The post Dreaded Data Sufficiency Questions That Will Test Your Knowledge of Number Properties appeared first on Veritas Prep Blog.

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

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

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

*Statement 1: b^a is negative*

*Statement 2: a^b is negative*

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

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

*Statement 1: b^a is negative*

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

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

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

Let’s look at Statement 2:

*Statement 2: a^b is negative*

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

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

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

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

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

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

If *a* = -1 and* b* = -1, then *a*^*b* = -1 (an integer)

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

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

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

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

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

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]]>The post 3 Ways to Solve a 750+ Level GMAT Question About Irregular Polygons appeared first on Veritas Prep Blog.

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

*(A) **7*

*(B)**10*

*(C) 12*

*(D) 15*

*(E) 16*

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

(n – 2)*180 = sum of all interior angles

(6 – 2)*180 = 720

Each of the 6 angles = 720/6 = 120 degrees

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

**Method 1: Visualization**

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

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

This means Side 10 is parallel to Side 18.

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

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

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

**Method 2: Using Right Triangles**

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

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

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

**Method 3: Using Equilateral Triangles**

First, extend the sides of the hexagon as shown to form a triangle:

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

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

PQ = 10 + 15 + 4 = 29

PR = 10 + BC + 7 = 29

BC = 12 (again, answer choice C)

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

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

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

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]]>The post Quarter Wit, Quarter Wisdom: Using Ingenuity on GMAT Remainder Questions appeared first on Veritas Prep Blog.

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

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

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

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

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

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

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

x = 6Q + 2

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

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

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

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

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

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

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

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

(z – 3)^2 = z^2 – 6z + 9

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

z^2 – 6z + 9 + 8z = z^2 + 2z + 9

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

Let’s look at the second statement:

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

2z = 8a + 2

z = 4a + 1

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

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

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

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

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

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]]>The post Using Parallel Lines and Transversals to Your Advantage on the GMAT appeared first on Veritas Prep Blog.

]]>**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.

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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]]>The post How to Answer GMAT Critical Reasoning Questions Involving Experiments appeared first on Veritas Prep Blog.

]]>The logic of these arguments is always rooted in the notion that we can only trust the results of the experiment if we have a legitimate control group, and there aren’t any other confounding variables that we’ve failed to account for. Spoiler alert: typically in GMAT questions, we will find such confounding variables tainting the experiment’s predictive value.

Imagine, for example, that you’re testing a drug designed to alleviate headaches. You have two groups of subjects: a control group that takes a placebo and an experimental group that receives the drug. The results of the experiment show that the control group has a higher rate of headaches than the group receiving the medication. Time to rejoice, notify the delighted shareholders, and move this drug to market as quickly as possible? Well, maybe.

But now imagine that the control group consisted largely of stressed-out, sleep-deprived college students living near construction sites, and the experiment group consisted of retired yoga instructors. Suddenly we’ve got other variables to contend with. Yes, it’s possible that the effectiveness of the drug is what accounts for the differential in headache incidence between the two groups. But it’s just as likely that other environmental factors are responsible. A good experiment would have controlled for these factors.

The upshot: whenever you see a question that involves an experiment with a control group, always ask yourself if there are variables that the experimenters have failed to account for.

Here’s a good example of such an argument:

*In Colorado subalpine meadows, nonnative dandelions co-occur with a native ﬂower, the larkspur. Bumblebees visit both species, creating the potential for interactions between the two species with respect to pollination. In a recent study, researchers selected 16 plots containing both species; all dandelions were removed from eight plots; the remaining eight control plots were left undisturbed. The control plots yielded significantly more larkspur seeds than the dandelion-free plots, leading the researchers to conclude that the presence of dandelions facilitates pollination (and hence seed production) in the native species by attracting more pollinators to the mixed plots. *

*Which of the following, if true, most seriously undermines the researchers’ reasoning? *

*A) Bumblebees preferentially visit dandelions over larkspurs in mixed plots.*

*B) In mixed plots, pollinators can transfer pollen from one species to another to augment seed production. *

*C) If left unchecked, nonnative species like dandelions q**u**i**c**kly crowd out native species. *

*D) Seed germination is a more reliable measure of a species’ ﬁtness than seed production.*

*E) Soil disturbances can result in fewer blooms, and hence lower seed production. *

This is a classic experiment argument. There are two populations: plots that contain both dandelions and larkspurs, and plots that have had all the dandelions removed, and thus contain only larkspurs. We’re told that the plots containing both types of flowers produced more larkspur seeds than the plots containing only larkspurs, thus validating the contention that the presence of dandelions has a positive benefit on larkspur seed yields.

Fortunately, the GMAT is pretty predictable. If we’re trying to weaken the conclusion derived from an experiment comparing two populations – a control group and an experimental group – we’re looking for a confounding variable. The initial hypothesis is that the presence of dandelions promotes seed production in larkspurs. An alternative hypothesis is that an environmental factor we haven’t yet considered accounts for the differential in larkspur seed production in the two groups, so that’s what we’re on the lookout for when we examine each of the answer choices.

A) Which flower bees prefer sheds no light on the validity of the experiment. A is out.

B) This answer option would be entirely consistent with the hypothesis that dandelions promote larkspur seed production. We’re trying to weaken the argument. B is also out.

C) This answer choice makes no sense. We’ve already been told that the plots containing both types of flower produce more larkspur seeds – we never want to contradict a premise. C is no good.

D) This tells us nothing about whether it is the presence of dandelions that’s helping promote larkspur seed production. D gets kicked to the curb.

E) If removing the dandelions *disrupts the soil*, perhaps it’s the disrupted soil, rather than the absence of dandelions, that accounts for the lower larkspur production in the plots where the dandelions have been removed. We’ve got our confounding variable – E is the answer.

Takeaway: On Critical Reasoning questions on the lookout for the tainted experiment. If you’re trying to weaken an argument regarding an experiment containing a control group and an experimental group, the key will be determining which answer choice provides a confounding variable, and thus, an alternative explanation for the conclusion given.

*Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on Facebook, YouTube, Google+ and Twitter!*

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

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]]>The post Learn How to Begin a GMAT Problem by Focusing on Keywords in the Question Stem appeared first on Veritas Prep Blog.

]]>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!

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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]]>The post The Pythagorean Triples Properties You’ll See on the GMAT appeared first on Veritas Prep Blog.

]]>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.

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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]]>The post Select Your Section Order on the New GMAT appeared first on Veritas Prep Blog.

]]>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.

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:

- Analytical Writing Assessment, Integrated Reasoning, Quantitative, Verbal (original order)
- Verbal, Quantitative, Integrated Reasoning, Analytical Writing Assessment
- 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.

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.

**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.)

**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.

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.

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 Facebook, YouTube, Google+, and Twitter for the latest in test prep and MBA admissions news.*

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]]>The post Using “Few” vs. “A Few” vs. “Quite a Few” in a GMAT Verbal Question appeared first on Veritas Prep Blog.

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

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

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

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

Here are a few other simple examples:

- A few people think that red wine is healthy.
- This implies some people think that red wine is healthy.

- Few people think that red wine is healthy.
- This implies only very few people, a very small number, think that red wine is healthy; most think that it is not.

- Quite a few people think that red wine is healthy.
- This implies many people, a large number, think that red wine is healthy.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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]]>The post How to Quickly Interpret Ranges of Variables in GMAT Questions appeared first on Veritas Prep Blog.

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

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

*(A) -8 < z < 4*

*(B) z > 2*

*(C) z > -8*

*(D) z < 4*

*(E) None of the above*

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

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

*(A) -8 < z < 4*

We know that z may be more than 4, so this range does not hold true for all possible values of z.

*(B) z > 2*

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

*(C) z > -8*

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

*(D) z < 4*

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

Our answer is C.

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

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

*(A) Her weight is 125 pounds.*

* (B) Her weight is more than 124 pounds.*

* (C) Her weight is less than 127 pounds.*

* (D) Her weight is more than 110 pounds.*

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

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

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

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

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

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

(i) 1/54

(ii) 1/23

(iii) 1/12

*(A) Only (i)*

* (B) Only (ii)*

* (C) (i) and (ii)*

* (D) (ii) and (iii)*

* (E) (i), (ii) and (iii)*

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

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

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

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

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

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

*(A) Her weight is 112 pounds.*

* (B) Her weight is 124 pounds.*

* (C) Her weight is 135 pounds.*

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

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

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

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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