we will learn a technique that can be employed by the test-taker at an advanced stage of preparation – requiring one to understand the situations in which one can use this simplifying technique.

(Before you continue reading, be sure to check out parts I, II, III, IV, and V of this series.)

We all love to use the plug-in method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one that satisfies the equation.

But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options:

If |4x−4|=|2x+30|, which of the following could be a value of x?

(A) –35/3
(B) −21/2
(C) −13/3
(D) 11/5 
(E) 47/5

This question is an ideal candidate for the “plug-in” method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than “plug-in”, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in.

Method 1:
|4x – 4| = |2x + 30|

4 * |x – 1| = 2 * |x + 15|

2 * |x – 1| = |x + 15|

This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from -15.

——————(-15) —————————————————(0)——(1)——————

There are two ways to find the value of x:

Case 1: x could be between -15 and 1 such that the distance between them is split in the ratio 2:1.

or

Case 2: x could be to the right of 1 such that the distance between x and -15 is twice the distance between x and 1.

Let’s examine both of these cases in further detail:

Case 1: The distance from -15 to 1 is of 16 units – this can be split into 3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from -15 two times 16/3, i.e. 32/3.

So, x should be at a point 16/3 away from 1 toward the left.

x = 1 – 16/3 = -13/3

This is one of our answer choices and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found -13/3 in the answer options, we would have moved on to Case 2:

Case 2: The distance between -15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and -15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17. If you would not have found -13/3 in the answer choices, then you would have found 17.

Now let’s move on to see how we can make the plug-in method work for us in this case by examining each answer choice we are given:

Method 2:
|4x – 4| = |2x + 30|

2 * |x – 1| = |x + 15|

(A) -35/3

It is difficult to solve for x = -35/3 to see if both sides match. Instead, let’s solve for the closest integer, -12.

2 * |-12 – 1| = |-12 + 15|

On the left-hand side, you will get 26, but on the right-hand side, you will get 3.

These values are far away from each other, so x cannot be -35/3.  As the value of x approaches the point where the equation holds – i.e. where the two sides are equal to each other – the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of -35/3 from -12 will bring the two sides to be equal.

(B) -21/2

For this answer choice, let’s solve for the nearest integer, x = -10.

2 * |-10 – 1| = |-10 + 15|

On the left-hand side, you will get 22; on the right-hand side, you will get 5.

Once again, these values are far away from each other and, hence, x will not be -21/2.

(C) -13/3

For this answer choice, let’s solve for x = -4.

2 * |-4 -1| = |-4 + 15|

On the left-hand side, you will get 10; on the right-hand side, you will get 11.

Here, there is a possibility that x can equal -13/3, as the two sides are so close to one another – plug in the actual value of -13/3 and you will see that the left-hand side of the equation does, in fact, equal the right-hand side. Therefore, C is the correct answer.

Basically, we approximated the answer choices we were given and shortlisted the one that gave us very close values. We checked for that and found that it is the answer.

We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have limited applications – only if you are given the actual values of x, can you use it. Method 1 is far more generic for absolute value questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Squares and Square Roots on the GMAT

Quarter Wit, Quarter WisdomIn today’s post, we will try to clear up your doubts regarding positive and negative solutions in the case of squares and square roots. We will explain the reasons behind each case, which will help you recall the fundamentals when you need to use them. While preparing for the GMAT, you have probably come across a discussion that says x^2 = 4 has two roots, 2 and -2, while √4 has only one value, 2.

Now, let’s try to understand why this is so:

1) x^2 = 4
Basic algebra tells us that quadratics have two roots. Here, x can be either 2 or -2; each, when squared, will give you 4.

x^2 – 4 = 0 and (x + 2)*(x – 2) = 0 when x equals -2 or 2.

2) √x is positive, only
Now this is odd, right? √4 must be 2. Why is that? Shouldn’t it be 2 or -2. After all, when we square both 2 and -2, we get 4 (as discussed above). So, √4 should be 2 or -2.

Here is the concept: √x denotes only the principal square root. x has two square roots – the positive square root (or principal square root) written as √x and the negative square root written as -√x. Therefore, when you take the square root of 4, you get two roots: √4 and -√4, which  is 2 and -2 respectively.

On a GMAT question, when you see √x, this is specifically referring to the positive square root of the number. So √4 is 2, only.

3) (√x)^2 = x
This is fairly straightforward – since x has a square root, it must be non-negative. When you square it, just the square root sign vanishes and you are left with x.

4) √(x^2) = |x|
Now this isn’t intuitive either. √(x^2) should simply be x – why do we have absolute value of x, then? Again, this has to do with the principal square root concept. First you will square x, and then when you write √, it is by default just the principal square root. The negative square root will be written as -√(x^2). So, irrespective of whether x was positive or negative initially, √(x^2) will definitely be positive x. Therefore, we will need to take the absolute value of x.

Here’s a quick recap with some examples:

  • √9 = 3
  • x^2 = 16 means x is either 4 or -4
  • √(5^2) = 5
  • √(-5^2) = 5
  • (√16)^2 = 16
  • √100 = 10

To see this concept in action, let’s take a look at a very simple official problem:

If x is not 0, then √(x^2)/x =

(A) -1
(B) 0
(C) 1
(D) x
(E) |x|/x

We know that √(x^2) is not simply x, but rather |x|. So, √(x^2)/x = |x|/x.

Depending on whether x is positive or negative, |x|/x will be 1 or -1 – we can’t say which one. Hence, there is no further simplification that we can do, and our answer must be E.

Now that you are all warmed up, let’s examine a higher-level question:

Is √[(x – 3)^2] = (3 – x)?

Statement 1: x is not 3
Statement 2: -x * |x| > 0

We know that √(x^2) = |x|, so √[(x – 3)^2] = |x – 3|.

This means that our question is basically:

Is |x – 3| = 3 – x?

Note that 3 – x can also be written as -(x – 3).

Is |x – 3| = -(x – 3)?

Recall the definition of absolute values: |a| = a if a is greater than or equal to 0, and -a if a < 0.

So, “Is |x – 3| = -(x – 3)?” depends on whether (x – 3) is positive or negative. If (x – 3) is negative (or 0), then |x – 3| is equal to -(x – 3).

So our question now boils down to:

Is (x – 3) negative (or 0)?

Statement 1: x is not 3

This means we know that (x – 3) is not 0, but we still don’t know whether it is negative or positive. This statement is not sufficient.

Statement 2: -x * |x| > 0

|x| is always non-negative, so for the product to be positive, “-x” must also be positive. This means x must be negative. If x is negative, x – 3 must be negative, too.

If (x – 3) is negative, |x – 3| is equal to -(x – 3). Hence, this statement alone is sufficient, and our answer is B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Why Critical Reasoning Needs Your Complete Attention on the GMAT!

Quarter Wit, Quarter WisdomLet’s look at a tricky and time consuming official Critical Reasoning question today. We will learn how to focus on the important aspects of the question and quickly evaluate our answer choices:

Tiger beetles are such fast runners that they can capture virtually any nonflying insect. However, when running toward an insect, a tiger beetle will intermittently stop and then, a moment later, resume its attack. Perhaps the beetles cannot maintain their pace and must pause for a moment’s rest; but an alternative hypothesis is that while running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop. 

Which of the following, if discovered in experiments using artificially moved prey insects, would support one of the two hypotheses and undermine the other? 

(A) When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately stops and runs away without its usual intermittent stopping. 
(B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.
(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insect’s direction, and it pauses equally frequently whether the chase is up or down an incline. 
(D) If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit. 
(E) The faster a beetle pursues an insect fleeing directly away from it, the more frequently the beetle stops.

First, take a look at the argument:

  • Tiger beetles are very fast runners.
  • When running toward an insect, a tiger beetle will intermittently stop and then, a moment later, resume its attack.

There are two hypotheses presented for this behavior:

  1. The beetles cannot maintain their pace and must pause for a moment’s rest.
  2. While running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.

We need to support one of the two hypotheses and undermine the other. We don’t know which one will be supported and which will be undermined. How will we support/undermine a hypothesis?

The beetles cannot maintain their pace and must pause for a moment’s rest.

Support: Something that tells us that they do get tired. e.g. going uphill they pause more.

Undermine: Something that says that fatigue plays no role e.g. the frequency of pauses do not increase as the chase continues.

While running, tiger beetles are unable to adequately process the resulting rapidly changing visual information and so quickly go blind and stop.

Support: Something that says that they are not able to process changing visual information e.g. as speed increases, frequency of pauses increases.

Undermine: Something that says that they are able to process changing visual information e.g. it doesn’t pause on turns.

Now, we need to look at each answer choice to see which one supports one hypothesis and undermines the other. Focus on the impact each option has on our two hypotheses:

(A) When a prey insect is moved directly toward a beetle that has been chasing it, the beetle immediately stops and runs away without its usual intermittent stopping.

This undermines both hypotheses. If the beetle is able to run without stopping in some situations, it means that it is not a physical ailment that makes him take pauses. He is not trying to catch his breath – so to say – nor is he adjusting his field of vision.

(B) In pursuing a swerving insect, a beetle alters its course while running and its pauses become more frequent as the chase progresses.

If the beetle alters its course while running, it is obviously processing changing visual information and changing its course accordingly while running. This undermines the hypothesis “it cannot process rapidly changing visual information”. However, if the beetle pauses more frequently as the chase progresses, it is tiring out more and more due to the long chase and, hence, is taking more frequent breaks. This supports the hypothesis, “it cannot maintain its speed and pauses for rest”.

Answer choice B strengthens one hypothesis and undermines the other. This must be the answer, but let’s check our other options, just to be sure:

(C) In pursuing a moving insect, a beetle usually responds immediately to changes in the insect’s direction, and it pauses equally frequently whether the chase is up or down an incline.

This answer choice undermines both hypotheses. If the beetle responds immediately to changes in direction, it is able to process changing visual information. In addition, if the beetle takes similar pauses going up or down, it is not the effort of running that is making it take the pauses (otherwise, going up, it would have taken more pauses since it takes more effort going up).

(D) If, when a beetle pauses, it has not gained on the insect it is pursuing, the beetle generally ends its pursuit.

This answer choice might strengthen the hypothesis that the beetle is not able to respond to changing visual information since it decides whether it is giving up or not after pausing (in case there is a certain stance that tells us that it has paused), but it doesn’t actually undermine the hypothesis that the beetle pauses to rest. It is very possible that it pauses to rest, and at that time assesses the situation and decides whether it wants to continue the chase. Hence, this option doesn’t undermine either hypothesis and cannot be our answer.

(E) The faster a beetle pursues an insect fleeing directly away from it, the more frequently the beetle stops.

This answer choice strengthens both of the hypotheses. The faster the beetle runs, the more rest it would need, and the more rapidly visual information would change causing the beetle to pause. Because this option does not undermine either hypothesis, it also cannot be our answer.

Only answer choice B strengthens one hypothesis and undermines the other, therefore, our answer must be B.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using Visual Symmetry to Solve GMAT Probability Problems

Quarter Wit, Quarter WisdomToday, let’s take a look at an official GMAT question involving visual skills. It takes a moment to understand the given diagram, but at close inspection, we’ll find that this question is just a simple probability question – the trick is in understanding the symmetry of the figure:

The figure shown represents a board with 4 rows of pegs, and at the bottom of the board are 4 cells numbered 1 to 4. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has the probability 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes through the first two pegs at the top it will end in cell 2?

qwqw pegs pic

 

 

 

 

(A) 1/16
(B) 1/8
(C) 1/4
(D) 3/8
(E) 1/2

First, understand the diagram. There are small pegs arranged in rows and columns. The ball falls between two adjacent pegs and hits the peg directly below. When it does, there are two ways it can go – either to the opening on the left or to the opening on the right. The probability of each move is equal, i.e. 1/2.

The arrow show the first path the ball takes. It is dropped between the top two pegs, hits the peg directly below it, and then either drops to the left side or to the right. The same process will be repeated until the ball falls into one of the four cells – 1, 2, 3 or 4.

Method 1: Using Symmetry
Now that we understand this process, let’s examine the symmetry in this diagram.

Say we flip the image along the vertical axis – what do we get? The figure is still exactly the same, but now the order of cells is reversed to be 4, 3, 2, 1. The pathways in which you could reach Cell 1 are now the pathways in which you can use to reach Cell 4.

OR think about it like this:

To reach Cell 1, the ball needs to turn left-left-left.

To reach Cell 4, the ball needs to turn right-right-right.

Since the probability of turning left or right is the same, the situations are symmetrical. This will be the same case for Cells 2 and 3. Therefore, by symmetry, we see that:

The probability of reaching Cell 1 = the probability of reaching Cell 4.

Similarly:

The probability of reaching Cell 2 = the probability of reaching Cell 3. (There will be multiple ways to reach Cell 2, but the ways of reaching Cell 3 will be similar, too.)

The total probability = the probability of reaching Cell 1 + the probability of reaching Cell 2 + the probability of reaching Cell 3 + the probability of reaching Cell 4 = 1

Because we know the probability of reaching Cells 1 and 4 are the same, and the probabilities of reaching Cells 2 and 3 are the same, this equation can be written as:

2*(the probability of reaching Cell 1) + 2*(the probability of reaching Cell 2) = 1

Let’s find the probability of reaching Cell 1:

After the first opening (not the peg, but the opening between pegs 1 and 2 in the first row), the ball moves left (between pegs 1 and 2 in second row) or right (between pegs 2 and 3 in second row). It must move left to reach Cell 1, and the probability of this = 1/2.

After that, the ball must move left again – the probability of this occurring is also 1/2, since probability of moving left or right is equal. Finally, the ball must turn left again to reach Cell 1 – the probability of this occurring is, again, 1/2. This means that the total probability of the ball reaching Cell 1 = (1/2)*(1/2)*(1/2) = 1/8

Plugging this value into the equation above:

2*(1/8) + 2 * probability of reaching Cell 2 = 1

Therefore, the probability of reaching Cell 2 = 3/8

Method 2: Enumerating the Cases
You can also answer this question by simply enumerating the cases.

At every step after the first drop between pegs 1 and 2 in the first row, there are two different paths available to the ball – either it can go left or it can go right. This happens three times and, hence, the total number of ways in which the ball can travel is 2*2*2 = 8

The ways in which the ball can reach Cell 2 are:

Left-Left-Right

Left-Right-Left

Right-Left-Left

So, the probability of the ball reaching Cell 2 is 3/8.

Note that here there is a chance that we might miss some case(s), especially in problems that involve many different probability options. Hence, enumerating should be the last option you use when tackling these types of questions on the GMAT.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using the Standard Deviation Formula on the GMAT

Quarter Wit, Quarter WisdomWe have discussed standard deviation (SD) in detail before. We know what the formula is for finding the standard deviation of a set of numbers, but we also know that GMAT will not ask us to actually calculate the standard deviation because the calculations involved would be way too cumbersome. It is still a good idea to know this formula, though, as it will help us compare standard deviations across various sets – a concept we should know well.

Today, we will look at some GMAT questions that involve sets with similar standard deviations such that it is hard to tell which will have a higher SD without properly understanding the way it is calculated. Take a look at the following question:

Which of the following distribution of numbers has the greatest standard deviation? 

(A) {-3, 1, 2} 
(B) {-2, -1, 1, 2} 
(C) {3, 5, 7} 
(D) {-1, 2, 3, 4} 
(E) {0, 2, 4}

At first glance, these sets all look very similar. If we try to plot them on a number line, we will see that they also have similar distributions, so it is hard to say which will have a higher SD than the others. Let’s quickly review their deviations from the arithmetic means:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2
For answer choice C, the mean = 5 and the deviations are 2, 0, 2
For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2
For answer choice E, the mean = 2 and the deviations are 2, 0, 2

We don’t need to worry about the arithmetic means (they just help us calculate the deviation of each element from the mean); our focus should be on the deviations. The SD formula squares the individual deviations and then adds them, then the sum is divided by the number of elements and finally, we find the square root of the whole term. So if a deviation is greater, its square will be even greater and that will increase the SD.

If the deviation increases and the number of elements increases, too, then we cannot be sure what the final effect will be – an increased deviation increases the SD but an increase in the number of elements increases the denominator and hence, actually decreases the SD. The overall effect as to whether the SD increases or decreases will vary from case to case.

First, we should note that answers C and E have identical deviations and numbers of elements, hence, their SDs will be identical. This means the answer is certainly not C or E, since Problem Solving questions have a single correct answer.

Let’s move on to the other three options:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2
For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2

Comparing answer choices A and D, we see that they both have the same deviations, but D has more elements. This means its denominator will be greater, and therefore, the SD of answer D is smaller than the SD of answer A. This leaves us with options A and B:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2
For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2

Now notice that although two deviations of answers A and B are the same, answer choice A has a higher deviation of 3 but fewer elements than answer choice B. This means the SD of A will be higher than the SD of B, so the SD of A will be the highest. Hence, our answer must be A.

Let’s try another one:

Which of the following data sets has the third largest standard deviation?

(A) {1, 2, 3, 4, 5} 
(B) {2, 3, 3, 3, 4} 
(C) {2, 2, 2, 4, 5} 
(D) {0, 2, 3, 4, 6} 
(E) {-1, 1, 3, 5, 7}

How would you answer this question without calculating the SDs? We need to arrange the sets in increasing SD order. Upon careful examination, you will see that the number of elements in each set is the same, and the mean of each set is 3.

Deviations of answer choice A: 2, 1, 0, 1, 2
Deviations of answer choice B: 1, 0, 0, 0, 1 (lowest SD)
Deviations of answer choice C: 1, 1, 1, 1, 2
Deviations of answer choice D: 3, 1, 0, 1, 3
Deviations of answer choice E: 4, 2, 0, 2, 4 (highest SD)

Obviously, option B has the lowest SD (the deviations are the smallest) and option E has the highest SD (the deviations are the greatest). This means we can automatically rule these answers out, as they cannot have the third largest SD.

Deviations of answer choice A: 2, 1, 0, 1, 2
Deviations of answer choice C: 1, 1, 1, 1, 2
Deviations of answer choice D: 3, 1, 0, 1, 3

Out of these options, answer choice D has a higher SD than answer choice A, since it has higher deviations of two 3s (whereas A has deviations of two 2s). Also, C is more tightly packed than A, with four deviations of 1. If you are not sure why, consider this:

The square of deviations for C will be 1 + 1+ 1 + 1  + 4 = 8
The square of deviations for A will be 4 + 1 + 0 + 1 + 4 = 10

So, A will have a higher SD than C but a lower SD than D. Arranging from lowest to highest SD’s, we get: B, C, A, D, E. Answer choice A has the third highest SD, and therefore, A is our answer

Although we didn’t need to calculate the actual SD, we used the concepts of the standard deviation formula to answer these questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: An Innovative Use of the Slope of a Line on the GMAT

Quarter Wit, Quarter WisdomLet’s continue our discussion on coordinate geometry today.

The concept of slope is extremely important on the GMAT – it is not sufficient to just know how to calculate it using (y2 – y1)/(x2 – x1).

In simple terms, the slope of a line specifies the units by which the y-coordinate changes and the direction in which it changes with each 1 unit increase in the x-coordinate. If the slope (m) is positive, the y-coordinate changes in the same direction as the x-coordinate. If m is negative, however, the y-coordinate changes in the opposite direction.

For example, if the slope of a line is 2, it means that every time the x-coordinate increases by 1 unit, the y-coordinate increases by 2 units. So if the point (3, 5) lies on a line with a slope of 2, the point (4, 7) will also lie on it. Here, when the x-coordinate increases from 3 to 4, the y-coordinate increases from 5 to 7 (by an increase of 2 units). Similarly,  the point (2, 3) will also lie on this same line – if the x-coordinate decreases by 1 unit (from 3 to 2), the y-coordinate will decrease by 2 units (from 5 to 3). Since the slope is positive, the direction of change of the x-coordinate will be the same as the direction of change of the y-coordinate.

Now, if we have a line where the slope is -2 and the point (3, 5) lies on it, when the x-coordinate increases by 1 unit, the y-coordinate DECREASES by 2 units – the point (4, 3) will also lie on this line. Similarly, if the x-coordinate decreases by 1 unit, the y-coordinate will increase by 2 units. So, for example, the point (2, 7) will also lie on this line.

This understanding of the concept of slope can be very helpful, as we will see in this GMAT question:

Line L and line K have slopes -2 and 1/2 respectively. If line L and line K intersect at (6,8), what is the distance between the x-intercept of line L and the y-intercept of line K? 

(A) 5
(B) 10
(C) 5√(5)
(D) 15
(E) 10√(5)

Method 1: The Traditional Approach
Traditionally, one would solve this question like this:

The equation of a line with slope m and constant c is given as y = mx + c. Therefore, the equations of lines L and K would be:

Line L: y = (-2)x + a
and
Line K: y = (1/2)x + b

As both these lines pass through (6,8), we would substitute x=6 and y=8 to get the values of a and b.

Line L: 8 = (-2)*6 + a
a = 20

Line K: 8 = (1/2)*6 + b
b = 5

Thus, the equations of the 2 lines become:

Line L: y = (-2)x + 20
and
Line K: y = (1/2)x + 5

The x-intercept of a line is given by the point where y = 0. So, the x-intercept of line L is given by:

0 = (-2)x + 20
x = 10

This means line L intersects the x-axis at the point (10, 0).

Similarly, the y-intercept of a line is given by the point where x = 0. So, y-intercept of line K is given by:

y = (1/2)*0 + 5
y = 5

This means that line K intersects the y-axis at the point (0, 5).

Looking back at our original question, the distance between these two points is given by √((10 – 0)^2 + (0 – 5)^2) = 5√(5). Therefore, our answer is C.

Method 2: Using the Slope Concept
Although the using the traditional method is effective, we can answer this question much quicker using the concept we discussed above.

Line L has a slope of -2, which means that for every 1 unit the x-coordinate increases, the y-coordinate decreases by 2. Line L also passes through the point (6, 8). We know the line must intersect the x-axis at y = 0, which is a decrease of 8 y-coordinates from the given point (6,8). If y increases by 8, according to our slope concept, x will increase by 4 to give 6 + 4 = 10. So the x-intercept of line L is at (10, 0).

Line K has slope of 1/2 and also passes through (6, 8). We know the this line must intersect the y-axis at x = 0, which is a decrease of 6 x-coordinates from the given point (6,8). This means y will decrease by 1/2 of that (6*1/2 = 3) and will become 8 – 3 = 5. So the y-intercept of line K is at (0, 5).

The distance between the two points can now be found using the Pythagorean Theorem – √(10^2 + 5^2) = 5√(5), therefore our answer is, again, C.

Using the slope concept makes solving this question much less tedious and saves us a lot of precious time. That is the advantage of using holistic approaches over the more traditional approaches in tackling GMAT questions.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Coordinate Geometry: Solving GMAT Problems With Lines Crossing Either the X-Axis or the Y-Axis

Quarter Wit, Quarter WisdomToday let’s learn about the cases in which lines on the XY plane cross, or do not cross, the x- or y-axis. Students often struggle with questions such as this:

Does the line with equation ax+by = c, where a,b and c are real constants, cross the x-axis?

What concepts will you use here? How will you find whether or not a line crosses the x-axis? What conditions should it meet? Think about this a little before you move ahead.

We know that most lines on the XY plane cross the x-axis as well as the y-axis. Even if it looks like a given line doesn’t cross either of these axes, eventually, it will if it has a slope other than 0 or infinity.

QWQW pic 1

 

 

 

 

 

Note that by definition, a line extends infinitely in both directions – it has no end points (otherwise it would be a line “segment”). We cannot depict a line extending infinitely, which is why we will only show a small section of it. Ideally, a line on the XY plane should be shown with arrowheads to depict that it extends infinitely on both sides, but we often omit them for our convenience. For instance, if we try to extend the example line above, we see that it does, in fact, cross the x-axis:

QWQW pic 2

 

 

 

 

 

So what kind of lines do not cross either the x-axis or the y-xis? We know that the equation of a line on the XY plane is given by ax + by  + c = 0. We also know that if we want to find the slope of a line, we can use the equation y = (-a/b)x – c/b, where the slope of the line is -a/b.

A line with a slope of 0 is parallel to the x-axis. For the slope (i.e. -a/b) to be 0, a must equal 0. So if a = 0, the line will not cross the x-axis – it is parallel to the x-axis. The equation of the line, in this case, will become y = k. In all other cases, a line will cross the x-axis at some point.

Similarly, it might appear that a line doesn’t cross the y-axis but it does at some point if its slope is anything other than infinity. A line with a slope of infinity is parallel to the y-axis. For -a/b to be infinity, b must equal 0. So if b = 0, the line will not cross the y-axis. The equation of the line in this case will become x = k. In all other cases, a line will cross the y-axis at some point.

Now, we can easily solve this official question:

Does the line with equation ax+by = c, where a, b and c are real constants, cross the x-axis?

Statement 1: b not equal to 0

Statement 2: ab > 0

As we discussed earlier, all lines cross the x-axis except lines which have a slope of 0, i.e. a = 0.

Statement 1: b not equal to 0

This tells statement us that b is not 0 – which means the line is not parallel to y-axis – but it doesn’t tell us whether or not a is 0, so we don’t know whether the line is parallel to the x-axis or crosses it. Therefore, this statement alone is not sufficient.

Statement 2: ab>0

If ab > 0, it means that neither a nor b is 0 (since any number times 0 will equal 0). This means the line is parallel to neither x-axis nor the y-xis, and therefore must cross the x-axis. This statement alone is sufficient and our answer is B.

Hopefully this has helped clear up some coordinate geometry concepts today.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Be Tolerant Towards Pronoun Ambiguity on the GMAT

Quarter Wit, Quarter WisdomWe encounter many different types of pronoun errors on the GMAT Verbal Section. Some of the most common errors include:

Using a pronoun without an antecedent. For example, the sentence, “Although Jack is very rich, he makes poor use of it,” is incorrect because “it” has no antecedent. The antecedent should instead be “money” or “wealth.”

Error in matching the pronoun to its antecedent in number and gender. For example, the sentence, “Pack away the unused packets, and save it for the next game,” is incorrect because the antecedent of “it” is referring to “unused packets,” which is plural.

Using a nominative/objective case pronoun when the antecedent is possessive. For example, the sentence, “The client called the lawyer’s office, but he did not answer,” is incorrect because the antecedent of “he” should be referring to “lawyer,” but it appears only in the possessive case. Official GMAT questions will not give you this rule as the only decision point between two options.

But note that the rules governing pronoun ambiguity are not as strict as other rules! Pronoun ambiguity should be the last decision point for eliminating an option after we have taken care of SV agreements, tenses, modifiers, parallelism etc.

Every sentence that has two nouns before a pronoun does not fall under the “pronoun ambiguity error” category. If the pronoun agrees with two nouns in number and gender, and both nouns could be the antecedent of the pronoun, then there is a possibility of pronoun ambiguity. But in other cases, logic can dictate that only one of the nouns can really perform (or receive) an action, and so it is logically clear to which noun the pronoun refers.

For example, “Take the bag out of the car and get it fixed.”

What needs to get fixed? The bag or the car? Either is possible. Here we have a pronoun ambiguity, but it is highly unlikely you will see something like this on the GMAT.

A special mention should be made here about the role nouns play in the sentence. Often, a pronoun which acts as the subject of a clause refers to the noun which acts as a subject of the previous clause. In such sentences, you will often find that the antecedent is unambiguous. Similarly, if the pronoun acts as the direct object of a clause, it could refer to the direct object of the  previous clause. If the pronoun and its antecedent play parallel roles, a lot of clarity is added to the sentence. But it is not necessary that the pronoun and its antecedent will play parallel roles.

Let’s look at a different example, “The car needs to be taken out of the driveway and its brakes need to get fixed.”

Here, obviously the antecedent of “its” must be the car since only it has brakes, not the driveway. Besides, the car is the subject of the previous clause and “its” refers to the subject. Hence, this sentence would be acceptable.

A good rule of thumb would be to look at the options. If no options sort out the pronoun issue by replacing it with the relevant noun, just forget about pronoun ambiguity. If there are options that clarify the pronoun issue by replacing it with the relevant noun, consider all other grammatical issues first and then finally zero in on pronoun ambiguity.

Let’s take a quick look at some official GMAT questions involving pronouns now:

Congress is debating a bill requiring certain employers provide workers with unpaid leave so as to care for sick or newborn children. 

(A) provide workers with unpaid leave so as to 
(B) to provide workers with unpaid leave so as to 
(C) provide workers with unpaid leave in order that they 
(D) to provide workers with unpaid leave so that they can 
(E) provide workers with unpaid leave and 

The answer is (D). Why? The correct sentence would use “to provide” (not “provide”) and “so that” (not “so as to”), and should read, “Congress is debating a bill requiring certain employers to provide workers with unpaid leave so that they can care for sick or newborn children.” In this sentence, “they” logically refers to “workers.” Even though “they” could refer to employers, too, after you sort out the rest of the errors, you are left with (D) only, hence answer must be (D).

Let’s look at another question:

While depressed property values can hurt some large investors, they are potentially devastating for homeowners, whose equity – in many cases representing a life’s savings – can plunge or even disappear.

(A) they are potentially devastating for homeowners, whose
(B) they can potentially devastate homeowners in that their
(C) for homeowners they are potentially devastating, because their
(D) for homeowners, it is potentially devastating in that their
(E) it can potentially devastate homeowners, whose

The correct answer is (A). The correct sentence should read, “While depressed property values can hurt some large investors, they are potentially devastating for homeowners, whose equity – in many cases representing a life’s savings – can plunge or even disappear.” The pronoun “they” logically refers to “depressed property values.” Both the pronoun and its antecedent serve as subjects in their respective clauses, so the pronoun antecedent is quite clear.

One more question:

Although Napoleon’s army entered Russia with far more supplies than they had in their previous campaigns, it had provisions for only twenty-four days. 

(A) they had in their previous campaigns 
(B) their previous campaigns had had 
(C) they had for any previous campaign 
(D) in their previous campaigns 
(E) for any previous campaign

The correct answer is (E). The correct sentence should read, “Although Napoleon’s army entered Russia with far more supplies than for any previous campaign, it had provisions for only twenty-four days.”

The pronoun “it” logically refers to “Napolean’s army” and not Russia. Both the pronoun and its antecedent serve as subjects in their respective clauses, so the pronoun antecedent is quite clear. Note that the pronoun and its antecedent are a part of the non-underlined portion of the sentence so we don’t need to worry about the usage here but it strengthens our understanding of pronoun ambiguity.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Using a Venn Diagram vs. a Double Set Matrix on the GMAT

Quarter Wit, Quarter WisdomCritics may have given a rotten rating to the recently released “Batman v. Superman” movie, but we sure can use it to learn a valuable GMAT lesson. A difficult decision point for GMAT test takers is picking the probable winner between Venn diagrams and Double Set Matrices for complicated sets questions. If that is true for you too, then the onscreen rivalry between Batman and Superman will help you remember this trick:

Venn diagrams are like Superman – all powerful. They can help you solve almost all questions involving either 2 or 3 overlapping sets. But then, there are some situations in which double set matrix method (aka Batman with his amazing weaponry) might be easier to use. It is possible to solve these questions using Venn diagrams, too, but it is more convenient to solve them using a Double Set Matrix.

We have discussed solving three overlapping sets using Venn diagrams here.

Today, we will look at the case in which using a Double Set Matrix is easier than using a Venn diagram – in instances where we have two sets of variables, such as English/Math and Middle School/High School, or Cake/Ice cream and Boys/Girls, etc.

Eventually, we will solve our question again using a Venn diagram, for those who like to use a single method for all similar questions. First, take a look at our question:

A business school event invites all of its graduate and undergraduate students to attend. Of the students who attend, male graduate students outnumber male undergraduates by a ratio of 7 to 2, and females constitute 70% of the group. If undergraduate students make up 1/6 of the group, which of the following CANNOT represent the number of female graduate students at the event?

(A) 18
(B) 27
(C) 36
(D) 72
(E) 180

To solve this problem using a Double Set Matrix, first jot down one set of variables as the row headings and the other as the column headings, as well as a row and column for “totals.” Now all you need to do is add in the information line by line as you read through the question.

“…male graduate students outnumber male undergraduates by a ratio of 7 to 2…
QWQW graph 1

 

 

“…females constitute 70% of the group.

Female students make up 70% of the group, which implies that male students (total of 9x) make up 30% of the group.

9x = (30/100)*Total Students

Total Students = 30x

Since 9x is the total number of male students while 30x is the total number of all students, the total number of female students must be 30x – 9x = 21x.

QWQW graph 2

 

 

If undergraduate students make up 1/6 of the group…

Undergrad students make 1/6 of the group, i.e. (1/6)*30x = 5x

If the total number of undergrad students is 5x and the number of male undergrad students is 2x, the number of female undergrad students must be 5x – 2x = 3x.

This implies that the number of graduate females must be 18x, since the total number of females is 21x.

QWQW graph 3

 

 

Therefore, the number of graduate females must be a multiple of 18. 27 is the only answer choice that is not a multiple of 18, so it cannot be the number of graduate females – therefore, our answer must be B.

Now, here is how Superman can rescue us in this question. An analysis similar to the one above will give us a Venn diagram which looks like this:

qwqw pic

 

 

 

 

 

 

Of course, we will get the same answer: the number of graduate females must be a multiple of 18. We know 27 is not a multiple of 18, so it cannot be the number of graduate females and therefore, our answer is still B.

Hopefully, next time you come across an overlapping sets question, you will know exactly who your superhero is!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

What Makes GMAT Quant Questions So Hard?

Quarter Wit, Quarter WisdomWe know that the essentials of the GMAT Quant section are pretty simple: advanced topics such as derivatives, complex numbers, matrices and trigonometry are not included, while fundamentals we all learned from our high school math books are included. So it would be natural to think that the GMAT Quant section should not pose much of a problem for most test-takers (especially for engineering students, who have actually covered far more advanced math during their past studies).

Hence, it often comes as a shock when many test-takers, including engineering students, receive a dismal Quant score on the first practice test they take. Of course, with practice, they usually wise up to the treachery of the GMAT, but until then, the Quant section is responsible for many a nightmare!

Today, let’s see what kind of treachery we are talking about – problems like this make some people laugh out loud and others pull at their hair!

Is the product pqr divisible by 12?
Statement 1: p is a multiple of 3
Statement 2: q is a multiple of 4

This seems like an easy C (Statements 1 and 2 together are sufficient, but alone are not sufficient), doesn’t it? P is a multiple of 3 and q is a multiple of 4, so together, p*q would be a multiple of 3*4 = 12. If p * q is already a multiple of 12, then obviously it would seem that p*q*r would be a multiple of 12, too.

But here is the catch – where is it mentioned that r must be an integer? Just because p and q are integers (multiples of 3 and 4 respectively), it does not imply that r must also be an integer.

If r is an integer, then sure, p*q*r will be divisible by 12. Imagine, however, that p = 3, q = 4 and r = 1/12. Now the product p*q*r = 3*4*(1/12) = 1. 1 is not divisible by 12, so in this case, pqr is not divisible by 12. Hence, both statements together are not sufficient to answer the question, and our answer is in fact E!

This question is very basic, but it still tricks us because we want to assume that p, q and r are clean integer values.

Along these same lines, let’s try the another one:

If 10^a * 3^b * 5^c = 450^n, what is the value of c?
Statement 1: a is 1.
Statement 2:  b is 2.

The first thing most of us will do here is split 450 into its prime factors:

450 = 2 * 3^2 * 5^2

450^n = 2^n * 3^2n * 5^2n

And do the same thing with the left side of the equation:

10^a * 3^b * 5^c = 2^a * 3^b * 5^(a+c)

Bringing the given equation back, we get:

2^a * 3^b * 5^(a+c) = 2^n * 3^2n * 5^2n

Statement 1: a is 1.

Equating the power of 2 on both sides, we see that a = n = 1.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

Statement 2:  b is 2.

Equating the power of 3 on both sides, we see that b = 2n = 2, so n = 1.

If n = 1, a = 1 by equating the powers of 2 on both sides.

a + c = 2n (equating the power of 5 on both sides)

1 + c = 2

c = 1

So it seems that both statements are separately sufficient. But hold on – again, the variables here don’t need to be cleanly fitting integers. The variables could pan out the way discussed in our first problem, or very differently.

Say, n = 1. When Statement 1 gives you that a = 1, you get 10^1 * 3^b * 5^c = 450^1.

3^b * 5^c = 45

Now note that value of c depends on the value of b, which needn’t be 2.

If b  = 3, then 3^3 * 5^c = 45.

5^c = 45/27

C will take a non-integer value here.

c = .3174

The question does not mention that all variables are integers, therefore there are infinite values that c can take depending on the values of b. Similarly, we can see that Statement 2 alone is also not sufficient. Using both statements together, you will get:

2^a * 3^b * 5^(a+c) = 450^n

2^1 * 3^2 * 5^(1 + c) = 450^n

5^(1 + c) = 450^n/18

By now, you’ve probably realized that depending on the value of n, c can take infinite different values. If n = 1, c = 1. If n = 2, c = 4.8. And so on… We don’t need to actually find these values – it is enough to know that different values of n will give different values of c.

With this in mind, we can see that both statements together are not sufficient, and therefore our answer must be E.

Hopefully, in future, this sneaky trick will not get you!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Dealing with Tangents on the GMAT

Quarter Wit, Quarter WisdomConsidering a two dimensional figure, a tangent is a line that touches a curve at a single point.  Here are some examples of tangents:
QWQW 1

 


 

In each of these cases, the line touches the curve at a single point. In the case of a circle, when you draw the radius of the circle from the center to the point of contact with the tangent, the radius is perpendicular to the tangent (as demonstrated in the figure on the right, above). A question discussing this concept is given in our post here.

Today, we will look at a question involving a tangent to a parabola:

If f(x) = 3x^2 – tx + 5 is tangent to the x-axis, what is the value of the positive number t?

(A) 2√15
(B) 4√15
(C) 3√13
(D) 4√13
(E) 6√15

Let’s first try to understand what the question is saying.

f(x) is a tangent to the x-axis. We know that the x-axis is a straight line, so f(x) must be a curve. A quadratic equation, such as our given equation of f(x) = 3x^2 -tx +5, gives a parabola. Since the x^2 term in the equation is positive, the parabola would be facing upwards and touching the x-axis at a single point, such as:

QWQW 2

 

 

 

 

 

Since the parabola touches the x-axis in only one point, it means the quadratic has only one root, or in other words, the quadratic must be a perfect square.

Therefore, f(x) = 3x^2 – tx + 5 = √3(x)^2 – tx + (√5)^2

To get f(x) in the form a^2 – 2ab + b^2 = (a – b)^2,

tx = 2ab = (2√3)x * √5

t = 2√15

Note that if t takes this value, the quadratic will have only one root.

Plugging this value of t back into our equation, we will get: f(x) = √3(x)^2 – 2(√15)(x) + (√5)^2

f(x) = (√3)x – (√5)^2

We know that the root of f(x) is the point where the value of the y coordinate is 0. Therefore:

(√3)x – (√5)^2  = 0

x = (√5)/(√3)

At this x co-ordinate, the parabola will touch the x axis.

[This calculation was shown only to help you completely understand the question. We could have easily stopped at t = 2(√15).]

Therefore, our answer is A.

The question can be solved in various other ways – think of how, and write your thoughts in the comments below!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Understanding Absolute Values with Two Variables

Quarter Wit, Quarter WisdomWe have looked at quite a few absolute value and inequality concepts. (Check out our discussion on the basics of absolute values and inequalities, here, and our discussion on how to handle inequalities with multiple absolute value terms in a single variable, here.) Today let’s look at an absolute value concept involving two variables. It is unlikely that you will see such a question on the actual GMAT, since it involves multiple steps, but it will help you understand absolute values better.

Recall the definition of absolute value:

|x| = x if x ≥ 0

|x| = -x if x < 0

So, to remove the absolute value sign, you will need to consider two cases – one when x is positive or 0, and another when it is negative.

Say, you are given an inequality, such as |x – y| < |x|. Here, you have two absolute value expressions: |x – y| and |x|. You need to get rid of the absolute value signs, but how will you do that?

You know that to remove the absolute value sign, you need to consider the two cases. Therefore:

|x – y| = (x – y) if (x – y) ≥ 0

|x – y| = – (x – y) if (x – y) < 0

But don’t forget, we also need to remove the absolute value sign that |x| has. Therefore:

|x| = x if x ≥ 0

|x| = -x if x < 0

In all we will get four cases to consider:

Case 1: (x – y) ≥ 0 and x ≥ 0

Case 2: (x – y) < 0 and x ≥ 0

Case 3: (x – y) ≥ 0 and x < 0

Case 4: (x – y) < 0 and x < 0

Let’s look at each case separately:

Case 1: (x – y) ≥ 0 (which implies x ≥ y) and x ≥ 0

|x – y| < |x|

(x – y) < x

-y < 0

Multiply by -1 to get:

y > 0

In this case, we will get 0 < y ≤ x.

Case 2: (x – y) < 0 (which implies x < y) and x ≥ 0

|x – y| < |x|

-(x – y) < x

2x > y

x > y/2

In this case, we will get 0 < y/2 < x < y.

Case 3: (x – y) ≥ 0 (which implies x ≥ y) and x < 0

|x – y| < |x|

(x – y) < -x

2x < y

x < y/2

In this case, we will get y ≤ x < y/2 < 0.

Case 4: (x – y) < 0 (which implies x < y) and x < 0

|x – y| < |x|

-(x – y) < -x

-x + y < -x

y < 0

In this case, we will get x < y < 0.

Considering all four cases, we get that both x and y are either positive or both are negative. Case 1 and Case 2 imply that if both x and y are positive, then x > y/2, and Case 3 and Case 4 imply that if both x and y are negative, then x < y/2. With these in mind, there is a range of values in which the inequality will hold. Both x and y should have the same sign – if they are both positive, x > y/2, and if they are both negative, x < y/2.

Here are some examples of values for which the inequality will hold:

x = 4, y = 5

x = 8, y = 2

x = -2, y = -1

x = -5, y = -6

etc.

Here are some examples of values for which the inequality will not hold:

x = 4, y = -5 (x and y have opposite signs)

x = 5, y = 15 (x is not greater than y/2)

x = -5, y = 9 (x and y have opposite signs)

x = -6, y = -14 (x is not less than y/2)

etc.

As said before, don’t worry about going through this method during the actual GMAT exam – if you do get a similar question, some strategies such as plugging in values and/or using answer choices to your advantage will work. Overall, this example hopefully helped you understand absolute values a little better.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: The Case of a Correct Answer Despite Incorrect Logic!

Quarter Wit, Quarter WisdomIt is common for GMAT test-takers to think in the right direction, understand what a question gives and what it is asking to be found out, but still get the wrong answer. Mistakes made during the execution of a problem are common on the GMAT, but what is rather rare is going with incorrect logic and still getting the correct answer! If only life was this rosy so often!

Today, we will look at a question in which exactly this phenomenon occurs – we will find the flaw in the logic that test-takers often come up with and then learn how to correct that flaw:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did had he driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100

(B) 120

(C) 140

(D) 150

(E) 160

This little gem (and it’s detailed algebra solution) is from our Advanced Word Problems book. We will post its solution here, too, for the sake of a comprehensive discussion:

Method 1: Algebra
Let’s start with the basic “Distance = Rate * Time” formula:

D = R*T ……….(I)

From here, the first theoretical trip can be represented as D + 70 = (R + 5)(T + 1), (the motorist travels for 1 extra hour at a rate of 5 mph faster), which can be expanded to D + 70 = RT + R + 5T +5.

We can then eliminate “D” by plugging in the value of “D” from our equation (I):

RT + 70 = RT + R + 5T + 5, which simplifies to 70 = R + 5T + 5 and then to 65 = R + 5T ……….. (II)

The second theoretical trip can be represented as (R+10)(T+2), which expands to RT + 2R + 10T + 20 (not that we only have an expression since we don’t know what the distance is).

The two middle terms (2R + 10T) can be factored to 2(R+5T), which allows us to use equation (II) here:

RT + 2(R+5T) + 20 = RT + 2(65) + 20 = RT + 150.

Since the original distance was RT, the additional distance is 150 more miles, or answer choice D.

We totally understand that this solution is a bit convoluted – algebra often is. So, understandably, students often look for a more direct logical solution.

Here is one they sometimes employ:

Method 2: Logic (Incorrect)
If the motorist had driven 1 hour longer at a rate 5 mph faster, then his original speed would be 70 miles subtracted by the extra 5 miles he drove in that hour to get 70 – 5 = 65 mph. If he drives at a rate 10 mph faster (i.e. at 65 + 10 = 75) * 2 for the extra hours, he/she would have driven 150 miles extra.

But here is the catch in this logic:

The motorist drove for an average rate of 5 mph extra. So the 70 includes not only the extra distance covered in the last hour, but also the extra 5 miles covered every hour for which he drove. Hence, his original speed is not 65. Now, let’s see the correct logical method of solving this:

Method 3: Logic (Correct)
Let’s review the original problem first. Say, speed is “S” mph – we don’t know the number of hours for which this speed was maintained.

STEP 1:

S + S + S + … + S + S = TOTAL DISTANCE COVERED

In the first hypothetical case, the motorist drove for an extra hour at a speed of 5 mph faster. This means he covered 5 extra miles every hour and then covered another S + 5 miles in the last hour. The underlined distances are the extra ones which all add up to 70.

STEP 2:

S + S + S + … + S + S = TOTAL DISTANCE COVERED

+5 +5 +5 + … + 5 + 5 = +70

In the second hypothetical case, in which the motorist drove for two hours longer at a speed of 10 mph faster,  he adds another 5 mph to his hourly speed and covers yet another distance of “S” in the second extra hour. In addition to S, he also covers another 10 miles in the second extra hour. The additional distances are shown in red  in the third case – every hour, the speed is 10 mph faster and he drove for two extra hours in this case (compared with Step 1).

STEP 3:

S + S + S + … + S + S + S + S = TOTAL DISTANCE COVERED

+5  +5  +5 + …  +5  +5  +5 = +70

+5  +5  +5 + …  +5  +5  +5 + 10 = +70 + 10

Note that the +5s and the S all add up to 70 (as seen in Step 2). We also separately add the extra 10 from the last hour. This is the logic of getting the additional distance of 70 + 70 + 10 = 150. It involves no calculations, but does require you to understand the logic. Therefore, our answer is still D.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: How to Find Composite Numbers on the GMAT

Quarter Wit, Quarter WisdomWe love to talk about prime numbers and their various properties for GMAT preparation, but composite numbers usually aren’t mentioned. Composite numbers are often viewed as whatever is leftover after prime numbers are removed from a set of positive integers (except 1 because 1 is neither prime, nor composite), but it is important to understand how these numbers are made, what makes them special and what should come to mind when we read “composite numbers.”

Principle: Every composite number is made up of 2 or more prime numbers. The prime numbers could be the same or they could be distinct.

For example:

2*2 = 4 (Composite number)

2*3*11 = 66 (Composite number)

5*23 = 115 (Composite number)

and so on…

Look at any composite number. You will always be able to split it into 2 or more prime numbers (not necessarily distinct). For example:

72 = 2*2*2*3*3

140 = 2*2*5*7

166 = 2*83

and so on…

This principle does look quite simple and intuitive at first, but when tested, we could face problems because we don’t think much about it. Let’s look at it with the help of one of our 700+ level GMAT questions:

x is the smallest integer greater than 1000 that is not prime and that has only one factor in common with 30!. What is x?

(A) 1009

(B) 1021

(C) 1147

(D) 1273

(E) 50! + 1

If we start with the answer choices, the way we often do when dealing with prime/composite numbers, we will get stuck. If we were looking for a prime number, we would use the method of elimination – we would find factors of all other numbers and the number that was left over would be the prime number.

But in this question, we are instead looking for a composite number – a specific composite number – and some of the answer choices are probably prime. Try as we might, we will not find a factor for them, and by the time we realize that it is prime, we will have wasted a lot of precious time. Let’s start from the question stem, instead.

We need a composite number that has only one factor in common with 30!. Every positive integer will have 1 as a factor, as will 30!, hence the only factor our answer and 30! will have in common is 1.

30! = 1*2*3*…*28*29*30

30! is the product of all integers from 1 to 30, so all prime numbers less than 30 are factors of 30!.

To make a composite number which has no prime factor in common with 30!, we must use prime numbers greater than 30. The first prime number greater than 30 is 31.

(As an aside, note that if we were looking for the smallest number with no factor other than 1 in common with 31!, we would skip to 37. All integers between 31 and 37 are composite and hence, would have factors lying between 1 and 31. Similarly, if we were looking for the smallest number with no factor other than 1 in common with 50!, 53 would be the answer.)

Let’s get back to our question. If we want to make a composite number without using any primes until 30, we must use two or more prime numbers greater than 30, and the smallest prime greater than 30 is 31. If we use two 31’s to get the smallest composite number, we get 31*31 = 961 But 961 is not greater than 1000, so it cannot be our answer.

So, let’s find the next prime number after 31 – it is 37. Multiplying 31 and 37, we get 31*37 = 1147. This is the smallest composite number greater than 1000 with no prime factors in common with 30! – the only factor it has in common with 30! is 1. Therefore, our answer is (C).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Ratios in GMAT Data Sufficiency

Quarter Wit, Quarter WisdomWe know that ratios are the building blocks for a lot of other concepts such as time/speed, work/rate and mixtures. As such, we spend a lot of time getting comfortable with understanding and manipulating ratios, so the GMAT questions that test ratios seem simple enough, but not always! Just like questions from all other test areas, questions on ratios can be tricky too, especially when they are formatted as Data Sufficiency questions.

Let’s look at two cases today: when a little bit of data is sufficient, and when a lot of data is insufficient.

When a little bit of data is sufficient!
Three brothers shared all the proceeds from the sale of their inherited property. If the eldest brother received exactly 5/8 of the total proceeds, how much money did the youngest brother (who received the smallest share) receive from the sale?

Statement 1: The youngest brother received exactly 1/5 the amount received by the middle brother.

Statement 2: The middle brother received exactly half of the two million dollars received by the eldest brother.

First impressions on reading this question? The question stem gives the fraction of money received by one brother. Statement 1 gives the fraction of money received by the youngest brother relative to the amount received by the middle brother. Statement 2 gives the fraction of money received by the middle brother relative to the eldest brother and an actual amount. It seems like the three of these together give us all the information we need. Let’s dig deeper now.

From the Question stem:

Eldest brother’s share = (5/8) of Total

Statement 1: Youngest Brother’s share = (1/5) * Middle brother’s share

We don’t have any actual number – all the information is in fraction/ratio form. Without an actual value, we cannot find the amount of money received by the youngest brother, therefore, Statement 1 alone is not sufficient.

Statement 2: Middle brother’s share = (1/2) * Eldest brother’s share, and the eldest brother’s share = 2 million dollars

Middle brother’s share = (1/2) * 2 million dollars = 1 million dollars

Now, we might be tempted to jump to Statement 1 where the relation between youngest brother’s share and middle brother’s share is given, but hold on: we don’t need that information. We know from the question stem that the eldest brother’s share is (5/8) of the total share.

So 2 million = (5/8) of the total share, therefore the total share = 3.2 million dollars.

We already know the share of the eldest and middle brothers, so we can subtract their shares out of the total and get the share of the youngest brother.

Youngest brother’s share = 3.2 million – 2 million – 1 million = 0.2 million dollars

Statement 2 alone is sufficient, therefore, the answer is B.

When a lot of data is insufficient!
A department manager distributed a number of books, calendars, and diaries among the staff in the department, with each staff member receiving x books, y calendars, and z diaries. How many staff members were in the department?

Statement 1: The numbers of books, calendars, and diaries that each staff member received were in the ratio 2:3:4, respectively.

Statement 2: The manager distributed a total of 18 books, 27 calendars, and 36 diaries.

First impressions on reading this question? The question stem tells us that each staff member received the same number of books, calendars, and diaries. Statement 1 gives us the ratio of books, calendars and diaries. Statement 2 gives us the actual numbers. It certainly seems that we should be able to obtain the answer. Let’s find out:

Looking at the question stem, Staff Member 1 recieved x books, y calendars, and z diaries, Staff Member 2 recieved x books, y calendars, and z diaries… and so on until Staff Member n (who also recieves x books, y calendars, and z diaries).

With this in mind, the total number of books = nx, the total number of calendars = ny, and the total number of diaries = nz.

Question: What is n?

Statement 1 tells us that x:y:z = 2:3:4. This means the values of x, y and z can be:

2, 3, and 4,

or 4, 6, and 8,

or 6, 9, and 12,

or any other values in the ratio 2:3:4.

They needn’t necessarily be 2, 3 and 4, they just need the required ratio of 2:3:4.

Obviously, n can be anything here, therefore, Statement 1 alone is not sufficient.

Statement 2 tell us that nx = 18, ny = 27, and nz = 36.

Now we know the actual values of nx, ny and nz, but we still don’t know the values of x, y, z and n.

They could be

2, 3, 4 and 9

or 6, 9, 12 and 3

Therefore, Statement 2 alone is also not sufficient.

Considering both statements together, note that Statement 2 tells us that nx:ny:nz = 18:27:36 = 2:3:4 (they had 9 as a common factor).

Since n is a common factor on left side, x:y:z = 2:3:4 (ratios are best expressed in the lowest form).

This is a case of what we call “we already knew that” – information given in Statement 1 is already a part of Statement 2, so it is not possible that Statement 2 alone is not sufficient but that together Statement 1 and 2 are. Hence, both statements together are not sufficient, and our answer must be E.

A question that arises often here is, “Why can’t we say that the number of staff members must be 9?”

This is because the ratio of 2:3:4 is same as the ratio of 6:9:12, which is same as 18:27:36 (when you multiply each number of a ratio by the same number, the ratio remains unchanged).

If 18 books, 27 calendars, and 36 diaries are distributed in the ratio 2:3:4, we could give them all to one person, or to 3 people (giving them each 6 books, 9 calendars and 12 diaries), or to 9 people (giving them each 2 books, 3 calendars and 4 diaries).

When we see 18, 27 and 36, what comes to mind is that the number of people could have been 9, which would mean that the department manager distributed 2 books, 3 calendars and 4 diaries to each person. But we know that 9 is divisible by 3, which should remind us that the number of people could also be 3, which would mean that the manager distributed 6 books, 9 calendars and 12 diaries to each person. As such, we still don’t know how many staff members there are, and our answer remians E.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Circular Reasoning in GMAT Critical Reasoning Questions

Quarter Wit, Quarter WisdomConsider this argument:

Anatomical bilateral symmetry is a common trait. It follows, therefore, that it confers survival advantages on organisms. After all, if bilateral symmetry did not confer such advantages, it would not be common.

What is the flaw here?

The argument restates rather than proves. The conclusion is a  premise, too – we start out by assuming that the conclusion is true and then state that the conclusion is true.

If A (bilateral symmetry) were not B (confer survival advantages), A (bilateral symmetry) would not be C (common).

A (bilateral symmetry) is C (common) so A (bilateral symmetry) is B (confer survival advantages).

Note that we did not try to prove that “A is C implies A is B”. We did not explain the connection between C and B. For our reasoning, all we said is that if A were not B, it would not be C, so we are starting out by taking the conclusion to be true.

This is called circular reasoning. It is a kind of logical fallacy – a flaw in the logic. You begin with what you are trying to prove, using your own conclusion as one of your premises.

Why is it good to understand circular reasoning for the GMAT? A critical reasoning question that asks you to mimic the reasoning argument could require you to identify such a flawed reasoning and find the argument that mimics it.

Continuing with the previous example:

Anatomical bilateral symmetry is a common trait. It follows, therefore, that it confers survival advantages on organisms. After all, if bilateral symmetry did not confer such advantages, it would not be common.

The pattern of reasoning in which one of the following arguments is most similar to that in the argument above?

(A) Since it is Sawyer who is negotiating for the city government, it must be true that the city takes the matter seriously. After all, if Sawyer had not been available, the city would have insisted that the negotiations be deferred.
(B) Clearly, no candidate is better qualified for the job than Trumbull. In fact, even to suggest that there might be a more highly qualified candidate seems absurd to those who have seen Trumbull at work.
(C) If Powell lacked superior negotiating skills, she would not have been appointed arbitrator in this case. As everyone knows, she is the appointed arbitrator, so her negotiating skills are, detractors notwithstanding, bound to be superior.
(D) Since Varga was away on vacation at the time, it must have been Rivers who conducted the secret negotiations. Any other scenario makes little sense, for Rivers never does the negotiating unless Varga is unavailable.
(E) If Wong is appointed arbitrator, a decision will be reached promptly. Since it would be absurd to appoint anyone other than Wong as arbitrator, a prompt decision can reasonably be expected.

We’ve established that the above pattern of reasoning has a circular reasoning flaw. Let’s consider each answer option to find the one which has similarly flawed reasoning.

(A) Since it is Sawyer who is negotiating for the city government, it must be true that the city takes the matter seriously. After all, if Sawyer had not been available, the city would have insisted that the negotiations be deferred.

Here is the structure of this argument:

If A (Sawyer) were not B (available), C (the city) would have D (insisted on deferring).

Since A (Sawyer) is B (available to the city), C (the city) does E (takes matter seriously).

Obviously, this argument structure is not the same as in the original argument.

(B) Clearly, no candidate is better qualified for the job than Trumbull. In fact, even to suggest that there might be a more highly qualified candidate seems absurd to those who have seen Trumbull at work.

Here is the structure of this argument:

A (people who have seen Trumbull at work) find B (Trumbull is not the best) absurd, therefore B (Trumbull is not the best) is false.

This is not circular reasoning. We have not assumed that B is false in our premises, we are simply saying that people think B is absurd. This is flawed logic too, but it is not circular reasoning.

(C) If Powell lacked superior negotiating skills, she would not have been appointed arbitrator in this case. As everyone knows, she is the appointed arbitrator, so her negotiating skills are, detractors notwithstanding, bound to be superior.

Here is the structure of this argument:

If A (Powell) were not B (had superior negotiating skills), A (Powell) would not have been C (appointed arbitrator).

A (Powell) is C (appointed arbitrator), therefore A (Powell) is B (had superior negotiating skills).

Note that the structure of the argument matches the structure of our original argument – this is circular reasoning, too. We are saying that if A were not B, A would not be C and concluding that since A is C, A is B. The conclusion is already taken to be true in the initial argument, so we can see it is is also an example of circular reasoning.

Hence (C) is the correct answer. Nevertheless, let’s look at the other two options and why they don’t work:

(D) Since Varga was away on vacation at the time, it must have been Rivers who conducted the secret negotiations. Any other scenario makes little sense, for Rivers never does the negotiating unless Varga is unavailable.

Here is the structure of this argument:

If A (Varga) is B (available), C (Rivers) does not do D (negotiate).

A (Varga) was not B (available), so C (Rivers) did D (negotiate).

This logic is flawed – the premise tells us what happens when A is B, however it does not tell us what happens when A is not B. We cannot conclude anything about what happens when A is not B. And because this is not circular reasoning, it cannot be the answer.

(E) If Wong is appointed arbitrator, a decision will be reached promptly. Since it would be absurd to appoint anyone other than Wong as arbitrator, a prompt decision can reasonably be expected.

Here is the structure of this argument:

If A (Wong) is B (appointed arbitrator), C (a decision) will be D (reached promptly).

A (Wong) not being B (appointed arbitrator) would be absurd, so C (a decision) will be D (reached promptly).

Again, this argument uses brute force, but it is not circular reasoning. “A not being B would be absurd” is not a convincing reason, so the argument is not strong as it is, but in any case, we don’t have to worry about it since it doesn’t use circular reasoning.

Take a look at this question for practice:

Dr. A: The new influenza vaccine is useless at best and possibly dangerous. I would never use it on a patient.
Dr. B: But three studies published in the Journal of Medical Associates have rated that vaccine as unusually effective.
Dr. A: The studies must have been faulty because the vaccine is worthless.

In which of the following is the reasoning most similar to that of Dr. A?

(A) Three of my patients have been harmed by that vaccine during the past three weeks, so the vaccine is unsafe.
(B) Jerrold Jersey recommends this milk, and I don’t trust Jerrold Jersey, so I won’t buy this milk.
(C) Wingz tennis balls perform best because they are far more effective than any other tennis balls.
(D) I’m buying Vim Vitamins. Doctors recommend them more often than they recommend any other vitamins, so Vim Vitamins must be good.
(E) Since University of Muldoon graduates score about 20 percent higher than average on the GMAT, Sheila Lee, a University of Muldoon graduate, will score about 20 percent higher than average when she takes the GMAT.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

A 750+ Level GMAT Geometry Question

Quarter Wit, Quarter WisdomToday we will discuss a pretty advanced GMAT question, because we can still use our basic GMAT concepts to find the answer. It may seem like we will need trigonometry to handle this question, but that is not so. In fact, the question will look familiar at first, but will present unforeseen problems later on.

While going through this exercise, we will learn a few tips and tricks which will be useful in our mainstream GMAT questions, hence, it will add value to our GMAT repertoire (especially in elimination techniques). Let’s go on to the question now:

In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = BC = X, what is the value of X?

(A) Sqrt(3) – 1

(B) Sqrt(3) + 2

(C) (Sqrt(3) – 1)/2

(D) (Sqrt(3) + 1)/2

(E) 2*(Sqrt(3) + 1)

What we see here is an isosceles triangle with one angle as 30 degrees and other two angles as (180 – 30)/2 = 75 degrees each.

The side opposite the 30 degrees angle is 2*sqrt(2). One simple observation is that X must be greater than 2*sqrt(2) because these sides are opposite the greater angles (75 degrees).

2*sqrt(2) is a bit less than 2*1.5 because Sqrt(2) = 1.414. So 2*sqrt(2) is a bit less than 3. Note that options (A), (C), and (D) are much smaller than 3, so these cannot be the value of X. We have already improved our chances of getting the correct answer by eliminating three options! Now we have to choose out of (B) and (E).
2Triangles

 

 

 

 

 

 

Here is what is given: Angle ABC = 30 degrees, and AC = 2*sqrt(2). We need to find the value of X. Now, our 30 degree angle reminds us of a 30-60-90 triangle in which we know the ratio of the sides – given one side, we can find the other two.

The problem is this: if we drop an altitude from angle B to AC, the angle 30 degrees will be split in half and we will actually get a 15-75-90 triangle, instead. We won’t have a 30-60-90 triangle anymore, so what do we do now? Let’s try to maintain the 30 degree angle as it is to get the 30-60-90 triangle, and drop an altitude from angle C to AB instead, calling it CE. Now we have a 30-60-90 triangle! Since BCE is a 30-60-90 triangle, its sides are in the ratio 1:sqrt(3):2. Side X corresponds to 2 on the ratio, so CE = x/2.

Area of triangle ABC = (1/2)*BD*AC = (1/2)*CE*AB

(1/2)*BD*2*sqrt(2) = (1/2)*(X/2)*X

BD = X^2/4*Sqrt(2)

Now DC = (1/2)AC = 2*sqrt(2)/2 = sqrt(2)

Let’s use the pythagorean theorem on triangle BDC:

BD^2 + DC^2 = BC^2

(X^2/4*Sqrt(2))^2 + (Sqrt(2))^2 = X^2

X^4/32 + 2 = X^2

X^4 – 32*X^2 + 64 = 0

X^4 – 16X^2 + 8^2 – 16X^2 = 0

(X^2 – 8)^2 – (4X)^2 = 0

(X^2 -8 + 4X) * (X^2 – 8 – 4X) = 0

Normally, this would require us to use the quadratic roots formula, but let’s not get that complicated. We can just plug in the the two shortlisted options and see if either of the factors is 0. If one of the factors becomes 0, the equation will be satisfied and we will have the root of the equation.

Since both options have both terms positive, it means the co-efficient corresponding to B in Ax^2 + Bx + C = 0 must be negative.

x = [-B +- Sqrt(B^2 – 4AC)]/2A

-B will give us a positive term if B is negative, so we will get the answer by plugging into (X^2 – 4X – 8):

Put X = Sqrt(3) + 2 in X^2 – 4X – 8 and you do not get 0.

Put X = 2*(Sqrt(3) + 1) in X^2 – 4X – 8 and you do get 0.

This means that X is 2*(Sqrt(3) + 1), so our answer must be (E).

To recap:

Tip 1: A greater side of a triangle is opposite a greater angle.

Tip 2: We can get the relation between sides and altitudes of a triangle by using the area of the triangle formula.

Tip 3: The quadratic formula can help identify the sign of the irrational roots.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Solving GMAT Critical Reasoning Questions Involving Rates

Quarter Wit, Quarter WisdomIn our “Quarter Wit, Quarter Wisdom” series, we have seen how to solve various rates questions – the basic ones as well as the complicated ones. But we haven’t considered critical reasoning questions involving rates, yet. In fact, the concept of rates makes these problems very difficult to both understand and explain. First, let’s look at what “rate” is.

Say my average driving speed is 60 miles/hr. Does it matter whether I drive for 2 hours or 4 hours? Will my average speed change if I drive more (theoretically speaking)? No, right? When I drive for more hours, the distance I cover is more. When I drive for fewer hours, the distance I cover is less. If I travel for a longer time, does it mean my average speed has decreased? No. For that, I need to know  what happened to the distance covered. If the distance covered is the same while time taken has increased, only then can I say that my speed was reduced.

Now we will look at an official question and hopefully convince you of the right answer:

The faster a car is traveling, the less time the driver has to avoid a potential accident, and if a car does crash, higher speeds increase the risk of a fatality. Between 1995 and 2000, average highway speeds increased significantly in the United States, yet, over that time, there was a drop in the number of car-crash fatalities per highway mile driven by cars.

Which of the following, if true about the United States between 1995 and 2000, most helps to explain why the fatality rate decreased in spite of the increase in average highway speeds?

(A) The average number of passengers per car on highways increased.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

(D) The average mileage driven on highways per car increased.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

Let’s break down the given argument:

  • The faster a car, the higher the risk of fatality.
  • In a span of 5 years, the average highway speed has increased.
  • In the same time, the number of car crash fatalities per highway mile driven by cars has reduced.

This is a paradox question. In last 5 years, the average highway speed has increased. This would have increased the risk of fatality, so we would expect the number of car crash fatalities per highway mile to go up. Instead, it actually goes down. We need to find an answer choice that explains why this happened.

(A) The average number of passengers per car on highways increased.

If there are more people in each car, the risk of fatality increases, if anything. More people are exposed to the possibility of a crash, and if a vehicle is in fact involved in an accident, more people are at risk. It certainly doesn’t explain why the rate of fatality actually decreases.

(B) There were increases in both the proportion of people who wore seat belts and the proportion of cars that were equipped with airbags as safety devices.

This option tells us that the safety features in the cars have been enhanced. That certainly explains why the fatality rate has gone down. If the cars are safer now, the risk of fatality would have reduced, hence this option does help us in explaining the paradox. This is the answer, but let’s double-check by looking at the other options too.

(C) The increase in average highway speeds occurred as legal speed limits were raised on one highway after another.

This option is irrelevant – why the average speed increased is not our concern at all. Our only concern is that average speed has, in fact, increased. This should logically increase the risk of fatality, and hence, our paradox still stands.

(D) The average mileage driven on highways per car increased.

This is the answer choice that troubles us the most. The rate we are concerned about is number of fatalities/highway mile driven, and this option tells us that mileage driven by cars has increased.

Now, let’s consider the parallel with our previous distance-rate-time example:

Rate = Distance/Time

We know that if I drive for more time, it doesn’t mean that my rate changes. Here, however:

Rate = Number of fatalities/highway mile driven

In this case, if more highway miles are driven, it doesn’t mean that the rate will change. It actually has no impact on the rate; we would need to know what happened to the number of fatalities to find out what happened to the rate. Hence this option does not explain the paradox.

(E) In most locations on the highways, the density of vehicles on the highway did not decrease, although individual vehicles, on average, made their trips more quickly.

This answer choice tells us that on average, the trips were made more quickly, i.e. the speed increased. The given argument already tells us that, so this option does not help resolve the paradox.

Our answer is, therefore, (B).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Should You Use the Permutation or Combination Formula?

Quarter Wit, Quarter WisdomA recurring question from many students who are preparing for GMAT is this: When should one use the permutation formula and when should one use the combination formula?

People have tried to answer this question in various ways, but some students still remain unsure. So we will give you a rule of thumb to follow in all permutation/combination questions:

You never NEED to use the permutation formula! You can always use the combination formula quite conveniently. First let’s look at what these formulas do:

Permutation: nPr = n!/(n-r)!
Out of n items, select r and arrange them in r! ways.

Combination: nCr = n!/[(n-r)!*r!]
Out of n items, select r.

So the only difference between the two formulas is that nCr has an additional r! in the denominator (that is the number of ways in which you can arrange r elements in a row). So you can very well use the combinations formula in place of the permutation formula like this:

nPr = nCr * r!

The nCr formula is far more versatile than nPr, so if the two formulas confuse you, just forget about nPr.

Whenever you need to “select,” “pick,” or “choose” r things/people/letters… out of n, it’s straightaway nCr. What you do next depends on what the question asks of you. Do you need to arrange the r people in a row? Multiply by r!. Do you need to arrange them in a circle? Multiply by (r-1)!. Do you need to distribute them among m groups? Do that! You don’t need to think about whether it is a permutation problem or a combination problem at all. Let’s look at this concept more in depth with the use of a few examples.

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can the presenters be chosen?

In this question, you simply have to choose 3 of the 8 teachers, and you know that you can do that in 8C3 ways. That is all that is required.

8C3 = 8*7*6/3*2*1 = 56 ways

Not too bad, right? Let’s look at another question:

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can all three presentations be done?

This question is a little different. You need to find the ways in which the presentations can be done. Here the presentations will be different if the same three teachers give presentations in different order. Say Teacher 1 presents, then Teacher 2 and finally Teacher 3 — this will be different from Teacher 2 presenting first, then Teacher 3 and finally Teacher 1. So, not only do we need to select the three teachers, but we also need to arrange them in an order. Select 3 teachers out of 8 in 8C3 ways and then arrange them in 3! ways:

We get 8C3 * 3! = 56 * 6 = 336 ways

Let’s try another one:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook. How many groups of photographs are possible?

For this problem, out of 7 photographs, we just have to select 3 to make a group. This can be done in 7C3 ways:

7C3 = 7*6*5/3*2*1 = 35 ways

Here’s another variation:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook, 1 each on the walls of his three best friends. In how many ways can he do that?

Here, out of 7 photographs, we have to first select 3 photographs. This can be done in 7C3 ways. Thereafter, we need to put the photographs on the walls of his three chosen friends. In how many ways can he do that? Now there are three distinct spots in which he will put up the photographs, so basically, he needs to arrange the 3 photographs in 3 distinct spots, which that can be done in 3! ways:

Total number of ways = 7C3 * 3! = (7*6*5/3*2*1) * 6= 35 * 6 = 210 ways

Finally, our last problem:

12 athletes will run in a race. In how many ways can the gold, silver and bronze medals be awarded at the end of the race?

We will start with selecting 3 of the 12 athletes who will win some position in the race. This can be done in 12C3 ways. But just selecting 3 athletes is not enough — they will be awarded 3 distinct medals of gold, silver, and bronze. Athlete 1 getting gold, Athlete 2 getting silver, and Athlete 3 getting bronze is not the same as Athlete 1 getting silver, Athlete 2 getting gold and Athlete 3 getting bronze. So, the three athletes need to be arranged in 3 distinct spots (first, second and third) in 3! ways:

Total number of ways = 12C3 * 3! ways

Note that some of the questions above were permutation questions and some were combination questions, but remember, we don’t need to worry about which is which. All we need to think about is how to solve the question, which is usually by starting with nCr and then doing any other required steps. Break the question down — select people and then arrange if required. This will help you get rid of the “permutation or combination” puzzle once and for all.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Keeping an Open Mind in Critical Reasoning

Quarter Wit, Quarter WisdomToday we will discuss why it is important to keep an open mind while toiling away on your GMAT studying. Don’t go into test day with biases expecting that if a question tells us this, then it must ask that. GMAC testmakers are experts in surprising you and taking advantage of your preconceived notions, which is how they confuse you and convert a 600-level question to a 700-level one.

We have discussed necessary and sufficient conditions before; we have also discussed assumptions before. This question from our own curriculum is an innovative take on both of these concepts. Let’s take a look.

All of the athletes who will win a medal in competition have spent many hours training under an elite coach. Michael is coached by one of the world’s elite coaches; therefore it follows logically that Michael will win a medal in competition.

The argument above logically depends on which of the following assumptions?

(A) Michael has not suffered any major injuries in the past year.

(B) Michael’s competitors did not spend as much time in training as Michael did.

(C) Michael’s coach trained him for many hours.

(D) Most of the time Michael spent in training was productive.

(E) Michael performs as well in competition as he does in training.

First we must break down the argument into premises and conclusions:

Premises:

  • All of the athletes who will win a medal in competition have spent many hours training under an elite coach.
  • Michael is coached by one of the world’s elite coaches.

Conclusion: Michael will win a medal in competition.

Read the argument carefully:

All of the athletes who will win a medal in competition have spent many hours training under an elite coach.

Are you wondering, “How does one know that all athletes who will win (in the future) would have spent many hours training under an elite coach?”

The answer to this is that it doesn’t matter how one knows – it is a premise and it has to be taken as the truth. How the truth was established is none of our business and that is that. If we try to snoop around too much, we will waste precious time. Also, what may seem improbable may have a perfectly rational explanation. Perhaps all athletes who are competing have spent many hours under an elite coach – we don’t know.

Basically, what this statement tells us is that spending many hours under an elite coach is a NECESSARY condition for winning. What you need to take away from this statement is that “many hours training under an elite coach” is a necessary condition to win a medal. Don’t worry about the rest.

Michael is coached by one of the world’s elite coaches.

It seems that Michael satisfies one necessary condition: he is coached by an elite coach.

Conclusion: Michael will win a medal in competition.

Now this looks like our standard “gap in logic”. To get this conclusion, the necessary condition has been taken to be sufficient. So if we are asked for the flaw in the argument, we know what to say.

Anyway, let’s check out the question (this is usually our first step):

The argument above logically depends on which of the following assumptions?

Note the question carefully – it is asking for an assumption, or a necessary premise for the conclusion to hold.

We know that “many hours training under an elite coach” is a necessary condition to win a medal. We also know that Michael has been trained by an elite coach. Note that we don’t know whether he has spent “many hours” under his elite coach. The necessary condition requires “many hours” under an elite coach.

If Michael has spent many hours under the elite coach then he satisfies the necessary condition to win a medal. It is still not sufficient for him to win the medal, but our question only asks for an assumption – a necessary premise for the conclusion to hold. It does not ask for the flaw in the logic.

Focus on what you are asked and look at answer choice (C):

(C) Michael’s coach trained him for many hours.

This is a necessary condition for Michael to win a medal. Hence, it is an assumption and therefore, (C) is the correct answer.

Don’t worry that the argument is flawed. There could be another question on this argument which asks you to find the flaw in it, however this particular question asks you for the assumption and nothing more.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: An Interesting Property of Exponents

Quarter Wit, Quarter WisdomToday, let’s take a look at an interesting number property. Once we discuss it, you might think, “I always knew that!” and “Really, what’s new here?” So let me give you a question beforehand:

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

Think about it for a few seconds – could you come up with the answer in the blink of an eye? If yes, great! Close this window and wait for the next week’s post. If no, then read on. There is much to learn today and it is an eye-opener!

Let’s start by jotting down some powers of numbers:

Power of 2: 1, 2, 4, 8, 16, 32 …

Power of 3: 1, 3, 9, 27, 81, 243 …

Power of 4: 1, 4, 16, 64, 256, 1024 …

Power of 5: 1, 5, 25, 125, 625, 3125 …

and so on.

Obviously, for every power of 2, when you multiply the previous power by 2, you get the next power (4*2 = 8).

For every power of 3, when you multiply the previous power by 3, you get the next power (27*3 = 81), and so on.

Also, let’s recall that multiplication is basically repeated addition, so 4*2 is basically 4 + 4.

This leads us to the following conclusion using the power of 2:

4 * 2 = 8

4 + 4 = 8

2^2 + 2^2 = 2^3

(2 times 2^2 gives 2^3)

Similarly, for the power of 3:

27 * 3 = 81

27 + 27 + 27 = 81

3^3 + 3^3 + 3^3 = 3^4

(3 times 3^3 gives 3^4)

And for the power of 4:

4 * 4 = 16

4 + 4 + 4 + 4 = 16

4^1 + 4^1 + 4^1 + 4^1 = 4^2

(4 times 4^1 gives 4^2)

Finally, for the power of 5:

125 * 5 = 625

125 + 125 + 125 + 125 + 125 = 625

5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 5^4

(5 times 5^3 gives 5^4)

Quite natural and intuitive, isn’t it? Take a look at the previous question again now.

For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?

A) 18

(B) 32

(C) 35

(D) 64

(E) 70

Which two powers when added will give 2^(36)?

From our discussion above, we know they are 2^(35) and 2^(35).

2^(35) + 2^(35) = 2^(36)

So x = 35 and y = 35 will satisfy this equation.

x + y = 35 + 35 = 70

Therefore, our answer is E.

One question arises here: Is this the only possible sum of x and y? Can x and y take some other integer values such that the sum of 2^x and 2^y will be 2^(36)?

Well, we know that no matter which integer values x and y take, 2^x and 2^y will always be positive, which means both x and y must be less than 36. Now note that no matter which two powers of 2 you add, their sum will always be less than 2^(36). For example:

2^(35) + 2^(34) < 2^(35) + 2^(35)

2^(2) + 2^(35) < 2^(35) + 2^(35)

etc.

So if x and y are both integers, the only possible values that they can take are 35 and 35.

How about something like this: 2^x + 2^y + 2^z = 2^36? What integer values can x, y and z take here?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Calculating the Probability of Intersecting Events

Quarter Wit, Quarter WisdomWe know our basic probability formulas (for two events), which are very similar to the formulas for sets:

P(A or B) = P(A) + P(B) – P(A and B)

P(A) is the probability that event A will occur.

P(B) is the probability that event B will occur.

P(A or B) gives us the union; i.e. the probability that at least one of the two events will occur.

P(A and B) gives us the intersection; i.e. the probability that both events will occur.

Now, how do you find the value of P(A and B)? The value of P(A and B) depends on the relation between event A and event B. Let’s discuss three cases:

1) A and B are independent events

If A and B are independent events such as “the teacher will give math homework,” and “the temperature will exceed 30 degrees celsius,” the probability that both will occur is the product of their individual probabilities.

Say, P(A) = P(the teacher will give math homework) = 0.4

P(B) = P(the temperature will exceed 30 degrees celsius) = 0.3

P(A and B will occur) = 0.4 * 0.3 = 0.12

2) A and B are mutually exclusive events

If A and B are mutually exclusive events, this means they are events that cannot take place at the same time, such as “flipping a coin and getting heads” and “flipping a coin and getting tails.” You cannot get both heads and tails at the same time when you flip a coin. Similarly, “It will rain today” and “It will not rain today” are mutually exclusive events – only one of the two will happen.

In these cases, P(A and B will occur) = 0

3) A and B are related in some other way

Events A and B could be related but not in either of the two ways discussed above – “The stock market will rise by 100 points” and “Stock S will rise by 10 points” could be two related events, but are not independent or mutually exclusive. Here, the probability that both occur would need to be given to you. What we can find here is the range in which this probability must lie.

Maximum value of P(A and B):

The maximum value of P(A and B) is the lower of the two probabilities, P(A) and P(B).

Say P(A) = 0.4 and P(B) = 0.7

The maximum probability of intersection can be 0.4 because P(A) = 0.4. If probability of one event is 0.4, probability of both occurring can certainly not be more than 0.4.

Minimum value of P(A and B):

To find the minimum value of P(A and B), consider that any probability cannot exceed 1, so the maximum P(A or B) is 1.

Remember, P(A or B) = P(A) + P(B) – P(A and B)

1 = 0.4 + 0.7 – P(A and B)

P(A and B) = 0.1 (at least)

Therefore, the actual value of P(A and B) will lie somewhere between 0.1 and 0.4 (both inclusive).

Now let’s take a look at a GMAT question using these fundamentals:

There is a 10% chance that Tigers will not win at all during the whole season. There is a 20% chance that Federer will not play at all in the whole season. What is the greatest possible probability that the Tigers will win and Federer will play during the season?

(A) 55%

(B) 60%

(C) 70%

(D) 72%

(E) 80%

Let’s review what we are given.

P(Tigers will not win at all) = 0.1

P(Tigers will win) = 1 – 0.1 = 0.9

P(Federer will not play at all) = 0.2

P(Federer will play) = 1 – 0.2 = 0.8

Do we know the relation between the two events “Tigers will win” (A) and “Federer will play” (B)? No. They are not mutually exclusive and we do not know whether they are independent.

If they are independent, then the P(A and B) = 0.9 * 0.8 = 0.72

If the relation between the two events is unknown, then the maximum value of P(A and B) will be 0.8 because P(B), the lesser of the two given probabilities, is 0.8.

Since 0.8, or 80%, is the greater value, the greatest possibility that the Tigers will win and Federer will play during the season is 80%. Therefore, our answer is E.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Basic Operations for GMAT Inequalities

Quarter Wit, Quarter WisdomWe know that we can perform all basic operations of addition, subtraction, multiplication and division on two equations:

a = b

c = d

When these numbers are equal, we know that:

a + c = b + d (Valid)

a – c = b – d (Valid)

a * c = b * d (Valid)

a / c = b / d (Valid assuming c and d are not 0)

When can we add, subtract, multiply or divide two inequalities? There are rules that we need to follow for those. Today let’s discuss those rules and the concepts behind them.

Addition:

We can add two inequalities when they have the same inequality sign.

a < b

c < d

a + c < b + d (Valid)

Conceptually, it makes sense, right? If a is less than b and c is less than d, then the sum of a and c will be less than the sum of b and d.

On the same lines:

a > b

c > d

a + c > b + d (Valid)

Case 2: What happens when the inequalities have opposite signs?

a > b

c < d

We need to multiply one inequality by -1 to get the two to have the same inequality sign.

-c > -d

Now we can add them.

a – c > b – d

Subtraction:

We can subtract two inequalities when they have opposite signs:

a > b

c < d

a – c > b – d (The result will take the sign of the first inequality)

Conceptually, think about it like this: from a greater number (a is greater than b), if we subtract a smaller number (c is smaller than d), the result (a – c) will be greater than the result obtained when we subtract the greater number from the smaller number (b – d).

Note that this result is the same as that obtained when we added the two inequalities after changing the sign (see Case 2 above). We cannot subtract inequalities if they have the same sign, so it is better to always stick to addition. If the inequalities have the same sign, we simply add them. If the inequalities have opposite signs, we multiply one of them by -1 (to get the same sign) and then add them (in effect, we subtract them).

Why can we not subtract two inequalities when they have the same inequality sign, such as when a > b and c > d?

Say, we have 3 > 1 and 5 > 1.

If we subtract these two, we get 3 – 5 > 1 – 1, or -2 > 0 which is not valid.

If instead it were 3 > 1 and 2 > 1, we would get 1 > 0 which is valid.

We don’t know how much greater one term is from the other and hence we cannot subtract inequalities when their inequality signs are the same.

Multiplication:

Here, the constraint is the same as that in addition (the inequality signs should be the same) with an extra constraint: both sides of both inequalities should be non-negative. If we do not know whether both sides are non-negative or not, we cannot multiply the inequalities.

If a, b, c and d are all non negative,

a < b

c < d

a*c < b*d (Valid)

When two greater numbers are multiplied together, the result will be greater.

Take some examples to see what happens in Case 1, or more numbers are negative:

-2 < -1

10 < 30

Multiply to get: -20 < -30 (Not valid)

-2 < 7

-8 < 1

Multiply to get: 16 < 7 (Not valid)

Division:

Here, the constraint is the same as that in subtraction (the inequality signs should be opposite) with an extra constraint: both sides of both inequalities should be non-negative (obviously, 0 should not be in the denominator). If we do not know whether both sides are positive or not, we cannot divide the inequalities.

a < b

c > d

a/c < b/d (given all a, b, c and d are positive)

The final inequality takes the sign of the numerator.

Think of it conceptually: a smaller number is divided by a greater number, so the result will be a smaller number.

Take some examples to see what happens in Case 1, or more numbers are negative.

1 < 2

10 > -30

Divide to get 1/10 < -2/30 (Not valid)

Takeaways: 

Addition: We can add two inequalities when they have the same inequality signs.

Subtraction: We can subtract two inequalities when they have opposite inequality signs.

Multiplication: We can multiply two inequalities when they have the same inequality signs and both sides of both inequalities are non-negative.

Division: We can divide two inequalities when they have opposite inequality signs and both sides of both inequalities are non-negative (0 should not be in the denominator).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Grammatical Structure of Conditional Sentences on the GMAT

Quarter Wit, Quarter WisdomToday, we will take a look at the various “if/then” constructions in the GMAT Verbal section. Let us start out with some basic ideas on conditional sentences (though I know that most of you will be comfortable with these):

A conditional sentence (an if/then sentence) has two clauses – the “if clause” (conditional clause) and the “then clause” (main clause).  The “if clause” is the dependent clause, meaning the verbs we use in the clauses will depend on whether we are talking about a real or a hypothetical situation.

Often, conditional sentences are classified into first conditional, second conditional and third conditional (depending on the tense and possibility of the actions), but sometimes we have a separate zero conditional for facts. We will follow this classification and discuss four types of conditionals:

1) Zero Conditional

These sentences express facts; i.e. implications – “if this happens, then that happens.”

  • If the suns shines, the clothes dry quickly.
  • If he eats bananas, he gets a headache.
  • If it rains heavily, the temperature drops.

These conditionals establish universally known facts or something that happens habitually (every time he eats bananas, he gets a headache).

2) First Conditional

These sentences refer to predictive conditional sentences. They often use the present tense in the “if clause” and future tense (usually with the word “will”) in the main clause.

  • If you come to  my place, I will help you with your homework.
  • If I am able to save $10,000 by year end, I will go to France next year.

3) Second Conditional

These sentences refer to hypothetical or unlikely situations in the present or future. Here, the “if clause” often uses the past tense and the main clause uses conditional mood (usually with the word “would”).

  • If I were you, I would take her to the dance.
  • If I knew her phone number, I would tell you.
  • If I won the lottery, I would travel the whole world.

4) Third Conditional

These sentences refer to hypothetical situations in the past – what could have been different in the past. Here, the “if clause” uses the past perfect tense and the main clause uses the conditional perfect tense (often with the words “would have”).

  • If you had told me about the party, I would have attended it.
  • If I had not lied to my mother, I would not have hurt her.

Sometimes, mixed conditionals are used here, where the second and third conditionals are combined. The “if clause” then uses the past perfect and the main clause uses  the word “would”.

  • If you had helped me then, I would be in a much better spot today.

Now that you know which conditionals to use in which situation, let’s take a look at a GMAT question:

Botanists have proven that if plants extended laterally beyond the scope of their root system, they will grow slower than do those that are more vertically contained.

(A) extended laterally beyond the scope of their root system, they will grow slower than do

(B) extended laterally beyond the scope of their root system, they will grow slower than

(C) extend laterally beyond the scope of their root system, they grow more slowly than

(D) extend laterally beyond the scope of their root system, they would have grown more slowly than do

(E) extend laterally beyond the scope of their root system, they will grow more slowly than do

Now that we understand our conditionals, we should be able to answer this question quickly. Scientists have established something here; i.e. it is a fact. So we will use the zero conditional here – if this happens, then that happens.

…if plants extend laterally beyond the scope of their root system, they grow more slowly than do…

So the correct answer must be (C).

A note on slower vs. more slowly – we need to use an adverb here because “slow” describes “grow,” which is a verb. So we must use “grow slowly”. If we want to show comparison, we use “more slowly”, so the use of “slower” is incorrect here.

Let’s look at another question now:

If Dr. Wade was right, any apparent connection of the eating of highly processed foods and excelling at sports is purely coincidental.

(A) If Dr. Wade was right, any apparent connection of the eating of

(B) Should Dr. Wade be right, any apparent connection of eating

(C) If Dr. Wade is right, any connection that is apparent between eating of

(D) If Dr. Wade is right, any apparent connection between eating

(E) Should Dr. Wade have been right, any connection apparent between eating

Notice the non-underlined part “… is purely coincidental” in the main clause. This makes us think of the zero conditional.

Let’s see if it makes sense:

If Dr. Wade is right, any connection … is purely coincidental.

This is correct. It talks about a fact.

Also, “eating highly processed foods and excelling at sports” is correct.

Hence, our answer must be (D).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions (Part 2)

Quarter Wit, Quarter WisdomLast week, we reviewed the concepts of cyclicity and remainders and looked at some basic questions. Today, let’s jump right into some GMAT-relevant questions on these topics:

 

 

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2.

The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1

So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

Since the exponent a remainder of 3, the new cycle ends at 3, 9, 7. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9.

This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E.

Not so bad; let’s try a data sufficiency problem:

If k is a positive integer, what is the remainder when 2^k is divided by 10?

Statement 1: k is divisible by 10

Statement 2: k is divisible by 4

With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k.

The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change:

If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6.

If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2.

If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4.

If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8.

If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case.

and so on…

So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4.

Statement 1: k is divisible by 10

With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient.

Statement 2: k is divisible by 4

This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B.

Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT:

What is the remainder of (3^7^11) divided by 5?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}.

We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4.

We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4.

For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”):

The remainder of 7^{11} divided by 4

= The remainder of (4 + 3)^{11} divided by 4

= The remainder of 3^{11} divided by 4

= The remainder of 3*3^{10} divided by 4

= The remainder of 3*9^5 divided by 4

= The remainder of 3*(8+1)^5 divided by 4

= The remainder of 3*1^5 divided by 4

= The remainder of 3 divided by 4, which itself = 3

So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4.

Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7

With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions

Quarter Wit, Quarter WisdomUsually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Today we will look at the cases in which it can, and we will see why it helps us in these cases.

First let’s look at a pattern:

 

20/10 gives us a remainder of 0 (as 20 is exactly divisible by 10)

21/10 gives a remainder of 1

22/10 gives a remainder of 2

23/10 gives a remainder of 3

24/10 gives a remainder of 4

25/10 gives a remainder of 5

and so on…

In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on…

Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5).

With this in mind:

20/5 gives a remainder of 0 (as 20 is exactly divisible by 5)

21/5 gives a remainder of 1

22/5 gives a remainder of 2

23/5 gives a remainder of 3

24/5 gives a remainder of 4

25/5 gives a remainder of 0 (as 25 is exactly divisible by 5)

26/5 gives a remainder of 1

27/5 gives a remainder of 2

28/5 gives a remainder of 3

29/5 gives a remainder of 4

30/5 gives a remainder of 0 (as 30 is exactly divisible by 5)

and so on…

So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10.

Let’s take a few questions now:

What is the remainder when 86^(183) is divided by 10?

Here, we need to find the last digit of 86^(183) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is (previously discussed in this post).

So the units digit of 86^(183) will be 6. So when we divide this by 10, the remainder will also be 6.

Next question:

What is the remainder when 487^(191) is divided by 5?

Again, when considering division by 5, the units digit can help us.

The units digit of 487 is 7.

7 has a cyclicity of 7, 9, 3, 1.

Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term.

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3

So the units digit of 487^(191) is 3, and the number would look something like ……………..3

As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5.

Therefore, the remainder of 487^(191) divided by 5 is 3.

Last question:

If x is a positive integer, what is the remainder when 488^(6x) is divided by 2?

Take a minute to review the question first. If you start by analyzing the expression 488^(6x), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether 488^(6x) is odd or even.

488^(6x) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even.

So 488^(6x) is even and will give remainder 0 when it is divided by 2.

That is all for today. We will look at some GMAT remainders-cyclicity questions next week!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Cyclicity of Units Digits on the GMAT (Part 2)

Quarter Wit, Quarter WisdomAs discussed last week, all units digits have a cyclicity of 1 or 2 or 4. Digits 2, 3, 7 and 8 have a cyclicity of 4, i.e. the units digit repeats itself every 4 digit:

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

Digits 4 and 9 have a cyclicity of 2, i.e. the units digit repeats itself every 2 digits:

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1

Digits 0, 1, 5 and 6 have a cyclicity of 1, i.e. the units digit is 0, 1, 5, or 6 respectively.

Now let’s take a look at how to apply these fundamentals:

What is the units digit of 813^(27)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 3.

Remember, our cyclicity of 3 is 3, 9, 7, 1 (four numbers total).

We need the units digit of 3^(27). How many full cycles of 4 will be there in 27? There will be 6 full cycles because 27 divided by 4 gives 6 as quotient and 3 will be the remainder. So after 6 full cycles of 4 are complete, a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

… (6 full cycles)

3, 9, 7 (new cycle for remainder of 3)

7 will be the units digit of 3^(27), so 7 will be the units digit of 813^(27).

Let’s try another question:

What is the units digit of 24^(1098)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 4.

Remember, our cyclicity of 4 is 4 and 6 (this time, only 2 numbers).

We need the units digit of 24^(1098) – every odd power of 24 will end in 4 and every even power of 24 will end in 6.

Since 1098 is even, the units digit of 24^(1098) is 6.

Not too bad; let’s try something a little harder:

What is the units digit of 75^(25)^5

Note here that you have 75 raised to power 25 which is further raised to the power of 5.

25^5 is not the same as 25*5 – it is 25*25*25*25*25 which is far more complicated. However, the simplifying element of this question is that the last digit of the base 75 is 5, so it doesn’t matter what the positive integer exponent is, the last digit of the expression will always be 5.

Now let’s take a look at a Data Sufficiency question:

Given that x and y are positive integers, what is the units digit of (5*x*y)^(289)?

Statement 1: x is odd.

Statement 2: y is even.

Here there is a new complication – we don’t know what the base is exactly because the base depends on the value of x and y. As such, the real question should be can we figure out the units digit of the base? That is all we need to find the units digit of this expression.

When 5 is multiplied by an even integer, the product ends in 0.

When 5 is multiplied by an odd integer, the product ends in 5.

These are the only two possible cases: The units digit must be either 0 or 5.

With Statement 1, we do not know whether y is odd or even, we only know that x is odd. If y is odd, x*y will be odd. If y is even, x*y will be even. Since we don’t know whether x*y is odd or even, we don’t know whether 5*x*y will end in 5 or 0, so this statement alone is not sufficient.

With Statement 2, if y is even, x*y will certainly be even because an even * any integer will equal an even integer. Therefore, it doesn’t matter whether x is odd or even – regardless, 5*x*y will be even, hence, it will certainly end in 0.

As we know from our patterns of cyclicity, 0 has a cyclicity of 1, i.e. no matter what the positive integer exponent, the units digit will be 0. Therefore, this statement alone is sufficient and the answer is B (Statement 2 alone is sufficient but Statement 1 alone is not sufficient).

Finally, let’s take a question from our own book:

If n and a are positive integers, what is the units digit of n^(4a+2) – n^(8a)?

Statement 1: n = 3

Statement 2: a is odd.

We know that the cyclicity of every digit is either 1, 2 or 4. So to know the units digit of n^{4a+2} – n^{8a}, we need to know the units digit of n. This will tell us what the cyclicity of n is and what the units digit of each expression will be individually.

Statement 1: n = 3

As we know from our patterns of cyclicity, the cyclicity of 3 is 3, 9, 7, 1

Plugging 3 into “n”, n^{4a+2} = 3^{4a+2}

In the exponent, 4a accounts for “a” full cycles of 4, and then a new cycle begins to account for 2.

3, 9, 7, 1

3, 9, 7, 1

3, 9

The units digit here will be 9.

Again, plugging 3 into “n”, n^{8a} = 3^{8a}

8a is a multiple of 4, so there will be full cycles of 4 only. This means the units digit of 3^{8a} will be 1.

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

Plugging these answers back into our equation: n^{4a+2} – n^{8a} = 9 – 1

The units digit of the combined expression will be 9 – 1 = 8.

Therefore, this statement alone is sufficient.

In Statement 2, we are given what the exponents are but not what the value of n, the base, is. Therefore, this statement alone is not sufficient, and our answer is A (Statement 1 alone is sufficient but Statement 2 alone is not sufficient).

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit Quarter Wisdom: Cyclicity of Units Digits on the GMAT

Quarter Wit, Quarter WisdomIn our algebra book, we have discussed finding and extrapolating patterns. In this post today, we will look at the patterns we get with various units digits.

The first thing you need to understand is that when we multiply two integers together, the last digit of the result depends only on the last digits of the two integers.

For example:

24 * 12 = 288

Note here: …4 * …2 = …8

So when we are looking at the units digit of the result of an integer raised to a certain exponent, all we need to worry about is the units digit of the integer.

Let’s look at the pattern when the units digit of a number is 2.

Units digit 2:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

2^9 = 512

2^10 = 1024

Note the units digits. Do you see a pattern? 2, 4, 8, 6, 2, 4, 8, 6, 2, 4 … and so on

So what will 2^11 end with? The pattern tells us that two full cycles of 2-4-8-6 will take us to 2^8, and then a new cycle starts at 2^9. 

2-4-8-6

2-4-8-6

2-4

The next digit in the pattern will be 8, which will belong to 2^11. 

In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base. 

So 652^(11) will end in 8,1896782^(11) will end in 8, and so on…

A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits. 

Units digit 3:

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on…

Units digit 4:

4^1 = 4

4^2 = 16

4^3 = 64

4^4 = 256

The pattern here is 4, 6, 4, 6, 4, 6, and so on… 

Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is:

1545^23 = ……..5

1650^19 = ……..0

161^28 = ………1

Hope you get the point.

Units digit 7:

7^1 = 7

7^2 = 49

7^3 = 343

7^4 = ….1 (Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit)

7^5 = ….7 (Now multiply 1 from above by 7 to get 7 as the units digit)

7^6 = ….9

The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on…

Units digit 8:

8^1 = 8

8^2 = 64

8^3 = …2

8^4 = …6

8^5 = …8

8^6 = …4

The pattern here is 8, 4, 2, 6, 8, 4, 2, 6, and so on…

Units digit 9: 

9^1 = 9

9^2 = 81

9^3 = 729

9^4 = …1

The pattern here is 9, 1, 9, 1, 9, 1, and so on…

Summing it all up:

1) Digits 2, 3, 7 and 8 have a cyclicity of 4; i.e. the units digit repeats itself every 4 digits.

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

2) Digits 4 and 9 have a cyclicity of 2; i.e. the units digit repeats itself every 2 digits.

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1 

3) Digits 0, 1, 5 and 6 have a cyclicity of 1.

Cyclicity of 0: 0

Cyclicity of 1: 1

Cyclicity of 5: 5

Cyclicity of 6: 6

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube, Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Permutation Involving Sum of Digits

Quarter Wit, Quarter WisdomWe have seen in previous posts how to deal with permutation and combination questions on the GMAT. There is a certain variety of questions that involve getting a bunch of numbers using permutation, and then doing some operations on the numbers we get. The questions can get a little overwhelming considering the sheer magnitude of the number of numbers involved! Let’s take a look at that concept today. We will explain it using an example and then take a question as an exercise:

 

What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 such that each digit is used exactly once in each integer?

First of all, we will use our basic counting principle to find the number of integers that are possible.

The first digit can be chosen in 4 ways. The next one in 3 ways since each digit can be used only once. The next one in 2 ways and there will be only one digit left for the last place.

This gives us a total of 4*3*2*1 = 24 ways of writing such a four digit number. This is what some of the numbers will look like:

1234

1243

1324

1342

2143

4321

Now we need to add these 24 integers to get their sum. Note that since each digit has an equal probability of occupying every place, out of the 24 integers, six integers will have 1 in the units place, six will have 2 in the units place, another six will have 3 in the units place and the rest of the six will have 4 in the units place. The same is true for all places – tens, hundreds and thousands.

Imagine every number written in expanded form such as:

1234 = 1000 + 200 + 30 + 4

2134 = 2000 + 100 + 30 + 4

…etc.

For the 24 numbers, we will get six 1000’s, six 2000’s, six 3000’s and six 4000’s.

In addition, we will get six 100’s, six 200’s, six 300’s and six 400’s.

For the tens place, will get six 10’s, six 20’s, six 30’s and six 40’s.

And finally, in the ones place we will get six 1’s, six 2’s, six 3’s and six 4’s.

Therefore, the total sum will be:

6*1000 + 6*2000 + 6*3000 + 6*4000 + 6*100 + 6*200 + … + 6*3 + 6*4

= 6*1000*(1 + 2 + 3 + 4) + 6*100*(1 + 2 + 3 + 4) + 6*10*(1 + 2 + 3 + 4) + 6*1*(1 + 2 + 3 + 4)

= 6*1000*10 + 6*100*10 + 6*10*10 + 6*10

= 6*10*(1000 + 100 + 10 + 1)

= 1111*6*10

= 66660

Note that finally, there aren’t too many actual calculations, but there is some manipulation involved. Let’s look at a GMAT question using this concept now:

What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A444440

B) 610000

C) 666640

D) 711040

E) 880000

Conceptually, this problem isn’t much different from the previous one.

Using the same basic counting principle to get the number of integers possible, the first digit can be chosen in 4 ways, the next one in 4 ways, the next one in again 4 ways and finally the last digit in 4 ways. This is what some of the numbers will look like:

1111

1112

1121

and so on till 4444.

As such, we will get a total of 4*4*4*4 = 256 different integers.

Now we need to add these 256 integers to get their sum. Since each digit has an equal probability of occupying every place, out of the 256 integers, 64 integers will have 1 in the units place, 64 will have 2 in the units place, another 64 integers will have 3 in the units place and the rest of the 64 integers will have 4 in the units place. The same is true for all places – tens, hundreds and thousands.

Therefore, the total sum will be:

64*1000 + 64*2000 + 64*3000 + 64*4000 + 64*100 + 64*200 + … + 64*3 + 64*4

= 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)

= (64*1 + 64*2 + 64*3 + 64*4) * (1000 + 100 + 10 + 1)

= 64*10*1111

= 711040

So our answer is D.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTube and Google+, and Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!