How to Prepare for Your Business School Interview

For many applicants the notification of an interview invite from your dream school is an exciting next step after an arduous application process. All of your hard work has finally boiled down to some initial success. However, typically the excitement soon turns to anxiety as candidates begin to realize they have no idea how to prepare for an admissions interview for business school. “Is it just like a regular job interview?” “What type of questions do they ask?” are just some of the common initial questions that can arise once an interview invitation is received.

The business school interview should not be viewed as anything new to you. It is more similar to the traditional job interview than you might expect. Just like a regular interview you are aiming to impress and the majority of the interview will be focused on YOU! The key difference with this interview is really just the goal, which in this case is admission to the MBA program of your dreams.

I would recommend preparing for your MBA interview the same way you prepare for any job interview, it starts with knowing your own personal background inside and out along with your motivations for that target business school. Then it’s researching your target school and identifying the aspects that make the school uniquely attractive to you. A nice way to do this is to pair up school-specific offerings of interest with an adjoining explanation for why that offering is uniquely attractive to you. This includes academic offerings, extracurricular activities/professional clubs, career support/recruiting strengths, etc.

Next I would identify common MBA questions like…

  • What Are Your Career Goals?
  • Why an MBA?
  • Why School X?
  • Walk Me Through Your Resume

As well as other common situational business school questions that address interpersonal skills like leadership, teamwork, and maturity. For the most part, these interviews have very few surprises, and you will know what’s coming, which makes the prep all the more important. Preparing conversational responses in a script format to each of the common interview questions can be a method for those that prefer a more structured approach to their interview prep. But make sure to incorporate elements of your personality into your script to avoid coming off as too rehearsed.

Also, breakthrough candidates will make sure to incorporate the “I” of what they accomplished into their script. Make sure to connect the dots with regards to the steps you’ve taken in your career, and remain structured in your responses. Utilizing the S.T.A.R format (Situation-Task-Action-Result) and talking in buckets – “There are 3 Reasons Why I Want to Go to Fuqua” are other tactics one can sneak into their preparation for the interview.

Finally, take particular note of how the interview style of certain schools can affect your responses. Some schools like Kellogg have “blind” interviews so the interviewer will not have seen your application, so they will not have access to important information like GPA, GMAT, essays etc. Other styles can be influenced by the type of interviewer (Alum vs. Student vs. Admissions) or the location (On Campus vs. Off Campus) which can dictate the type of information you are prepared to share as well as list on your resume for the interview.

Don’t let the interview be the end of your business school journey, prepare accordingly and come decision day you will be all smiles!

Want to craft a strong application? Call us at 1-800-925-7737 and speak with an MBA admissions expert today, or sign up for a free admissions consultation. Let’s get started!

Dozie A. is a Veritas Prep Head Consultant for the Kellogg School of Management at Northwestern University. His specialties include consulting, marketing, and low GPA/GMAT applicants.

All About That on the GMAT

The word “that” is often found in the daymares of many a GMAT student! The reason for that is that “that” could play various different roles in a sentence:

  1. Demonstrative Determiner
  2. Demonstrative Pronoun
  3. Relative Pronoun

Demonstrative Determiner – In this role, “that” specifies the specific person/thing about which we are talking. It is followed by a noun.

Can I have some of that cake, please?

I have never been to that part of Italy.

When we are talking about a plural noun, “that” becomes “those”.

Demonstrative Pronoun – In this role, “that” replaces a noun.

That is beautiful.

Look at that!

When we replace a plural noun, “that” becomes “those”.

Relative Pronoun – “that” introduces a defining/restrictive clause. This clause is essential to the sentence.

Loki is on the team that lost.

The produce that is sourced locally is environment-friendly.

There is no “that”/“those” distinction in this case. The clause is always introduced by “that”.

Hope these simple examples clarified the various roles “that” can play in a sentence. Not understanding this distinction could lead to a lot of confusion. The words around “that” will help you understand exactly what role it is playing in each case.

Let’s take a look at one of our own questions in which knowing this distinction comes in handy.

Question: In nests across North America, the host mother tries to identify their own eggs and weed out the fakes, but the brown-headed cowbird – a brood parasite that sneaks its eggs into other birds’ nests – produces eggs that look very similar to those of the host, making that task surprisingly difficult.

(A) the host mother tries to identify their own eggs and weed out the fakes, but the brown-headed cowbird – a brood parasite that sneaks its eggs into other birds’ nests – produces eggs that look very similar to those of the host, making that task surprisingly difficult

B) the host mother tries to identify its own eggs and weed out the fakes, but the brown-headed cowbird – a brood parasite that sneaks its eggs into other birds’ nests – produces eggs that look very similar to that of the host, making it surprisingly difficult

C) host mothers try to identify their own eggs and weed out the fakes, but the brown-headed cowbird – a brood parasite that sneaks its eggs into other birds’ nests – produces eggs that look very similar to the host’s, making that task surprisingly difficult

D) host mothers try to identify their own eggs and weed out the fakes, but the brown-headed cowbird – a brood parasite that sneaks its eggs into other birds’ nests – produces eggs that look very similar to that of the host’s, making it surprisingly difficult

E) host mothers try to identify its own eggs and weed out the fakes, but the brown-headed cowbird – a brood parasite that sneaks its eggs into other birds’ nests – produces eggs that look very similar to those of the host’s, making that task surprisingly difficult

Solution:

This is a complicated sentence and unfortunately, almost the entire sentence is underlined. That just makes it harder and more time consuming.

  1. … the host mother tries to identify their own eggs…

In the beginning itself, we see that the subject is “host mother” which is singular and the pronoun that refers back to it – “those” – is plural. Hence this sentence is incorrect. We just move on.

(B) … produces eggs that look very similar to that of the host …

We have two instances of the use of “that” here. The first “that” is used as a relative pronoun to introduce the clause “that look very similar to ….”

The second “that” is  used as a placeholder for “eggs” hence we need to use “those” – the plural form – here.

(C) All correct

(D) … produces eggs that look very similar to that of the host’s…

The explanation is the same as that of (B). The second “that” is  used as a placeholder for “eggs” hence we need to use “those” – the plural form – here.

Also, the correct comparison is:

either

“A’s eggs look very similar to those of B” (where “those” stands for eggs)

or

“A’s eggs look very similar to B’s” (where eggs is implied at the end).

But “A’s eggs look very similar to those of B’s” is incorrect since it implies

“A’s eggs look very similar to eggs of B’s eggs”

(E) … host mothers try to identify its own eggs…

The subject is “host mothers”, which is plural, but the pronoun is “its”, which is singular.

Hope this clarifies the various ways in which “that” can be used.

Getting ready to take the GMAT? Check out one of our many free GMAT resources to get a jump start on your GMAT prep. And as always, be sure to follow us on FacebookYouTubeGoogle+, and Twitter for more helpful tips like this one!

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!

The Third Type of GMAT Quant Question

We all know that GMAT Quant questions are of two types: Problem Solving and Data Sufficiency.

But, if we look carefully, we will see a third type of question – combination of the two. There are a few statements given in them (like in Data Sufficiency questions) and five options to choose from (like in Problem Solving questions). But since we know how to solve both these question types, we shouldn’t really have a problem in solving this third type, or so one would think!

In any GMAT question, it is very important to know two things:

  1. What is given
  2. What is asked

Now, one might think that it is a very obvious distinction and why are we even trying to discuss it in a post. In this third question type, this exact distinction is far harder to explain because here the statements do NOT represent the data given. Here the statements actually ask “Is this true?” and many test-takers find it hard to make that switch. To clarify, let’s discuss the structure of the three question types.

Problem Solving Question:

Question: A and B are given, what is X?

(A) X is …

(B) X is …

(C) X is …

(D) X is …

(E) X is …

Data Sufficiency Question:

Question: A and B are given, what is X?

I. We are given that X and Y are related.

II. We are given that X and Z are related.

“Which of the following must be true?” Question:

Question: A and B are given, which of the following must be true about X?

I. Is this true about X?

II. Is this true about X?

III. Is this true about X?

(A) I is true

(B) I and II are true

and so on…

We hope you see that the statements in a Data Sufficiency question are different from the statements in this third type of question.

We will elaborate with the help of an example now:

Question: If |x| > 3, which of the following must be true?

I. x > 3

II. x^2 > 9

III. |x – 1| > 2

(A) I only

(B) II only

(C) I and II only

(D) II and III only

(E) I, II, and III

Solution:

We are given that |x| > 3

This implies that x is a point at a distance of more than 3 from 0. So x could be greater than 3 or less than -3. Before we go any further, let’s think about the values x can take: 3.00001, 3.5, 4.2, 5.7, 67, 1000, -3.45, -4, -8, -100 etc. The only values it cannot take are -3 <= x <= 3

Which of the following must be true?

I. x > 3

This is a question even though it looks like a statement.

Is it necessary that x > 3?

For every value that x can take, must x be greater than 3? No. As discussed above, x could take values such as 3.00001, 3.5, 4.2, 5.7, 67, 1000 but it could also take values such as  -3.45, -4, -8, -100.

So this is not necessarily true.

II. x^2 > 9

Again, this is a question even though it looks like a statement.

Taking square root on both sides since they are positive, we get

Sqrt(x^2) > Sqrt(9)

|x| > 3

This is what we are given, hence it certainly is true.

III. |x-1|>2

Yet again, we are asked: Is |x – 1| > 2?

What does |x – 1|> 2 imply?

The distance of x from 1 must be greater than 2. So x is either greater than 3 or less than -1. Now, recall all the values that x can take.

So this is the question now: Is every value that x can take greater than 3 or less than -1?

Recall the values that x can take (discussed above)

3.00001 : x is greater than 3

3.5 : x is greater than 3

4.2 : x is greater than 3

5.7 : x is greater than 3

67 : x is greater than 3

1000 : x is greater than 3

-3.45 : x is less than -1

-4 : x is less than -1

-8 : x is less than -1

-100 : x is less than -1

For every value that x can take, x will be either greater than 3 or less than -1. Note that we are not saying that every value less than -1 must be valid for x. We are saying that every value that is valid for x (found by using |x| > 3) will be either greater than 3 or less than -1 since any value less than -3 is obviously less than -1 too. Hence |x-1|>2 must be true for every value that x can take.

Answer (D)

We hope you are quite clear about how to handle this third question type now!

Getting ready to take the GMAT? Check out one of our many free GMAT resources to get a jump start on your GMAT prep. And as always, be sure to follow us on FacebookYouTubeGoogle+, and Twitter for more helpful tips like this one!

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!

The Importance of Context in Verb Tenses

Quarter Wit, Quarter WisdomIn our last post, we looked at verb tenses and noted that there is no restriction on how many tenses we can use and mix within a sentence, as long as they are appropriate for the context of the sentence. The problem is that sometimes the context can be a bit complicated to crack. We may think that a tense shift is required when it is actually not.

Let’s take a look at an official GMAT question to better understand this concept:

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

So the first decision point is “have” vs. “had”. What is correct here? We know that we use past perfect tense when there are two actions in the past. So do we have two actions in the past here – “finding” and “electing” – of which, it may seem, “electing” would have happened before “finding?” Sure, we have two actions but here is the catch – we use past perfect only when the previous action takes place completely before the recent past action. Here, we know that “within the past few years” implies the recent years. The study shows that most probably, doctors are still electing early retirement. So the use of past perfect is incorrect here. In this context, we will use present perfect only.

The other error that helps us to arrive at the right answer is lack of parallelism. “retire” and “face” need to be parallel while rising should not be parallel to them because it is a sub-list under “face”.

They elected to retire … rather than face A and B.

A – the threats
B – the rising costs

“[R]ising” is a present participle that is modifying the noun “costs” in the non underlined part. So our verbs “retire” and “face” should not be in the -ing form. Answer choice E satisfies all these criteria and hence is the right answer.

Note that the correct answer uses present perfect for both verbs since the context requires us to.

Let’s look at a rewrite of this question:

A recent article in The Economic Times reported that many recent MBA graduates had decided on taking a job rather than face the uncertainty of entrepreneurship.

(A) had decided on taking a job rather than face
(B) had decided on taking a job instead of facing
(C) have decided to take a job instead of facing
(D) had decided to take a job rather than facing
(E) have decided to take a job rather than face

How does the solution change now? Again we have two verbs “report” and “decide”. The reporting has already happened so the simple past “reported” has been used in the non-underlined part. Which tense will we use with “decide”? Again, the concept is still the same. We are talking about recent MBA graduates and it shows a trend. It is something that is not completely over, hence the use of past perfect is not justified. We should use the present perfect tense only though it may seem a bit counterintuitive since “report” is in the past tense.

“take” and “face” should be parallel to each other so out of (C) and (E), (E) fits. This is the reason making sweeping statements in grammar is dangerous – a lot depends on the context.

Getting ready to take the GMAT? Check out one of our many free GMAT resources to get a jump start on your GMAT prep. And as always, be sure to follow us on FacebookYouTubeGoogle+, and Twitter for more helpful tips like this one!

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!

Is It Incorrect to Use Multiple Verb Tenses in a Sentence?

Quarter Wit, Quarter WisdomSome GMAT test-takers wonder whether it is grammatically correct to use multiple tenses in a single sentence. Today we will discuss the cases in which this is acceptable and those in which this is not. The bottom line is this: there is no restriction on what tenses we can use and mix within a sentence, as long as they are appropriate for the context.

Take a look at this example sentence:

I have heard that Mona left Manchester this morning, and has already arrived in London, where she will be for the next three weeks.

Here, we have present perfect tense, simple past tense and simple future tense all in the same sentence, but they all make sense together to create a logical sequence of events.

The confusion over using multiple verb tenses in one sentence probably arises because we have heard that we need to maintain verb tense consistency. These two things are different.

Tense Consistency – We do not switch one tense to another unless the timing of the action demands that we do. We do not switch tenses when there is no time change for the actions.

Let’s take a look at some examples to understand this:

Example 1: During the match, my dad stood up and waved at me.

These two actions (“stood” and “waved”) happen at the same time and hence, need to have the same tense. This sentence could take place in the present or future tense too, but both verbs will still need to take on the same tense. For example:

Example 2: During my matches, my dad stands up and waves at me.
Example 3: During the match tomorrow, my dad will stand up and wave at me.

On the other hand, a sentence such as…

Example 4: During the match, my dad stood up and waves at me.

This sentence is grammatically incorrect. Since both actions (“stood” and “waves”) happen at the same time, we need them to be in the same tense, as shown in the variations of this sentence above. Consider this case, however:

Example 5: My dad reached for the sandwich after he had already eaten a whole pizza.

Here, the two actions (“reached” and “eaten”) happen at different times in the past, so we use both the simple past and past perfect tenses. The shift in tense is correct in this context.

Takeaway: The tenses of verbs in a sentence must be consistent when the actions happen at the same time. When dealing with actions that occur at different points in time, however, we can use multiple tenses in the same sentence.

Let’s look at an official GMAT question now to see how multiple tenses can be a part of the same sentence:

For the farmer who takes care to keep them cool, providing them with high-energy feed, and milking them regularly, Holstein cows are producing an average of 2,275 gallons of milk each per year.

(A) providing them with high-energy feed, and milking them regularly, Holstein cows are producing
(B) providing them with high-energy feed, and milked regularly, the Holstein cow produces
(C) provided with high-energy feed, and milking them regularly, Holstein cows are producing
(D) provided with high-energy feed, and milked regularly, the Holstein cow produces
(E) provided with high-energy feed, and milked regularly, Holstein cows will produce

This is a very tricky question. Let’s first shortlist our options based on the obvious errors.

The non-underlined part of the sentence uses the pronoun “them” to refer to the cows, so using “the Holstein cow” (singular) as the antecedent will be incorrect. The antecedent must be “Holstein cows” (plural) – this means answer choices B and D are out.

Also, we know for sure that “provide” and “milk” are parallel elements in the sentence, so they should take the same verb tense. Hence, answer choice C is also out.

Let’s look at A now. If we assume this option is correct, “providing” and “milking” act as modifiers to “keep them cool”. That certainly does not make sense since “providing with high energy feed” and “milking regularly” are not ways of keeping cows cool.

This means the correct answer is E, but we need to see how.

For the farmer who takes care to keep them cool, provided with high-energy feed, and milked regularly, Holstein cows will produce an average of 2,275 gallons of milk each per year.

Let’s break down the sentence:

For the farmer who takes care to keep them…

  • cool,
  • provided with high-energy feed,
  • milked regularly,

…Holstein cows will produce an average of 2,275 gallons of milk each per year.

Note that we use two different tenses here: “For the farmer who takes care…” and “cows will produce…”. The word “takes” is the present tense while “will produce” is the future, but that does not make this sentence incorrect. The context of the author could very well justify the use of the future tense. Perhaps the farmers have obtained Holstein cows recently, and hence, will see the produce of 2,275 gallons in the future, only.

A shift in the tense certainly doesn’t make the sentence incorrect. When you’re presented with multiple verbs in various tenses in a problem, check to determine whether the verbs convey a logical sequence of events.

Getting ready to take the GMAT? Check out one of our many free GMAT resources to get a jump start on your GMAT prep. And as always, be sure to follow us on FacebookYouTubeGoogle+, and Twitter for more helpful tips like this one!

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!

Dreaded Data Sufficiency Questions That Will Test Your Knowledge of Number Properties

Quarter Wit, Quarter WisdomHere is an often-repeated complaint we hear from test takers – Data Sufficiency questions that deal with number properties are very difficult to handle (even for people who find problem-solving number properties questions manageable)! They feel that such questions are time consuming and often involve too many cases.

Here is our advice – when solving number properties questions, imagine a number line. It reminds us that numbers behave differently “between 0 and 1”, “between -1 and 0”, “less than -1”, and “more than 1”, and that integers occur only at regular intervals and that there are infinite numbers in between them. The integers are, in turn, even and odd. Also, 0, 1 and -1 are special numbers, hence it is always a good idea to consider cases with them.

Let’s see how thinking along these lines can help us on a practice Data Sufficiency question:

If a and b are non-zero integers, is a^b an integer?

Statement 1: b^a is negative
Statement 2: a^b is negative

The answer to this problem does not lie in actually drawing a number line. The point is that we need to think along these lines: -1, 0, 1, ranges between them, integers, negatives-positives, even-odd, decimals and how each of these comes into play in this case.

What we know from the question stem is that a and b are non-zero integers, which means they occur at regular intervals on the number line. To answer the question, “Is a^b an integer?”, let’s first look at Statement 1:

Statement 1: b^a is negative

For a number to be negative, its base must be negative. But that is not enough – the exponent should not be an even integer. If the exponent is an even integer, the negative signs will cancel out. Since a and b are integers, if a is not an even integer, it must be an odd integer.

We know that the sign of the exponent is immaterial as far as the sign of the result is concerned (since a^(-n) is just 1/a^n). For b^a to be negative, then we know that b must be a negative integer and a must be an odd integer. Does this help us in deducing whether a^b is an integer? Not necessarily!

If b is negative, say -2, a^(-2) = 1/a^2. a could be 1, in which case 1/a^2 = 1 (an integer), or a could be 3, in which case 1/a^2 = 1/9 (not an integer). Because there are two possible answers, this statement alone is not sufficient.

Let’s look at Statement 2:

Statement 2: a^b is negative

Again, the logic remains the same – for a number to be negative, its base must also be negative and the exponent should not be an even integer. If the exponent is an even integer, the negative signs will cancel out. Since a and b are integers, if b is not an even integer, it must be an odd integer. Again, we know that the sign of the exponent is immaterial as far as the sign of the result is concerned (since a^(-n) is just 1/a^n).

For a^b to be negative, then we know that a must be a negative integer and b must be an odd integer. a could be -1/-2/-3/-4… etc, and b could be 1/3/5… or -1/-3/-5.

If a = -1 and b = 1, then a^b = -1 (an integer). If a = -2 and b = -3, then a^b = (-2)^(-3) = 1/(-2)^3 = -1/8 (not an integer). This statement alone is also not sufficient.

We hope you see how we are using values of 1 and -1 to enumerate our cases. Now, let’s consider using both statements together:

a is a negative, odd integer, so it can take values such as -1, -3, -5, -7, …
b is a negative, odd integer too, so it can also take values such as -1, -3, -5, -7, …

If a = -1 and b = -1, then a^b = -1 (an integer)
If a = -3 and b = -3, then a^b = (-3)^(-3) = -1/27 (not an integer)

Even using both statements together, we do not know whether a^b is an integer or not. therefore, our answer is E.

Thinking of a number line and knowing what it represents will help you tackle many Data Sufficiency questions that are about number properties.

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!

3 Ways to Solve a 750+ Level GMAT Question About Irregular Polygons

Quarter Wit, Quarter WisdomWe have examined how to deal with polygons when you encounter them on a GMAT question in a previous post. Today, we will look at a relatively difficult polygon question, however we would like to remind you here that the concepts being tested in this question are still very simple (although we won’t give away exactly which concepts they are yet). First, take a look at the question itself:

The hexagon above has interior angles whose measures are all equal. As shown, only five of the six side lengths are known: 10, 15, 4, 18, and 7. What is the unknown side length?

 

 

 

 

(A) 7
(B)10
(C) 12
(D) 15
(E) 16

There are various ways to solve this question, but each takes a bit of effort. Note that the polygon we are given is not a regular polygon, since the side lengths are not all equal. The angles, however, are all equal. Let’s first find the measure of each one of those angles using the formula discussed in this previous post.

(n – 2)*180 = sum of all interior angles
(6 – 2)*180 = 720
Each of the 6 angles = 720/6 = 120 degrees

Though we would like to point out here that if you see a question such as this one on the actual GMAT exam, you should already know that if each angle of a hexagon is equal, each angle must be 120 degrees, so performing the above calculation would not be necessary.

Method 1: Visualization
This is a very valid approach to obtaining the correct answer on this GMAT question since we don’t need to explain the reasoning or show our steps, however it may be hard to comprehend for the beginners. We will try to explain it anyway, since it requires virtually no work and will help build your math instinct.

Note that in the given hexagon, each angle is 120 degrees – this means that each pair of opposite sides are parallel. Think of it this way: Side 4 turns on Side 18 by 120 degrees. Then Side 15 turns on Side 4 by another 120 degrees. And finally, Side 10 turns on Side 15 by another 120 degrees. So Side 10 has, in effect, turned by 360 degrees on Side 18.

This means Side 10 is parallel to Side 18.

Now, think of the 120 degree angle between Side 4 and Side 15 – it has to be kept constant. Plus, the angles of the legs must also stay constant at 120 degrees with Sides 10 and 18. Since the slopes of each leg of that angle are negatives of each other (√3 and -√3), when one leg gets shorter, the other gets longer by the same length (use the image below as a visual of what we’re talking about).

 

 

 

 

Hence, the sum of the sides will always be 15 + 4 = 19. This means 7 + Unknown = 19, so Unknown = 12. Our answer is C.

If you struggled to understand the approach above, you’re not alone. This method involves a lot of intuition, and struggling to figure it out may not be the best use of your time on the GMAT, so let’s examine a couple of more tangible solutions!

Method 2: Using Right Triangles
As we saw in Method 1 above, AB and DE are parallel lines. Since each of the angles A, B, C, D, E and F are 120 degrees, the four triangles we have made are all 30-60-90 triangles. The sides of a 30-60-90 triangle can be written using the ratio 1:√(3):2.

 

 

 

 

 

AT = 7.5*√3 and ME = 2*√3, so the distance between the sides of length 10 and 18 is 9.5*√3. We know that DN = 3.5*√3, so BP = (9.5*√3) – (3.5*√3) = 6*√3.

Since the ratios of our sides should be 1:√(3):2, side BC = 2*6 = 12. Again, the answer is C. Let’s look at our third and final method for solving this problem:

Method 3: Using Equilateral Triangles
First, extend the sides of the hexagon as shown to form a triangle:

 

 

 

 

 

 

 

Since each internal angle of the hexagon is 120 degrees, each external angle will be 60 degrees. In that case, each angle between the dotted lines will become 60 degrees too, and hence, triangle PAB becomes an equilateral triangle. This means PA = PB = 10. Triangle QFE  and triangle RDC also become equilateral triangles, so QF = QE = 4, and RD = RC = 7.

Now note that since angles P, Q, and R are all 60 degrees, triangle PQR is also equilateral, and hence, PQ = PR.

PQ = 10 + 15 + 4 = 29
PR = 10 + BC + 7 = 29
BC = 12 (again, answer choice C)

Note the geometry concepts that we used to solve this problem: regular polygon, parallel lines, angles, 30-60-90 right triangles, and equilateral triangles. We know all of these concepts very well individually, but applying them to a GMAT question can take some ingenuity!

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 Ingenuity on GMAT Remainder Questions

Quarter Wit, Quarter WisdomWe have looked at various types of GMAT remainder questions and discussed how to tackle them in a few previous posts. Specifically, we have examined the concepts of general divisibility, divisibility as applied to GMAT questions, and divisibility specifically applied to remainders. There is one concept, however, that we haven’t discussed yet, and that is using ingenuity on remainder questions.

Say “x” gives you a remainder of 2 when divided by 6. What will be the remainder when x + 1 is divided by 6?

Go back to the divisibility concepts discussed above. When x balls are split into groups of 6, we will have 2 balls leftover. If we are given 1 more ball, it will join the 2 balls and now we will have 3 balls leftover. The remainder will be 3.

What happens in the case of x + 6 – what will be the remainder when this is divided by 6? This additional 6 balls will just make an extra group of 6, so we will still have 2 balls leftover.

What about the case of x + 9? Now, of the extra 9 balls, we will make one group of 6 and will have 3 balls leftover. These 3 balls will join the 2 balls leftover from x, giving us a remainder of 5.

Now, what about the case of 2x? Recall that 2x = x + x. The number of groups will double and so will the remainder, so 2x will give us a remainder of 2*2 = 4.

On the other hand, if x gives us a remainder of 4 when divided by 6, then 2x divided by 6 will have a remainder of 2*4 = 8, which gives us a remainder of 2 (since another group of 6 will be formed from the 8 balls).

Let’s consider the tricky case of x^2 now. If x gives us a remainder of 2 when it is divided by 6, it means:

x = 6Q + 2
x^2 = (6Q + 2)*(6Q + 2) = 36Q^2 + 24Q + 4

Note here that the first and the second terms are divisible by 6. The remainder when you divide this by 6 will be 4.

We hope you understand how to deal with these various cases of remainders. Let’s take a look at a GMAT sample question now:

If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r?

Statement 1: When (z−3)^2 is divided by 8, the remainder is 4.
Statement 2: When 2z is divided by 8, the remainder is 2.

This is not our typical, “When z is divided by 8, r is the remainder” type of question. Instead, we are given a quadratic equation in the form of z that, when divided by 8, gives us a remainder of r. We need to find r. This question might feel complicated, but look at the statements – at least one of them gives us data on a quadratic! Looks promising!

Statement 1: When (z−3)^2 is divided by 8, the remainder is 4

(z – 3)^2 = z^2 – 6z + 9

We know that when z^2 – 6z + 9 is divided by 8, the remainder is 4. So no matter what z is, z^2 – 6z + 9 + 8z, when divided by 8, will only give us a remainder of 4 (8z is a multiple of 8, so will give remainder 0).

z^2 – 6z + 9 + 8z = z^2 + 2z + 9

z^2 + 2z + 9 when divided by 8, gives remainder 4. This means z^2 + 2z + 5 is divisible by 8 and would give remainder 0, further implying that z^2 + 2z + 4 would be 1 less than a multiple of 8, and hence, would give us a remainder of 7 when divided by 8. This statement alone is sufficient.

Let’s look at the second statement:

Statement 2: When 2z is divided by 8, the remainder is 2

2z = 8a + 2
z = 4a + 1
z^2 = (4a + 1)^2 = 16a^2 + 8a + 1

When z^2 is divided by 8, the remainder is 1. When 2z is divided by 8, the remainder is 2. So when z^2 + 2z is divided by 8 the remainder will be 1+2 = 3.

When z^2 + 2z + 4 is divided by 8, remainder will be 3 + 4 = 7. This statement alone is also sufficient. Because both statements alone are sufficient, 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!

Using Parallel Lines and Transversals to Your Advantage on the GMAT

Quarter Wit, Quarter WisdomToday, we will look at a Geometry concept involving parallel lines and transversals (a line that cuts through two parallel lines). This is the property:

The ratios of the intercepts of two transversals on parallel lines is the same.

Consider the diagram below:

 

 

 

 

 

 

Here, we can see that:

  • “a” is the intercept of the first transversal between L1 and L2.
  • “b” is the intercept of the first transversal between L2 and L3.
  • “c” is the intercept of the second transversal between L1 and L2.
  • “d” is the intercept of the second transversal between L2 and L3.

Therefore, the ratios of a/b = c/d. Let’s see how knowing this property could be useful to us on a GMAT question. Take a look at the following example problem:

In triangle ABC below, D is the mid-point of BC and E is the mid-point of AD. BF passes through E. What is the ratio of AF:FC ?

 

 

 

 

 

 

(A) 1:1
(B) 1:2
(C) 1:3
(D) 2:3
(E) 3:4

Here, the given triangle is neither a right triangle, nor is it an equilateral triangle. We don’t really know many properties of such triangles, so that will probably not help us. We do know, however, that AD is the median and E is its mid-point, but again, we don’t know any properties of mid-points of medians.

Instead, we need to think outside the box – parallel lines will come to our rescue. Let’s draw lines parallel to BF passing through the points A, D, and C, as shown in the diagram below:

 

 

 

 

 

 

Now we have four lines parallel to each other and two transversals, AD and AC, passing through them.

Consider the three parallel lines, “line passing through A”, “BF”, and “line passing through D”. The ratio of the intercepts of the two transversals on them will be the same.

AE/ED = AF/FP

We know that AE = ED since E is the mid point of AD. Hence, AE/ED = 1/1. This means we can say:

AE/ED = 1/1 = AF/FP
AF = FP

Now consider these three parallel lines: “BF”, “line passing through D”, and “line passing through C”. The ratio of the intercepts of the two transversals on them will also be the same.

BD/DC = FP/PC

We know that BD = DC since D is the mid point of BC. Hence, BD/DC = 1/1. This means we can also say:

BD/DC = 1/1 = FP/PC
FP = PC

From these two calculations, we will get AF = FP = PC, and hence, AF:FC = 1:(1+1) = 1:2.

Therefore, the answer is B. We hope you see that Geometry questions on the GMAT can be easily resolved once we bring in parallel lines.

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!

Learn How to Begin a GMAT Problem by Focusing on Keywords in the Question Stem

Quarter Wit, Quarter WisdomToday, we will not begin our post as we usually do by introducing the topic we intend to discuss. Instead, we will directly ask you to think about a question. The reason is this – when faced with similar questions on the GMAT without any preface, we often struggle to identify the concept being tested, which is the starting point of our efforts. Our post today focuses on how to observe the keywords in the question stem, and how to know where to go from there.

Take a look at this example Quant question:

The length and width of a rectangle are integer values. What is the area of the smallest such rectangle that can be inscribed in a circle whose radius is also an integer?

(A) 12 
(B) 24 
(C) 36 
(D) 48 
(E) 60

Now here is the problem – the question stem does not give us any numbers! We don’t know any dimensions of the rectangle or the circle, yet the answer choice options are very specific numbers! So how do we begin? The smallest positive integer is 1, so should we start by testing the radius of the circle as 1, and then try to go on from there? And if 1 doesn’t work, then move on to 2, 3, 4… etc?

No – we are not a computer algorithm and on top of that, the GMAT only gives us around 2 minutes to figure out the answer. With this in mind, the question should enough clues to make all all of that trial and error testing unnecessary. So if plugging in numbers isn’t the way to go, how should we start solving this problem?

Now, the moment we read “rectangle inscribed in a circle”, what comes to mind is that a rectangle has 90 degree angles, and hence, the diagonal of the rectangle is the diameter of the circle (an arc that subtends a 90 degree angle at the circumference is a semicircle). The rectangle inside of the circle will look something like this:

 

 

 

 

Now we can see that we have a circle with a diameter (AB) and 90 degree angles subtended in each semicircle (angle AMB and angle ANB).

Essentially then, we have two right triangles (triangle AMB and triangle ANB) that share the hypotenuse AB. Also, it’s important to note that each side of these triangles is an integer – since we know the radius of the circle is an integer, the diameter has to be an integer too. This should make us think of Pythagorean triples!

Whenever all three sides of a right triangle are integers, they will form a Pythagorean triple. Can you have a right triangle with all integer sides such that the length of one side is 1? No. There are no Pythagorean triples with 1 as a side. The smallest Pythagorean triple we know of is 3, 4, 5 (so there can be no right triangle with all integer sides such that the length of one side is 2, either).

We already know Pythagorean triples are the lengths of the sides of right triangles where all sides are integers. What we need to internalize is that ONLY Pythagorean triples are the lengths of sides of right triangles where all sides are integers. You cannot have a right triangle with all integer sides but whose sides are not a Pythagorean triple.

This means that the smallest right triangle with all integer sides is a 3, 4, 5 triangle.

Now note that in the given question, the hypotenuse is the diameter of the circle. We are given that the radius of the circle is an integer, so the diameter will be twice an integer, i.e. an even integer.

So we know the hypotenuse is an even integer, but as we discussed last week, the hypotenuse of a primitive Pythagorean right triangle must be odd. So this triangle must be a non-primitive Pythagorean triple. The smallest such triple will be twice of 3, 4, 5, i.e. the triangle will have sides with lengths 6, 8, 10.

This means the sides of the rectangle must be 6 and 8, while its diagonal must have a length of length 10. The area of the rectangle, then, must be 6*8 = 48. The answer is D.

Finally, at the end of the post we have figured out that this post is a continuation of last week’s post on properties of Pythagorean triples! We hope you enjoyed it!

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!

The Pythagorean Triples Properties You’ll See on the GMAT

Quarter Wit, Quarter WisdomToday, let’s discuss a few useful properties of primitive Pythagorean triples. A primitive Pythagorean triple is one in which a, b and c (the length of the two legs and the hypotenuse, respectively) are co-prime. So, for example, (3, 4, 5) is a primitive Pythagorean triple while its multiple, (6, 8, 10), is not.

Now, without further ado, here are the properties of primitive Pythagorean triples that you’ll probably encounter on the GMAT:

I. One of a and b is odd and the other is even.
II. From property I, we can then say that c is odd.
III. Exactly one of a, b is divisible by 3.
IV. Exactly one of a, b is divisible by 4.
V. Exactly one of a, b, c is divisible by 5.

If you keep in mind the first primitive Pythagorean triple that we used as an example (3, 4, 5), it is very easy to remember all these properties.

If we look at some other examples:

(3, 4, 5), (5, 12, 13), (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73), etc.

we will see that these properties hold for all primitive Pythagorean triples.

Now, let’s take a look at an example GMAT question which can be easily solved if we know these properties:

The three sides of a triangle have lengths p, q and r, each an integer. Is this triangle a right triangle?

Statement 1: The perimeter of the triangle is an odd integer.
Statement 2: If the triangle’s area is doubled, the result is not an integer.

We know that the three sides of the triangle are all integers. So if the triangle is a right triangle, the three sides will represent a Pythagorean triple. Given that p, q and r are all integers, let’s use the properties of primitive Pythagorean triples to break down each of the statements.

Statement 1: The perimeter of the triangle is an odd integer.

Looking at the properties above, we know that a primitive Pythagorean triple can be represented as:

(Odd, Even, Odd) (The first two are interchangeable.)

Non-primitive triples are made by multiplying each member of the primitive triple by an integer n greater than 1. Depending on whether n is odd or even, the three sides can be represented as:

(Odd*Odd, Even*Odd, Odd*Odd) = (Odd, Even, Odd)
or
(Odd*Even, Even*Even, Odd*Even) = (Even, Even, Even)

However, the perimeter of a right triangle can never be odd because:

Odd + Even + Odd = Even
Even + Even + Even = Even

Hence, the perimeter will be even in all cases. (If the perimeter of the given triangle is odd, we can say for sure that it is not a right triangle.) This statement alone is sufficient.

Statement 2: If the triangle’s area is doubled, the result is not an integer.

If p, q and r are the sides of a right triangle such that r is the hypotenuse (the hypotenuse could actually be either p, q, or r but for the sake of this example, let’s say it’s r), we can say that:

The area of this triangle = (1/2)*p*q
and
Double of area of this triangle = p*q

Double the area of the triangle has to be an integer because we are given that both p and q are integers, but this statement tells us that this is not an integer. In that case, this triangle cannot be a right triangle. If the triangle is not a right triangle, double the area would be the base * the altitude, and the altitude would not be an integer in this case.

This statement alone is sufficient, too. Therefore, our answer is D.

As you can see, understanding the special properties of primitive Pythagorean triples can come in handy on the GMAT – especially in tackling complicated geometry 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!

Using “Few” vs. “A Few” vs. “Quite a Few” in a GMAT Verbal Question

Quarter Wit, Quarter WisdomOn quite a few occasions, we at Veritas Prep find ourselves explaining the difference between the terms “few” and “a few” – a subtle, but very important distinction which has, on occasion, completely changed the meaning of a sentence. Hence, we realized that a post on this difference is warranted.

“Few”, when used without a preceding “a”, means “very few” or “none at all”. “Few” is a negative, which puts the quantity of what you are describing near zero.

On the other hand, “a few” is used to indicate “not a large number”. “A few” also indicates a small approximate number, but it is positive nonetheless.

The difference between the two is subtle, yet there are instances where the two can mean completely opposite things. For example, “I have a few friends” is the same as saying “I have some friends”. “I have few friends”, however, implies that I have only very few friends (as opposed to many). It can also imply that I don’t feel very well about it, and I wish I had more friends.

Also, note that there is a very common expression, “quite a few”, which looks like it could mean “rather few” or “very few”, but it does not. It actually means the exact opposite: “a large or significant number” or “many”. So saying, “I have quite a few friends,” is the same as saying “I have quite a lot of friends”.

Here are a few other simple examples:

  • A few people think that red wine is healthy.
    • This implies some people think that red wine is healthy.
  • Few people think that red wine is healthy.
    • This implies only very few people, a very small number, think that red wine is healthy; most think that it is not.
  • Quite a few people think that red wine is healthy.
    • This implies many people, a large number, think that red wine is healthy.

Let’s examine an official Critical Reasoning question in which confusion among these terms could lead to an incorrect answer:

Until now, only injectable vaccines against influenza have been available. They have been primarily used by older adults who are at risk for complications from influenza. A new vaccine administered in a nasal spray form has proven effective in preventing influenza in children. Since children are significantly more likely than adults to contract and spread influenza, making the new vaccine widely available for children will greatly reduce the spread of influenza across the population. 

Which of the following, if true, most strengthens the argument?

(A) If a person receives both the nasal spray and the injectable vaccine, they do not interfere with each other. 
(B) The new vaccine uses the same mechanism to ward off influenza as injectable vaccines do. 
(C) Government subsidies have kept the injectable vaccines affordable for adults. 
(D) Of the older adults who contract influenza, relatively few contract it from children with influenza. 
(E) Many parents would be more inclined to have their children vaccinated against influenza if it did not involve an injection. 

Let’s break down the argument of this passage first. We are given following premises:

  • Until now, only injections of the influenza vaccine were available.
  • These injections were primarily used by older adults.
  • Now nasal sprays are available that prevent influenza in children.
  • Children are more likely to contract and spread influenza.
  • Conclusion: If nasal sprays are made available for children, it will greatly reduce the spread of influenza across the population.

Does something come to mind when you read this conclusion? What initially came to my mind was that if children are most likely to contract and spread influenza, they should have just been given the injections and that would have prevented the spread of disease across the population. Why is it that the availability of a nasal spray will prevent the spread of influenza but injections have not been able to do this?

We need to strengthen the argument, so we should focus on our conclusion and find out what will strengthen it the most. Let’s go through each of the answer choices:

(A) If a person receives both the nasal spray and the injectable vaccine, they do not interfere with each other.

If a person has already been given an injection, he or she is immune to influenza – taking the nasal spray on top of this will not have any impact on his or her immunity. This option is irrelevant to the argument, thus A cannot be our answer.

(B) The new vaccine uses the same mechanism to ward off influenza as injectable vaccines do.

This answer choice only says that the nasal sprays work in the same way the injections do. We are not told exactly why injections could not prevent the spread of influenza while the nasal spray will, so this option is also not correct.

(C) Government subsidies have kept the injectable vaccines affordable for adults.

This option tells us that the subsidies have kept injections affordable for all older adults, but it doesn’t say anything about the cost of the nasal spray. If, instead, this option stated, “Injections are very expensive but nasal spray is a cheap alternative”, it might have made a stronger contender, however we do not know whether cost is a factor that parents consider at all when getting their children vaccinate (to make this option the correct answer, we might even have to add something like, “Parents are not willing to get their kids immunized if the vaccine is very expensive”). As is, however, this answer choice is not correct.

(D) Of the older adults who contract influenza, relatively few contract it from children with influenza.

Here is the trick – many test takers feel that this option is like an assumption, and hence, it certainly strengthens the conclusion. “Few” is assumed to be “some”, so it seems to them that this option is saying, “Some older adults do contract influenza from children”. It certainly seems to be an assumption, since that is how the spread of influenza will reduce across the population of older adults.

We know, however, that “few” actually means “hardly any” or “near zero”. If few (near zero) older adults catch flu from children, it doesn’t strengthen the conclusion. If anything, it has the opposite effect since the older adults will be unaffected, and hence, it is unlikely that the spread of influenza will reduce across the population. Because of this, option D is not correct.

(E) Many parents would be more inclined to have their children vaccinated against influenza if it did not involve an injection.

Now this is what we are looking for – a reason why parents don’t give influenza shots to their kids but will be willing to give them nasal sprays. Parents don’t like to give shots to their kids (could be due pain associated with a shot or whatever, the reason why doesn’t really matter here), but now that a nasal spray version of the vaccine is available, they will be more inclined to get their kids vaccinated. This will probably help prevent the spread of influenza across the population. The correct answer, therefore, 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 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!

How to Quickly Interpret Ranges of Variables in GMAT Questions

Quarter Wit, Quarter WisdomSometimes a GMAT Quant question will give us multiple ranges of values that apply to a single variable, and when this happens it can really take us for a ride. Evaluating these ranges to arrive at deductions can extremely confusing, so today we will look at some strategies for how to deal with such problems.

To start off, let’s take a look at an example problem:

If it is true that z < 8 and 2z > -4, which of the following must be true?

(A) -8 < z < 4
(B) z > 2
(C) z > -8
(D) z < 4
(E) None of the above

Given that z < 8 and 2z > -4, we know that z > -2. This means -2 < z < 8. z must lie within that range, hence z can take values such as -1, 0, 5, 7.4, etc.

Now, which of the given answer choices would hold true for ALL such values? Let’s examine each option and see:

(A) -8 < z < 4
We know that z may be more than 4, so this range does not hold true for all possible values of z.

(B) z > 2
We know that z may be less than 2, so this also does not hold true for all possible values of z.

(C) z > -8
No matter what value z will take, it will always be more than -8. This range holds true for all values of z.

(D) z < 4
We know that z may be greater than 4, so this does not hold for all possible values of z.

Our answer is C.

To understand this concept more clearly, let’s use a real life example:

We know that Anna’s weight is more than 120 pounds but less than 130 pounds. Which of the following is definitely true about her weight?

(A) Her weight is 125 pounds.
(B) Her weight is more than 124 pounds.
(C) Her weight is less than 127 pounds.
(D) Her weight is more than 110 pounds.

Can we say that her weight is 125 pounds? No – we just know that it is more than 120 but less than 130. It could be anything in this range, such as 122, 125, 127.5, etc.

Can we say that her weight is more than 124 pounds? This may be true, but it might not be true. Knowing our given range, her weight could very well be 121 pounds, instead.

Can we say her weight is less than 127 pounds? Again, this might not necessarily be true. Her weight could be 128 pounds.

Now, can we say that her weight is more than 110 pounds? Yes – since we know Anna’s weight is between 120 and 130 pounds, it must be more than 110 pounds.

This question uses the same concept as the first question! If you look at that question again, it will hopefully make much more sense. Now try solving this example problem:

If 1/55 < x < 1/22 and 1/33 < x < 1/11, then which of the following could be the value of x?

(i) 1/54
(ii) 1/23
(iii) 1/12

(A) Only (i)
(B) Only (ii)
(C) (i) and (ii)
(D) (ii) and (iii)
(E) (i), (ii) and (iii)

In this problem, we are given two ranges of x. We know that 1/55 < x < 1/22 and 1/33 < x < 1/11, so x is greater than 1/55 AND it is greater than 1/33. Since 1/33 is greater than 1/55 (the smaller the denominator, the larger the number), we just need to know that x will be greater than 1/33.

We are also given that x is less than 1/22 AND it is less than 1/11. Since 1/22 is less than 1/11, we really just need to know that x is less than 1/22.

Hence, the range for x should be 1/33 < x < 1/22. x could take all values that lie within this range, such as 1/32, 1/31, 1/24, 1/23, etc.

Looking at the answer choices, we can see that 1/54 and 1/12 (i and iii) are both out of this range. Therefore, our answer is B.

If we go back to our real life example, this is what the question would look like now:

We know that Anna’s weight is more than 110 pounds but less than 130 pounds. We also know that her weight is more than 115 pounds but less than 140 pounds. Which of the following is definitely true about her weight?

(A) Her weight is 112 pounds.
(B) Her weight is 124 pounds.
(C) Her weight is 135 pounds.

We are given that Anna’s weight is more than 110 pounds and also more than 115 pounds. Since 115 is more than 110, we just need to know that her weight is more than 115 pounds. We are also given that Anna’s weight is less than 130 pounds and also less than 140 pounds. Since 130 is less than 140, we just need to know that her weight is less than 130 pounds.

Now we have the following range: 115 pounds < Anna’s weight < 130 pounds. Only answer choice B lies within this range, so that is our answer.

We hope you see that evaluating ranges of numbers on GMAT questions is not difficult when we consider them in terms of a real life example. The same logic that we use in the simple weight problem is also applicable when algebraic data is given.

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: Ignore the Diagram in That GMAT Geometry Question!

Quarter Wit, Quarter WisdomIf you follow the Veritas Prep blog, you have probably heard us talk about the importance of diagrams in many GMAT Quant questions  – coordinate geometry, races, time-speed-distance problems, sets, etc. We even suggest you to make diagrams when they are not given on such questions.

But sometimes, the GMAT Testmakers give such diagrams that we wish we were not given the diagram at all. In fact, the addition of a diagram – something that often simplifies our questions – can take the difficulty of the question to a whole new level. By now you are probably thinking that I am surely exaggerating, so I will proceed with an example.

Try to figure this out: when the figure given below is cut along the solid lines, folded along the dashed lines, and then taped along the solid lines, the result is a model of a geometric solid.

Now, can you use your imagination and figure out what kind of a geometric solid you will get in this case? Don’t go ahead just yet – first, give it a shot for a few minutes:

To be honest, I have given it a try and it is certainly not easy. I will know for sure only when I actually carry out the aforementioned steps – cut the paper along the solid lines, fold along the dashed lines and then tape up along the solid lines. Without carrying out the steps I am not sure exactly what kind of a figure I will get.

So the test maker comes to our rescue here. Here is the complete question:

When the figure above is cut along the solid lines, folded along the dashed lines, and taped along the solid lines, the result is a model of a geometric solid. This geometric solid consists of two pyramids each with a square base that they share. What is the sum of number of edges and number of faces of this geometric solid?

(A) 10
(B) 18
(C) 20
(D) 24
(E) 25

The Testmaker specifies what kind of a figure we get – two pyramids, each with a square base that they share. Figuring this out in one minute without an actual paper and scissor at hand would need extraordinary skill. Many test-takers spend precious minutes trying to make sense of the given diagram, but in problems like this, it should be completely ignored because we already know what it will look like – two pyramids with a common square base.

This, we understand! We know what a pyramid looks like – triangular faces converge to a single point at the top with a polygon (often a square) base. We need two pyramids joined together at the base.

This is what the solid will look like:

Just the 4 triangular faces of each of the two pyramids (8 triangles total) will be visible.  Since they will share the square base, the base will not be visible. Hence, the figure will have 8 faces.

Now let’s see how many edges there will be: to make the top pyramid, four triangular faces join to give four edges. To make the bottom pyramid, another four triangular faces join to give four more edges. The two pyramids join on the square base to give yet another four edges.

So all in all, we have 4 + 4 + 4 = 12 edges

When we sum up the faces and edges, we get 8 + 12 = 20

The question is much more manageable now. All we had to do was ignore the diagram given to us!

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 3-Step Method to Solving Complex GMAT Algebra Problems

Quarter Wit, Quarter WisdomIf you have been practicing GMAT questions for a while, you will realize that not every question can be solved using pure algebra, especially at higher levels. There will be questions that will require logic and quite a bit of thinking on your part.  These questions tend to throw test-takers off – students often complain, “Where do I start from? Thinking through the question takes too much time!” Unfortunately, there is no getting away from such questions.

Today, let’s see how to handle such questions step-by-step by looking at an example problem:

N and M are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either N or M. What is the smallest possible positive difference between N and M?

(A) 29
(B) 49
(C) 58
(D) 113
(E) 131

This is not a simple algebra question, where we are asked to make equations and solve them.

We are given 6 digits: 1, 2, 3, 6, 7, 8. Each digit needs to be used to form two 3-digit numbers. This means that we will use each of the digits only once and in only one of the numbers.

We also need to minimize the difference between the two numbers so they are as close as possible to each other. Since the numbers cannot share any digits, they obviously cannot be equal, and hence, the smaller number needs to be as large as possible and the greater number needs to be as small as possible for the numbers to be close to each other.

Think of the numbers  of a number line. You need to reduce the difference between them. Then, under the given constraints, push the smaller number to the right on the number line and the greater number to the left to bring them as close as possible to each other.

STEP 1:
The first digit (hundreds digit) of both numbers should be consecutive integers – i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3** (the difference between the latter will certainly be more than 100).

We get lots of options for hundreds digits: (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**). All of these options could satisfy our purpose.

STEP 2:
Now let’s think about what the next digit (the tens digit) should be. To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1, if possible) and the tens digit of the smaller number should be as large as possible (8, if possible). So let’s not use 1 or 8 in the hundreds places and reserve them for the tens places instead, since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?

Let’s try to make a pair of numbers in the form of 2** and 3**. We need to make the 2** number as large as possible and make the 3** number as small as possible. As discussed above, the tens digit of the smaller number should be 8 and the tens digit of the greater number should be 1. We now have 28* and 31*.

STEP 3:
Now let’s use the same logic for the units digit – make the units digit of the smaller number as large as possible and the units digit of the greater number as small as possible. We have only two digits left over – 6 and 7.

The two numbers could be 287 and 316 – the difference between them is 29.

Let’s try the same logic on another pair of hundreds digits, and make the pair of numbers in the form of 6** and 7**. We need the 6** number to be as large as possible and the 7** number to be as small as possible. Using the same logic as above, we’ll get 683 and 712. The difference between these two is also 29.

The smallest of the given answer choices is 29, so we need to think no more. The answer must be A.

Note that even if you try to express the numbers algebraically as:

N = 100a + 10b + c
M = 100d + 10e + f

a lot of thought will still be needed to find the answer, and there is no real process that can be followed.

Assuming N is the greater number, we need to minimize N – M.

N – M = 100 (a – d) + 10( b – e) + (c – f)

Since a and d cannot be the same, the minimum value a – d can take is 1. (a – d) also cannot be negative because we have assumed that N is greater than M. With this in mind, a and d must be consecutive (2 and 1, or 3 and 2, or 7 and 6, etc). This is another way of completing STEP 1 above.

Next, we need to minimize the value of (b – e). From the available digits, 1 and 8 are the farthest from each other and can give us a difference of -7. So b = 1 and e = 8. This leaves the consecutive pairs of 2, 3 and 6, 7 for hundreds digits. This takes care of our STEP 2 above.

(c – f) should also have a minimum value. We have only one pair of digits left over and they are consecutive, so the minimum value of (c – f) is -1. If the hundreds digits are 3 and 2, then c = 6 and f = 7. This is our STEP 3.

So, the pair of numbers could be 316 and 287 – the difference between them is 29. The pair of numbers could also be 712 and 683 – the difference between them is also 29.

In either case, note that you do not have a process-oriented approach to solving this problem. A bit of higher-order thinking is needed to find the correct 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!

Which is Worse to Encounter on a GMAT Question: Median or Mean?

Quarter Wit, Quarter WisdomHypothetically speaking, given a choice between a question on median and one on mean, which would you choose? (Not that we are fortunate enough to have a choice on test day, but no harm in dreaming!) I would certainly pick the question testing on median, and here is why:

Median is the value at a point – to be precise, the point which divides the increasing data set into two equal halves. You don’t care what is on the left and what is on the right of this point, so an outlier will do nothing to the median. The mean, however depends on every value in the set. If you increase one element of data, the mean of the set changes – outliers can drastically change the value of the mean. Hence, every element has to be kept in mind! With the median, there is a lot less to worry about.

Let’s illustrate this with an example data sufficiency question:

Question on Median:
At a bakery, cakes are sold every day for a certain number of days. If 6 or more cakes were sold for 20% of the total number of days, is the median number of cakes sold less than 4?

Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.
Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.

The following is the number of cakes sold on any of the days mentioned in the question:

 

Say there were 100 days (since all figures are in terms of percentages, we can assume a number to simplify our understanding).

The question stem tells us that 6 or more cakes were sold for 20% of the days, so for 20 days, 6 or more cakes were sold. Then for 80 days, 1/2/3/4/5 cakes were sold.

With this information in mind, is the median number of cakes sold in one day less than 4?

We know how to get the median. When we arrange all figures in increasing order, the median will be the average of the 50th and the 51st terms. We need to know if the average of the 50th and 51st term is less than 4. Let’s tackle the statements one at a time:

Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.

The number of days that less than 6 cakes were sold = 80. 75% of these 80 days will be 60 days. In 60 days, less than 4 cakes were sold. So the 50th and 51st terms will be less than 4 and so will their average. Hence, the median will be less than 4. This statement alone is sufficient.

Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.

In 20 days, 6 or more cakes were sold. This constitutes 50% of the days during which 4 or more cakes were sold, so in another 20 days, 4 or 5 cakes were sold. Hence, during the leftover 60 days, less than 4 cakes were sold. The 50th and 51st terms will be less than 4 and so will their average. Hence, the median will be less than 4. This statement alone is also sufficient, so our answer is D.

All we needed to worry about here were the 50th and 51st terms, however the whole problem changes when we talk about mean instead of median.

Same Question on Mean:
At a bakery, cakes are sold every day for a certain number of days. If 6 or more cakes were sold for 20% of the total number of days, is the average number of cakes sold less than 4?

Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.
Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.

Again, the question stem tells us that 6 or more cakes were sold for 20% of the days, so for 20 days, 6 or more cakes were sold. Then for 80 days, 1/2/3/4/5 cakes were sold.

We now need to ask ourselves is the average number of cakes sold in one day less than 4?

This question asks us about the average. – that is far more complicated than the median. Every value matters when we talk about the average. We need to know the number of cakes sold on each of these 100 days to get the average.

6 or more cakes were sold in 20 days. Note that the number of cakes sold during these 20 days could be any number greater than 6, such as 20 or 50 or 120, etc. The minimum number of cakes sold on these 20 days would be 6*20 = 120. There is no limit to the maximum number of cakes sold.

With this in mind, let’s examine the statements:

Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.

In 80 days, less than 6 cakes were sold. Of this number, 75% is 60 days. In 60 days, less than 4 cakes were sold.

So in 60 days, you have a minimum of 1*60 = 60 cakes sold and a maximum of 3*60 = 180 cakes sold. During the leftover 20 days 4 or 5 cakes were sold, so you have a minimum of 4*20 = 80 cakes and a maximum of 5*20 = 100 cakes.

The minimum value of the average is (120 + 60 + 80)/ 100 = 2.6 cakes, but the maximum average could be anything. Therefore, this statement alone is not sufficient.

Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.

The 20 days when 6 or more cakes were sold make up 50% of the days when 4 or more cakes were sold. So for another 20 days, 4 or 5 cakes were sold. This gives us a minimum of 4*20 = 80 cakes and a maximum of 5*20 = 100 cakes. For 60 days, 1/2/3 cakes were sold. So in 60 days, you have minimum of 1*60 = 60 cakes sold and a maximum of 3*60 = 180 cakes sold.

The minimum value of the average is (120 + 60 + 80)/ 100 = 2.6 cakes, but again, the maximum average could be anything. This statement alone is also not sufficient.

Note that both statements give you the same information, so if they are not sufficient independently, they are not sufficient together. The answer of this modified question would be E.

Here, we had to assume the minimum and maximum value for each data point to get the range of the average – we couldn’t just rely on one or two data points. Finding the mean during a GMAT question requires much more information than finding the median!

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!

An Interesting Right Triangle Property You’ll Need to Know for the GMAT

Quarter Wit, Quarter WisdomIn a previous post, we discussed medians, altitudes and angle bisectors of isosceles and equilateral triangles. Today, we will discuss an interesting property of perpendicular bisectors and circumcenter of right triangles.

Property: The circumcenter of a right triangle is the mid point of the hypotenuse.

Let’s prove this first and then we will see its application.

 

 

 

 

 

 

Say, we have a right triangle ABC right angled at B. Let’s draw the perpendicular bisector of AB which intersects AB at its mid point M. Say this line intersects the hypotenuse AC at N. We need to prove that AN = CN. Note that triangle AMN and triangle ABC are similar triangles using the AA property (angle AMN = angle ABC = 90 degrees and angle A is common to both triangles). So the ratio of the sides of the two triangle is the same. Since MN is the perpendicular bisector of line AB, AM = MB which means that AM is half of AB.

So AM/AB = 1/2 = AN/AC

Hence AN = NC

So N is the mid point of AC.

Using the exact same logic for side BC, we will see that its perpendicular bisector also bisects the hypotenuse. So N would be the circumcenter of triangle ABC and the mid point of AC.

Using an official question, let’s see how this property can be useful to us:

In the rectangular coordinate system shown above, points O, P, and Q represent the sites of three proposed housing developments. If a fire station can be built at any point in the coordinate system, at which point would it be equidistant from all three developments?

(A) (3,1)
(B) (1,3)
(C) (3,2)
(D) (2,2)
(E) (2,3)

 

 

 

 

 

First, let’s see how we will solve this question without knowing this property and using co-ordinate geometry instead.

Method 1:
Points O and Q lie on the X axis and are 4 units apart. We need a point equidistant from both O and Q. All such points will lie on the line lying in the middle of O and Q and perpendicular to the X axis. The equation of such a line will be x = 2. The fire station should be somewhere on this line.

Points O and P lie on the Y axis and are 6 units apart. We need a point equidistant from both O and P. All such points will lie on the line lying in the middle of O and P and perpendicular to the Y axis. The equation of such a line will be y = 3. The fire station should be somewhere on this line too.

Any two lines on the XY plane intersect at most at one point (if they are not overlapping). Since the fire station must lie on both these lines, it must be on their intersection i.e. at (2, 3).

This point (2,3) will be equidistant from O, Q and P. Therefore, the answer is E.

Method 2:
Think of the question in terms of the perpendicular bisectors of triangle OPQ. Their point of intersection will be equidistant from all three vertices.

We know that the circumcenter lies on the mid point of the hypotenuse. The end points of the hypotenuse are (4, 0) and (0, 6). The mid point will be

x = (4 + 0)/2 = 2
y = (0 + 6)/2 = 3

As in Method 1, the point (2, 3) will be equidistant from all three points, O, P and Q. Again, the 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!

Tackling GMAT Critical Reasoning Boldface Questions

Quarter Wit, Quarter WisdomFor some reason, GMAT test takers automatically associate boldface questions with the 700 level, but this fear is unfounded, honestly!

We have often found that one strategy, which is very helpful in other question types too, helps sort out most questions of this type, though not in the same way. That strategy is – ‘find the conclusion(s)’

The conclusion of the argument is the position taken by the author.

Boldface questions (and others too) sometimes have more than one conclusion – One would be the conclusion of the argument i.e. the author’s conclusion. The argument could mention another conclusion which could be the conclusion of a certain segment of people/ some scientists/ some researchers/ a politician etc. We need to segregate these two and how each premise supports/opposes the various conclusion. Once this structure is in place, we automatically find the answer. Let’s see how with an example.

Question: Recently, motorists have begun purchasing more and more fuel-efficient economy and hybrid cars that consume fewer gallons of gasoline per mile traveled. There has been debate as to whether we can conclude that these purchases will actually lead to an overall reduction in the total consumption of gasoline across all motorists. The answer is no, since motorists with more fuel-efficient vehicles are likely to drive more total miles than they did before switching to a more fuel-efficient car, negating the gains from higher fuel-efficiency.

Which of the following best describes the roles of the portions in bold?

(A)The first describes a premise that is accepted as true; the second introduces a conclusion that is opposed by the argument as a whole.

(B)The first states a position taken by the argument; the second introduces a conclusion that is refuted by additional evidence.

(C)The first is evidence that has been used to support a position that the argument as a whole opposes; the second provides information to undermine the force of that evidence.

(D)The first is a conclusion that is later shown to be false; the second is the evidence by which that conclusion is proven false.

(E)The first is a premise that is later shown to be false; the second is a conclusion that is later shown to be false.

Solution: As our first step, let’s try to figure out the conclusion of the argument:

The author’s view is that “purchases of fuel efficient vehicles will NOT lead to an overall reduction in the total consumption of gasoline across all motorists.”

This is the position the argument (and author) takes.

The argument gives us another conclusion: these purchases will actually lead to an overall reduction in the total consumption of gasoline across all motorists.

Some people take this position (implied by the use of “there has been debate”)

This is our second bold statement. It introduces the opposing conclusion.

Let’s look at our options now.

(A) The first describes a premise that is accepted as true; the second introduces a conclusion that is opposed by the argument as a whole.

The first bold statement: Recently, motorists have begun purchasing more and more fuel-efficient economy and hybrid cars that consume fewer gallons of gasoline per mile traveled.

This is a premise and has been accepted as true. We know it has been accepted as true since the last line ends with – “…negating the gains from higher fuel-efficiency”

We have seen above that the second bold statement tells us about a conclusion that the argument opposes.

So (A) is correct. We have found our answer but let’s look at the other options too.

(B) The first states a position taken by the argument; the second introduces a conclusion that is refuted by additional evidence.

The first bold statement is a premise. It is not the position taken by the argument. Let’s move on.

(C) The first is evidence that has been used to support a position that the argument as a whole opposes; the second provides information to undermine the force of that evidence.

This option often confuses test-takers.

The evidence is – “Recently, motorists have begun purchasing more and more fuel-efficient economy and hybrid cars that consume fewer gallons of gasoline per mile traveled.”

That is, “the motorists have begun purchasing fuel efficient cars that give better mileage.”

The second bold statement does not undermine this evidence at all. In fact, it builds up on it with – “This brings up a debate on whether it will lead to overall decreased fuel consumption?”

Hence (C) is not correct.

(D)The first is a conclusion that is later shown to be false; the second is the evidence by which that conclusion is proven false.

The first bold statement is not a conclusion. So no point dwelling on this option.

(E)The first is a premise that is later shown to be false; the second is a conclusion that is later shown to be false.

The premise is taken to be true. The argument ends with “… the gains from higher fuel-efficiency”. Hence, this option doesn’t stand a chance either.

We hope you see how easy it is to break down the options once we identify the conclusion(s).

Keep practicing!

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 Complex Word Problems

Quarter Wit, Quarter WisdomIn studying for the GMAT, we often come across a strategy for how to handle complex questions – simplify them until they become a problem that we know how to solve. But how exactly does one simplify a complicated GMAT question? Let’s try to understand this with an example today:

Twenty-four men can complete a job in sixteen days. Thirty-two women can complete the same job in twenty-four days. Sixteen men and sixteen women started working on the job for twelve days. How many more men must be added to complete the job in 2 days?

(A) 16
(B) 24
(C) 36
(D) 48
(E) 54

Here, we are dealing with two groups of people: men and women. These two groups have different rates of completing a job. We are also told that a certain number of men and women do a part of the job, and we are asked to find the number of additional “men” required to finish the job in a shorter amount of time.

Recall that we have already come across questions where workers start some work and then more workers join in to complete the work before time.

The problem with this question is that we have two types of workers, not just one. So let’s try to simplify the question to a form that we know how to easily solve.

We’ll start by finding the relation between the rate of work done by men and the rate of work done by women. Let’s make the number of men and women the same to find the number of days it will take each group to complete 1 job.

Given: 24 men complete 1 job in 16 days

Given: 32 women complete 1 job in 24 days

So how many days will 24 women take to complete 1 work? (Why 24 women? Because we know how many days 24 men take)

We know how to solve this problem. (It has already been discussed in a past post).

32 women ……………. 1 work ………………. 24 days

24 women ……………. 1 work ………………. ?? days

No. of days taken = 24 * (32/24) = 32 days

Now this is what we have: 24 men take 16 days while 24 women take 32 days

So women take twice the time taken by men to do the same work (32 days vs 16 days). This means the rate of work of women is half the rate of work of men. This means 2 women are equivalent to 1 man i.e. 2 women will do the same work as 1 man does in the same time.

So now, let us replace all women by men so that we have only one type of worker.

Now this is our regular work rate question –

Given: 24 men complete the work in 16 days

Given: 16 men and 16 women work for 12 days

This means that we have 16 men and 8 men work for 12 days

which implies 24 men work for 12 days

We know that 24 men complete the work in 16 days. If they work for 12 days, there are 4 more days to go. But the work has to be completed in 2 days.

24 men …………… 4 days

?? men ……………. 2 days

No of men needed = 24 * (4/2) = 48

So we need 24 additional men to complete the work in 2 days.

Or looking at it another way, 24 men need 16 days to complete the work, so they need another 4 days to complete. But if we want them to complete the work in half the time (2 days), we will need twice the work force. So we need another 24 men.

Answer (B)

Basically, the question involved solving two smaller work-rate problems. Doesn’t seem daunting now, right?

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: Beware of Assumption in GMAT Critical Reasoning Options

Quarter Wit, Quarter WisdomSometimes, while evaluating the answer choices in in strengthen/weaken questions, we unknowingly go beyond the options and make assumptions about what they may imply if we were to have additional pieces of data. What we have to remember is that we do not have this additional information – we have to judge each option on its own merits, only. Let’s discuss this in detail with one of our own practice GMAT questions:

In 2009, a private school spent $200,000 on a building which housed classrooms, offices, and a library. In 2010, the school was unable to turn a profit. Therefore, the principal should be fired.

Each of the following, if true, weakens the author’s conclusion EXCEPT:

(A) The principal was hired primarily for her unique ability to establish a strong sense of community, which many parents cited as a quality that kept children enrolled in the school longer.
(B) The new library also features a seating area big enough for all students to participate in cultural arts performances, which the head of school intends to schedule more frequently now.
(C) The principal was hired when the construction of the new building was almost completed.
(D) A significant number of families left the school in 2010 because a favourite teacher retired.
(E) More than half of the new families who joined the school in 2010 cited the beautiful new school facility as an important factor in their selection of the school.

This is a weaken/exception question, so four of the five answer choices will weaken the argument, while the fifth option (which will be the correct answer) will either not have any impact on the argument or it might even strengthen it. As we know, such questions require a bit more effort to answer, since four of the five options will definitely be relevant to the argument. The important thing is to focus on what we are given and not assume what the various answer options may or may not lead to. Let’s understand this:

The gist of the argument:

  • Last year, a lot of money was spent to construct a new building with many amenities.
  • This year, the school did not see a profit.
  • Hence, fire the principal.

Based on the two given facts – “a lot of money was spent to make the building in 2009” and “the school did not see a profit in 2010” – the author has decided to fire the principal. Many pieces of information could weaken his stance. For example:

  • It was not the principal’s decision to construct the building.
  • The school’s revenue in 2010 took a hit because of some other factor.
  • The school’s losses reduced by a huge amount in 2010 and the probability of it seeing a profit in 2011 is high.

Information such as this could improve the principal’s case to stay. We know that for this particular question, there will only be one option that does not help the principal.

You will have to choose the answer choice which, with the given information, does not help the principal’s case. Let’s look at the options now:

(A) The principal was hired primarily for her unique ability to establish a strong sense of community, which many parents cited as a quality that kept children enrolled in the school longer.

With this answer choice, we see that the principal was hired not to increase school profits, but for another critical purpose. Perhaps the school’s finance department is in charge of worrying about profits, and so the head of that department needs to be fired! This answer choice makes a strong case for keeping the principal, and hence, weakens the author’s argument.

(B) The new library also features a seating area big enough for all students to participate in cultural arts performances, which the head of school intends to schedule more frequently now.

If true, this statement would have no impact on whether or not the principal should be fired. It describes an amenity provided by the new building and how it will be used – it neither strengthens nor weakens the principal’s case to stay, hence, this is the correct answer choice. But let’s look at the rest of the options too, just to be safe:

(C) The principal was hired when the construction of the new building was almost completed.

This tells us that the new building was not her decision. So if it did not have the desired effect, she cannot be blamed for it. So it again helps her case.

(D) A significant number of families left the school in 2010 because a favourite teacher retired.

This answer choice shows that there was another reason behind the school’s loss in profit. The construction of the building could still be a good idea that leads to future profits, which the principal’s case and weakens the author’s argument.

(E) More than half of the new families who joined the school in 2010 cited the beautiful new school facility as an important factor in their selection of the school.

For some reason, this is the answer choice that often trips up students. They feel that it doesn’t help the principal’s case – that because the new building attracts students, if there are losses, it means that the loss is due to a fault with the new building, and thus, the principal is at fault. But note that we are assuming a lot to arrive at that conclusion. All we are told is that the new building is attracting students. This means the new building is serving its purpose – it is generating extra revenue. The fact that the school is still experiencing losses could be explained by many different reasons.

Since the author’s decision to fire the principal is based solely on the premise that a lot of money was spent to construct the new building, which now seems to serve no purpose (because the school experienced losses), this answer choice certainly weakens the argument. The option tells us that the principal’s decision to make the building was justified, so it helps her case to stay with the school.

After examining each answer choice, we can see that the answer is clearly B. Remember, in Critical Reasoning questions it is crucial to come to conclusions only based on the facts that are given – creating assumptions based on information that is not given can lead you to fall in a Testmaker trap.

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: When Can You Divide by a Variable?

Quarter Wit, Quarter WisdomWe have often come across test takers confused about division by a variable. When is it allowed, when is it not allowed? Why is it allowed in some cases and not in others? What are the constraints we need to look out for?

For example:

Is division by x allowed here: x^2 = 10x?
Is division by x allowed here: y = 4x?
Is division by x allowed here: x^2 < 4x?

Let’s take a detailed look at all these questions today.

The basic guidelines:

  1. Division by 0 is not allowed, hence you cannot divide by a variable until and unless we know that it cannot be 0.
  2. In the case of an inequality, when you divide by a negative number, the sign of the inequality flips. So we cannot divide by a variable until and unless we know that it cannot be 0 AND whether it is positive or negative.

Let’s look at the three questions given above and try to solve them using these guidelines:

Is division by x allowed here: x^2 = 10x?

The first thing to find out here is whether or not x can equal 0.

Case 1: If no other information has been given, then x can be 0 and we cannot divide by it. This is how we proceed in that case:

x^2 – 10x = 0
x(x – 10) = 0
x = 0 or 10

Case 2: If the question stem tells us that x is not 0, then we can divide by x.

x^2/x = 10x/x
x = 10

Obviously, we don’t get the second solution (x = 0) in this case, as we already know that x cannot be 0. Now let’s look at the second problem:

Is division by x allowed here: y = 4x?

Again, this is an equation and we need to know whether or not x can equal 0.

Case 1: If x can be 0, you cannot divide by it. In this case, x = 0 and y = 0 is one of the infinite possible solutions.

Case 2: If the question stem states that x cannot be 0, then we can do the following:

y/x = 4

Now let’s look at the final question:

Is division by x allowed here: x^2 > -4x?

Here, we have an inequality. Before deciding whether we can divide by x or not, we need to know not only whether x can be equal to 0, but also whether x is positive or negative.

Case 1: If we know nothing about the possible values that x can take, then this is how we proceed:

x^2 + 4x > 0
x(x + 4) > 0

Now we can use the method discussed in the first problem to arrive at the range of x.

x > 0 or x < -4

Case 2: If we know that x is positive, then we can proceed like this:

x^2/x > -4x/x
x > -4

Since we are given that x is positive, we know that that x > 0 (looking at the two options above).

Case 3: If we know that x is negative, then this is how we will proceed:

x^2/x < -4x/x (we flip the sign of the inequality because we divide by x, which is negative)
x < -4

The results obtained are logical, right? When x can be anywhere on the number line, we get the range as x > 0 or x < -4.

If x has to be positive, the range is x > 0.
If x has to be negative, the range is x < -4.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTubeGoogle+, 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: When a Little Information is Enough to Solve a GMAT Problem

Quarter Wit, Quarter WisdomWe have reviewed what standard deviation is in a past post. We know what data is necessary to calculate the standard deviation of a set, but in some cases, we could actually do with a lot less information than the average test-taker may think they need.

Let’s explore this idea through an example GMAT data sufficiency question:

What is the standard deviation of a set of numbers whose mean is 20?

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.
Statement 2: The sum of the squares of the differences from the mean is greater than 100.

We need to determine whether the information we have been given is sufficient to get us the exact value of the standard deviation of a particular set of numbers. To find the standard deviation of a set, we need to know the deviation of each term from the mean so that we can square those deviations, sum the squares, divide them by the number of terms, and then find the square root.

Essentially, to find the standard deviation we either need to know each element of the set, or we need to know the deviation of each element from the mean (which will also give us the number of terms), or we need to know the sum of the square of deviations and the number of terms in the set.

The question stem here tells us that the mean of the set is 20. We have no other information about any of the actual elements of the set or the number of elements. With this in mind, let’s examine each of the statements:

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.

With this statement, we don’t actually know what the absolute value of the difference is. We also don’t know how many elements there are. The set could be something like:

19, 21 (each term is exactly 1 away from the mean 20)
or
18, 18, 22, 22 (each term is exactly 2 away from the mean 20)
etc.

The standard deviation in each case will be different. We don’t know the elements of the set and we don’t know the number of elements in the set. Because of this, there is no way for us to know the value of the standard deviation – this statement alone is not sufficient.

Statement 2: The sum of the squares of the differences from the mean is greater than 100.

“Greater than 100” encompasses a large range of numbers – it could be any value larger than 100. Again, we cannot find the exact standard deviation of the set, so this statement is also not sufficient alone.

Using both statements together, we still do not have any idea of what the elements of the set are or what the sum of the squares of the differences from the mean is. We also still don’t know the number of elements. Hence, both statements together are not sufficient, so the answer is E.

Now, let us add just one more piece of information to the problem in this similar question:

What is the standard deviation of a set of 7 numbers whose mean is 20?

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.
Statement 2: The sum of the squares of the differences from the mean is greater than 100.

What would you expect the answer to be? Still E, right? The sum of the deviations are still unknown and the exact elements of the set are still unknown – all we know is the number of elements. Actually, this information is already too much. All we need to know is that the number of elements is odd and suddenly we can find the standard deviation.

Here is why:

Statement 1 is quite tricky.

If we have an odd number of elements, in which case can the absolute values of the differences of each number in the set from the mean be equal?

Think about it – the mean of the set is 20. What could a possible set look like such that the mean is 20 and the absolute values of the differences of each number in the set from the mean are equal. Try to think of such a set with just 3 elements. Can you come up with one?

19, 19, 21? No, the mean is not 20

19, 20, 21? No, the absolute value of the difference of each number in the set from the mean is not equal. 19 is 1 away from mean but 20 is 0 away from mean.

Note that in this case, the only possible set that could fit the given criteria is one consisting of just an odd number of 20s (all elements in this set must be 20). Only then can each number be equidistant from the mean, i.e. each number would be 0 away from mean. If the numbers of the set all have equal elements, then obviously the standard deviation of the set is 0. It doesn’t matter how many elements it has; it doesn’t matter what the mean is! In this case, Statement 1 alone is sufficient so the answer would be A.

Takeaway:
If a set has an even number of distinct terms, the absolute values of the distances of each term from the mean could be equal. But if a set has an odd number of terms and the absolute values of the distances of each term from the mean are equal, all the terms in the set must be the same and will be equal to the mean.

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 Read GMAT Questions Carefully

Quarter Wit, Quarter WisdomWe all know that we need to be very careful while reading GMAT questions – that every word is important. Even small oversights can completely change an answer for you. This is what happens with many test takers who try to tackle this official question. Even though the question looks very simple, the way it is worded causes test-takers to often ignore one word, which changes the solution entirely. Let’s look at this question now:

Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month. The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save. If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Let’s consider the question stem sentence by sentence:

“Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month.”

Say Alice’s take-home pay last year was $100 each month. She saves a fraction of this every month – let the amount saved be x.

“The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save.”

What would be “the total amount of money that she had saved at the end of the year”? Since Alice saves x every month, she would have saved 12x by the end of the year.

What would be “the amount of that portion of her monthly take-home pay that she did NOT save”? Note that this is going to be (100 – x). Many test takers end up using (100 – x)*12, however this equation is not correct. The key word here is “monthly” – we are looking for how much Alice does not save each month, not how much she does not save during the whole year.

The total amount of money that Alice saved at the end of the year is 3 times the amount of that portion of her MONTHLY take-home pay that she did not save. Now we know we are looking for:

12x = 3*(100 – x)
x = 20

“If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?”

From our equation, we have determined that Alice saved $20 out of every $100 she earned every month, so she saved 20/100 = 1/5 of her take-home pay.

Therefore, the answer is D.

Often, test-takers make the mistake of writing the equation as:

12x = 3*(100 – x)*12
x = 300/4

However, this will give them the fraction (300/4)/100 = 3/4, and that’s when they will wonder what went wrong.

Be extra careful when reading GMAT questions so that precious minutes are not wasted on such avoidable errors.

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 the Fuel-Up Puzzles

Quarter Wit, Quarter WisdomWe hope you are enjoying the puzzles we have been putting up in the last few weeks. Though all of them may not be directly convertible to GMAT questions, they are great mathematical brain teasers!

(Before we tackle today’s puzzle, first take a look at our posts on how to solve pouring water puzzles, weighing puzzles, and hourglass puzzles.)

Another variety of puzzle involves distributing fuel among vehicles to reach a destination. Let’s look at this type of question today:

A military car carrying an important letter must cross a desert. There is no petrol station in the desert, and the car’s fuel tank is just enough to take it halfway across. There are other cars with the same fuel capacity that can transfer their petrol to one another. There are no canisters to carry extra fuel or rope to tow the cars.

How can the letter be delivered?

Here, we are given that a single car can only reach the midpoint of the desert on its own tank of gas. Since there are no canisters, the car cannot carry extra fuel, so it will need to be fueled up by other cars traveling along with it.

Let’s fill up 4 cars and get them to start crossing the desert together. By the time they cover a quarter of the desert, half of their fuel tanks will be empty. Hence, we will have 4 cars with half tanks, and the status of their fuel tanks will be:

(0.5, 0.5, 0.5, 0.5)

If we transfer the fuel from two of the cars into two other cars, we will have:

(1, 1, 0, 0)

The two cars with fuel in their tanks will continue to cross the desert and cover another quarter of it. Now both of the cars will have half tanks again, and they will have reached the middle of the desert:

(0.5, 0.5, 0, 0)

Now one car will transfer all of its fuel to the other car, allowing that car to have one full tank:

(1, 0, 0, 0)

That car can then carry the letter through the remaining half of the desert.

For this problem, we didn’t really care about the stalled cars in the middle of the desert since we are not required to bring them back. The only important thing is to get the letter completely across the desert. Now, how do we handle a puzzle that asks us to get all of the vehicles back, too? Let’s look at an example question with those constraints:

A distant planet “X” has only one airport located at the planet’s North Pole. There are only 3 airplanes and lots of fuel at the airport. Each airplane has just enough fuel capacity to get to the South Pole (which is diametrically opposite the North Pole). The airplanes can land anywhere on the planet and transfer their fuel to one another.

The mission is for at least one airplane to fly completely around the globe and stay above the South Pole; in the end, all of the airplanes must return to the airport at the North Pole.

For this problem, we are given that a plane with a full tank of fuel can only reach the South Pole, i.e. cover half the distance it needs to travel for the mission. We need it to take a full trip around the planet – from the North Pole, to the South pole, and back again to North Pole. Obviously, we will need more than one plane to fuel the plane which will fly above the South pole.

Let’s divide the distance from pole to pole into thirds (from the North Pole to the South Pole we have three thirds, and from the South Pole to the North Pole we have another three thirds).

Step #1: 2 airplanes will fly to the first third. A third of their fuel will be used, so the status of their fuel tanks will be:

(2/3, 2/3)

One airplane will then fuel up the other plane and go back to the airport. Now the status of their tanks is:

(3/3, 1/3)

Step #2: 2 airplanes will again fly from the airport to the first third – one airplane will fuel up the other plane and go back to the airport. So the status of these two airplanes is this:

(3/3, 1/3)

Step #3: Now there are two airplanes at the first third mark with their tanks full. They will now fly to the second third point, giving us:

(2/3, 2/3)

One of the airplanes will fuel up the second one (until its tank is full) and go back to the first third, where it will meet the third airplane (which has just come back from the airport to support it with fuel) so that they both can return to the airport.

In the meantime, the airplane at the second third, with a full tank of fuel, will fly as far as it can – over the South Pole and towards the North pole, to the last third before the airport.

Step #4: One of the two airplanes from the airport can now go to the first third (on the opposite side of the North pole as before), and share its 1/3 fuel so that both airplanes safely land back at the airport.

And that is how we can have one plane travel completely around the planet and still have all airplanes arrive back safely!

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 the Weighing Puzzle (Part 2)

Quarter Wit, Quarter WisdomA couple of weeks back, we discussed how to handle puzzles involving a two pan balance. In those problems, we learned how to tackle problems that ask you to measure items against one another.

Today, we will look at some puzzles that require the use of a traditional weighing scale. When we put an object on this scale, it shows us the weight of the object.

This is what such a scale looks like:

Puzzles involving a weighing scale can be quite tricky! Let’s take a look at a couple of examples:

You have 10 bags with 1000 coins in each. In one of the bags, all of the coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 grams. 

If you have an accurate weighing scale, which you can use only once, how can you identify the bag with the forgeries?

We are allowed only a single weighing, so we cannot weigh all 10 bags on the scale individually to measure which one has counterfeit coins. We need to find the bag in only one weighing, so we need to somehow make the coins in the bags distinctive.

How do we do that? We can take out one coin from the first bag, two coins from the second bag, three coins from the third bag and so on. Finally, we will have 1 + 2 + 3 + … + 10 = 10*11/2 = 55 coins.

Let’s weigh these 55 coins now.

If all coins were true, the total weight would have been 55 grams. But since some coins are counterfeit, the total weight will be more. Say, the total weight comes out to be 55.2 grams. What can we deduce from this? We can deduce that there must be two counterfeit coins (because each counterfeit coin weighs 0.1 gram extra). So the second bag must be the bag of counterfeit coins.

Let’s try one more:

A genuine gummy bear has a mass of 10 grams, while an imitation gummy bear has a mass of 9 grams. You have 7 cartons of gummy bears, 4 of which contain real gummy bears while the others contain imitation bears. 

Using a scale only once and the minimum number of gummy bears, how can you determine which cartons contain real gummy bears?

Now this has become a little complicated! There are three bags with imitation gummy bears. Taking a cue from the previous question, we know that we should take out a fixed number of gummy bears from each bag, but now we have to ensure that the sum of any three numbers is unique. Also, we have to keep in mind that we need to use the minimum number of gummy bears.

So from the first bag, take out no gummy bears.

From the second bag, take out 1 gummy bear.

From the third bag, take out 2 gummy bears (if we take out 1 gummy bear, the sum will be the same in case the second bag has imitation gummy bears or in case third bag has imitation gummy bears.

From the fourth bag, take out 4 gummy bears. We will not take out 3 because otherwise 0 + 3 and 1 + 2 will give us the same sum. So we won’t know whether the first and fourth bags have imitation gummy bears or whether second and third bags have imitation gummy bears.

From the fifth bag, take out 7 gummy bears. We have obtained this number by adding the highest triplet: 1 + 2 + 4 = 7. Note that anything less than 7 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 6 = 7 and 1 + 2 + 4 = 7
or
0 + 1 + 5 = 6 and 0 + 2 + 4 = 6

But we need the sum to be obtainable in only one way so that we can find out which three bags contain the imitation gummy bears.

At this point, we have taken out 0, 1, 2, 4, and 7 gummy bears.

From the sixth bag, take out 13 gummy bears. We have obtained this number by adding the highest triplet: 2 + 4 + 7 = 13. Note that anything less than 13 will, again, give us a sum that can be made in multiple ways, such as:

12 + 1 + 0 = 13 and 2 + 4 + 7 = 13
or
0 + 1 + 9 = 10 and 1 + 2 + 7 = 10
…etc.

Note that this way, we are also ensuring that we measure only the minimum number of gummy bears, which is what the question asks us to do.

From the seventh bag, take out 24 gummy bears. We have obtained this number by adding the highest triplet again: 4 + 7 + 13 = 24. Again, anything less than 24 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 15 = 16 and 1 + 2 + 13 = 16
or
0 + 1 + 19 = 20 and 0 + 7 + 13 = 20
or
0 + 1 + 23 = 24 and 4 + 7 + 13 = 24
…etc.

Thus, this is the way we will pick the gummy bears from the 7 bags: 0, 1, 2, 4, 7, 13, 24.

In all, 51 gummy bears will be weighed. Their total weight should be 510 grams (51*10 = 510) but because three bags have imitation gummy bears, the weight obtained will be less.

Say the weight is less by 8 grams. This means that the first bag (which we pulled 0 gummy bears from), the second bag (which we pulled 1 gummy bear from) and the fifth bag (which we pulled 7 gummy bears from) contain the imitation gummy bears. This is because 0 + 1 + 7 = 8 – note that we will not be able to make 8 with any other combination.

We hope this tricky little problem got you thinking. Work those grey cells and the GMAT will not seem hard at 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: Solving the Hourglass Puzzle

Quarter Wit, Quarter WisdomLet’s continue our puzzles discussion today with another puzzle type – time measurement using an hourglass. (Before you continue reading this article, check out our posts on how to solve pouring water puzzles and weighing and balancing puzzles)

First, understand what an hourglass is – it is a mechanical device used to measure the passage of time. It is comprised of two glass bulbs connected vertically by a narrow neck that allows a regulated trickle of sand from the upper bulb to fall into the lower one. The sand also takes a fixed amount of time to fall from the upper bulb to the lower bulb. Hourglasses may be reused indefinitely by inverting the bulbs once the upper bulb is empty.

This is what they look like:

Say a 10-minute hourglass will let us measure time in intervals of 10 minutes. This means all of the sand will flow from the upper bulb to the lower bulb in exactly 10 minutes. We can then flip the hourglass over – now sand will start flowing again for the next 10 minutes, and so on. We cannot measure, say, 12 minutes using just a 10-minute hourglass, but we can measure more time intervals when we have two hourglasses of different times. Let’s look at this practice problem to see how this can be done:

A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He used a 7-minute and an 11-minute hourglass. During the whole time, he turned the hourglasses only 3 times (turning both hourglasses at once counts as one flip). Explain how the teacher measured out 15 minutes.

Here, we have a 7-minute hourglass and an 11-minute hourglass. This means we can measure time in intervals of 7 minutes as well as in intervals of 11 minutes. But consider this: if both hourglasses start together, at the end of 7 minutes, we will have 4 minutes of sand leftover in the top bulb of the 11-minute hourglass. So we can also measure out 4 minutes of time.

Furthermore, if we flip the 7-minute hourglass over at this time and let it flow for that 4 minutes (until the sand runs out of the top bulb of the 11-minute hourglass), we will have 3 minutes’ worth of sand leftover in the 7-minute hourglass. Hence, we can measure a 3 minute time interval, too, and so on…

Now, let’s see how we can measure out 15 minutes of time using our 7-minute and 11-minute hourglasses.

First, start both hourglasses at the same time. After the top bulb of the 7-minute hourglass is empty, flip it over again. At this time, we have 4 minutes’ worth of sand still in the top bulb of the 11-minute hourglass. When the top bulb of the 11-minute hourglass is empty, the bottom bulb of 7-minute hourglass will have 4 minutes’ worth of sand in it. At this point, 11 minutes have passed

Now simply flip the 7-minute hourglass over again and wait until the sand runs to the bottom bulb, which will be in 4 minutes.

This is how we measure out 11 + 4 = 15 minutes of time using a 7-minute hourglass and an 11-minute hourglass.

Let’s look at another problem:

Having two hourglasses, a 7-minute one and a 4-minute one, how can you correctly time out 9 minutes?

Now we need to measure out 9 minutes using a 7-minute hourglass and a 4-minute hourglass. Like we did for the last problem, begin by starting both hourglasses at the same time. After 4 minutes pass, all of the sand in the 4-minute hourglass will be in the lower bulb. Now flip this 4-minute hourglass back over again. In the 7-minute hourglass, there will be 3 minutes’ worth of sand still in the upper bulb.

After 3 minutes, all of the sand from the 7-minute hourglass will be in the lower bulb and 1 minute’s worth of sand will be in the upper bulb of the 4-minute hourglass.

This is when we will start our 9-minute interval.

The 1 minute’s worth of sand will flow to the bottom bulb of the 4-minute hourglass. Then we just need to flip the 4-minute hourglass over and let all of the sand flow out (which will take 4 minutes), and then flip the hourglass over to let all of the sand flow out again (which will take another 4 minutes).

In all, we have measured out a 1 + 4 + 4 = 9-minute interval, which is what the problem has asked us to find.

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 the Weighing and Balancing Puzzle

Quarter Wit, Quarter WisdomLet’s continue the discussion on puzzles that we began last week. Today we look at another kind of puzzle – weighing multiple objects using a two-pan balance while we are given a limited number of times to weight the objects against each other.

First of all, do we understand what a two-pan balance looks like?

Here is a picture.

Law School Images

 

 

 

 

 

As you can see, it has two pans that will be even if the weights in them are equal. If one pan has heavier objects in it, that pan will go down due to the weight. With this in mind, let’s try our first puzzle:

One of twelve coins is a bit lighter than the other 11 (which have the same weight). How would you identify this lighter coin if you could use a two-pan balance scale only 3 times? (You can only balance one set of coins against another, i.e. you have no weight measurements.)

There are various ways in which we can solve this.

We are given 12 coins, all of same weight, except one which is a bit lighter.

Let’s split the coins into two groups of 6 coins each and put them in the two pans. Since there is one lighter coin, one pan will be lighter than the other and will rise higher. So now we know that one of these 6 coins is the lighter coin.

Now split these 6 coins into another two groups of 3 coins each. Again, one pan will rise higher since it will have the lighter coin on it. Now we know that one of these three coins is the lighter coin.

Now what do we do? We have 3 coins and we cannot split them equally. What we can do is put one coin in each pan. What happens if the pans are not balanced? Then we know the pan that rises higher has the lighter coin on it (and thus, we have identified our coin). But what if both pans are balanced? The catch is that then the leftover coin is the lighter one! In any case, we would be able to identify the lighter coin using this strategy.

We hope you understand the logic here. Now let’s try another puzzle:

One of 9 coins is a bit lighter than the other 8. How would you identify this lighter coin if you could use a two-pan balance scale only 2 times?

Now we can use the balance only twice, and we are given an odd number of coins so we cannot split them evenly. Recall what we did in the first puzzle when we had an odd number of coins – we put one coin aside. What should we do here? Can we try putting 1 coin aside and splitting the rest of the 8 coins into two groups of 4 each? We can but once we have a set of 4 coins that contain the lighter coin, we will still need 2 more weighings to isolate the light coin, and we only have a total of 2 weighings to use.

Instead, we should split the 9 coins into 3 groups of 3 coins each. If we put one group aside and put the other two groups into the two pans of the scale, we will be able to identify the group which has the lighter coin. If one pan rises up, then that pan is holding the lighter coin; if the pans weight the same, then the group put aside has the lighter coin in it.

Now the question circles back to the strategy we used in the first puzzle. We have 3 coins, out of which one is lighter than the others, and we have only one chance left to weigh the coins. Just like in the first puzzle, we can put one coin aside and weigh the other two against each other – if one pan rises, it is holding the lighter coin, otherwise the coin put aside is the lighter coin! Thus, we were able to identify the lighter coin in just two weighings. Can you use the same method to answer the first puzzle now?

We will leave you with a final puzzle:

On a Christmas tree there were two blue, two red, and two white balls. All seemed the same, however in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls weighed the same. Using a 2-pan balance scale only twice, identify the lighter balls.

Can you solve this problem using the strategies above? Let us know in the comments!

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 the Pouring Water Puzzle

Quarter Wit, Quarter WisdomSome time back, we came across a GMAT Data Sufficiency word problem question based on the “pouring water puzzle”. That made us think that it is probably a good idea to be comfortable with the various standard puzzle types. From this week on, we will look at some fundamental puzzles to acquaint ourselves with these mind benders in case we encounter them on test day.

Today, we will look at the popular “pouring water puzzle”. You may remember a similar puzzle from the movie Die Hard with a Vengeance, where Bruce Willis and Samuel L. Jackson had to diffuse a bomb by placing a 4 gallon jug of water on a set of scales.

Here is the puzzle:

You have a 3- and a 5-liter water container – each container has no markings except for that which gives us its total volume. We also have a running tap. We must use the containers and the tap in such a way that we measure out exactly 4 liters of water. How can this be done?

Don’t worry that this question is not written in a traditional GMAT format! We need to worry only about the logic behind the puzzle – we can then answer any question about it that is given in any GMAT format.

Let’s break down what we are given. We have only two containers – one of 3-liter and the other of 5-liter capacity. The containers have absolutely no markings on them other than those which give us the total volumes, i.e. the markings for 3 liters and 5 liters respectively. There is no other container. We also have a tap/faucet of running water, so basically, we have an unlimited supply of water. Environmentalists may not like my saying this, but this fact means we can throw out water when we need to and just refill again.

Now think about it:

STEP 1: Let’s fill up the 5-liter container with water from the tap. Now we are at (5, 0), with 5 being the liters of water in the 5-liter container, and 0 being the liters of water in the 3-liter container.

STEP 2: Now, there is nothing we can do with this water except transfer it to the 3-liter container (there is no other container and throwing out the water will bring us back to where we started). After we fill up the 3-liter container, we are left with 2 liters of water in the 5-liter container. This brings us to (2, 3).

STEP 3: We gain nothing from transferring the 3 liters of water back to 5-liter container, so let’s throw out the 3 liters that are in the 3-liter container. Because we just threw out the water from the 3-liter container, we will gain nothing by simply refilling it with 3 liters of water again. So now we are at (2, 0).

STEP 4: The next logical step is to transfer the 2 liters of water we have from the 5-liter container to the 3-liter container. This means the 3-liter container has space for 1 liter more until it reaches its maximum volume mark. This brings us to (0, 2).

STEP 5: Now fill up the 5-liter container with water from the tap and transfer 1 liter to the 3-liter container (which previously had 2 liters of water in it). This means we are left with 4 liters of water in the 5-liter container. Now we are at (4, 3).

This is how we are able to separate out exactly 4 liters of water without having any markings on the two containers. We hope you understand the logic behind solving this puzzle. Let’s take a look at another question to help us practice:

We are given three bowls of 7-, 4- and 3-liter capacity. Only the 7-liter bowl is full of water. Pouring the water the fewest number of times, separate out the 7 liters into 2, 2, and 3 liters (in the three bowls).

This question is a little different in that we are not given an unlimited supply of water. We have only 7 liters of water and we need to split it into 2, 2 and 3 liters. This means we can neither throw away any water, nor can we add any water. We just need to work with what we have.

We start off with (7, 0, 0) – with 7 being the liters of water in the 7-liter bowl, the first 0 being the liters of water in the 4-liter bowl, and the second 0 being the liters of water in the 3-liter bowl – and we need to go to (2, 2, 3). Let’s break this down:

STEP 1: The first step would obviously be to pour water from the 7-liter bowl into the 4-liter bowl. Now you will have 3 liters of water left in the 7-liter bowl. We are now at (3, 4, 0).

STEP 2: From the 4-liter bowl, we can now pour water into the 3-liter bowl. Now we have 1 liter in the 4-liter bowl, bringing us to (3, 1, 3).

STEP 3: Empty out the 3-liter bowl, which is full, into the 7-liter bowl for a total of 6 liters – no other transfer makes sense [if we transfer 1 liter of water to the 7-liter bowl, we will be back at the (4, 0, 3) split, which gives us nothing new]. This brings us to (6, 1, 0).

STEP 4: Shift the 1 liter of water from the 4-liter bowl to the 3-liter bowl. We are now at (6, 0, 1).

STEP 5: From the 7-liter bowl, we can now shift 4 liters of water into the 4-liter bowl. This leaves us with with 2 liters of water in the 7-liter bowl. Again, no other transfer makes sense – pouring 1 liter of water into some other bowl takes us back to a previous step. This gives us (2, 4, 1).

STEP 6: Finally, pour water from the 4-liter bowl into the 3-liter bowl to fill it up. 2 liters will be shifted, bringing us to (2, 2, 3). This is what we wanted.

We took a total of 6 steps to solve this problem. At each step, the point is to look for what helps us advance forward. If our next step takes us back to a place at which we have already been, then we shouldn’t take it.

Keeping these tips in mind, we should be able to solve most of these pouring water puzzles in the future!

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 Answer GMAT Questions That are About an Unfamiliar Topic

Quarter Wit, Quarter WisdomUsually, GMAT questions that are based on your field of work or interests are simple to comprehend and relatively easy to answer correctly. But what happens when, say, an engineer gets a question based on psychiatry? Is he or she bound to fail? No.

Remember that the GMAT offers a level playing field for test takers from different backgrounds – it doesn’t matter whether your major was literature or physics. If you feel lost on a question about renaissance painters, remember that the guy next to you is lost on the problem involving planetary systems.

So how can you successfully handle GMAT questions on any topic? By sticking to the basics. The logic and reasoning required to answer these questions will stay the same no matter which field the information in the question stem comes from.

To give an example of this, let’s today take a look at a GMAT question involving psychoanalysis:

Studies in restaurants show that the tips left by customers who pay their bill in cash tend to be larger when the bill is presented on a tray that bears a credit-card logo. Consumer psychologists hypothesize that simply seeing a credit-card logo makes many credit-card holders willing to spend more because it reminds them that their spending power exceeds the cash they have immediately available.

Which of the following, if true, most strongly supports the psychologists’ interpretation of the studies? 

(A) The effect noted in the studies is not limited to patrons who have credit cards. 
(B) Patrons who are under financial pressure from their credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.
(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.
(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.
(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

Let’s break down the argument:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo.

Why would that be? Why would there be a difference in customer behavior when the tray has no logo from when the tray has a credit card logo? Psychologists’ hypothesize that seeing a credit-card logo reminds people of the spending power given by the credit card they carry (and that their spending power exceeds the actual cash they have right now).

The question asks us to support the psychologists’ interpretation. And what is the psychologists’ interpretation of the studies? It is that seeing a logo reminds people of their own credit card status. Say we change the argument a little by adding a line:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo. Patrons under financial pressure from credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.

Now, does the psychologists’ interpretation make even more sense? The psychologists’ interpretation is only that “seeing a logo reminds people of their own credit card status.” The fact “that their spending power exceeds the cash they have right now” explains the higher tips. If we are given that some customers tip more upon seeing that card logo and some tip less upon seeing it, it makes sense, right? Different people have different credit card obligation status, hence, people are reminded of their own card obligation status and they tip accordingly.

Answer choice B increases the probability that the psychologists’ interpretation is true because it tells you that in the cases of very high credit card obligations, customers tip less. This is what you would expect if the psychologists’ interpretation were correct.

In simpler terms, the logic here is similar to the following situation:

A: After 12 hours of night time sleep, I can’t study.
B: Yeah, because your sleep pattern is linked to your level of concentration. After a long sleep, your mind is still muddled and lazy so you can’t study.
A: After only 4 hrs of night time sleep, I can’t study either.

Does B’s theory make sense? Sure! B’s theory is that “sleep pattern is linked to level of concentration.” If A sleeps too much, her concentration is affected. If she sleeps too little, again her concentration is affected. So B’s theory certainly makes more sense.

Let’s now review answer choice E since it tends to confuse people:

(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

This option supports the hypothesis that credit card logos remind people of their own card – not of their card obligations. The psychologists’ interpretation talks about the logo reminding people of their card status (high spending power or high obligations). Hence, this option is not correct.

Now let’s examine the rest of the answer choices to see why they are also incorrect:

(A) The effect noted in the studies is not limited to patrons who have credit cards.

This argument is focused only on credit cards, not on credit cards and their logos, so this is irrelevant.

(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.

This option questions the validity of the psychologists’ interpretation. Hence, this is also not correct.

(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.

This argument deals with people who have credit cards but are tipping by cash, hence this is also irrelevant.

Therefore, our answer is B.

We hope you see that if you approach GMAT questions logically and stick to the basics, it is not hard to interpret and solve them, even if they include information from an unfamiliar field.

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 Find the Maximum Distance Between Points on a 3D Object

Quarter Wit, Quarter WisdomHow do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?

A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)

There are various different diagonals in a rectangular solid. Look at the given figure:
MaxDistRectangularSolid

 

 

 

 

 

BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?

The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.

Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.

In our question we know that:

l = 10 inches
w = 10 inches
h = 5 inches

Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.

DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:

DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125

Going back to triangle DHB, we can now say that:

BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15

Thus, our answer to this question is A.

Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:

The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?

(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15

Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:
MaxDistanceCylinder

 

 

 

 

 

 

The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:

Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5

Thus, our answer is C.

In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.

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 VII

Quarter Wit, Quarter WisdomWe have seen a number of posts on divisibility, odd-even concepts and perfect squares. Individually, each topic has very simple concepts but when they all come together in one GMAT question, it can be difficult to wrap one’s head around so many ideas. The GMAT excels at giving questions where multiple concepts are tested. Let’s take a look at one such Data Sufficiency question today:

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

1) When p is divided by 8, the remainder is 5.
2) x – y = 3

This Data Sufficiency question has a lot of information in the question stem.  First, we need to sort through this information before we move on to the statements.

We know that p, x and y are positive integers. y is an unknown odd number, so it can be written in the form 2n + 1. We also know that p = x^2 + y^2.

Because y is written in the form 2n + 1, y^2 can be written as:

y^2 =(2n + 1)^2

y^2 = 4n^2 + 4n + 1

y^2 = 4n(n + 1) + 1

An interesting thing to note here is that one case of n and (n+1) will be odd and the other will be even. In every case, n(n + 1) is even. Therefore, y^2 is 1 more than a multiple of 8. In other words, we can write it as y^2 = 8m + 1.

Now we can say p = x^2 + 8m + 1.

With this in mind, is x divisible by 4? Let’s examine the statements to find out:

Statement 1: When p is divided by 8, the remainder is 5.

Because y^2 = 8m + 1, we can see that when y^2 is divided by 8, the remainder will be 1. Therefore, to get a remainder of 5 when p is divided by 8, when x^2 is divided by 8, we should get a remainder of 4.

Now we know that x^2 can be written in the form 8a + 4 (i.e. we can make “a’” groups of 8 each and have 4 leftover).

x^2 = 4*(2a + 1)

So x = 2 * √(an odd number)

Note that square root of an odd number will be an odd number only. If there is no 2 in the perfect square, obviously there was no 2 in the number, too.

So, x = 2 * some other odd number, which means x will be a multiple of 2, but not of 4 definitely. This statement alone is sufficient.

Now let’s look at the next statement:

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (an even – an odd = an odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since we have no constraints on p.

This statement alone is not sufficient to answer the question. Therefore, our answer is A.

Test takers might feel that not every step in this solution is instinctive. For example, how do we know that we should put y^2 in the form 4n(n+1) + 1? Keep the target in mind – we know that we need to find whether x is divisible by 4. Hence, try to get everything in terms of multiples of 4 + a remainder.

See you next week!

(For more advanced number properties on the GMAT, check out Parts IIIIIIIV, V and VI 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!

How NOT to Write the Equation of a Line on the GMAT

Quarter Wit, Quarter WisdomA question brought an interesting situation to our notice. Let’s start by asking a question: How do we write the equation of a line? There are two formulas:

y = mx + c (where m is the slope and c is the y-intercept)
and
yy1 = m * (xx1) [where m is the slope and (x1,y1) is a point on the line]

We also know that m = (y2y1)/(x2x1) – this is how we find the slope given two points that lie on a line. The variables are x1, y1 and x2, y2, and they represent specific values.

But think about it, is m = (y2y)/(xx1) really the equation of a line? Let’s further clarify this idea using a GMAT practice question:

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x + y =

(A) 3.5
(B) 7 
(C) 8
(D) 10
(E) 14

We have been given that the line passes through (0, 0) and has a slope of 2. We can find the equation of the line from this information.

y = mx + c
y = 2x + 0 (Since the line passes through (0, 0), its y-intercept is 0 – when x is 0, y is also 0.)
y = 2x

Since we are given two other points, (3, y) and (x, 4), on the line and we have a slope of 2, many test-takers will be tempted to make another equation for the line using this information.

(4 – y)/(x – 3) = 2
(4 – y) = 2*(x – 3)
Thus, 2+ y = 10

Here, test-takers will use the two equations to solve for x and y and get x = 5/2 and y = 5.

After adding x and y together, they then wonder why 7.5 is not one of the answer choices. If this were an actual GMAT question, it is quite likely that 7.5 would have been one of the options. So all in all, the test-taker would not even have realized that he or she made a mistake, and would choose 7.5 as the (incorrect) answer.

The error is conceptual here. Note that the equation of the line, 2x + y = 10, is not the same as the equation we obtained above, y = 2x. They represent two different lines, but we have only a single line in the question. So which is the actual equation of that line?

To get the second equation, we have used m = (y2y)/(xx1). But is this really the equation of a line? No. This formula doesn’t have y and x, the generic variables for the x– and y-coordinates in the equation of a line.

To further clarify, instead of x and y, try using the variables a and b in the question stem and see if it makes sense:

“In the coordinate plane, line k passes through the origin and has slope 2. If points (3, a) and (b, 4) are on line k, then a + b =”

You can write (4 – a)/(b – 3) = 2 and this would be correct. But can we solve for both a and b here? No – we can write one of them in terms of the other, but we can’t get their exact values.

We know a and b must have specific values. (3, a) is a point on the line y = 2x. For x = 3, the value of of the y-coordinate, a, will be y = 2*3 = 6. Therefore, a = 6.

(b, 4) is also on the line y = 2x. So if the y-coordinate is 4, the x-coordinate, b, will be 4 = 2b, i.e. b = 2. Thus, a + b = 6 + 2 = 8, and our answer is C.

This logic remains the same even if the variables used are x and y, although test-takers often get confused because of it. Let’s solve the question in another way using the variables as given in the original question.

Recall what we have learned about slope in the past. If the slope of the line is 2 and the point (0, 0) lies on the line, the value of y – if point (3, y) also lies on the line – will be 6 (a slope of 2 means a 1-unit increase in x will lead to a 2-unit increase in y).

Again, if point (x, 4) lies on the line too, an increase of 4 in the y-coordinate implies an increase of 2 in the x-coordinate. So x will be 2, and again, x + y = 2 + 6 = 8.

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!

3 Formats for GMAT Inequalities Questions You Need to Know

Quarter Wit, Quarter WisdomAs if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

  1. Must Be True
  2. Could Be True
  3. Complete Range

Case 1: Must Be True
If |-x/3 + 1| < 2, which of the following must be true?
(A) x > 0
(B) x < 8
(C) x > -4
(D) 0 < x < 3
(E) None of the above

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2
(1/3) * |x – 3| < 2
|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0
Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8
Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4
Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3
Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Case 2: Could Be True
If −1 < x < 5, which is the following could be true?
(A) 2x > 10
(B) x > 17/3
(C) x^2 > 27
(D) 3x + x^2 < −2
(E) 2x – x^2 < 0

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10
x > 5
No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3
x > 5.67
No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27
x^2 – 27 > 0
x > 3*√(3) or x < -3*√(3)
√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

(Details on how to solve such inequalities are discussed here.)

(D) 3x + x^2 < −2
x^2 + 3x + 2 < 0
(x + 1)(x + 2) < 0
-2 < x < -1
No values of x will lie between -2 and -1, so this also cannot be true.

(E) 2x – x^2 < 0
x * (x – 2) > 0
x > 2 or x < 0
If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.

Case 3: Complete Range
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?
(A) 0 < |x| < ½
(B) |x| > ½
(C) -½ < x < 0 or ½ < x
(D) x < -½ or 0 < x < ½
(E) x < -½ or x > 0

We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice.

We are given that x^3 – 4x^5 < 0. This inequality can be solved to:

x^3 ( 1 – 4x^2) < 0
x^3*(2x + 1)*(2x – 1) > 0
> 1/2 or -1/2 < x < 0

This is our universe of the values of x. It is given that all values of x lie in this range.

Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.

We hope these practice problems will help you become able to distinguish between the three cases 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!

Using Special Formats on GMAT Variable Problems

Quarter Wit, Quarter WisdomIn today’s post, we will discuss some special formats when we assume variables on the GMAT. These will allow us to minimize the amount of manipulations and calculations that are required to solve certain Quant problems.

Here are some examples:

An even number: 2a
Logic: It must be a multiple of 2.

An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.

Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).

Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.

Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.

A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.

Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.

Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.

Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:

The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Let’s have the four consecutive odd numbers be the following, where “a” is any integer: (2a – 3), (2a – 1), (2a + 1), (2a + 3)

The sum of these numbers is: (2a – 3) + (2a – 1) + (2a + 1) + (2a + 3) = 8a

Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)

The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b

Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.

We are given that the sum 8a is equal to the sum 6b.

8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.

With this in mind, possible solutions for a and b are:

a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
etc.

We are also given that the middle term of the even numbers is greater than 101 and less than 200.

So 101 < 2b < 200, i.e. 50.5 < b < 100.

B must be an integer, hence, 51 ≤ b ≤ 99.

Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96

The number of values b can take = (Last term – First term)/Common Difference + 1 = (96 – 52)/4 + 1 = 12

For each of these 12 values of b, there will be a corresponding value of a and, hence, we will get 12 such sequences. Therefore, the answer to our question is A.

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: Beware of Sneaky Answer Choices on the GMAT!

Quarter Wit, Quarter WisdomTest-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

QWQW_11_21

 

 

 

 

 

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that  there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

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!

The Holistic Approach to Absolute Values – Part V

Quarter Wit, Quarter WisdomWe will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

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

Let’s look at the following GMAT question:

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

In this question, we are given the inequality |x – 6| > 3*|x + 2|

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.
QWQW image 2

 

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

QWQW image 1

 

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

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!

The Holistic Approach to Absolute Values – Part IV

Quarter Wit, Quarter WisdomLast week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

  • |x – 3| will be the distance covered by the friend at 3 to reach x.
  • |x + 1| will be the distance covered by the friend at -1 to reach x.
  • |x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.

Today, let’s look at how to solve a more advanced GMAT question using the same logic:

For how many integer values of x, is |x – 5| + |x + 1| + |x|  + |x – 7| < 15?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x|  + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

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!

The Holistic Approach to Absolute Values – Part III

Quarter Wit, Quarter WisdomA while back, we discussed some holistic approaches to answering absolute value questions. Today, we will enhance our understanding of absolute values with some variations that you might see on the GMAT.

Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.

A quick review:

  • |x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.
  • |x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or  -3.
  • |x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.

Let’s move ahead now and see how we can use these concepts to solve inequalities:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:

QWQW

 

 

 

What will be the total distance at any value of x between these two points?

Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4

Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7

In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).

Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.

Along the same lines, consider a slight variation of this question:

For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.

We hope this logic is clear. We will look at some other variations of this concept 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 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!

Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient

Quarter Wit, Quarter WisdomToday we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?

Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.

A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?

(1) The daily rental rate for a luxury car was $15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday

We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.

There are four variables to consider here:

  1. Number of compact cars rented (this is what we need to find)
  2. Number of luxury cars rented
  3. Daily rental rate of compact cars
  4. Daily rental rate of luxury cars

Let’s examine the information given to us by the statements:

Statement 1: The daily rental rate for a luxury car was $15 higher than the rate for a compact car.

This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:

Statement 2: The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday.

This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.

Now, let’s try to tackle both statements together:

The daily rate for luxury cars is $15 higher than it is for compact cars, and the total rental rates for luxury cars is $105 higher than it is for compact cars. What constitutes this $105? It is the higher rental cost of each luxury car (the extra $15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more  luxury cars were rented, 105 would account for the $15 higher rent of each luxury car and also for the rent of the extra luxury cars.

Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.

We start with what we know:

The total extra money collected by renting luxury cars is $105.

105/15 = 7

Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is $0 (theoretically), the rent of luxury cars is $15 and the extra rent charged will be $105 (7*15 = 105) – this is a valid case.

Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.

Let’s make a slight change to our current numbers to see if they still fit:

Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be $15*8 = $120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is $105 – the $15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be $15/9 = $5/3.

This would mean that the daily rental rate of each luxury car is $5/3 + $15 = $50/3

The total rental cost of luxury cars in this case would be 8 * $50/3 = $400/3

The total rental cost of compact cars in this case would be 17 * $5/3 = $85/3

The difference between the two total rental costs is $400/3 – $85/3 = 315/3 = $105

Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.

The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:

Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be 10*$15 = $150 extra, but it is actually only $105 extra, a difference of $45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be $45/5 = $9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, $105. This is a valid case, too.

Hence, there are two strategies we saw in action today:

  • Tweak the numbers slightly to see if you will get the same results
  • Go for the easy split when choosing numbers to plug in

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 Solve “Unsolvable” Equations on the GMAT

Quarter Wit, Quarter WisdomThe moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Statement 2: x^2 – x – 2 < 0

This, we can easily solve:

x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0

We know how to solve this inequality using the method discussed here.

This this will give us -1 < x < 2.

Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.

To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.

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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!