The post Quarter Wit, Quarter Wisdom: Solving the Pouring Water Puzzle appeared first on Veritas Prep Blog.

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

Here is the puzzle:

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

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

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

Now think about it:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on 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 How to Answer GMAT Questions That are About an Unfamiliar Topic appeared first on Veritas Prep Blog.

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

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

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

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

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

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

*(B) Patrons who are under financial pressure from their credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.*

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

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

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

Let’s break down the argument:

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

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

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

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

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

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

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

A: After 12 hours of night time sleep, I can’t study.

B: Yeah, because your sleep pattern is linked to your level of concentration. After a long sleep, your mind is still muddled and lazy so you can’t study.

A: After only 4 hrs of night time sleep, I can’t study either.

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

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

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

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

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

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

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

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

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

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

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

Therefore, our answer is B.

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

*Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on 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 Solving GMAT Geometry Problems That Involve Infinite Figures appeared first on Veritas Prep Blog.

]]>*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 = 2*s*^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).

*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 How to Find the Maximum Distance Between Points on a 3D Object appeared first on Veritas Prep Blog.

]]>*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

Thus, our answer is C.

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

**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 Advanced Number Properties on the GMAT – Part VII appeared first on Veritas Prep Blog.

]]>*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 **I**, **II**, **III**, **IV,** **V **and** VI** of this series.)

**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 How NOT to Write the Equation of a Line on the GMAT appeared first on Veritas Prep Blog.

]]>*y* = *mx* + *c* (where *m* is the slope and *c* is the *y*-intercept)

and

*y* – *y*_{1} = *m* * (*x* – *x*_{1}) [where *m* is the slope and (*x*_{1},*y*_{1}) is a point on the line]

We also know that *m* = (*y*_{2} – *y*_{1})/(*x*_{2} – *x*_{1}) – this is how we find the slope given two points that lie on a line. The variables are *x*_{1}, *y*_{1} and *x*_{2}, *y*_{2,} and they represent specific values.

But think about it, is *m* = (*y*_{2} – *y*)/(*x* – *x*_{1}) 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* = 2*x* + 0 (Since the line passes through (0, 0), its *y*-intercept is 0 – when *x* is 0, *y* is also 0.)

*y* = 2*x*

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*x *+ *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, 2*x* + *y* = 10, is not the same as the equation we obtained above, *y* = 2*x*. 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* = (*y*_{2} – *y*)/(*x* – *x*_{1}). 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* = 2*x*. 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* = 2*x*. So if the *y*-coordinate is 4, the *x*-coordinate, *b,* will be 4 = 2*b*, 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.

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]]>The post 3 Formats for GMAT Inequalities Questions You Need to Know appeared first on Veritas Prep Blog.

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

- Must Be True
- Could Be True
- 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 – 4*x*^5 < 0 and the other will be the correct answer choice.

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

*x*^3 ( 1 – 4*x*^2) < 0

*x*^3*(2*x* + 1)*(2*x* – 1) > 0

*x *> 1/2 or -1/2 < *x* < 0

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

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

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

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

The post 3 Formats for GMAT Inequalities Questions You Need to Know appeared first on Veritas Prep Blog.

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]]>Here are some examples:

**An even number: 2a**

Logic: It must be a multiple of 2.

**An odd number: (2a + 1) or (2a – 1)**

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

**Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a**

Logic: One number will be even and the other will be the next odd number (or the other way around).

**Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)**

In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.

**Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)**

In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.

**A prime number: (6a+1) / (6a – 1)**

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

**Three consecutive numbers:**

If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).

Logic: They add up to give 6a.

In a more generic case, we will have: 3a, (3a+1), (3a+2).

This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.

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

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

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

*(A) 12*

*(B) 17*

*(C) 25*

*(D) 33*

*(E) 50*

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

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

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

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

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

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

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

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

a = 3, b = 4;

a = 6, b = 8;

a = 9, b = 12

etc.

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

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

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

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

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

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

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **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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]]>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!

*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 The Holistic Approach to Absolute Values – Part V appeared first on Veritas Prep Blog.

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

Let’s look at the following GMAT question:

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

*(A) 1*

*(B) 3*

*(C) 5*

*(D) 7*

*(E) Infinite*

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

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

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

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

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

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

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

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

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

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

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

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

*For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?*

*(A) 1*

*(B) 3*

*(C) 5*

*(D) 7*

*(E) Infinite*

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a

a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b

b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

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

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

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