The post A 750+ Level GMAT Geometry Question appeared first on Veritas Prep Blog.

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

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

*(A) Sqrt(3) – 1*

*(B) Sqrt(3) + 2*

*(C) (Sqrt(3) – 1)/2*

*(D) (Sqrt(3) + 1)/2*

*(E) 2*(Sqrt(3) + 1)*

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

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

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

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

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

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

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

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

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

Let’s use the pythagorean theorem on triangle BDC:

BD^2 + DC^2 = BC^2

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

X^4/32 + 2 = X^2

X^4 – 32*X^2 + 64 = 0

X^4 – 16X^2 + 8^2 – 16X^2 = 0

(X^2 – 8)^2 – (4X)^2 = 0

(X^2 -8 + 4X) * (X^2 – 8 – 4X) = 0

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

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

x = [-B +- Sqrt(B^2 – 4AC)]/2A

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

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

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

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

To recap:

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

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

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

Getting ready to take the GMAT? We have **free online GMAT seminars** running all the time. And, be sure to follow us on **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: Solving GMAT Critical Reasoning Questions Involving Rates appeared first on Veritas Prep Blog.

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

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

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

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

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

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

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

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

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

Let’s break down the given argument:

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

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

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

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

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

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

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

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

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

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

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

Rate = Distance/Time

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

Rate = Number of fatalities/highway mile driven

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

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

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

Our answer is, therefore, (B).

Getting ready to take the GMAT? We have **free online GMAT seminars** running all the time. And, be sure to follow us on **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: Should You Use the Permutation or Combination Formula? appeared first on Veritas Prep Blog.

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

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

**Permutation:** nPr = n!/(n-r)!

Out of n items, select r and arrange them in r! ways.

**Combination:** nCr = n!/[(n-r)!*r!]

Out of n items, select r.

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

nPr = nCr * r!

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

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

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

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

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

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

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

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

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

Let’s try another one:

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

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

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

Here’s another variation:

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

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

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

Finally, our last problem:

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

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

Total number of ways = 12C3 * 3! ways

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

Getting ready to take the GMAT? We have **free online GMAT seminars** running all the time. And, be sure to follow us on **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 Quarter Wit, Quarter Wisdom: Should You Use the Permutation or Combination Formula? appeared first on Veritas Prep Blog.

]]>The post Quarter Wit, Quarter Wisdom: Keeping an Open Mind in Critical Reasoning appeared first on Veritas Prep Blog.

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

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

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

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

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

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

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

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

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

**Premises:**

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

**Conclusion:** Michael will win a medal in competition.

Read the argument carefully:

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

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

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

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

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

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

**Conclusion:** Michael will win a medal in competition.

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

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

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

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

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

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

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

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

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

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

**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 Quarter Wit, Quarter Wisdom: An Interesting Property of Exponents appeared first on Veritas Prep Blog.

]]>*For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?*

Think about it for a few seconds – could you come up with the answer in the blink of an eye? If yes, great! Close this window and wait for the next week’s post. If no, then read on. There is much to learn today and it is an eye-opener!

Let’s start by jotting down some powers of numbers:

**Power of 2:** 1, 2, 4, 8, 16, 32 …

**Power of 3:** 1, 3, 9, 27, 81, 243 …

**Power of 4:** 1, 4, 16, 64, 256, 1024 …

**Power of 5:** 1, 5, 25, 125, 625, 3125 …

and so on.

Obviously, for every power of 2, when you multiply the previous power by 2, you get the next power (4*2 = 8).

For every power of 3, when you multiply the previous power by 3, you get the next power (27*3 = 81), and so on.

Also, let’s recall that multiplication is basically repeated addition, so 4*2 is basically 4 + 4.

This leads us to the following conclusion using the power of 2:

4 * 2 = 8

4 + 4 = 8

2^2 + 2^2 = 2^3

(2 times 2^2 gives 2^3)

Similarly, for the power of 3:

27 * 3 = 81

27 + 27 + 27 = 81

3^3 + 3^3 + 3^3 = 3^4

(3 times 3^3 gives 3^4)

And for the power of 4:

4 * 4 = 16

4 + 4 + 4 + 4 = 16

4^1 + 4^1 + 4^1 + 4^1 = 4^2

(4 times 4^1 gives 4^2)

Finally, for the power of 5:

125 * 5 = 625

125 + 125 + 125 + 125 + 125 = 625

5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 5^4

(5 times 5^3 gives 5^4)

Quite natural and intuitive, isn’t it? Take a look at the previous question again now.

*For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?*

*A) 18*

*(B) 32*

*(C) 35*

*(D) 64*

*(E) 70*

Which two powers when added will give 2^(36)?

From our discussion above, we know they are 2^(35) and 2^(35).

2^(35) + 2^(35) = 2^(36)

So x = 35 and y = 35 will satisfy this equation.

x + y = 35 + 35 = 70

Therefore, our answer is E.

One question arises here: Is this the only possible sum of x and y? Can x and y take some other integer values such that the sum of 2^x and 2^y will be 2^(36)?

Well, we know that no matter which integer values x and y take, 2^x and 2^y will always be positive, which means both x and y must be less than 36. Now note that no matter which two powers of 2 you add, their sum will always be less than 2^(36). For example:

2^(35) + 2^(34) < 2^(35) + 2^(35)

2^(2) + 2^(35) < 2^(35) + 2^(35)

etc.

So if x and y are both integers, the only possible values that they can take are 35 and 35.

How about something like this: 2^x + 2^y + 2^z = 2^36? What integer values can x, y and z take here?

**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 Quarter Wit, Quarter Wisdom: Calculating the Probability of Intersecting Events appeared first on Veritas Prep Blog.

]]>**P(A or B) = P(A) + P(B) – P(A and B)**

P(A) is the probability that event A will occur.

P(B) is the probability that event B will occur.

P(A or B) gives us the union; i.e. the probability that at least* one* of the two events will occur.

P(A and B) gives us the intersection; i.e. the probability that *both* events will occur.

Now, how do you find the value of P(A and B)? The value of P(A and B) depends on the relation between event A and event B. Let’s discuss three cases:

**1) ****A and B are independent events**

If A and B are independent events such as “the teacher will give math homework,” and “the temperature will exceed 30 degrees celsius,” the probability that both will occur is the product of their individual probabilities.

Say, P(A) = P(the teacher will give math homework) = 0.4

P(B) = P(the temperature will exceed 30 degrees celsius) = 0.3

P(A and B will occur) = 0.4 * 0.3 = 0.12

**2) A and B are mutually exclusive events**

If A and B are mutually exclusive events, this means they are events that cannot take place at the same time, such as “flipping a coin and getting heads” and “flipping a coin and getting tails.” You cannot get both heads and tails at the same time when you flip a coin. Similarly, “It will rain today” and “It will not rain today” are mutually exclusive events – only one of the two will happen.

In these cases, P(A and B will occur) = 0

**3) A and B are related in some other way**

Events A and B could be related but not in either of the two ways discussed above – “The stock market will rise by 100 points” and “Stock S will rise by 10 points” could be two related events, but are not independent or mutually exclusive. Here, the probability that both occur would need to be given to you. What we can find here is the range in which this probability must lie.

Maximum value of P(A and B):

The maximum value of P(A and B) is the lower of the two probabilities, P(A) and P(B).

Say P(A) = 0.4 and P(B) = 0.7

The maximum probability of intersection can be 0.4 because P(A) = 0.4. If probability of one event is 0.4, probability of both occurring can certainly not be more than 0.4.

Minimum value of P(A and B):

To find the minimum value of P(A and B), consider that any probability cannot exceed 1, so the maximum P(A or B) is 1.

Remember, P(A or B) = P(A) + P(B) – P(A and B)

1 = 0.4 + 0.7 – P(A and B)

P(A and B) = 0.1 (at least)

Therefore, the actual value of P(A and B) will lie somewhere between 0.1 and 0.4 (both inclusive).

Now let’s take a look at a GMAT question using these fundamentals:

*There is a 10% chance that Tigers will not win at all during the whole season. There is a 20% chance that Federer will not play at all in the whole season. What is the greatest possible probability that the Tigers will win and Federer will play during the season?*

*(A) 55%*

*(B) 60%*

*(C) 70%*

*(D) 72%*

*(E) 80%*

Let’s review what we are given.

P(Tigers will not win at all) = 0.1

P(Tigers will win) = 1 – 0.1 = 0.9

P(Federer will not play at all) = 0.2

P(Federer will play) = 1 – 0.2 = 0.8

Do we know the relation between the two events “Tigers will win” (A) and “Federer will play” (B)? No. They are not mutually exclusive and we do not know whether they are independent.

If they are independent, then the P(A and B) = 0.9 * 0.8 = 0.72

If the relation between the two events is unknown, then the maximum value of P(A and B) will be 0.8 because P(B), the lesser of the two given probabilities, is 0.8.

Since 0.8, or 80%, is the greater value, the greatest possibility that the Tigers will win and Federer will play during the season is 80%. Therefore, our answer is E.

**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 Quarter Wit, Quarter Wisdom: Basic Operations for GMAT Inequalities appeared first on Veritas Prep Blog.

]]>a = b

c = d

When these numbers are equal, we know that:

a + c = b + d (Valid)

a – c = b – d (Valid)

a * c = b * d (Valid)

a / c = b / d (Valid assuming c and d are not 0)

When can we add, subtract, multiply or divide two inequalities? There are rules that we need to follow for those. Today let’s discuss those rules and the concepts behind them.

**Addition**:

We can add two inequalities when they have the same inequality sign.

a < b

c < d

a + c < b + d (Valid)

Conceptually, it makes sense, right? If a is less than b and c is less than d, then the sum of a and c will be less than the sum of b and d.

On the same lines:

a > b

c > d

a + c > b + d (Valid)

Case 2: What happens when the inequalities have opposite signs?

a > b

c < d

We need to multiply one inequality by -1 to get the two to have the same inequality sign.

-c > -d

Now we can add them.

a – c > b – d

**Subtraction**:

We can subtract two inequalities when they have opposite signs:

a > b

c < d

a – c > b – d (The result will take the sign of the first inequality)

Conceptually, think about it like this: from a greater number (a is greater than b), if we subtract a smaller number (c is smaller than d), the result (a – c) will be greater than the result obtained when we subtract the greater number from the smaller number (b – d).

Note that this result is the same as that obtained when we added the two inequalities after changing the sign (see Case 2 above). We cannot subtract inequalities if they have the same sign, so it is better to always stick to addition. If the inequalities have the same sign, we simply add them. If the inequalities have opposite signs, we multiply one of them by -1 (to get the same sign) and then add them (in effect, we subtract them).

Why can we not subtract two inequalities when they have the same inequality sign, such as when a > b and c > d?

Say, we have 3 > 1 and 5 > 1.

If we subtract these two, we get 3 – 5 > 1 – 1, or -2 > 0 which is not valid.

If instead it were 3 > 1 and 2 > 1, we would get 1 > 0 which is valid.

We don’t know how much greater one term is from the other and hence we cannot subtract inequalities when their inequality signs are the same.

**Multiplication**:

Here, the constraint is the same as that in addition (the inequality signs should be the same) with an extra constraint: both sides of both inequalities should be non-negative. If we do not know whether both sides are non-negative or not, we cannot multiply the inequalities.

If a, b, c and d are all non negative,

a < b

c < d

a*c < b*d (Valid)

When two greater numbers are multiplied together, the result will be greater.

Take some examples to see what happens in Case 1, or more numbers are negative:

-2 < -1

10 < 30

Multiply to get: -20 < -30 (Not valid)

-2 < 7

-8 < 1

Multiply to get: 16 < 7 (Not valid)

**Division**:

Here, the constraint is the same as that in subtraction (the inequality signs should be opposite) with an extra constraint: both sides of both inequalities should be non-negative (obviously, 0 should not be in the denominator). If we do not know whether both sides are positive or not, we cannot divide the inequalities.

a < b

c > d

a/c < b/d (given all a, b, c and d are positive)

The final inequality takes the sign of the numerator.

Think of it conceptually: a smaller number is divided by a greater number, so the result will be a smaller number.

Take some examples to see what happens in Case 1, or more numbers are negative.

1 < 2

10 > -30

Divide to get 1/10 < -2/30 (Not valid)

**Takeaways: **

**Addition:** We can add two inequalities when they have the same inequality signs.

**Subtraction:** We can subtract two inequalities when they have opposite inequality signs.

**Multiplication:** We can multiply two inequalities when they have the same inequality signs and both sides of both inequalities are non-negative.

**Division:** We can divide two inequalities when they have opposite inequality signs and both sides of both inequalities are non-negative (0 should not be in the denominator).

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

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]]>A conditional sentence (an if/then sentence) has two clauses – the “**if clause**” (conditional clause) and the “**then clause**” (main clause). The “if clause” is the dependent clause, meaning the verbs we use in the clauses will depend on whether we are talking about a real or a hypothetical situation.

Often, conditional sentences are classified into first conditional, second conditional and third conditional (depending on the tense and possibility of the actions), but sometimes we have a separate zero conditional for facts. We will follow this classification and discuss four types of conditionals:

**1) Zero Conditional**

These sentences express facts; i.e. implications – “if this happens, then that happens.”

*If the suns shines, the clothes dry quickly.**If he eats bananas, he gets a headache.**If it rains heavily, the temperature drops.*

These conditionals establish universally known facts or something that happens habitually (every time he eats bananas, he gets a headache).

**2) First Conditional**

These sentences refer to predictive conditional sentences. They often use the present tense in the “if clause” and future tense (usually with the word “will”) in the main clause.

*If you come to my place, I will help you with your homework.**If I am able to save $10,000 by year end, I will go to France next year.*

**3) Second Conditional**

These sentences refer to hypothetical or unlikely situations in the present or future. Here, the “if clause” often uses the past tense and the main clause uses conditional mood (usually with the word “would”).

*If I were you, I would take her to the dance.**If I knew her phone number, I would tell you.**If I won the lottery, I would travel the whole world.*

**4) Third Conditional**

These sentences refer to hypothetical situations in the past – what could have been different in the past. Here, the “if clause” uses the past perfect tense and the main clause uses the conditional perfect tense (often with the words “would have”).

*If you had told me about the party, I would have attended it.**If I had not lied to my mother, I would not have hurt her.*

Sometimes, mixed conditionals are used here, where the second and third conditionals are combined. The “if clause” then uses the past perfect and the main clause uses the word “would”.

*If you had helped me then, I would be in a much better spot today.*

Now that you know which conditionals to use in which situation, let’s take a look at a GMAT question:

*Botanists have proven that if plants extended laterally beyond the scope of their root system, they will grow slower than do those that are more vertically contained.*

*(A) extended laterally beyond the scope of their root system, they will grow slower than do*

*(B) extended laterally beyond the scope of their root system, they will grow slower than*

*(C) extend laterally beyond the scope of their root system, they grow more slowly than*

*(D) extend laterally beyond the scope of their root system, they would have grown more slowly than do*

*(E) extend laterally beyond the scope of their root system, they will grow more slowly than do*

Now that we understand our conditionals, we should be able to answer this question quickly. Scientists have established something here; i.e. it is a fact. So we will use the zero conditional here – if this happens, then that happens.

…if plants extend laterally beyond the scope of their root system, they grow more slowly than do…

So the correct answer must be (C).

A note on slower vs. more slowly – we need to use an adverb here because “slow” describes “grow,” which is a verb. So we must use “grow slowly”. If we want to show comparison, we use “more slowly”, so the use of “slower” is incorrect here.

Let’s look at another question now:

*If Dr. Wade was right, any apparent connection of the eating of highly processed foods and excelling at sports is purely coincidental.*

*(A) If Dr. Wade was right, any apparent connection of the eating of*

*(B) Should Dr. Wade be right, any apparent connection of eating*

*(C) If Dr. Wade is right, any connection that is apparent between eating of*

*(D) If Dr. Wade is right, any apparent connection between eating*

*(E) Should Dr. Wade have been right, any connection apparent between eating*

Notice the non-underlined part “… is purely coincidental” in the main clause. This makes us think of the zero conditional.

Let’s see if it makes sense:

If Dr. Wade is right, any connection … is purely coincidental.

This is correct. It talks about a fact.

Also, “eating highly processed foods and excelling at sports” is correct.

Hence, our answer must be (D).

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

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

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*If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?*

*(A) 0*

*(B) 1*

*(C) 2*

*(D) 3*

*(E) 4*

In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2.

The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1

So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

…

3, 9, 7, 1

3, 9, 7

Since the exponent a remainder of 3, the new cycle ends at 3, 9, **7**. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9.

This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E.

Not so bad; let’s try a data sufficiency problem:

*If k is a positive integer, what is the remainder when 2^k is divided by 10?*

*Statement 1: k is divisible by 10*

*Statement 2: k is divisible by 4*

With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k.

The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change:

If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6.

If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2.

If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4.

If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8.

If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case.

and so on…

So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4.

*Statement 1: k is divisible by 10*

With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient.

*Statement 2: k is divisible by 4*

This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B.

Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT:

*What is the remainder of (3^7^11) divided by 5?*

*(A) 0*

*(B) 1*

*(C) 2*

*(D) 3*

*(E) 4*

For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}.

We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4.

We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4.

For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”):

The remainder of 7^{11} divided by 4

= The remainder of (4 + 3)^{11} divided by 4

= The remainder of 3^{11} divided by 4

= The remainder of 3*3^{10} divided by 4

= The remainder of 3*9^5 divided by 4

= The remainder of 3*(8+1)^5 divided by 4

= The remainder of 3*1^5 divided by 4

= The remainder of 3 divided by 4, which itself = **3**

So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4.

Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining:

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

…

3, 9, 7, 1

3, 9, **7**

With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C.

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

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]]>First let’s look at a pattern:

**20/10** gives us a remainder of** 0** (as 20 is exactly divisible by 10)

**21/10** gives a remainder of **1**

**22/10** gives a remainder of **2**

**23/10** gives a remainder of **3**

**24/10** gives a remainder of **4**

**25/10** gives a remainder of **5**

and so on…

In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on…

Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5).

With this in mind:

**20/5** gives a remainder of** 0** (as 20 is exactly divisible by 5)

**21/5** gives a remainder of **1**

**22/5** gives a remainder of **2**

**23/5** gives a remainder of **3**

**24/5** gives a remainder of **4**

**25/5** gives a remainder of **0** (as 25 is exactly divisible by 5)

**26/5** gives a remainder of **1**

**27/5** gives a remainder of **2**

**28/5** gives a remainder of **3**

**29/5** gives a remainder of **4**

**30/5** gives a remainder of** 0** (as 30 is exactly divisible by 5)

and so on…

So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10.

Let’s take a few questions now:

*What is the remainder when 86^(183) is divided by 10?*

Here, we need to find the last digit of 86^(183) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is (**previously discussed in this post**).

So the units digit of 86^(183) will be 6. So when we divide this by 10, the remainder will also be 6.

Next question:

*What is the remainder when 487^(191) is divided by 5?*

Again, when considering division by 5, the units digit can help us.

The units digit of 487 is 7.

7 has a cyclicity of 7, 9, 3, 1.

Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term.

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

7, 9, 3, 1

…

7, 9, 3

So the units digit of 487^(191) is 3, and the number would look something like ……………..3

As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5.

Therefore, the remainder of 487^(191) divided by 5 is 3.

Last question:

*If x is a positive integer, what is the remainder when 488^(6x) is divided by 2?*

Take a minute to review the question first. If you start by analyzing the expression 488^(6x), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether 488^(6x) is odd or even.

488^(6x) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even.

So 488^(6x) is even and will give remainder 0 when it is divided by 2.

That is all for today. We will look at some GMAT remainders-cyclicity questions next week!

**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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