# Why We Need to Redraw GMAT Geometry Figures Many test takers, though good at Math find Data Sufficiency difficult. They are much more used to the straight forward Problem Solving pattern. The very principles behind the two question types are very different.

In Problem Solving questions, our target is to find just one solution. For example, when we have questions involving percentages, we assume some values and get the answer. No matter what values we assume, we will always get the same answer as long as the integrity of the data is maintained.

In Data Sufficiency questions, our target is to find multiple possible solutions  after using all the given data and arrive at answer (E). If we are unable to find more than 1 solution using either statement (1) and/or statement (2), we arrive at answers (A), (B), (C) or (D).

The aim is diametrically opposite in the two cases. Therefore, our strategies in the two cases would also be different and they are. Consider Geometry questions with figures in them. In Problem Solving questions, we try to make the figures as symmetrical as possible under the given constraints. With symmetrical figures, it is easier to get an answer. One answer is all we need.

In Data Sufficiency questions, we try to make the figures as extreme as possible. Only the given data should hold in such a figure and no symmetry should exist in the other dimensions. Only then will we be able to really figure out whether the given information is enough to arrive at a unique answer.

Let’s explain this using two examples:

Problem Solving Question

PSvsDSQuesPS1.jpg ********************************

In the figure above, the area of square PQRS is 64. What is the area of triangle QRT?

(A) 48
(B) 32
(C) 24
(D) 16
(E) 8

This is a Problem Solving question.

All we are given is that PQRS is a square. Note that the location of point T is not defined. It is just any point on side PS. We can place it anywhere we like as long as it is on PS. At what point will it be easy for us to calculate the area of triangle QRT? Of course, T could be the middle point of PS (bringing in symmetry) and we could calculate the area of the triangle or we could make it coincide with S so that QRT is a right triangle half of square PQRS. Then, the area of triangle QRT will simply be half of 64, i.e. 32.

Note that we don’t necessarily need to do this. We can assume T to be a random point, drop an altitude from T to QR, find that the length of the altitude will be same as the side of the square, find that side of the square will be √(64) = 8 and area of triangle QRT will be (1/2)*8*8 = 32

We will arrive at the same answer of course! But, assuming a better position for point T (but only because it is not defined) will cut the calculations and help us arrive directly at 32 from 64.

Data Sufficiency Question

PSvsDSQuesDS1.jpg ********************************

If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

Statement 1: Angle ABD = 60°
Statement 2: AC = 12

Looking at the figure, many test takers are tempted to think that the altitude AD will bisect BC. Note that that may not be the case.

According to the data given in the question stem alone, the figure could very well look something like this:

PSvsDSQuesDS2.jpg ********************************

All we know is that ADC is a right angle and the length of the altitude is 6. We don’t know whether any of the sides are equal, etc. Hence, it is a good idea to redraw the figure with extreme proportions – one side much greater than the other.

Now we can use the given statements to re-adjust the proportions.

Area of triangle ABC = (1/2)*AD*BC

We know that AD is 6. But we don’t know BC. Let’s examine each of the statements separately.

Statement 1: Angle ABD = 60°

This statement tells us that triangle ABD is a 30-60-90 triangle. Knowing the length of AD will give us the length of the other two sides too. But here is the problem – to know BC, we need to know length of CD too. That we cannot find from this statement alone. This statement alone is not sufficient to answer the question.

Statement 2: AC = 12

We know that ADC is a right angled triangle. Knowing AC and AD, we can find the length of CD using Pythagorean Theorem. But we cannot find BD using this statement and that is needed to get the length of BC. This statement alone is also not sufficient to answer the question.

Using both statements, we can find the lengths of both BD and CD, and hence, can find the length of BC. This will give us the area of the triangle. Therefore, our answer is C.

Note here that if we mistakenly assume that D is the mid point of BC, we might come to the conclusion that each statement alone is sufficient and might mark the answer as D, instead of C. Hence, it is a good idea to redraw the given figure in a Data Sufficiency question to ensure that it has as little symmetry as possible.

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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!

# Breaking Down the Scale Method for Weighted Averages Before you dive into this post, make sure you are are familiar with the Scale Method for weighted averages, which we have discussed in previous posts.

We know that the scale formula of weighted averages is the following:

w1/w2 = (A2 – Aavg)/(Aavg – A1)

One point of confusion for many test takers regarding this formula is figuring out what A1, A2, w1 and w2 actually are.

Here is the simple answer: they can be anything. You can choose to set up the solution as you want. The only thing is that it must be consistent across. A1 and w1 could be the parameters of either solution; A2 and w2 will be the parameters of the other solution. We could also work with the concentration of either ingredient of the solution. We will illustrate this point with an example GMAT question:

A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

(A) 4/3
(B) 5/3
(C) 7/3
(D) 8/3
(E) 10/3

Now, we have been given two solutions that we have to mix:

1. A container holding 4 quarts of alcohol and 4 quarts of water
2. Water (which means it has no alcohol in it)

When these solutions are mixed together, they give us a mixture that is 3 parts alcohol to 5 parts water by volume.

So, what are A1, w1, A2, w2 and Aavg? We can work with the concentration of either alcohol or water. Let’s first see how we can work with the concentration of water:

Method 1:
A1 is the concentration of water in the solution of 4 quarts of alcohol and 4 quarts of water. So A1 = 4/8.
w1 is the volume of this solution.
A2 is the concentration of water in the solution of water only. So A2 = 8/8 (we want to write this in the same format that we write A1 in.)
w2 is the volume of this solution.
Aavg is the concentration of water in the final solution i.e. 5/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (8/8 – 5/8)/(5/8 – 4/8)
w1/w2 = 3/1

So 3 parts of the solution with alcohol and water should be mixed with 1 part of pure water.

Method 2:
A1 is the concentration of water in pure water. So A1 is 8/8
w1 is the volume of this solution.
A2 is the concentration of water in the solution of 4 quarts alcohol and 4 quarts water. So A2 is 4/8
w2 is the volume of this solution.
Aavg is the concentration of water in the final solution i.e. 5/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (4/8 – 5/8)/(5/8 – 8/8)
w1/w2 = 1/3

So 1 part of water should be mixed with 3 parts of the solution with alcohol and water (same result as above).

Now we will see how to work with the concentration of alcohol. Of course the result will be the same.

Method 3:
A1 is the concentration of alcohol in the solution of 4 quarts alcohol and 4 quarts water. So A1 is 4/8.
w1 is the volume of this solution.
A2 is the concentration of alcohol in the solution of water only. So A2 is 0/8 (to write in the same way as above)
w2 is the volume of this solution.
Aavg is the concentration of alcohol in the final solution i.e. 3/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (0/8 – 3/8)/(3/8 – 4/8)
w1/w2 = 3/1

So 3 parts of the solution with alcohol and water should be mixed with 1 part of pure water (same as above).

Method 4:
A1 is the concentration of alcohol in pure water. So A1 is 0/8
w1 is the volume of this solution.
A2 is the concentration of alcohol in the solution of 4 quarts alcohol and 4 quarts water. So A2 is 4/8.
w2 is the volume of this solution.
Aavg is the concentration of alcohol in the final solution i.e. 3/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (4/8 – 3/8)/(3/8 – 0/8)
w1/w2 = 1/3

So 1 part of pure water should be mixed with 3 parts of the solution with alcohol and water (same result as above).

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!

# 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.

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

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.

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

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 Here 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.

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 We 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!

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 We 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.

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 Today, 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.

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 Today, 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!

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 Today, 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.

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!

# Select Your Section Order on the New GMAT Good news! Starting July 11, 2017 the GMAT will allow you to in which you take the sections of the test (from a menu of three options). This new “Select Section Order” feature gives you more control over your test-day experience and an opportunity to play to your strengths.

The bad news? Now in addition to the 37 Quant questions, 41 Verbal questions, 12 Integrated Reasoning questions, and Essay, you have one more question you have to answer. But don’t stress – here’s an analysis of how to make this important decision:

Most importantly: statistically, the order of the sections on the GMAT does not matter. GMAC ran a pilot program last year and concluded that reordering the sections of the exam had no impact on scores. So there is no way you can make this decision “wrong” – choosing Quant first vs. Verbal first (or vice versa) doesn’t put you at a disadvantage (or give you an advantage). The only impact that this option will have on your score is a psychological one: which order makes you feel like you’re giving yourself the best shot.

Also hugely important: make sure you have a plan well before test day. Select Section Order has great potential to give you confidence on test day, but you don’t want the added stress of one more “big” decision on test day or even the day before. Make your plan at least a week before test day, take your final practice test(s) in the exact order you’ll use on the real thing, and save your decision-making capacity for test questions. A great option for this is the Veritas Prep practice tests, which are currently the only GMAT practice tests in the industry that let you customize the order of your test like the real exam.

### THE ANALYSIS

And now for the ever-important question on everyone’s mind: in what order should I take the sections? Make sure that you recognize that you only have three options:

1. Analytical Writing Assessment, Integrated Reasoning, Quantitative, Verbal (original order)
2. Verbal, Quantitative, Integrated Reasoning, Analytical Writing Assessment
3. Quantitative, Verbal, Integrated Reasoning, Analytical Writing Assessment

Note that you don’t have the option to split up the AWA and IR sections, and that the AWA/IR block comes either first or last: Quant and Verbal will remain adjacent no matter what order you choose, so you can’t plan yourself a nice “break” in between the two.

Also, recognize that all test-takers are different. As there is no inherent, universal advantage to one order versus the other, your decision isn’t so much “Quant vs. Verbal” but rather “stronger subject vs. not-as-strong subject.” You can fill in the names “Quant” and “Verbal” based on your own personal strengths. For this analysis, we’ll use “Stronger” and “Not as Strong” to refer to your choice between Quant/Verbal, and “AWA/IR” as the third category.

### YOU SHOULD TAKE THE AWA LAST

Traditionally, one of the biggest challenges of the GMAT has been related to stamina and fatigue: it’s a long test, and by the end people are worn out. And over the last 5 years, the fast-paced Integrated Reasoning section has also proven a challenge – very few people comfortably finish the IR section, so it’s quite common to be a combination of tired and demoralized heading into the Quant section. Plus, let’s be honest: the IR and AWA scores just don’t matter as much as the Quant/Verbal scores, so if stamina and confidence are potentially limited quantities, you want to use as much of them as possible on the sections that b-schools care about most.

Who should take AWA/IR first?
Non-native speakers for whom the essay will be important. The danger of waiting until all the way at the end of the test to write the essay is that doing so increases the difficulty of writing clearly and coherently: you’ll just be really tired. If you need your AWA to shine and you’re a bit concerned about it as it is, you may want to attack it first.
Not-morning-people with first-thing-in-the-morning test appointments. If you got stuck with a test appointment that’s much earlier than the timeframe when you feel alert and capable, AWA/IR is a good opportunity to spend an hour of extended warmup getting into the day. If you have a later test appointment and still want a warmup, though, you’re better served doing a few practice problems before you head to the test center.

### REASONS TO DO YOUR STRONGER SECTION (Q vs. V) FIRST

1) You like a good “warmup” to get started on a project. At work you typically start the day by responding to casual emails or reading industry news, because you know your most productive/creative/impactful work will come after you’ve taken a bit of time to get your head in the game. Playing to your strength first will let you experience early success so that your mind is primed for the tougher section to come.

2) You want to start with a confidence booster. Test-taking is very psychological – for example, studies show that test results are significantly impacted when examinees are prompted beforehand with reasons that they should perform well or poorly. Getting started with a section that reminds you that “you’re good at this!” is a great way to prime your mind for success and confidence.

3) You need your stronger section to carry your overall score. Those with specific score targets often find that the easiest way to hit them is to max out on their better score, gaining as many points as possible there and then hoping to scrounge up enough on the other section to hit that overall threshold. Doing your strength first may help you hit it while you’re fresh and gather up all those points before you get worn down by other sections. (Be careful, though: elite schools tend to prefer balanced scores to imbalanced scores, so make sure you consider that.)

### REASONS TO DO YOUR WEAKER SECTION (Q vs. V) FIRST

1) You’re a fast starter. If like to hit the ground running on projects or workdays, you may want to deal with your biggest challenge first while you’re freshest and before fatigue sets in.

2) You hate having stress looming on the horizon. Similarly, if you’re the type who always did your homework immediately after school and always pays your bills the day you get them, there mere presence of the challenge waiting you could add stress through the earlier sections. Why not confront it immediately and get it over with?

3) Your test appointment is late in the day. If you’ve been waiting all day to get the test started, you’ve likely been anxious knowing that you have a major event in front of you. Warm up with some easier problems and review in the hour before the test and attack it quickly.

4) You’re retaking the test to specifically improve that section. In some cases, students are told that they can get off the waitlist or will only be considered if they get a particular section score to a certain threshold. If that’s you, turn that isolated section into a 75-minute test followed by a couple hours of formality, instead of forcing yourself to wait for the important part.

5) You crammed for it. We’ve all been there: your biology midterm is at 11am but you have to go to a history class from 9-10:30, and all the while you’re sitting there worried that you’re losing the information you memorized last night. If you’re worried about remembering certain formulas, rules, or strategies, you might as well use them immediately before you get distracted. Note: this does not mean you should cram for the GMAT! But if you did, you may want to apply that short-term memory as quickly as possible.

### CAN’T DECIDE? THE CASE FOR DOING VERBAL FIRST

If the above reasons leave you conflicted, Veritas Prep recommends doing the Verbal section first. The skills required on the Verbal section are largely about focus – noting precision in wording, staying engaged in bland reading passages, switching between a variety of different topics – and focus is something that naturally fades over the course of the test. The ability to take the Verbal section when you’re most alert and able to concentrate is a terrific luxury.
Ultimately it’s best that you choose the order that makes you personally feel most confident, but if you can’t decide, most experts report that they would personally choose Verbal first.

### SUMMARY

Because, statistically, the order of the sections doesn’t really matter, the only thing that matters with Select Section Order is doing what makes you feel most confident and comfortable. So recognize that you cannot make a bad decision! What’s important is that you don’t let this decision add stress or fatigue to your test day. Make your decision at least 2 practice tests before the real thing, considering the advice above, and then don’t look back. The section selection option is a great way to ensure that your test experience feels as comfortable as possible, so, whatever you choose, believe in your decision and then go conquer the GMAT.

Getting ready to take the GMAT? Prepare for the exam with a computer-adaptive Veritas Prep practice test – the only test in the industry that allows you to practice section selection like the real exam! And as always, be sure to follow us on FacebookYouTubeGoogle+, and Twitter for the latest in test prep and MBA admissions news.

# How to Quickly Interpret Ranges of Variables in GMAT Questions Sometimes 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.

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.

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! If 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!

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 If 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.

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!

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

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 In a, 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.

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 Use Ratios in GMAT Verbal Questions I’ve written in the past about the GMAT’s tendency to use simple math concepts in the context of a Critical Reasoning question. One instance of this phenomenon is the test’s predilection for incorporating ratios in the Verbal section. It makes sense for the question-writers to do this. If we think about the types of core concepts you’re likely to encounter in your future MBA program: output/worker or price/earnings, etc., simple ratios are inescapable.

Here’s all we need to know:

• If the numerator increases and the denominator remains constant, the ratio will increase.
• If the denominator increases and the numerator remains constant, the ratio will decrease.

From this, we can also intuit that if the ratio doubled and the denominator remained constant, the numerator must have doubled. And if the ratio doubled and the numerator remained constant, the denominator must have been halved. Pretty simple, right? For whatever reason, these concepts tend not to produce any difficulty in the Quantitative section when test-takers are expecting them, but cause all sorts of problems when they crop up in Verbal questions. Let’s see an example.

That the application of new technology can increase the productivity of existing coal mines is demonstrated by the case of Tribnia’s coal industry. Coal output per miner in Tribnia is double what it was five years ago, even though no new mines have opened.

Which of the following can be properly concluded from the statement about coal output per miner in the passage?

A) If the number of miners working in Tribnian coal mines has remained constant in the past five years, Tribnia’s total coal production has doubled in that period of time.
B) Any individual Tribnian coal mine that achieved an increase in overall output in the past five years has also experienced an increase in output per miner.
C) If any new coal mines had opened in Tribnia in the past five years, then the increase in output per miner would have been even greater than it actually was.
D) If any individual Tribnian coal mine has not increased its output per miner in the past five years, then that mine’s overall output has declined or remained constant.
E) In Tribnia the cost of producing a given quantity of coal has declined over the past five years.

As soon as we see “per” we know we’re dealing with a ratio problem. In this case, we’re discussing coal output per miner. As a ratio, or fraction, this can be expressed as follows: Total Coal Output/Total Number of Miners. Further, we know that this ratio has doubled over the last five years. Employing the logic we used earlier, we now know that because the ratio doubled, if the number of miners (the denominator) remained constant, then the coal output (the numerator) doubled. And we also know that if the coal output (the numerator) remained constant, then the number of miners (the denominator) must have been halved. If we recognize this relationship, the correct answer is going to leap out at us.

1. This is a restatement of the relationship we’ve already documented – namely that if the denominator remained constant, the numerator must have doubled. Clearly, we’ve got our answer. (But it’s still helpful to evaluate why all the wrong answer choices are incorrect, something you should be doing with every practice problem you attempt.)
2. We can’t deduce what any individual coal mine has achieved based on the output per worker of all the mines in aggregate.
3. Again, there’s no way to know what the productivity level of any mine might have been, let alone a hypothetical new one.
4. If we understand how ratios work, we can see that this is not necessarily true. If the ratio has not increased, there are two possible explanations. First, the numerator has not increased. (This is what’s stated in the answer choice.) Second, the denominator has increased by more than the numerator has increased. Therefore we don’t know that output has declined or remained constant. It could be the case that the number of miners has gone up.
5. This is out of scope. We don’t know what’s happened to the cost of producing coal.

Takeaway: You see plenty of ratios in Critical Reasoning, so make sure you understand that when a ratio changes, it means that either the numerator or denominator (or both) has changed. If you treat these questions as simple Quant problems rather than as abstruse Verbal questions, you’re far less likely to be tripped up.

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By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles by him here.

# Quarter Wit, Quarter Wisdom: Dealing with Complex Word Problems In 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.

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

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!

# GMAT Tip of the Week Is Not A Player, It Just Crushes A Lot On this last day of Hip Hop Month in the GMAT Tip of the Week space, let’s talk about the big picture related to your GMAT score with a nod to one of hip hop’s most notorious B-I-Gs: Big Punisher. The “Big Punisher” of your GMAT score – the item that can take what would have been a great day and leave you walking away from the test sobbing “It’s So Hard” (another Big Pun hit…look it up – he had more than one!) – is poor time management.

On a test-taker’s route to a strong section score, there lie a handful of questions that tempt you to devote several fruitless minutes playing around with equations, calculations, and techniques that aren’t working. A few questions later you look at the clock and realize that even though 90% of the problems have gone well for you, you’re several minutes off your target pace…all because of that one big punisher, the question you should have left alone.

Fortunately, Big Punisher has a mantra for you to keep in mind on test day:

“I’m not a player, I just crush a lot”

Meaning, of course, that you’re not the kind of test-taker who aimlessly plays around with the 3-4 “big punisher” questions that will ruin the time you have left for the others. You quickly identify that no one question is worth taking your whole pacing strategy on (as Snoop would say, “I’m too swift on my toes to get caught up with you hos,” hos, of course, being short for “horribly involved problems that I’ll probably get wrong anyway) and bank that time for the many other problems that you’ll crush…a lot.

Functionally that means this: when you realize that you’re more likely wasting time than progressing toward a right answer, cut your losses and move on so that you save the time for the problems that you will undoubtedly get right…as long as you have a reasonable amount of time for them. You might consider paying homage to Big Pun by using his name as a quick mnemonic for your strategic options:

P: Pick Numbers. If the calculations or algebra you’re performing seems to either be going in circles or getting worse, look back and see if you could simply pick numbers instead. This often works when you’re dealing with variables as parts of the answer choices.

U: Use Answer Choices. Again, if you feel like you’re running in circles, check and see if there are clues in the answer choices or if you can plug them in and backsolve directly.

N: Not Worth My Time. And if a quick assessment tells you that you can’t pick numbers or use answer choices, recognize that this problem simply isn’t worth your time, and blow in a guess. Remember: you’re not a player – you won’t let the test bait you into playing with a single crazy question for more than a minute without a direct path to the finish line – so save the time to focus on crushing a lot of problems that you know you can crush.

On your journey to completing entire GMAT sections on time, heed Big Pun’s warning: don’t stop (to play around with questions you already know you’re not getting right), get it, get it – meaning pick up the pace to have meaningful time to spend on the questions you can get. The biggest punisher of what should be high GMAT scores is poor time management, almost always caused by spending far too long on just a few problems. So remember: you’re not a player on those problems…go out there and crush a lot of the problems you know you can crush.

By Brian Galvin.

# GMAT Tip of the Week: Make J. Cole One of Your Critical Reasoning Role Modelz Today, we’re going to discuss how a seemingly random hip-hop lyric relates to boosting your GMAT Score: “Don’t save her; she don’t want to be saved.” – J. Cole, “No Role Modelz”

One of the most common misconceptions that GMAT examinees have about the exam is that, while on quantitative questions, only one answer can be correct and everything else is wrong, on verbal questions “my wrong answer was good, but maybe not the best.” It is critical to realize that on GMAT verbal questions, exactly one answer is right and the other four are fatally flawed and 100% wrong! Visit a GMAT classroom or a GMAT Club forum thread discussing a Critical Reasoning problem, and you’re almost certain to see/hear students protesting for why their wrong answer could be right. “Well but what if the argument said X, would I be right?” “Well but what if instead of “some” it said “most” would it be right then?”

But students love trying to save an incorrect answer to verbal questions, and in particular Critical Reasoning questions. And to an extent that’s understandable: in high school and college, math was always black and white but in “verbal” classes (literature and the arts, history, philosophy…) as long as you could defend your stance or opinion you could be considered “right” even if that opinion differed from that of your professor. You could “save” an incorrect or unpopular position on an issue by finding a way to justify your stance, and in some cases you were even rewarded for proposing and defending an unorthodox, contrarian viewpoint. But on Critical Reasoning problems, remember this important mantra about incorrect answers:

Don’t save her; she don’t want to be saved.

Consider an example from the Veritas Prep Question Bank:

According to a recent study, employees who bring their own lunches to work take fewer sick days and and are, on average, more productive per hour spent at work than those who eat at the workplace cafeteria. In order to minimize the number of sick days taken by its staff, Boltech Industries plans to eliminate its cafeteria.

Which of the following, if true, provides the most reason to believe that Boltech Industries’ strategy will not accomplish its objective?

A) Boltech’s cafeteria is known for serving a diverse array of healthy lunch options.
B) Because of Boltech’s location, employees who choose to visit a nearby restaurant for lunch will seldom be able to return within an hour.
C) Employees have expressed concern about the cost of dining at nearby restaurants compared with the affordability of the Boltech cafeteria.
D) Employees who bring their lunch from home tend to lead generally healthier lifestyles than those of employees who purchase lunch.
E) Many Boltech employees chose to work for the company in large part because of the generous benefits, such as an on-site cafeteria and fitness center, that Boltech offers.

Less than half of all test-takers get this problem right, in large part because they try to “save” wrong answer choices. The goal of this plan is very clearly stated as “to minimize the number of sick days” but students very frequently pick choices B and E. With B, they try to save it by thinking “but isn’t being away from your desk a long time for a lunch really bad, too?” And the answer may very well be “yes” but the question specifically asks for a reason to think that the strategy will not achieve its objective, and that objective is very clearly stated as pertaining only to sick days.

So as you study, and especially on test day, heed the wisdom of J. Cole. If you fall into the trap of saving answers, tell the GMAT “fool me one time, shame on you; fool me twice can’t put the blame on you.” But most importantly, as you look at Critical Reasoning answer choices, don’t save her. She don’t want to be saved.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on Facebook, YouTubeand Twitter!

By Brian Galvin.

# GMAT Tip of the Week: The Song Remains the Same Welcome back to hip hop month in the GMAT Tip of the Week space, where we’re constantly asking ourselves, “Wait, where have I heard that before?” If you listen to enough hip hop, you’ll recognize that just about every beat or lyric you hear either samples from or derives from another track that came before it (unless, of course, the artist is Ol’ Dirty Bastard, for whom, as his nickname derives,

Biggie’s “Hypnotize” samples directly from “La Di Da Di” (originally by Doug E. Fresh – yep, he’s the one who inspired “The Dougie” that Cali Swag District wants to teach you – and Slick Rick). “Biggie Biggie Biggie, can’t you see, sometimes your words just hypnotize me…” was originally “Ricky, Ricky, Ricky…” And right around the same time, Snoop Dogg and 2Pac just redid the entire song just about verbatim, save for a few brand names.

The “East Coast edit” of Chris Brown’s “Loyal”? French Montana starts his verse straight quoting Jay-Z’s “I Just Wanna Love U” (“I’m a pimp by blood, not relation, I don’t chase ’em, I replace ’em…”), which (probably) borrowed the line “I don’t chase ’em I replace ’em” from a Biggie track, which probably got it from something else. And these are just songs we heard on the radio this morning driving to work…

The point? Hip hop is a constant variation on the same themes, one of the greatest recycling centers the world has ever known.

And so is the GMAT.

Good test-takers – like veteran hip hop heads – train themselves to see the familiar within what looks (or sounds) unique. A hip hop fan often says, “Wait, where I have heard that before?” and similarly, a good test-taker sees a unique, challenging problem and says, “Wait, where have I seen that before?”

And just like you might recite a lyric back and forth in your mind trying to determine where you’ve heard it before, on test day you should recite the operative parts of the problem or the rule to jog your memory and to remind yourself that you’ve seen this concept before.

Is it a remainder problem? Flip through the concepts that you’ve seen during your GMAT prep about working with remainders (“the remainder divided by the divisor gives you the decimals; when the numerator is smaller then the denominator the whole numerator is the remainder…”).

Is it a geometry problem? Think of the rules and relationships that showed up on tricky geometry problems you have studied (“I can always draw a diagonal of a rectangle and create a right triangle; I can calculate arc length from an inscribed angle on a circle by doubling the measure of that angle and treating it like a central angle…”).

Is it a problem that asks for a seemingly-incalculable number? Run through the strategies you’ve used to perform estimates or determine strange number properties on similar practice problems in the past.

The GMAT is a lot like hip hop – just when you think they’ve created something incredibly unique and innovative, you dig back into your memory bank (or click to a jazz or funk station) and realize that they’ve basically re-released the same thing a few times a decade, just under a slightly different name or with a slightly different rhythm.

The lesson?

You won’t see anything truly unique on the GMAT. So when you find yourself stumped, act like the old guy at work when you tell him to listen to a new hip hop song: “Oh I’ve heard this before…and actually when I heard it before in the ’90s, my neighbor told me that she had heard it before in the ’80s…” As you study, train yourself to see the similarities in seemingly-unique problems and see though the GMAT’s rampant plagiarism of itself.

The repetitive nature of the GMAT and of hip hop will likely mean that you’re no longer so impressed by Tyga, but you can use that recognition to be much more impressive to Fuqua.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: When Can You Divide by a Variable? We 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.

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!

# GMAT Tip of the Week: Evolving Your GMAT Quant Score with Help from The Evolution Of Rap If it’s March, it must be Hip Hop Month at the GMAT Tip of the Week space, where this year we’ve been transfixed by Vox’s video on the evolution of rhyme schemes in the rap world.

And if you don’t have the study break time right now, we’ll summarize. While a standard rhyme might have a one-syllable rhyme at the end of each bar (do you like green eggs and HAM, yes I like them Sam I AM), rappers have continued to evolve to the point where nowadays each bar can contain multiple rhyme schemes. Consider Eminem’s “Lose Yourself”:

Snap back to reality, oh there goes gravity
Oh there goes Rabbit he choked, he’s so mad but he won’t
Give up that easy, nope, he won’t have it he knows
His whole back’s to these ropes, it don’t matter he’s dope
He knows that but he’s broke, he’s so stagnant he knows…

Where “gravity,” “Rabbit, he,” “mad but he,” “that easy,” “have it he,” “back’s to these,” “matter he’s,” “that but he’s,” and “stagnant, he” all rhyme with one another, the list of goes/goes/choked/so/won’t/knows/whole/ropes/don’t/dope… keeps that hard “O” sound rhyming consistently throughout, too. And that was 15 years ago…since them, Eminem, Kendrick, and others have continued to build elaborate rhyme schemes that reward those listeners who don’t just listen for the simple rhyme at the end of each bar, but pick up the subtle rhyme flows that sometimes don’t come back until a few lines later.

So what does this have to do with your GMAT score?

One of the most common study mistakes that test-takers make is that they study skills as individual, standalone entities, and don’t look for the subtle ways that the GMAT testmaker can layer in those sophisticated Andre-3000-style combinations. Consider an example of an important GMAT skill, the “Difference of Squares” rule that (x + y)(x – y) = x^2 – y^2. A standard (think early 1980s Sugarhill Gang or Grandmaster Flash) GMAT question might test it in a relatively “obvious” way:

What is the value of (x + y)?

(1) x^2 – y^2 = 0
(2) x does not equal y

Here if you factor Statement 1 you’ll get (x + y)(x – y) = 0, and then Statement 2 tells you that it’s not (x – y) that equals zero, so it must be x + y. This Data Sufficiency answer is C, and the test is essentially just rewarding you for knowing the Difference of Squares.

The GMAT it cares
’bout the Difference of Squares
When there’s squares and subtraction
Put this rule into action

A slightly more sophisticated question (think late 1980s/early 1990s Rob Bass or Run DMC) won’t so obviously show you the Difference of Squares. It might “hide” that behind a square that few people tend to see as a square, the number 1:

If y = 2^(16) – 1, the greatest prime factor of y is:

(A) Less than 6
(B) Between 6 and 10
(C) Between 10 and 14
(D) Between 14 and 18
(E) Greater than 18

Here, many people don’t recognize 1 as a perfect square, so they don’t see that the setup is 2^(16) – 1^(2), which can be factored as:

(2^8 + 1)(2^8 – 1)

And that 2^8 – 1 can be factored again, since 1 remains 1^2:

(2^8 + 1)(2^4 + 1)(2^4 – 1)

And that ultimately you could do it again with 2^4 – 1 if you wanted, but you should know that 2^4 is 16 so you can now get to work on smaller numbers. 2^8 is 256 and 2^4 is 16, so you have:

257 * 17 * 15

And what really happens now is that you have to factor out 257 to see if you can break it into anything smaller than 17 as a factor (since, if not, you can select “greater than 18”). Since you can’t, you know that 257 must have a prime factor greater than 18 (it turns out that it’s prime) and correctly select E.

The lesson here? This problem directly tests the Difference of Squares (you don’t want to try to calculate 2^16, then subtract 1, then try to factor out that massive number) but it does so more subtly, layering it inside the obvious “prime factor” problem like a rapper might embed a secondary rhyme scheme in the middle of each bar.

But in really hard problems, the testmaker goes full-on Greatest of All Time rapper, testing several things at the same time and rewarding only the really astute for recognizing the game being played. Consider:

The size of a television screen is given as the length of the screen’s diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

(A) 2
(B) 4
(C) 16
(D) 38
(E) 40

Now here you KNOW you’re dealing with a geometry problem, and it also looks like a word problem given the television backstory. As you start calculating, you’ll know that you have to take the diagonal of each square TV and use that to determine the length of each side, using the 45-45-90 triangle ratio, where the diagonal = x√2. So the length of a side of the smaller TV is 19/√2 and the length of a side of the larger TV is 21/√2.

Then you have to calculate the area, which is the side squared, so the area of the smaller TV is (19/√2)^2 and the area of the larger TV is (21/√2)^2. This is starting to look messy (Who knows the squares for 21 and 19 offhand? And radicals in denominators never look fun…) UNTIL you realize that you have to subtract the two areas. Which means that your calculation is:

(21/√2)^2 – (19/√2)^2

This fits perfectly in the Difference of Squares formula, meaning that you can express x^2 – y^2 as (x + y)(x – y). Doing that, you have:

[(21 + 19)/√2][(21-19)/√2]

Which is really convenient because the math in the numerators is easy and leaves you with:

(40/√2) * (2/√2)

And when you multiply them, the √2 terms in the denominators square out to 2, which factors with the 2 in the numerator of the right-side fraction, and everything simplifies to 40. And then, in classic “oh this guy’s effing GOOD” hip-hop style (like in the Eminem lyric “you’re witnessing a massacre like you’re watching a church gathering take place” and you realize that he’s using “massacre” and “mass occur” – the church gathering taking place – simultaneously), you realize that you should have seen it coming all along. Because when you subtract the area of one square minus the area of another square you’re LITERALLY taking the DIFFERENCE of two SQUARES.

So what’s the point?

Too often people study for the GMAT like they’d listen to 1980s rap. They expect the Difference of Squares to pair nicely at the end of an Algebra-with-Exponents bar, and the Isosceles Right Triangle formula to pair nicely with a Triangle question. They learn skills in distinct silos, memorize their flashcards in nice, tidy sets, and then go into the test and realize that they’re up against an exam that looks a lot more like a 2017 mixtape with layers of rhyme schemes and motives.

You need to be prepared to use skills where they don’t seem to obviously belong, to jot down and rearrange your scratchwork, label your unknowns, etc., looking for how you might reposition the math you’re given to help you bring in a skill or concept that you’ve used countless times, just in totally different contexts. The GMAT testmaker has a much more sophisticated flow than the one you’re likely studying for, so pay attention to that nuance when you study and you’ll have a much better chance of keeping your score 800.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on Facebook, YouTubeand Twitter!

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: When a Little Information is Enough to Solve a GMAT Problem We 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.

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 We 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.

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.

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 We 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!

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) A 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!

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!

# GMAT Tip of the Week: Keep Your GMAT Score Safe from the Bowling Green Massacre The hashtag of the day is #bowlinggreenmassacre, inspired by an event that never happened. Whether intentionally or accidentally (we’ll let you and your news agency of choice decide which), White House staffer Kellyanne Conway referenced the “event” in an interview, inspiring an array of memes and references along the way.

Whatever Ms. Conway’s intentions (or lack thereof; again we’ll let you decide) with the quote, she is certainly guilty of inadvertently doing one thing: she didn’t likely intend to help you avoid a disaster on the GMAT, but if you’re paying attention she did.

Your GMAT test day does not have to be a Bowling Green Massacre!

Here’s the thing about the Bowling Green Massacre: it never happened. But by now, it’s lodged deeply enough in the psyche of millions of Americans that, to them, it did. And the same thing happens to GMAT test-takers all the time. They think they’ve seen something on the test that isn’t there, and then they act on something that never happened in the first place. And then, sadly, their GMAT hopes and dreams suffer the same fate as those poor souls at Bowling Green (#thoughtsandprayers).

Here’s how it works:

The Quant Section’s Bowling Green Massacre
On the Quant section, particularly with Data Sufficiency, your mind will quickly leap to conclusions or jump to use a rule that seems relevant. Consider the example:

What is the perimeter of isosceles triangle LMN?

(1) Side LM = 4
(2) Side LN = 4√2

A. Statement (1) ALONE is sufficient, but statement (2) alone is insufficient
B. Statement (2) ALONE is sufficient, but statement (1) alone is insufficient
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient

When people see that square root of 2, their minds quickly drift back to all those flash cards they studied – flash cards that include the side ratio for an isosceles right triangle: x, x, x√2. And so they then leap to use that rule, inferring that if one side is 4 and the other is 4√2, the other side must also be 4 to fit the ratio and they can then calculate the perimeter. With both statements together, they figure, they can derive that perimeter and select choice C.

But think about where that side ratio comes from: an isosceles right triangle. You’re told in the given information that this triangle is, indeed, isosceles. But you’re never told that it’s a right triangle. Much like the Bowling Green Massacre, “right” never happened. But the mere suggestion of it – the appearance of the √2 term that is directly associated with an isosceles, right triangle – baits approximately half of all test-takers to choose C here instead of the correct E (explanation: “isosceles” means only that two sides match, so the third side could be either 4, matching side LM, or 4√2, matching side LN).

Your mind does this to you often on Data Sufficiency problems: you’ll limit the realm of possible numbers to integers, when that wasn’t defined, or to positive numbers, when that wasn’t defined either. You’ll see symptoms of a rule or concept (like √2 leads to the isosceles right triangle side ratio) and assume that the entire rule is in play. The GMAT preys on your mind’s propensity for creating its own story when in reality, only part of that story really exists.

The Verbal Section’s Bowling Green Massacre
This same phenomenon appears on the Verbal section, too – most notably in Critical Reasoning. Much like what many allege that Kellyanne Conway did, your mind wants to ascribe particular significance to events or declarations, and it will often exaggerate on you. Consider the example:

About two million years ago, lava dammed up a river in western Asia and caused a small lake to form. The lake existed for about half a million years. Bones of an early human ancestor were recently found in the ancient lake-bottom sediments that lie on top of the layer of lava. Therefore, ancestors of modern humans lived in Western Asia between two million and one-and-a-half million years ago.

Which one of the following is an assumption required by the argument?

A. There were not other lakes in the immediate area before the lava dammed up the river.
B. The lake contained fish that the human ancestors could have used for food.
C. The lava that lay under the lake-bottom sediments did not contain any human fossil remains.
D. The lake was deep enough that a person could drown in it.
E. The bones were already in the sediments by the time the lake disappeared.

The key to most Critical Reasoning problems is finding the conclusion and knowing EXACTLY what the conclusion says – nothing more and nothing less. Here the conclusion is the last sentence, that “ancestors of modern humans lived” in this region at this time. When people answer this problem incorrectly, however, it’s almost always for the same reason. They read the conclusion as “the FIRST/EARLIEST ancestors of modern humans lived…” And in doing so, they choose choice C, which protects against humans having come before the ones related to the bones we have.

“First/earliest” is a classic Bowling Green Massacre – it’s a much more noteworthy event (“scientists have discovered human ancestors” is pretty tame, but “scientists have discovered the FIRST human ancestors” is a big deal) that your brain wants to see. But it’s not actually there! It’s just that, in day to day life, you’d rarely ever read about a run-of-the-mill archaeological discovery; it would only pop up in your social media stream if it were particularly noteworthy, so your mind may very well assume that that notoriety is present even when it’s not.

In order to succeed on the GMAT, you need to become aware of those leaps that your mind likes to take. We’re all susceptible to:

• Assuming that variables represent integers, and that they represent positive numbers
• Seeing the symptoms of a rule and then jumping to apply it
• Applying our own extra superlatives or limits to conclusions

So when you make these mistakes, commit them to memory – they’re not one-off, silly mistakes. Our minds are vulnerable to Bowling Green Massacres, so on test day #staywoke so that your score isn’t among those that are, sadly, massacred.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: Solving the Hourglass Puzzle Let’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.

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 Let’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. 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!

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 Some 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.

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!

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!

# GMAT Tip of the Week: Taking the Least Amount of Time to Solve “At Least” Probability Problems In its efforts to keep everyone from getting perfect 800s, the GMAT has two powerful tools to stop you from perfection. For one, it can bait you into wrong answers (with challenging content, tempting trap answers, or a combination thereof). And secondly, it can waste your time, making it look like you need to do a lot of work when there’s a much simpler way.

Fortunately, and contrary to popular belief, the GMAT isn’t “pure evil.” Wherever it provides opportunities for less-savvy examinees to waste their time, it also provides a shortcut for those who have put in the study time to learn it or who have the patience to look for the elevator, so to speak, before slogging up the stairs. And one classic example of that comes with the “at least one” type of probability question.

To illustrate, let’s consider an example:

In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?

(A) 5/17
(B) 12/17
(C) 25/81
(D) 56/81
(E) 4/9

Here, you can first streamline the process along the lines of one of those “There are two types of people in the world: those who _______ and those who don’t _______” memes. Your goal is to determine whether you get a yellow marble, so you don’t care as much about “blue” and “black”…those can be grouped into “not yellow,” thereby giving you only two groups: 8 yellow marbles and 10 not-yellow marbles. Fewer groups means less ugly math!

But even so, trying to calculate the probability of every sequence that gives you one or two yellow marbles is labor intensive. You could accomplish that “not yellow” goal several ways:

First marble: Yellow; Second: Not Yellow
First: Not Yellow; Second: Yellow
First: Yellow; Second: Yellow

That’s three different math problems each involving fractions and requiring attention to detail. There ought to be an easier way…and there is. When a probability problem asks you for the probability of “at least one,” consider the only situation in which you WOULDN’T get at least one: if you got none. That’s a single calculation, and helpful because if the probability of drawing two marbles is 100% (that’s what the problem says you’re doing), then 100% minus the probability of the unfavorable outcome (no yellow) has to equal the probability of the favorable outcome. So if you determine “the probability of no yellow” and subtract from 1, you’re finished. That means that your problem should actually look like:

PROBABILITY OF NO YELLOW, FIRST DRAW: 10 non-yellow / 18 total
PROBABILITY OF NO YELLOW, SECOND DRAW: 9 remaining non-yellow / 17 remaining total

10/18 * 9/17 reduces to 10/2 * 1/17 = 5/17. Now here’s the only tricky part of using this technique: 5/17 is the probability of what you DON’T want, so you need to subtract that from 1 to get the probability you do want. So the answer then is 12/17, or B.

More important than this problem is the lesson: when you see an “at least one” probability problem, recognize that the probability of “at least one” equals 100% minus the probability of “none.” Since “none” is always a single calculation, you’ll always be able to save time with this technique. Had the question asked about three marbles, the number of favorable sequences for “at least one yellow” would be:

Yellow Yellow Yellow
Yellow Not-Yellow Not-Yellow
Yellow Not-Yellow Yellow
Yellow Yellow Not-Yellow
Not-Yellow Yellow Yellow

(And note here – this list is not yet exhaustive, so under time pressure you may very well forget one sequence entirely and then still get the problem wrong even if you’ve done the math right.)

Whereas the probability of No Yellow is much more straightforward: Not-Yellow, Not-Yellow, Not-Yellow would be 10/18 * 9/17 * 8/16 (and look how nicely that last fraction slots in, reducing quickly to 1/2). What would otherwise be a terrifying slog, the “long way” becomes quite quick the shorter way.

So, remember, when you see “at least one” probability on the GMAT, employ the “100% minus probability of none” strategy and you’ll save valuable time on at least one Quant problem on test day.

By Brian Galvin.

# Solving GMAT Geometry Problems That Involve Infinite Figures Sometimes, we come across GMAT geometry questions that involve figures inscribed inside other figures. One shape inside of another shape may not be difficult to work with, but how do we handle problems that involve infinite figures inscribed inside one another? Such questions can unsettle even the most seasoned test takers. Let’s take a look at one of them today:

A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in this way and this process is continued indefinitely. If a side of the first square is 4 cm. Determine the sum of areas of all squares?

(A) 18
(B) 32
(C) 36
(D) 64
(E) None

Now the first thing that might come to our mind is this – how do we mathematically, in the time limit of approximately 2 minutes, calculate areas of infinite squares?

There has to be a formula for this. Recall that we do, in fact, have a formula that calculates the sum of infinite terms – the geometric progression formula! Let’s see if we can use that to find the areas of the squares mentioned in this problem. First, we’ll see if we can find a pattern in the areas of the squares:

Say the side of the outermost square is “s“. The area of the outermost square will be s^2 and half of the side will be s/2. The side of the next square inside this outermost square (the second square) forms the hypotenuse of a right triangle with legs of length s/2 each. Using the Pythagorean Theorem:

Hypotenuse^2 = (s/2)^2 + (s/2)^2 = s^2/2
Hypotenuse = s/√(2)

So now we know the sides of the second square will each equal s/√(2), and the area of the second square will be s^2/2.

Our calculations will be far easier if we note that the diagonal of the second square will be the same length as the side of the outer square. We know that area of a square given diagonal d is d^2/2, so that would directly bring us to s^2/2 as the area of the second square.

The second square and the square inscribed further inside it (the third square) will have the same relation. The area of the third square will be (s^2/2)*(1/2) = s^2/4.

Now we know the area of every subsequent square will be half the area of the outside square. So the total area of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …Each term is half the previous term.

Therefore, the sum of an infinite Geometric Progression where the common ratio is less than 1 is:

Total Sum = a/(1-r)
a: First Term
r: Common Ratio

Sum of areas of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …
Sum of areas of all squares = s^2/(1 – 1/2)
Sum of areas of all squares = 2s^2

Since s is the length of the side of the outermost square, and s = 4 (this fact is given to us by the questions stem), the sum of the areas of all the squares = 2*4^2 = 32 cm^2. Therefore, our answer is B.

We hope you understand how we have used the geometric progression formula to get to our answer. To recap, the sum of an infinite geometric progression is a/(1 – r).

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 How 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: 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: 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

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.

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 We 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.)

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!

# GMAT Tip of the Week: 3 Guiding Principles for Exponent Problems If you’re like many GMAT examinees, you’ve found yourself in this familiar situation. You KNOW the rules for exponents. You know them cold. When you’re multiplying the same base and different exponents, you add the exponents. When you’re taking one exponent to another power, you multiply those exponents. A negative exponent? Flip that term into the denominator. A number to the zero power? You’ve got yourself a 1.

But as thoroughly and quickly as you know those rules, this exponent-based problem in front of you has you stumped. You know what you need to KNOW, but you’re not quite sure what you need to DO. And that’s an ever-important part about taking the GMAT – it’s necessary to know the core rules, facts, and formulas, but it’s also every bit as important to have action items for how you’ll apply that knowledge to tricky problems.

For exponents, there are three “guiding principles” that you should keep in mind as your action items. Any time you’re stuck on an exponent-based problem, look to do one (or more) of these things:

1) Find Common Bases
Most of the exponent rules you know only apply when you’re dealing with two exponents of the same base. When you multiply same-base exponents, you add the exponents; when you divide two same-base exponents, you subtract. And if two exponents of the same base are set equal, then you know that the exponents are equal. But keep in mind – these major rules all require you to be using exponents with the same base! If the GMAT gives you a problem with different bases, you have to find ways to make them common, usually by factoring them into their prime bases.

So for example, you might see a problem that says that:

2^x * 4^2x = 8^y. Which of the following must be true?

(A) 3x = y
(B) x = 3y
(C) y = (3/5)x
(D) x = (3/5)y
(E) 2x^2 = y

In order to apply any rules that you know, you must get the bases in a position where they’ll talk to each other. Since 2, 4, and 8 are all powers of 2, you should factor them all in to base 2, rewriting as:

2^x * (2^2)^2x = (2^3)^y

Which simplifies to:

2^x * 2^4x = 2^3y

Now you can add together the exponents on the left:

2^5x = 2^3y

And since you have the same base set equal with two different exponents, you know that the exponents are equal:

5x = 3y

This means that you can divide both sides by 5 to get x = (3/5)y, making answer choice D correct. But more importantly in a larger context, heed this lesson – when you see an exponent problem with different bases for multiple exponents, try to find ways to get the bases the same, usually by prime-factoring the bases.

2) Factor to Create Multiplication
Another important thing about exponents is that they represent recurring multiplication. x^5, for example, is x * x * x * x * x…it’s a lot of x’s multiplied together. Naturally, then, pretty much all exponent rules apply in cases of multiplication, division, or more exponents – you don’t have rules that directly apply to addition or subtraction. For that reason, when you see addition or subtraction in an exponent problem, one of your core instincts should be to factor common terms to create multiplication or division so that you’re in a better position to leverage the rules you know. So, for example, if you’re given the problem:

2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

(A) 18
(B) 20
(C) 21
(D) 22
(E) 24

You should see that in order to do anything with the left-hand side of the equation, you’ll need to factor the common 2^x in order to create multiplication and be in a position to divide and cancel terms from the right. Doing so leaves you with:

2^x(1 + 2^3) = (6^2)(2^18)

Here, you can simplify the 1 + 2^3 parenthetical: 2^3 = 8, so that term becomes 9, leaving you with:

9(2^x) = (6^2)(2^18)

And here, you should heed the wisdom from above and find common bases. The 9 on the left is 3^2, and the 6^2 on the right can be broken into 3^2 * 2^2. This gives you:

(3^2)(2^x) = (3^2)(2^2)(2^18)

Now the 3^2 terms will cancel, and you can add the exponents of the base-2 exponents on the right. That means that 2^x = 2^20, so you know that x = 20. And a huge key to solving this one was factoring the addition into multiplication, a crucial exponent-based action item on test day.

3) Test Small Numbers and Look For Patterns
Remember: exponents are a way to denote repetitive, recurring multiplication. And when you do the same thing over and over again, you tend to get similar results. So exponents lend themselves well to finding and extrapolating patterns. When in doubt – when a problem involves too much abstraction or too large of numbers for you to get your head around – see what would happen if you replaced the large or abstract terms with smaller ones, and if you find a pattern, then look to extrapolate it. With this in mind, consider the problem:

What is the tens digit of 11^13?

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

Naturally, calculating 11^13 without a calculator is a fool’s errand, but you can start by taking the first few steps and seeing if you establish a pattern:

11^1 = 11 –> tens digit of 1

11^2 = 121 –> tens digit of 2

11^3 = 1331 –> tens digit of 3

And depending on how much time you have you could continue:

11^4 = 14641 –> tens digit of 4

But generally feel pretty good that you’ve established a recurring pattern: the tens digit increases by 1 each time, so by 11^13 it will be back at 3. So even though you’ll never know exactly what 11^13 is, you can be confident in your answer.

Remember: the GMAT is a test of how well you apply knowledge, not just of how well you can memorize it. So for any concept, don’t just know the rules, but also give yourself action items for what you’ll do when problems get tricky. For exponent problems, you have three guiding principles:

1) Find Common Bases
2) Factor to Create Multiplication
3) Test Small Numbers to Find a Pattern

By Brian Galvin.

# How NOT to Write the Equation of a Line on the GMAT A 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.

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 As 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.

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 In 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.

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! Test-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! 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!