# Quarter Wit, Quarter Wisdom: Solving GMAT Critical Reasoning Questions Involving Rates

In our “Quarter Wit, Quarter Wisdom” series, we have seen how to solve various rates questions – the basic ones as well as the complicated ones. But we haven’t considered critical reasoning questions involving rates, yet. In fact, the concept of rates makes these problems very difficult to both understand and explain. First, let’s look at what “rate” is.

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.

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

# Quarter Wit, Quarter Wisdom: Should You Use the Permutation or Combination Formula?

A recurring question from many students who are preparing for GMAT is this: When should one use the permutation formula and when should one use the combination formula?

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.

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

# Quarter Wit, Quarter Wisdom: Keeping an Open Mind in Critical Reasoning

Today we will discuss why it is important to keep an open mind while toiling away on your GMAT studying. Don’t go into test day with biases expecting that if a question tells us this, then it must ask that. GMAC testmakers are experts in surprising you and taking advantage of your preconceived notions, which is how they confuse you and convert a 600-level question to a 700-level one.

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.

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.

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

# Quarter Wit, Quarter Wisdom: An Interesting Property of Exponents

Today, let’s take a look at an interesting number property. Once we discuss it, you might think, “I always knew that!” and “Really, what’s new here?” So let me give you a question beforehand:

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

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?

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

# Quarter Wit, Quarter Wisdom: Calculating the Probability of Intersecting Events

We know our basic probability formulas (for two events), which are very similar to the formulas for sets:

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.

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

# Quarter Wit, Quarter Wisdom: Basic Operations for GMAT Inequalities

We know that we can perform all basic operations of addition, subtraction, multiplication and division on two equations:

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.

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

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).

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

# Quarter Wit, Quarter Wisdom: Grammatical Structure of Conditional Sentences on the GMAT

Today, we will take a look at the various “if/then” constructions in the GMAT Verbal section. Let us start out with some basic ideas on conditional sentences (though I know that most of you will be comfortable with these):

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 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).

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

# Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions (Part 2)

Last week,. Today, let’s jump right into some GMAT-relevant questions on these topics:

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.

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

# Quarter Wit, Quarter Wisdom: Cyclicity in GMAT Remainder Questions

Usually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Today we will look at the cases in which it can, and we will see why it helps us in these cases.

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!

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

# Quarter Wit, Quarter Wisdom: Cyclicity of Units Digits on the GMAT (Part 2)

As discussed last week, all units digits have a cyclicity of 1 or 2 or 4. Digits 2, 3, 7 and 8 have a cyclicity of 4, i.e. the units digit repeats itself every 4 digit:

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

Digits 4 and 9 have a cyclicity of 2, i.e. the units digit repeats itself every 2 digits:

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1

Digits 0, 1, 5 and 6 have a cyclicity of 1, i.e. the units digit is 0, 1, 5, or 6 respectively.

Now let’s take a look at how to apply these fundamentals:

What is the units digit of 813^(27)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 3.

Remember, our cyclicity of 3 is 3, 9, 7, 1 (four numbers total).

We need the units digit of 3^(27). How many full cycles of 4 will be there in 27? There will be 6 full cycles because 27 divided by 4 gives 6 as quotient and 3 will be the remainder. So after 6 full cycles of 4 are complete, a new cycle will start:

3, 9, 7, 1

3, 9, 7, 1

… (6 full cycles)

3, 9, 7 (new cycle for remainder of 3)

7 will be the units digit of 3^(27), so 7 will be the units digit of 813^(27).

Let’s try another question:

What is the units digit of 24^(1098)?

To get the desired units digit here, all we need to worry about is the units digit of the base, which is 4.

Remember, our cyclicity of 4 is 4 and 6 (this time, only 2 numbers).

We need the units digit of 24^(1098) – every odd power of 24 will end in 4 and every even power of 24 will end in 6.

Since 1098 is even, the units digit of 24^(1098) is 6.

Not too bad; let’s try something a little harder:

What is the units digit of 75^(25)^5

Note here that you have 75 raised to power 25 which is further raised to the power of 5.

25^5 is not the same as 25*5 – it is 25*25*25*25*25 which is far more complicated. However, the simplifying element of this question is that the last digit of the base 75 is 5, so it doesn’t matter what the positive integer exponent is, the last digit of the expression will always be 5.

Now let’s take a look at a Data Sufficiency question:

Given that x and y are positive integers, what is the units digit of (5*x*y)^(289)?

Statement 1: x is odd.

Statement 2: y is even.

Here there is a new complication – we don’t know what the base is exactly because the base depends on the value of x and y. As such, the real question should be can we figure out the units digit of the base? That is all we need to find the units digit of this expression.

When 5 is multiplied by an even integer, the product ends in 0.

When 5 is multiplied by an odd integer, the product ends in 5.

These are the only two possible cases: The units digit must be either 0 or 5.

With Statement 1, we do not know whether y is odd or even, we only know that x is odd. If y is odd, x*y will be odd. If y is even, x*y will be even. Since we don’t know whether x*y is odd or even, we don’t know whether 5*x*y will end in 5 or 0, so this statement alone is not sufficient.

With Statement 2, if y is even, x*y will certainly be even because an even * any integer will equal an even integer. Therefore, it doesn’t matter whether x is odd or even – regardless, 5*x*y will be even, hence, it will certainly end in 0.

As we know from our patterns of cyclicity, 0 has a cyclicity of 1, i.e. no matter what the positive integer exponent, the units digit will be 0. Therefore, this statement alone is sufficient and the answer is B (Statement 2 alone is sufficient but Statement 1 alone is not sufficient).

Finally, let’s take a question from our own book:

If n and a are positive integers, what is the units digit of n^(4a+2) – n^(8a)?

Statement 1: n = 3

Statement 2: a is odd.

We know that the cyclicity of every digit is either 1, 2 or 4. So to know the units digit of n^{4a+2} – n^{8a}, we need to know the units digit of n. This will tell us what the cyclicity of n is and what the units digit of each expression will be individually.

Statement 1: n = 3

As we know from our patterns of cyclicity, the cyclicity of 3 is 3, 9, 7, 1

Plugging 3 into “n”, n^{4a+2} = 3^{4a+2}

In the exponent, 4a accounts for “a” full cycles of 4, and then a new cycle begins to account for 2.

3, 9, 7, 1

3, 9, 7, 1

3, 9

The units digit here will be 9.

Again, plugging 3 into “n”, n^{8a} = 3^{8a}

8a is a multiple of 4, so there will be full cycles of 4 only. This means the units digit of 3^{8a} will be 1.

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

3, 9, 7, 1

Plugging these answers back into our equation: n^{4a+2} – n^{8a} = 9 – 1

The units digit of the combined expression will be 9 – 1 = 8.

Therefore, this statement alone is sufficient.

In Statement 2, we are given what the exponents are but not what the value of n, the base, is. Therefore, this statement alone is not sufficient, and our answer is A (Statement 1 alone is sufficient but Statement 2 alone is not sufficient).

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

# Quarter Wit Quarter Wisdom: Cyclicity of Units Digits on the GMAT

In our algebra book, we have discussed finding and extrapolating patterns. In this post today, we will look at the patterns we get with various units digits.

The first thing you need to understand is that when we multiply two integers together, the last digit of the result depends only on the last digits of the two integers.

For example:

24 * 12 = 288

Note here: …4 * …2 = …8

So when we are looking at the units digit of the result of an integer raised to a certain exponent, all we need to worry about is the units digit of the integer.

Let’s look at the pattern when the units digit of a number is 2.

Units digit 2:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

2^9 = 512

2^10 = 1024

Note the units digits. Do you see a pattern? 2, 4, 8, 6, 2, 4, 8, 6, 2, 4 … and so on

So what will 2^11 end with? The pattern tells us that two full cycles of 2-4-8-6 will take us to 2^8, and then a new cycle starts at 2^9.

2-4-8-6

2-4-8-6

2-4

The next digit in the pattern will be 8, which will belong to 2^11.

In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base.

So 652^(11) will end in 8,1896782^(11) will end in 8, and so on…

A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits.

Units digit 3:

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on…

Units digit 4:

4^1 = 4

4^2 = 16

4^3 = 64

4^4 = 256

The pattern here is 4, 6, 4, 6, 4, 6, and so on…

Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is:

1545^23 = ……..5

1650^19 = ……..0

161^28 = ………1

Hope you get the point.

Units digit 7:

7^1 = 7

7^2 = 49

7^3 = 343

7^4 = ….1 (Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit)

7^5 = ….7 (Now multiply 1 from above by 7 to get 7 as the units digit)

7^6 = ….9

The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on…

Units digit 8:

8^1 = 8

8^2 = 64

8^3 = …2

8^4 = …6

8^5 = …8

8^6 = …4

The pattern here is 8, 4, 2, 6, 8, 4, 2, 6, and so on…

Units digit 9:

9^1 = 9

9^2 = 81

9^3 = 729

9^4 = …1

The pattern here is 9, 1, 9, 1, 9, 1, and so on…

Summing it all up:

1) Digits 2, 3, 7 and 8 have a cyclicity of 4; i.e. the units digit repeats itself every 4 digits.

Cyclicity of 2: 2, 4, 8, 6

Cyclicity of 3: 3, 9, 7, 1

Cyclicity of 7: 7, 9, 3, 1

Cyclicity of 8: 8, 4, 2, 6

2) Digits 4 and 9 have a cyclicity of 2; i.e. the units digit repeats itself every 2 digits.

Cyclicity of 4: 4, 6

Cyclicity of 9: 9, 1

3) Digits 0, 1, 5 and 6 have a cyclicity of 1.

Cyclicity of 0: 0

Cyclicity of 1: 1

Cyclicity of 5: 5

Cyclicity of 6: 6

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

# Quarter Wit, Quarter Wisdom: Permutation Involving Sum of Digits

We have seen in previous posts how to deal with permutation and combination questions on the GMAT. There is a certain variety of questions that involve getting a bunch of numbers using permutation, and then doing some operations on the numbers we get. The questions can get a little overwhelming considering the sheer magnitude of the number of numbers involved! Let’s take a look at that concept today. We will explain it using an example and then take a question as an exercise:

What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 such that each digit is used exactly once in each integer?

First of all, we will use our basic counting principle to find the number of integers that are possible.

The first digit can be chosen in 4 ways. The next one in 3 ways since each digit can be used only once. The next one in 2 ways and there will be only one digit left for the last place.

This gives us a total of 4*3*2*1 = 24 ways of writing such a four digit number. This is what some of the numbers will look like:

1234

1243

1324

1342

2143

4321

Now we need to add these 24 integers to get their sum. Note that since each digit has an equal probability of occupying every place, out of the 24 integers, six integers will have 1 in the units place, six will have 2 in the units place, another six will have 3 in the units place and the rest of the six will have 4 in the units place. The same is true for all places – tens, hundreds and thousands.

Imagine every number written in expanded form such as:

1234 = 1000 + 200 + 30 + 4

2134 = 2000 + 100 + 30 + 4

…etc.

For the 24 numbers, we will get six 1000’s, six 2000’s, six 3000’s and six 4000’s.

In addition, we will get six 100’s, six 200’s, six 300’s and six 400’s.

For the tens place, will get six 10’s, six 20’s, six 30’s and six 40’s.

And finally, in the ones place we will get six 1’s, six 2’s, six 3’s and six 4’s.

Therefore, the total sum will be:

6*1000 + 6*2000 + 6*3000 + 6*4000 + 6*100 + 6*200 + … + 6*3 + 6*4

= 6*1000*(1 + 2 + 3 + 4) + 6*100*(1 + 2 + 3 + 4) + 6*10*(1 + 2 + 3 + 4) + 6*1*(1 + 2 + 3 + 4)

= 6*1000*10 + 6*100*10 + 6*10*10 + 6*10

= 6*10*(1000 + 100 + 10 + 1)

= 1111*6*10

= 66660

Note that finally, there aren’t too many actual calculations, but there is some manipulation involved. Let’s look at a GMAT question using this concept now:

What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A444440

B) 610000

C) 666640

D) 711040

E) 880000

Conceptually, this problem isn’t much different from the previous one.

Using the same basic counting principle to get the number of integers possible, the first digit can be chosen in 4 ways, the next one in 4 ways, the next one in again 4 ways and finally the last digit in 4 ways. This is what some of the numbers will look like:

1111

1112

1121

and so on till 4444.

As such, we will get a total of 4*4*4*4 = 256 different integers.

Now we need to add these 256 integers to get their sum. Since each digit has an equal probability of occupying every place, out of the 256 integers, 64 integers will have 1 in the units place, 64 will have 2 in the units place, another 64 integers will have 3 in the units place and the rest of the 64 integers will have 4 in the units place. The same is true for all places – tens, hundreds and thousands.

Therefore, the total sum will be:

64*1000 + 64*2000 + 64*3000 + 64*4000 + 64*100 + 64*200 + … + 64*3 + 64*4

= 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)

= (64*1 + 64*2 + 64*3 + 64*4) * (1000 + 100 + 10 + 1)

= 64*10*1111

= 711040

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

# Quarter Wit, Quarter Wisdom: The Tricky Critical Reasoning Conclusion

As discussed previously, the most important aspect of a strengthen/weaken question on the GMAT is “identifying the conclusion,” but sometimes, that may not be enough. Even after you identify the conclusion, you must ensure that you have understood it well. Today, we will discuss the “tricky conclusions.”

First let’s take a look at some simple examples:

Conclusion 1: A Causes B.

We can strengthen the conclusion by saying that when A happens, B happens.

We can weaken the conclusion by saying that A happened but B did not happen.

How about a statement which suggests that “C causes B,” or, “B happened but A did not happen”?

Do these affect the conclusion? No, they don’t. The relationship here is that A causes B. Whether there are other factors that cause B too is not our concern, so whether B can happen without A is none of our business.

Conclusion 2: Only A Causes B.

This is an altogether different conclusion. It is apparent that A causes B but the point of contention is whether A is the only cause of B.

Now here, a statement suggesting, “C causes B,” or, “B happened but A did not happen,” does affect our conclusion. These weaken our conclusion – they suggest that A is not the only cause of B.

This distinction can be critical in solving the question. We will now illustrate this point with one of our own GMAT practice questions:

Two types of earthworm, one black and one red-brown, inhabit the woods near the town of Millerton. Because the red-brown worm’s coloring affords it better camouflage from predatory birds, its population in 1980 was approximately five times that of the black worm. In 1990, a factory was built in Millerton and emissions from the factory blackened much of the woods. The population of black earthworms is now almost equal to that of the red-brown earthworm, a result, say local ecologists, solely stemming from the blackening of the woods.

Which of the following, if true, would most strengthen the conclusion of the local ecologists?

(A) The number of red-brown earthworms in the Millerton woods has steadily dropped since the factory began operations.

(B) The birds that prey on earthworms prefer black worms to red-brown worms.

(C) Climate conditions since 1990 have been more favorable to the survival of the red-brown worm than to the black worm.

(D) The average life span of the earthworms has remained the same since the factory began operations.

(E) Since the factory took steps to reduce emissions six months ago, there has been a slight increase in the earthworm population.

Let’s look at the argument.

Premises:

• There are two types of worms – Red and Black.
• Red has better camouflage from predatory birds, hence its population was five times that of black.
• The factory has blackened the woods and now the population of both worms is the same.

Conclusion:

From our premises, we can determine that the blackening of the woods is solely responsible for equalization of the population of the two earthworms.

We need to strengthen this conclusion. Note that there is no doubt that the blackening of the woods is responsible for equalization of populations; the question is whether it is solely responsible.

(A) The number of red-brown earthworms in the Millerton woods has steadily dropped since the factory began operations.

Our conclusion is that only the blackening of the woods caused the numbers to equalize (either black worms are able to hide better or red worms are not able to hide or both), therefore, we need to look for the option that strengthens that there is no other reason. Option A only tells us what the argument does anyway – the population of red worms is decreasing (or black worm population is increasing or both) due to the blackening of the woods. It doesn’t strengthen the claim that only blackening of the woods is responsible.

(B) The birds that prey on earthworms prefer black worms to red-brown worms.

The fact that birds prefer black worms doesn’t necessarily mean that they get to actually eat black worms. Even if we do assume that they do eat black worms over red worms when they can, this strengthens the idea that “the blackening of the woods is responsible for equalization of population,” but does not strengthen the idea that “the blackening of the woods is solely responsible for equalization,” hence, this is not our answer.

(C) Climate conditions since 1990 have been more favorable to the survival of the red-brown worm than to the black worm.

Option C tells us that another factor that could have had an effect on equalization (i.e. climate) is not responsible. This strengthens the conclusion that better camouflage is solely responsible – it doesn’t prove the conclusion beyond doubt, since there could be still another factor that could be responsible, but it does discard one of the other factors. Therefore, it does improve the probability that the conclusion is true.

(D) The average life span of the earthworms has remained the same since the factory began operations.

This option does not distinguish between the two types of earthworms. It just tells us that as a group, the average lifespan of the earthworms has remained the same. Hence, it doesn’t affect our conclusion, which is based on the population of two different earthworms.

(E) Since the factory took steps to reduce emissions six months ago, there has been a slight increase in the earthworm population.

Again, this option does not distinguish between the two types of earthworms. It just tells us that as a group, the earthworm population has increased, so it also does not affect our conclusion, which is based on the population of two different earthworms.

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 Critical Role of Reading in GMAT Critical Reasoning Questions!

Most non-native English users have one question: How do I improve my Verbal GMAT score?  There are lots of strategies and techniques we discuss in our books, in our class and on our blog. But one thing that we seriously encourage our students to do (that they need to do on their own) is read more – fiction, non fiction, magazines (mind you, good quality), national dailies, etc. Reading high quality material helps one develop an ear for correct English. It is also important to understand the idiomatic usage of English, which no one can teach in the class. At some time, most of us have thought how silly some things are in the English language, haven’t we?

For example:

Fat chance” and “slim chance’”mean the same thing – Really? Shouldn’t they mean opposite things?

But “wise man” and “wise guy” are opposites – Come on now!

A house burns up as it burns down and you fill in a form by filling it out?

And let’s not even get started on the multiple unrelated meanings many words have – The word on the top of the page, “critical,” could mean “serious” or “important” or “inclined to find fault” depending on the context!

Well, you really must read to understand these nuances or eccentricities, if you may, of the English language. Let’s look at an official question today which many people get wrong just because of the lack of familiarity with the common usage of phrases in English. But before we do that, some quick statistics on this question – 95% students find this question hard and more than half answer it incorrectly. And, on top of that, it is quite hard to convince test takers of the right answer.

Some species of Arctic birds are threatened by recent sharp increases in the population of snow geese, which breed in the Arctic and are displacing birds of less vigorous species. Although snow geese are a popular quarry for hunters in the southern regions where they winter, the hunting season ends if and when hunting has reduced the population by five percent, according to official estimates. Clearly, dropping this restriction would allow the other species to recover.

Which of the following, if true, most seriously undermines the argument?

(A) Hunting limits for snow geese were imposed many years ago in response to a sharp decline in the population of snow geese.

(B) It has been many years since the restriction led to the hunting season for snow geese being closed earlier than the scheduled date.

(C) The number of snow geese taken by hunters each year has grown every year for several years.

(D) As their population has increased, snow geese have recolonized wintering grounds that they had not used for several seasons.

(E) In the snow goose’s winter habitats, the goose faces no significant natural predation.

As usual, let’s start with the question stem – “… most seriously undermines the argument”

This is a weaken question. The golden rule is to focus on the conclusion and try to weaken it.

Let’s first understand the argument:

Snow geese breed in the Arctic and fly south for the winter. They are proliferating, and that is bad for other birds. Southern hunters reduce the number of geese when they fly south. There is a restriction in place that if the population of the geese that came in reduces by 5%, hunting will stop. So if 1000 birds flew south and 50 were hunted, hunting season will be stopped. The argument says that we should drop this restriction to help other Arctic birds flourish (conclusion), then hunters will hunt many more geese and reduce their numbers.

What is the conclusion here? It is: “Clearly, dropping this restriction would allow the other species to recover.”

You have to try to weaken it, i.e. give reasons why even after dropping this restriction, it is unlikely that other species will recover. Even if this restriction of “not hunting after 5%” is dropped and hunters are allowed to hunt as much as they want, the population of geese will still not reduce.

Now, first look at option (B);

(B) It has been many years since the restriction led to the hunting season for snow geese being closed earlier than the scheduled date.

What does this option really mean?

Does it mean the hunting season has been closing earlier than the scheduled date for many years? Or does it mean the exact opposite, that the restriction came into effect many years ago and since then, it has not come into effect.

It might be obvious to the native speakers and to the avid readers, but many non-native test takers actually fumble here and totally ignore option (B) – which, I am sure you have guessed by now, is the correct answer.

The correct meaning is the second one – the restriction has not come into effect for many years now. This means the restriction doesn’t really mean much. For many years, the restriction has not caused the hunting season to close down early because the population of geese hunted is less than 5% of the population flying in. So if the hunting season is from January to June, it has been closing in June, only, so even if hunters hunt for the entire hunting season, they still do not reach the 5% of the population limit (Southern hunters hunt less than 50 birds when 1000 birds fly down South).

Whether you have the restriction or not, the number of geese hunted is the same. So even if you drop the restriction and tell hunters that they can hunt as much as they want, it will not help as they will not want to hunt geese much anyway. This implies that even if the restriction is removed, it is likely that there will be no change in the situation. This definitely weakens our conclusion that dropping the restriction will help other species to recover.

So when people ignore (B), on which option do they zero in? Some fall for (C) but many fall for (D). Let’s look at all other options now:

(A) Hunting limits for snow geese were imposed many years ago in response to a sharp decline in the population of snow geese.

This is out of scope to our argument. It doesn’t really matter when and why the limits were imposed.

(C) The number of snow geese taken by hunters each year has grown every year for several years.

This doesn’t tell us how dropping the restriction would impact the population of geese, it just tells us what has happened in the past – the number of geese hunted has been increasing. If anything, it might strengthen our conclusion if the number of geese hunted is close to 5% of the population. When the population decreases by 5%, if the restriction is dropped, chances are that more geese will be hunted and other species will recover. We have to show that even after dropping the restriction, the other species may not recover.

(D) As their population has increased, snow geese have recolonised wintering grounds that they had not used for several seasons.

With this answer choice, “wintering grounds” implies the southern region (where they fly for winter). In the South, they have recolonised regions they had not occupied for a while now, which just tells you that the population has increased a lot and the geese are spreading. It doesn’t say that removing the restrictions and letting hunters hunt as much as they want will not help. In fact, if anything, it may make the argument a little stronger. If the geese are occupying more southern areas, hunting grounds may become easily accessible to more hunters and dropping hunting restrictions may actually help more!

(E) In the snow goose’s winter habitats, the goose faces no significant natural predation.

We are concerned about the effect of hunting, thus natural predation is out of scope.

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!

# Complicated GMAT Work-Rate Questions Made Easy!

Today, we will take up a gem of a work-rate question from our own curriculum. Its basics lie in a post on joint variation that we discussed many weeks ago. Here is a quick recap of the actual methodology:

If 10 workers complete a work in 5 days working 8 hours a day, how much work will be done by 6 workers in 10 days working 2 hours a day?

Here is what it looks like:

10 workers……………..5 days …………….. 8 hours ……………1 work

6 workers………………10 days …………… 2 hours …………… ? work

We need to find the amount of work done, so we start with the work done in the first case and then multiply it by the respective ratios:

Work done = 1 * (6/10) * (10/5) * (2/8) = 3/10

We multiply by 6/10 because number of men decreases from 10 to 6. The work done will reduce, so we multiply by 6/10 (the fraction less than 1).

We also multiply by 10/5 because number of days increases from 5 to 10. Because of this, the work done will increase, so we multiply by 10/5 (the fraction more than 1).

We also multiply by 2/8 because number of hours decreases from 8 to 2. Because of this, the work done will decrease, hence, we multiply by 2/8 (the fraction less than 1).

So the process is super simple – start with what you need to find out, say x, and multiply it by the ratio of each thing that changes from A to B. Whether you multiply by A/B or B/A depends on whether with this change increases or reduces x. If x increases, you will multiply by the fraction that is greater than 1, however if x decreases, you will multiply by the fraction that is less than 1.

On this same concept, let’s look at the question:

16 horses can haul a load of lumber in 24 minutes. 12 horses started hauling a load and after 14 minutes, 12 mules joined the horses. Will it take less than a quarter-hour for all of them together to finish hauling the load?

Statement 1: Mules work more slowly than horses.

Statement 2: 48 mules can haul the same load of lumber in 16 minutes.

Let’s see what data we have in the question stem:

16 horses …….. 24 mins ………. 1 work

12 horses …….. 14 mins ………. ? work

Work done = 1*(14/24)*(12/16) = (7/16)th of the work

We multiply by 14/24 because if the time taken to do the work decreases, the work done will also decrease. 14/24 is less than 1 so it will decrease the work done.

We also multiply by 12/16 because if the number of horses decreases, the work done will also decrease. 12/16 is less than 1 so it will decrease the work done.

All in all, we now know that 12 horses complete 7/16th of the work in 14 mins. So there is still 1 – 7/16 = 9/16 of the work left to do.

Now let’s review the two statements.

Statement 1: Mules work more slowly than horses.

This statement doesn’t give us any figures, so how can we analyse it mathematically? What we can do is find the range in which the time taken by all the horses and mules together will lie according to this statement.

Case 1: When mules work at a rate that is infinitesimally smaller than the rate of horses.

In this case, 12 mules are equivalent to 12 horses. So we have a total of 12 + 12 = 24 horses working together to complete (9/16)th of the work.

16 horses …….. 24 mins ………. 1 work

24 horses ……… ? mins ………. 9/16 work

Time taken = 24*(16/24)*(9/16) = 9 mins

Since the mules are slower than the horses, the time taken to complete the work will be more than 9 minutes. How much more than 9 minutes, we do not know. Now look at the flip side:

Case 2: When the mules work at a rate close to 0.

If the mules work slower, time taken will be more till the point when mules work so slowly that they do almost no work.

16 horses …….. 24 mins ………. 1 work

12 horses ……… ? mins ………. 9/16 work

Time taken = 24*(16/12)*(9/16) = 18 minutes

Therefore, depending on how fast/slow the mules are, the time taken to do the rest of the work could be anywhere from 9 minutes to 18 minutes. Therefore the time taken could be either less or more than 15 minutes – this statement alone is not sufficient.

Statement 2: 48 mules can haul the same load of lumber in 16 minutes.

We now know exactly how fast the mules are, so this must be sufficient to say whether the time taken to do the rest of the work was less or more than 15 minutes – we don’t need to actually find the time taken here – therefore, the answer is B, Statement 2 alone is sufficient.

However, if you would like to find out for practice, just find the equivalence between the horses and the mules first.

To haul the load in 16 minutes, we need 48 mules

To haul the load in 24 minutes, we need 48 * (16/24) = 32 mules

So 32 mules are equivalent to 16 horses (because 16 horses haul the load in 24 minutes). This means that 2 mules are equivalent to 1 horse, and 12 mules are, therefore, equivalent to 6 horses.

So now, in effect we have a total of 12 + 6 = 18 horses, and the situation now becomes this:

16 horses …….. 24 mins ………. 1 work

18 horses ……… ? mins ………. 9/16 work

Time taken = 24*(16/18)*(9/16) = 12 minute – less than a quarter-hour to finish the work.

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!

# 7 Formulas for Tackling Three Overlapping Sets on the GMAT

In a previous post, we saw how to solve three overlapping sets questions using venn diagrams. Today, we will look at all of the various formulas floating around on three overlapping sets. Most of these are self explanatory but we will look into the details of some of them.

There are two basic formulas that we already know:

1) Total = n(No Set) + n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets)

2) Total = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) + n(No Set)

From these two formulas, we can derive all others.

n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets) gives us n(At least one set). So we get:

3) Total = n(No Set) + n(At least one set)

From (3), we get n(At least one set) = Total – n(No Set)

Plugging this into (2), we then get:

4) n(At least one set) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)

Now let’s see how we can calculate the number of people in exactly two sets. There is a reason we jumped to n(Exactly two sets) instead of following the more logical next step of figuring out n(At least two sets) – it will be more intuitive to get n(At least two sets) after we find n(Exactly two sets).

n(A and B) includes people who are in both A and B and it also includes people who are in A, B and C. Because of this, we should remove n(A and B and C) from n(A and B) to get n(A and B only). Similarly, you get n(B and C only) and n(C and A only), so adding all these three will give us number of people in exactly 2 sets.

n(Exactly two sets) = n(A and B) – n(A and B and C) + n(B and C) – n(A and B and C) + n(C and A) – n(A and B and C). Therefore:

5) n(Exactly two sets) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)

Now we can easily get n(At least two sets):

6) n(At least two sets) = n(A and B) + n(B and C) + n(C and A) – 2*n(A and B and C)

This is just n(A and B and C) more than n(Exactly two sets). That makes sense, doesn’t it? Here, you include the people who are in all three sets once and n(Exactly two sets) converts to n(At least two sets)!

Now, we go on to find n(Exactly one set). From n(At least one set), let’s subtract n(At least two sets); i.e. we subtract (6) from (4)

n(Exactly one set) = n(At least one set) – n(At least two sets), therefore:

7) n(Exactly one set) = n(A) + n(B) + n(C) – 2*n(A and B) – 2*n(B and C) – 2*n(C and A) + 3*n(A and B and C)

You don’t need to learn all these formulas. Just focus on first two and know how you can arrive at the others if required. Let’s try this in an example problem:

Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185

(B) 180

(C) 175

(D) 190

(E) 195

You are given that:

n(At least one channel) = 250

n(Exactly two channels) = 50

So we know that n(At least one channel) = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels) = 250

250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let’s find the value of n(Exactly 3 channels) = x

We also know that n(At least one channel) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) = 250

Also, n(Exactly two channels) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)

So n(A and B) + n(B and C) + n(C and A) = n(Exactly two channels) + 3*n(A and B and C)

Plugging this into the equation above:

250 = n(A) + n(B) + n(C) – n(Exactly two channels) – 3*x + x

250 = 116 + 127 + 107 – 50 – 2x

x = 25

250 = n(Exactly 1 channel) + 50 + 25

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!

# Manipulating Standard Formulas on the GMAT

We know the formula we need to use to find the sum of n consecutive positive integers starting from 1. The formula is given as n(n+1)/2.

So the sum of first four positive integers is 4 * (4 + 1)/2 = 10.

This might seem a bit cumbersome, since it is easy to see that 1 + 2 + 3 + 4 = 10, but we know that the formula comes in very handy when n is a large number. For example, the sum of first 50 positive integers = 50 * 51/2 = 1275. Obviously, this will be a lot harder when done the “1 + 2 + 3 + 4 … + 49 + 50” way.

Now the question is, how do we adjust the same formula to find the sum of consecutive integers which do not start from 1?

Say, how do we find the sum of all positive integers from 8 to 20? The formula assumes a starting point of 1, so then we insert only the last number, n. How do we manage the 8? Let’s try to figure it out

Say the sum of first 20 positive integers = 1 + 2 + 3 + 4 + …. + 19 + 20 = 20 * 21/2

(1 + 2 + 3 +… + 7) + (8 + 9 +10 + … + 19 + 20) = 20 * 21/2

We need the value of the part in red, let’s call it the required sum.

(1 + 2 + 3 +… + 7) + The Required Sum = 20 * 21/2

Note here that we know the sum of 1 + 2 + 3 + … + 7 = 7 * 8/2

So, 7*8/2 + The Required Sum = 20 * 21/2, therefore the Required Sum = 20*21/2 – 7*8/2

To get the sum of consecutive integers from 8 to 20, we find the sum of all integers from 1 to 20 (using the formula we know) and subtract the sum of integers from 1 to 7 out of it (using the same formula).

To generalize, the sum of all positive integers from m to n is given as:

n(n+1)/2 – (m-1)*m/2

Let’s look at a question based on this concept:

If the sum of the consecutive integers from –40 to n inclusive is 356, what is the value of n?

(A) 47

(B) 48

(C) 49

(D) 50

(E) 51

If you are thinking that we haven’t gone over how to adjust the formula for negative numbers, you are right, but what we have discussed is enough to solve this question.

Numbers around 0 are symmetrical. So 1 and -1 add up to equal 0. Similarly, 2 and -2 add up to equal 0, and so on…

-40, -39 … 0 … 39, 40, 41, 42, 43, 44, 45 …

The sum of all numbers from -40 to 40 will be 0. Or another way to look at it is that 0 is the mean of all numbers from -40 to 40. So the total sum of these numbers will also be 0.

The given sum is actually the sum of numbers from 41 to n only.

We know how to calculate that:

n(n+1)/2 – 40*41/2 = 356

n(n+1) = 2352

From the options, we see that n cannot be 49 or 50 because the product of 49*50 or 50*51 will end in 0, so plug in n = 48 to check whether 48*49 is equal to 2352. It is, therefore our answer is B

(Had we obtained a lower product than required, we could have said that n must be 51. Had we obtained a higher product than was required, we could have said that n is 47.)

Another method:

Use the concept of arithmetic mean and ballpark. The mean of numbers from 41 to 47 or 48 or 49… will be somewhere between 44 and 46.

Let’s estimate the number of integers we need to get the sum of about 356. Each additional integer is quite large (more than 45) therefore, a difference of about 10-15 in the sum due to the various possible values of the mean will be immaterial.

45*7 = 315

45*8 = 360

This brings us very close to the value of 356.

Assuming there are 8 integers, their values will be from 41 to 48. The average of these 8 numbers will be 44.5. The total sum will be 44.5 * 8 = 356. It matches, so our answer is still B.

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!

Today, we will discuss the concept of sampling. People with a statistics background will be very comfortable with it, but if you have not studied statistics, a little bit of knowledge will be helpful. You are not required to know this for the GMAT, however there could be questions framed on the sampling premise, and you will be far more comfortable solving them with some understanding in place. A sample is a selection made from a larger group (the “population”) which helps you examine certain characteristics of the larger group using limited resources.

For example:

In a large population, say all the people in a state, it is difficult to find the number of people with a certain trait, such as red hair. So you pick up 100 people at random (from different families, different areas, different backgrounds) and find the number of people who have red hair in this selection of 100.

Let’s say 12 have red hair. You can then generalize that approximately 12% of the whole population has red hair. The more unbiased your sample, the better the approximation.

In this example, you found something about the entire population (12% has red hair) based on a small sample and hence, using few resources. To find the actual percentage of people who have red hair in the entire population, you would need far more effort, time and money. Usually the use of fewer resources justifies the use of sampling even though it comes with some error.

So that is a bit of background on sampling. It will help you make sense of the  official question given below:

In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate number of fish in the pond?

A) 400

B) 625

C) 1,250

D) 2,500

E) 10,000

This is what took place: From a pond, 50 fish were caught, tagged and returned to the pond. Then 50 were caught again and 2 of those were found to be tagged.

The total number of fish in the pond is the population of the pond. It is unknown. Since counting the total number of fish in the pond was hard, they tagged 50 of them and let them disperse evenly in the population. This means they gave a certain trait to a known number of fish in the pond – they tagged 50 fish.

Then they caught 50 fish again and these fish became the sample. Out of these 50, 2 were found to be tagged. So 2 of the 50 fish caught were found to have the trait given (tagged) – 4% of our sample was tagged.

The question tells us that “… the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond …” that is, the question tells us that the sample is representative of the population. This implies that 50 (the number of fish we tagged) is 4% of the entire fish population of the pond.

50 = 4% of Total Fish Population, therefore, we can calculate that the Total Fish Population = 50 * 100/4 = 1250. Our answer is then C.

Using sampling, we were able to calculate the total population of the pond without actually counting each fish. For increased accuracy, often the exercise of taking samples is repeated many times and then some kind of average is used to get the best approximation.

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

# Advanced Exponent Properties for the GMAT

Today, let’s discuss the relative placements of exponents on the number line.

We know what the graph of 2^x looks like:

It shows that when x is positive, with increasing value of x, 2^x increases very quickly (look at the first quadrant), but we don’t know exactly how it increases.

It also shows that when x is negative, 2^x stays very close to 0. As x decreases, the value of 2^x decreases by a very small amount.

Now note the spacing of the powers of 2 on the number line:

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

and so on…

2^1 = 2 * 2^0 = 2^0 + 2^0

2^2 = 2 * 2^1 = 2^1 + 2^1

2^3 = 2 * 2^2 = 2^2 + 2^2

2^4 = 2 * 2^3 = 2^3 + 2^3

So every power of 2 is equidistant from 0 and the next power. This means that a power of 2 would be much closer to 0 than the next higher powers. For example, 2^2 is at the same distance from 0 as it is from 2^3.

But 2^2 is much closer to 0 than it is to 2^4, 2^5 etc.

Let’s look at a question based on this concept. Most people find it a bit tough if they do not understand this concept:

Given that x = 2^b – (8^30 + 16^5), which of the following values for b yields the lowest value for |x|?

A) 35

B) 90

C) 91

D) 95

E) 105

We need the lowest value of |x|. We know that the smallest value any absolute value function can take is 0. So 2^b should be as close as possible to (8^30 + 16^5) to get the lowest value of |x|.

Let’s try to simplify:

(8^30 + 16^5)

= (2^3)^30 + (2^4)^5

= 2^90 + 2^20

Which value should b take such that 2^b is as close as possible to 2^90 + 2^20?

2^90 + 2^20 is obviously larger than 2^90. But is it closer to 2^90 or 2^91 or higher powers of 2?

Let’s use the concept we have learned today – let’s compare 2^90 + 2^20 with 2^90 and 2^91.

2^90 = 2^90 + 0

2^91 = 2^90 + 2^90

So now if we compare these two with 2^90 + 2^20, we need to know whether 2^20 is closer to 0 or closer to 2^90.

We already know that 2^20 is equidistant from 0 and 2^21, so obviously it will be much closer to 0 than it will be to 2^90.

Hence, 2^90 + 2^20 is much closer to 2^90 than it is to 2^91 or any other higher powers.

We should take the value 90 to minimize |x|, therefore the answer is B.

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!

# Know the Concept of Cost Price for the GMAT

Most of us are quite comfortable with the concepts of percentages, cost price and sale price, but when we come across a toughie from these topics, we feel lost. Then we go back to the theory but there seems to be nothing new there – nothing new that could potentially help us tackle such questions with ease in the future. The point is, the basic theory of these topics is quite simple – there isn’t anything else to it – but it’s application to GMAT questions is an altogether different deal. There are small but critical things that you need to keep in mind, one of which we will discuss today: what is the cost price?

Let’s take a look at this with an official question:

A photography dealer ordered 60 Model X cameras to be sold for \$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer’s initial cost. What was the dealer’s approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

(A) 7% loss

(B) 13% loss

(C) 7% profit

(D) 13% profit

(E) 14% profit

Solution:

Here are the various data points:

• 60 cameras bought at 20% markup.
• Selling Price = \$250
• 6 not sold and 50% of initial cost refunded
• Profit/Loss = ?

Now look at the solution:

The cost price per camera = 250/1.2 = 1250/6

The total cost price = (1250/6)*60 = \$12,500

50% of the cost of 6 cameras was returned.

The cost price of 6 cameras = (1250/6)*6 = \$1250

50% of this = 1250/2 = \$625

This means the effective cost price = 12,500 – 625 = \$11,875

If the selling price per camera = \$250, the total selling price = 54 * 250 = \$13500 (only 54 cameras were sold)

Hence, the profit % = [(13500 – 11875) / 11875] x 100 = (1625/11875) x 100 = 13.684%

This gives us approximately 14% as the answer (rounding up). But that is not correct. Before you move ahead, try to figure out the problem with this solution. If you are able to, it means you do understand this topic very well.

Here is the problem with the solution:

The cost price is the total initial cost price. You cannot subtract the refund out of it. The refund is effectively the price at which the 6 cameras were sold. You cannot cancel off your cost price with your sale price and have a smaller cost price. Your initial investment in the transaction is your cost price. When you reduce it by cancelling off some sale price (or refund), you are artificially increasing your profit percentage.

Say, we buy a few thing for \$100. While selling them off, we get \$50 for half of them. We reduce our cost price by \$50 and get \$50 as cost price. For the other half, we sell them for \$60. We say that \$50 is out cost price and \$60 is our selling price. The profit we made is \$10, which is fine. The issue is that our profit percentage is not (10/50) * 100 = 20%. Rather, our profit percentage will be (10/100) * 100 = 10% only, so \$100 would be our actual cost price.

Keeping this in mind, here is the correct algebra solution:

The total cost price = (1250/6)*60 = \$12,500

The total selling price = 54 * 250 + \$625 = \$13,500 + \$625 = \$14,125 (60 cameras were sold, 54 at \$250 each and 6 at 50% of cost price)

The profit = 14,125 – 12,500 = \$1625 (same as before)

The profit percentage = (1625/12,500) * 100 = 13%

Obviously, we can always use our trusted weighted averages formula here for a quick and efficient solution:

Weighted Averages

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. The ratio of the cost price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Average Profit/Loss percentage = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit.

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

# Advanced Averages Concepts for the GMAT

Let’s discuss an advanced averages concept today.

Say, you have the following set of consecutive integers: 2, 3, 4, 5, 6, 7, 8

What is the average of this set? There are 7 consecutive integers here and the average is 5, the middle number.

Say the set is changed to: 2, 3, 4, 5, 6, 7, 8, 9 (another consecutive number is added to the extreme right). Now what is the average? It is the average of the two middle numbers (5+6)/2 = 5.5.

Let’s edit the set one more time: 1, 2, 3, 4, 5, 6, 7, 8, 9 (another consecutive number is added to the extreme left). The average now is 5 again.

Whenever you add a number on either side of a set of consecutive integers, the average changes by 0.5. This is obvious because odd number of consecutive integers have the middle number as the average and an even number of consecutive integers have the average of two middle numbers as the average. Since every time you add an integer, the number of integers changes from odd to even or from even to odd, the average changes by 0.5.

By the same logic, what happens when you remove an integer from either extreme?

Given a set 3, 4, 5, 6, 7, 8, 9, how will its average change if you remove 3?

The average of 3, 4, 5, 6, 7, 8, 9 is 6, and the average of 4, 5, 6, 7, 8, 9 is 6.5 — the average increases to 6.5 because you removed a small number.

Now how will the average change if you remove 9 instead of 3?

The average of 3, 4, 5, 6, 7, 8, 9 is 6, and the average of 3, 4, 5, 6, 7, 8 is 5.5 — here, the average decreases to 5.5 because you removed a large number.

So, every time you add or remove a number from one of the extremes, the average will move by 0.5.

What happens if you remove a number from somewhere in the middle?

The average changes but by how much? When you remove the greatest or the least number, the average changes by 0.5. So when you remove some other number, the average will change by something less than 0.5. For example, from the set 3, 4, 5, 6, 7, 8, 9, if you remove 8, the average changes from 6 to 5.667. If instead, you remove 7, the average changes to 5.833.

A few takeaways:

1. When you remove an integer very close to the average, the average changes by very little. If you remove the average, the average doesn’t change (changes by 0). When you remove a number close to the extreme, the average changes by a larger number (up to a maximum of 0.5).
2. When you remove a number less than the average, the average increases. When you remove a number more than the average, the average decreases.
3. When you remove the smallest number, the average increases by 0.5. When you remove the greatest number, the average decreases by 0.5.

Now, a question based on this concept:

In a class, the teacher wrote a set of consecutive integers beginning with 1 on the blackboard. A student erased one number. The average of the remaining numbers was 29(14/19). What was the number that the student erased?

(A) 13

(B) 16

(C) 28

(D) 36

(E) 50

Solution:

The numbers on the board: 1, 2, 3, 4, …

The new average is 29(14/19). Since the average changes by not more than 0.5 when you remove an integer from a set of consecutive integers, the original average was either 29.5 or 30. So originally there were either 58 numbers (average 29.5) or 59 numbers (average 30).

When you remove a number, you are left with either 57 numbers or with 58 numbers. Now, the new average will tell you whether you are left with 57 numbers or 58 numbers. The denominator is 19 in the fraction, so when you divide the sum of all remaining integers by the number of integers, the number of integers (denominator) is 19 or a multiple of 19 — 57 is a multiple of 19, 58 is not. So you must have been left with 57 integers and the original number of integers must be 58. This means the original average must have been 29.5.

The original average of 29(1/2) increases to 29(14/19), i.e. an increase of 14/19 – 1/2 = 9/38.

When an integer was removed, the average increased by 9/38 so the integer must be less than the original average. Now use the concept of average that we have learned. One integer was bringing the rest of the numbers down by 9/38 each so the integer must have been (9/38)*57 = 13.5, which is less than the original average of 29.5.

This means the integer that was removed must have been (29.5 – 13.5) = 16, so the answer is B.

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!

# A Closer Look at Set and Ratio GMAT Quant Questions

Writing this post on Teacher’s Day made me dedicate this post to questions on teachers! Considering that all GMAT questions are written by teachers, oddly enough, I found very few questions actually involving them. Looks like we are a humble bunch! Today, we will discuss two GMAT Quant questions on two different topics of discussion – sets and ratios. Both questions are official and of higher difficulty.

Question 1: Of the 1400 college teachers surveyed, 42% said they considered engaging in research an essential goal. How many of the college teacher surveyed were women?

Statement I: In the survey 36% of men and 50% of women said that they consider engaging in research activity an essential goal.

Statement II: In the survey 288 men said that they consider engaging in research activity an essential goal.

Solution:

On reading the question stem we realise that this question involves two variables:

Research Essential – Not Essential

Men – Women

This should immediately make us think about a matrix. Not that we cannot solve the question without one, but you know that I am a huge proponent of visual approaches.

We are given that 42% of total teachers (1400) considered research essential. So this means that 58% did not consider it essential. No need to actually calculate the number right now, let’s wait and see what else we know (anyway, we love to procrastinate calculations in Data Sufficiency questions).

Statement I: In the survey 36% of men and 50% of women said that they consider engaging in research activity an essential goal.

Say the number of women is W. We need the value of W. The number of men must be ‘Total – W’ = 1400 – W. 36% of men and 50% of women consider research essential. Knowing this, we see that we get:

36% * (1400 – W) + 50% * W = 42% * 1400

This is a linear equation in W so we can solve it to get the value of W. Therefore, this statement alone is sufficient.

Statement II: In the survey 288 men said that they consider engaging in research activity an essential goal.

This statement doesn’t tell us the number of women who consider research essential, so it is not sufficient alone, therefore the answer is A, Statement I alone is sufficient but Statement II is not.

Question 2: If the ratio of the number of teachers to the number of students is the same in School District A and School District B, what is the ratio of the number of students in School District A to the number of students in School District B?

Statement I: There are 10,000 more students in School District A than there are in School District B.

Statement II: The ratio of the number of teachers to the number of students in School District A is 1 to 20.

Solution:

In both schools, the ratio of the number of teachers : the number of students is the same.

Statement I: There are 10,000 more students in School District A than there are in School District B.

We don’t know the number of students in either school district, so it is not informative enough to know that School District A has 10,000 more students. Therefore, this statement alone is not sufficient.

Statement II: The ratio of the number of teachers to the number of students in School District A is 1 to 20.

With this statement, we know that the ratio of the number of teachers : the number of students in School District A = 1:20.

Say the number of teachers in A = a; the number of students in A = 20a. We also know the ratio of the number of teachers : the number of students in School District B = 1:20.

Say the number of teachers in B = b; the number of students in B = 20b. Mind you, we don’t know the value of a and b. All we know is that the teacher student ratio is 1:20 in both.

The ratio of the number of students in A: the number of students in B = 20a : 20b = a:b. With this ratio, we don’t know a:b (even using both statements, we just know that a – b = 10,000). Therefore, the answer is E, Statements 1 and 2 together are not sufficient.

Were you able to solve both questions effortlessly? No? Don’t worry, that’s what we are here for! (Ignore the preposition at the end. It sounds most natural this way.)

Not so humble anymore, eh? :)

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

# How to Compare Effectively During the GMAT

A lot of GMAT test takers complain about insufficient time. This is understandable as far as the Verbal section is concerned. We all have different reading speeds and that itself accounts for a lot of time issues in the Verbal section. Obviously then there are other factors – your comfort with the language, your comprehension skills, your conceptual understanding of the Verbal question types, etc.

However, timing issues should not arise in the Quant section. Your reading speed has very little effect on the overall timing scheme because most of the time during the Quant section is spent in solving the question. So if you are falling short on time, it means the methods you are using are not appropriate. We have said it before and will say it again – most GMAT Quant questions can be done in under one minute if you just look for the right thing.

For example, of the four listed numbers below, which number is the greatest and which is the least?

2/3

2^2/3^2

2^3/3^3

Sqrt(2)/Sqrt(3)

Now, how much time you take to solve this depends on how you approach this problem. If you get into ugly calculations, you will end up wasting a ton of time.

2/3 = .667

2^2/3^2 = 4/9 = .444

2^3/3^3 = 8/27 = .296

Sqrt(2)/Sqrt(3) = 1.414/1.732 = .816

So we know that the greatest is Sqrt(2)/Sqrt(3) and the least is 2^3/3^3. We got the answer but we wasted at least 2-3 mins in getting it.

We can do the same thing very quickly. We know that the squares/cubes/roots etc of numbers vary according to where the numbers lie on the number line.

2/3 lies in between 0 and 1, as does 1/4.

The Sqrt(1/4) = 1/2, which is greater than 1/4, so we know that the Sqrt(2/3) will be greater than 2/3 as well.

Also, the square and cube of 1/4 is less than 1/4, so the square and cube of 2/3 will also be less than 2/3. So the comparison will look like this:

(2/3)^3 < (2/3)^2 < 2/3 < Sqrt(2/3)

That is all you need to do! We arrived at the same answer using less than 30 secs.

Using this technique, let’s solve a question:

Which of the following represents the greatest value?

(A) Sqrt(3)/Sqrt(5) + Sqrt(5)/Sqrt(7) + Sqrt(7)/Sqrt(9)

(B) 3/5 + 5/7 + 7/9

(C) 3^2/5^2 + 5^2/7^2 + 7^2/9^2

(D) 3^3/5^3 + 5^3/7^3 + 7^3/9^3

(E) 3/5 + 1 – 5/7 + 7/9

Such a question can baffle someone who believes in calculating everything. We know better than that!

Note that the base values in all the options are 3/5, 5/7 and 7/9. This should hint that we need to compare term to term and not the entire expressions. Also, all values lie between 0 and 1 so they will behave the same way.

Sqrt(3)/Sqrt(5) is the same as Sqrt(3/5). The square root of a number between 0 and 1 is greater than the number itself.

3^2/5^2 is the same as (3/5)^2. The square (and cube) of a number between 0 and 1 is less than the number itself.

So, the comparison will look like this:

(3/5)^3 < (3/5)^2 < 3/5 < Sqrt(3/5)

(5/7)^3 < (5/7)^2 < 5/7 < Sqrt(5/7)

(7/9)^3 < (7/9)^2 < 7/9 < Sqrt(7/9)

This means that out of (A), (B), (C) and (D), the greatest one is (A).

Now we just need to analyse (E) and compare it with (B).

The first term is the same, 3/5.

The last term is the same, 7/9.

The only difference is that (B) has 5/7 in the middle and (E) has 1 – 5/7 = 2/7 in the middle. So (E) is certainly less than (B).

We already know that (A) is greater than (B), so we can say that (A) must be the greatest value.

A quick recap of important number properties:

Case 1: N > 1

N^2, N^3, etc. will be greater than N.

The Sqrt(N) and the CubeRoot(N) will be less than N.

The relation will look like this:

… CubeRoot(N) < Sqrt(N) < N < N^2 < N^3 …

Case II: 0 < N < 1

N^2, N^3 etc will be less than N.

The Sqrt(N) and the CubeRoot(N) will be greater than N.

The relation will look like this:

… N^3 < N^2 < N < Sqrt(N) < CubeRoot(N)  …

Case III: -1 < N < 0

Even powers will be greater than N and positive; Odd powers will be greater than N but negative.

The square root will not be defined, and the cube root of N will be less than N.

CubeRoot(N) < N < N^3 < 0 < N^2

Case IV: N < -1

Even powers will be greater than N and positive; Odd powers will be less than N.

The square root will not be defined, and the cube root of N will be greater than N.

N^3 < N < CubeRoot(N) < 0 < N^2

Note that you don’t need to actually remember these relations, just take a value in each range and you will know how all the numbers in that range behave.

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

# Finding the Product of Factors on GMAT Questions

We have discussed how to find the factors of a number and their properties in these two posts:

Writing Factors of an Ugly Number

Factors of Perfect Squares

Today let’s discuss the concept of ‘product of the factors of a number’.

From the two posts above, we know that the factors equidistant from the centre multiply to give the number. We also know that the behaviour is a little different for perfect squares. Let’s take two examples to understand this.

Example 1: Say N = 6

Factors of 6 are 1, 2, 3, 6

1*6 = 6 (first factor * last factor)

2*3 = 6 (second factor and second last factor)

Product of the four factors of 6 is given by 1*6 * 2*3 = 6*6 = 6^2 = [Sqrt(N)]^4

Example 2: Say N = 25 (a perfect square)

Factors of 25 are 1, 5, 25

1*25 = 25 (first factor * last factor)

5*5 = 25 (middle factor multiplied by itself)

Product of the three factors of 25 is given by 1*25 * 5 = 5^3 = [Sqrt(N)]^3

If a number, N, can be expressed as: 2^a * 3^b * 5^c *…

The total number of factors f = (a+1)*(b+1)*(c+1)…

The product of all factors of N is given by [Sqrt(N)]^f i.e. N^(f/2)

Let’s look at a couple of questions based on this principle:

Question 1: If the product of all the factors of a positive integer, N, is

2^(18) * 3^(12), how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(18) * 3^(12)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(18) * 3^(12)

a*(a+1)*(b+1)/2 = 18

b*(a+1)*(b+1)/2 = 12

Dividing the two equations, we get a/b = 3/2

Smallest values: a = 3, b = 2. It satisfies our two equations.

Can we have more values for a and b? Can a = 6 and b = 4? No. Then the product a*(a+1)*(b+1)/2 would be much larger than 18.

So N = 2^3 * 3^2

There is only one such value of N.

Question 2: If the product of all the factors of a positive integer, N, is 2^9 * 3^9, how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(9) * 3^(9)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(9) * 3^(9)

a*(a+1)*(b+1)/2 = 9

b*(a+1)*(b+1)/2 = 9

Dividing the two equations, we get a/b = 1/1

Smallest values: a = 1, b = 1 – Does not satisfy our equation

Next set of values: a = 2, b = 2 – Satisfies our equations

All larger values will not satisfy our equations.

Note that we can easily use hit and trial in these questions without actually working through the equations.

This is how we will do it:

N^(f/2) = 2^(18) * 3^(12)

Case 1: Assume values of f/2 from common factors of 18 and 12 – say 2

[2^9 * 3^6]^2

Can f/2 = 2 i.e. can f = 4?

If N = 2^9 * 3^6, total number of factors f = (9+1)*(6+1) = 70

This doesn’t work.

Case 2: Assume f/2 is 6

[2^3 * 3^2]^6

Can f/2 = 6 i.e. can f = 12?

If N = 2^3 * 3^2, total number of factors f = (3+1)*(2+1) = 12

This works.

The reason hit and trial isn’t a bad idea is that there will be only one such set of values. If we can quickly find it, we are done.

Why should we then bother to find it at all. Shouldn’t we just answer with option ‘B’ in both cases? Think of a case in which the product of all factors is given as 2^(16) * 3^(14). Will there be any value of N in such a case?

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

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

# How to Solve Relative Rate of Work Questions on the GMAT

Today, we look at the relative rate concept  of work, rate and time – the parallel of relative speed of distance, speed and time.

But before we do that, we will first look at one fundamental principle of work, rate and time (which has a parallel in distance, speed and time).

Say, there is a straight long track with a red flag at one end. Mr A is standing on the track 100 feet away from the flag and Mr B is standing on the track at a distance 700 feet away from the flag. So they have a distance of 600 feet between them. They start walking towards each other. Where will they meet? Is it necessary that they will meet at 400 feet from the red flag – the mid point of the distance between them? Think about it – say Mr A walks very slowly and Mr B is super fast. Of the 600 feet between them, Mr A will cover very little distance and Mr B will cover most of the distance. So where they meet depends on their rate of walking. They will not necessarily meet at the mid point. When do they meet at the mid point? When their rate of walking is the same. When they both cover equal distance.

Now imagine that you have two pools of water. Pool A has 100 gallons of water in it and the Pool B has 700 gallons. Say, water is being pumped into pool A and water is being pumped out of pool B. When will the two pools have equal water level? Is it necessary that they both have to hit the 400 gallons mark to have equal amount of water? Again, it depends on the rate of work on the two pools. If water is being pumped into pool A very slowly but water is being pumped out of pool B very fast, at some point, they both might have 200 gallons of water in them. They will both have 400 gallons at the same time only when their rate of pumping is the same. This case is exactly like the case above.

Now let’s go on to the question from the GMAT Club tests which tests this understanding and the concept of  relative rate of work:

Question: Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

(A) 5/(M+K) hours

(B) 6/(M+K) hours

(C) 300/(M+K) hours

(D) 300/(M−K) hours

(E) 60/(M−K) hours

Solution: There are two tanks with different water levels. Note that the rate of pumping is given as K gallons per min and M gallons per min i.e. they are different. So we cannot say that they both will have equal amount of water when they have 350 gallons. They could very well have equal amount of water at 300 gallons or 400 gallons etc. So when one expects that water in both tanks will be at 350 gallon level, one is making a mistake. The two tanks are working for the same time to get their level equal but their rates are different. So the work done is different. Note here that equal level does not imply equal work done. The equal level could be achieved at 300 gallons when work done would be different – 200 gallons removed from tank X and 100 gallons added to tank Y. The equal level could be achieved at 400 gallons when work done would be different again – 100 gallons removed from tank X and 200 gallons added to tank Y.

To achieve the ‘equal level,’ tank Y needs to gain water and tank X needs to lose water. Total 300 gallons (500 gallons – 200 gallons) of work needs to be done. Which tank will do how much depends on their respective rates.

Work to be done together = 300 gallons

Relative rate of work = (K + M) gallons/minute

The rates get added because they are working in opposite directions – one is removing water and the other is adding water. So we get relative rate (which is same as relative speed) by adding the individual rates.

Note here that rate is given in gallons per minute. But the options have hours so we must convert the rate to gallons per hour.

Relative rate of work = (K + M) gallons/minute = (K + M) gallons/(1/60) hour = 60*(K + M)  gallons/hour

Time taken to complete the work = 300/60(K+M) hours = 5/(K+M) hours

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!

# Identifying the Paradox on GMAT Critical Reasoning Questions

Let’s take a look at a very tricky GMAT Prep critical reasoning problem today. Problems such as these make CR more attractive than RC and SC to people who have a Quantitative bent of mind. It’s one of the “explain the paradox” problems, which usually tend to be easy if you know exactly how to tackle them, but the issue here is that it is hard to put your finger on the paradox.

Once you do, then the problem is quite easy.

Question: Technological improvements and reduced equipment costs have made converting solar energy directly into electricity far more cost-efficient in the last decade. However, the threshold of economic viability for solar power (that is, the price per barrel to which oil would have to rise in order for new solar power plants to be more economical than new oil-fired power plants) is unchanged at thirty-five dollars.

Which of the following, if true, does most to help explain why the increased cost-efficiency of solar power has not decreased its threshold of economic viability?

(A) The cost of oil has fallen dramatically.

(B) The reduction in the cost of solar-power equipment has occurred despite increased raw material costs for that equipment.

(C) Technological changes have increased the efficiency of oil-fired power plants.

(D) Most electricity is generated by coal-fired or nuclear, rather than oil-fired, power plants.

(E) When the price of oil increases, reserves of oil not previously worth exploiting become economically viable.

Solution: We really need to understand this \$35 figure that is given. The argument calls it “the threshold of economic viability for solar plant.” It is further explained as price per barrel to which oil would have to rise in order for new solar power plants to be more economical than new oil-fired power plants.

Note the exact meaning of this “threshold of economic viability”. It is the price TO WHICH oil would have to rise to make solar power more economical i.e. the price to which oil would have to rise to make electricity generated out of oil power plants more expensive than electricity generated out of solar power plants. So this is a hypothetical price of oil. It is not the price BY WHICH oil would have to rise. So this number 35 has nothing to do with the actual price of oil right now – it could be \$10 or \$15. The threshold of economic viability will remain 35.

So what the argument tells us is that tech improvements have made solar power cheaper but the price to which oil should rise has stayed the same. If you are not sure where the paradox is, let’s take some numbers to understand:

Previous Situation:

– Sunlight is free. Infrastructure needed to convert it to electricity is expensive. Say for every one unit of electricity, you need to spend \$50 in a solar power plant.

– Oil is expensive. Infrastructure needed to convert it to electricity, not so much. Say for every one unit of electricity, the oil needed costs \$25 and cost of infrastructure to produce a unit of electricity is \$15. So total you spend \$40 for a unit of electricity in an oil fired plant.

Oil based electricity is cheaper. If the cost of oil rises by \$10 and becomes \$35 from \$25 assumed above, solar power will become viable. Electricity produced from both sources will cost the same.

Again, note properly what the \$35 implies.

Raw material cost in solar plant + Infrastructure cost in solar plant = Raw material cost in oil plant + Infrastructure cost in oil plant

0 + 50 = Hypothetical cost of oil + 15

Hypothetical cost of oil = 50 – 15

That is, this \$35 = Infra price per unit in solar plant – Infra price per unit in oil plant

This threshold of economic viability for solar power is the hypothetical price per barrel to which oil would have to rise (mind you, this isn’t the actual price of oil) to make solar power viable.

What happens if you need to spend only \$45 in a solar power plant for a unit of electricity? Now, for solar viability, ‘cost of oil + cost of infrastructure in oil power plant’ should be only \$45. If ‘cost of infrastructure in oil power plant’ = 15, we need the oil to go up to \$30 only. That will make solar power plants viable. So the threshold of economic viability will be expected to decrease.

Now here lies the paradox – The argument tells you that even though the cost of production in solar power plant has come down, the threshold of economic viability for solar power is still \$35! It doesn’t decrease. How can this be possible? How can you resolve it?

One way of doing it is by saying that ‘Cost of infrastructure in oil power plant’ has also gone down by \$5.

Raw material cost in solar plant + Infrastructure cost in solar plant = Raw material cost in oil plant + Infrastructure cost in oil plant

0 + \$45 = \$35 + Infrastructure cost in oil plant

Infrastructure cost in oil plant = \$10

Current Situation:

– Sunlight is free. Infrastructure needed to convert it to electricity is expensive. For every one unit of electricity, you need to spend \$45 in a solar power plant.

– Oil is expensive. Infrastructure needed to convert it to electricity, not so much. For every one unit of electricity, you need to spend \$25 + \$10 = \$35 in an oil fired power plant.

You still need the oil price to go up to \$35 so that cost of electricity generation in oil power plant is \$45.

So you explained the paradox by saying that “Technological changes have increased the efficiency of oil-fired power plants.” i.e. price of infrastructure in oil power plant has also decreased.

Hence, option (C) is correct.

The other option which seems viable to many people is (A). But think about it, the actual price of the oil has nothing to do with ‘the threshold of economic viability for solar power’. This threshold is \$35 so you need the oil to go up to \$35. Whether the actual price of oil is \$10 or \$15 or \$20, it doesn’t matter. It still needs to go up to \$35 for solar viability. So option (A) is irrelevant.

We hope the paradox and its solution make sense.

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

# 3 Important Concepts for Statistics Questions on the GMAT

We have discussed these three concepts of statistics in detail:

– Arithmetic mean is the number that can represent/replace all the numbers of the sequence. It lies somewhere in between the smallest and the largest values.

– Median is the middle number (in case the total number of numbers is odd) or the average of two middle numbers (in case the total number of numbers is even).

– Standard deviation is a measure of the dispersion of the values around the mean.

A conceptual question is how these three measures change when all the numbers of the set are varied is a similar fashion.

For example, how does the mean of a set change when all the numbers are increased by say, 10? How does the median change? And what about the standard deviation? What happens when you multiply each element of a set by the same number?

Let’s discuss all these cases in detail but before we start, we would like to point out that the discussion will be conceptual. We will not get into formulas though you can arrive at the answer by manipulating the respective formulas.

When you talk about mean or median or standard deviation of a list of numbers, imagine the numbers lying on the number line. They would be spread on the number line in a certain way. For example,

——0—a———b—c———————d———e————————f—g———————

Case I:

When you add the same positive number (say x) to all the elements, the entire bunch of numbers moves ahead together on the number line. The new numbers a’, b’, c’, d’, e’, f’ and g’ would look like this

——0——————a’———b’—c’———————d’———e’————————f’—g’——————

The relative placement of the numbers does not change. They are still at the same distance from each other. Note that the numbers have moved further to the right of 0 now to show that they have moved ahead on the number line.

The mean lies somewhere in the middle of the bunch and will move forward by the added number. Say, if the mean was d, the new mean will be d’ = d + x.

So when you add the same number to each element of a list,

New mean = Old mean + Added number.

On similar lines, the median is the middle number (d in this case) and will move ahead by the added number. The new median will be d’ = d + x

So when you add the same number to each element of a list,

New median = Old median + Added number

Standard deviation is a measure of dispersion of the numbers around the mean and this dispersion does not change when the whole bunch moves ahead as it is. Standard deviation does not depend on where the numbers lie on the number line. It depends on how far the numbers are from the mean. So standard deviation of 3, 5, 7 and 9 is the same as the standard deviation of 13, 15, 17 and 19. The relative placement of the numbers in both the cases will be the same. Hence, if you add the same number to each element of a list, the standard deviation will stay the same.

Case II:

Let’s now move on to the discussion of multiplying each element by the same positive number.

The original placing of the numbers on the number line looked like this:

——0—a———b—c———————d———e————————f—g———————

The new placing of the numbers on the  number line will look something like this:

——0———a’——————b’———c’————————————d’—————————e—- etc

The numbers spread out. To understand this, take an example. Say, the initial numbers were 10, 20 and 30. If you multiply each number by 2, the new numbers are 20, 40 and 60. The difference between them has increased from 10 to 20.

If you multiply each number by x, the mean also gets multiplied by x. So, if d was the mean initially, d’ will be the new mean which is x*d.

New mean = Old mean * Multiplied number

Similarly, the median will also get multiplied by x.

New median = Old median * Multiplied number

What happens to standard deviation in this case? It changes! Since the numbers are now further apart from the mean, their dispersion increases and hence the standard deviation also increases. The new standard deviation will be x times the old standard deviation. You can also establish this using the standard deviation formula.

New standard deviation = Old standard deviation * Multiplied number

The same concept is applicable when you increase each number by the same percentage. It is akin to multiplying each element by the same number. Say, if you increase each number by 20%, you are, in effect, multiplying each number by 1.2. So our case II applies here.

Now, think about what happens when you subtract/divide each element by the same number.

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!

# What to Do When Math Fails You on the GMAT!

They say Mathematics is a perfect Science. There is a debate over this among scientists but we can definitely say that Mathematical methods are not perfect so we cannot use them blindly. We could very well use the standard method for some given numbers and get stranded with “no solution.” The issue is what do we do when that happens?

For example, review this post on averages.

Here we saw that:

Average Speed = 2ab/(a + b)

Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds.

So now, say if we have a question which looks like this:

Question: In the morning, Chris drives from Toronto to Oakville and in the evening he drives back from Oakville to Toronto on the same road. Was his average speed for the entire round trip less than 100 miles per hour?

Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville.

Statement 2: In the evening, Chris drove at an average speed which was no more than 50 miles per hour while travelling from Oakville to Toronto.

Solution: We know that the question involves average speed. The case involves travelling at a particular average speed for one half of the journey and at another average speed for the other half of the journey.

So average speed of the entire trip will be given by 2ab/(a+b)

But the first problem is that we are given a range of speeds. How do we handle ‘at least 10’ and ‘no more than 50’ in equation form? We have learnt that we should focus on the extremities so let’s analyse the problem by taking the numbers are the extremities:10 and 50

Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville.

What if Chris drives at an average speed of 10 mph in the morning and averages 100 mph for the entire journey? What will be his average speed in the evening? Perhaps around 200, right? Let’s see.

100 = 2*10*b/(10 + b)

1000 + 100b = 20b

1000 = -80b

b = – 1000/80

How can speed be negative?

Let’s hold on here and try the same calculation for statement 2 too.

Statement 2: In the evening, Chris drove at an average speed which no more than 50 miles per hour while travelling from Oakville to Toronto.

If Chris drives at an average speed of 50 mph in the evening, and averages 100 mph, let’s find his average speed in the morning.

100 = 2a*50/(a + 50)

100a + 5000 = 100a

5000 = 0

This doesn’t make any sense either!

What is going wrong? Look at it conceptually:

Say, Toronto is 100 miles away from Oakville. If Chris wants his average speed to be 100 mph over the entire trip, he should cover 100+100 = 200 miles in 2 hrs.

What happens when he travels at 10 mph in the morning? He takes 100/10 = 10 hrs to reach Oakville in the morning. He has already taken more time than what he had allotted for the entire round trip. Now, no matter what his speed in the evening, his average speed cannot be 100mph. Even if he reaches Oakville to Toronto in the blink of an eye, he would have taken 10 hours and then some time to cover the total 200 miles distance. So his average speed cannot be equal to or more than 200/10 = 20 mph.

Similarly, if he travels at 50 mph in the evening, he takes 2 full hours to travel 100 miles (one side distance). In the morning, he would have taken some time to travel 100 miles from Toronto to Oakville. Even if that time is just a few seconds, his average speed cannot be 100 mph under any circumstances.

But statement 1 says that his speed in morning was at least 10 mph which means that he could have traveled at 10 mph in the morning or at 100 mph. In one case, his average speed for the round trip cannot be 100 mph and in the other case, it can very well be. Hence statement 1 alone is not sufficient.

On the other hand, statement 2 says that his speed in the evening was 50 mph or less. This means he would have taken AT LEAST 2 hours in the morning. So his average speed for the round trip cannot be 100 mph under any circumstances. So statement 2 alone is sufficient to answer this question with ‘No’.

Takeaway: If your average speed is s for a certain trip, your average speed for half the distance must be more than s/2.

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!

# Expression vs Equation on GMAT

Today, we want to take up a conceptual discussion on expressions and equations and the differences between them. The concept is quite simple but a discussion on these is warranted because of the similarity between the two.

An expression contains numbers, variables and operators.

For example

x + 4

2x – 4x^2

5x^2 + 4x -18

and so on…

These are all expressions. We CANNOT equate these expressions to 0 by default. We cannot solve for x in these cases. As the value of x changes, the value of the expression changes.

For example, given x + 4, if x is 1, value of the expression is 5. If x is 2, value of the expression is 6. If value of the expression is given to be 10, x is 6 and so on.

We cannot say, “Solve x + 4.”

If we set an algebraic expression equal to something, with an “=“ sign, we have an equation.

So here are some ways of converting the above expressions into equations:

I. x + 4 = -3

II. 2x – 4x^2 = 0

III. 5x^2 + 4x -18 = 3x

Now the equation can be solved. Note that the right hand side of the equation needn’t always be 0. It might be something other than 0 and you might need to make it 0 by bringing whatever is on the right hand side to the left hand side or by segregating the variable if possible:

I. x + 4 = -3

x + 7 = 0

x = -7

II. 2x – 4x^2 = 0

2x(1 – 2x) = 0

x = 0 or 1/2

III. 5x^2 + 4x -18 – 3x = 0

5x^2 + x – 18 = 0

5x^2 + 10x – 9x – 18 = 0

5x(x + 2) -9(x + 2) = 0

(x + 2)(5x – 9) = 0

x = -2, 9/5

In each of these cases, we get only a few values for x because we were given equations.

Think about what you mean by “solving an equation”. Let’s take a particular type of equation – a quadratic.

This is how you usually depict a quadratic:

f(x) = ax^2 + bx + c

or

y = ax^2 + bx + c

This is a parabola – upward facing if a is positive and downward facing if a is negative.

When we solve ax^2 + bx + c = 0 for x, it means, when y = 0, what is the value of x? So you are looking for x intercepts.

When we solve ax^2 + bx + c = d for x, it means, when y = d, what is the value of x? Depending on the values of a, b, c and d, you may or may not get values for x.

Let’s take an example:

x^2 – 2x – 3 = 0

(x + 1)(x – 3) = 0

x = -1 or 3

This is what it looks like:

When y is 0, x can take two values: -1 and 3.

So what do we do when we have x^2 – 2x -3 = -3?

We solve it in the same way:

x^2 – 2x -3 + 3 = 0

x(x – 2) = 0

x = 0 or 2

So when y is -3, x is 0 or 2. It has 2 values for y = -3 as is apparent from the graph too.

Similarly, you can solve for it when y = 5 and get two values for x.

What happens when you put y = -5? x will have no value for y = -5 so the equation x^2 – 2x – 3 = – 5 has no real solutions (so ‘no solutions’ as far as we are concerned).

We hope you understand the difference between an expression and an equation now and also that you cannot equate any given expression to 0 and solve it.

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

# When Not to Use Number Plugging on the GMAT

A few weeks back we discussed the kind of questions which beg you to think of the process of elimination – a strategy probably next only to number plugging in popularity.

Today we discuss the kind of questions which beg you to stay away from number plugging (but somehow, people still insist on using it because they see variables).

Not every question with variables is suitable for number plugging. If there are too many variables, it can be confusing and error prone. Then there are some other cases where number plugging is not suitable. Today we discuss an official question where you face two of these problems.

Question: If m, p , s and v are positive, and m/p < s/v, which of the following must be between m/p and s/v?

I. (m+s)/(p+v)

II. ms/pv

III. s/v – m/p

(A) None

(B) I only

(C) II only

(D) III only

(E) I and II both

Solution: The moment people see m, p, s and v variables, they jump to m = 1, p = 2 etc.

But two things should put you off number plugging here:

– There are four variables – just too many to plug in and manage.

– The question is a “must be true” question. Plugging in numbers is not the best strategy for ‘must be true’ questions. If you know that say, statement 1 holds for some particular values of m, p, s and v (say, 1, 2, 3 and 4), that’s fine but how do you know that it will be true for every set of valid values of m, p, s and v? You cannot try every set because the variables can take an infinite variety of values. If you find a set of values for the variables such that statement 1 does not hold, then you know for sure that it may not be true. In this case, number plugging does have some use but it may be a while before you can arrive at values which do not satisfy the conditions. In such questions, it is far better to take the conceptual approach.

We can solve this question using some number line and averaging concepts.

We are given that m/p < s/v

This means, this is how they look on the number line:

…………. 0 ……………….. m/p …………………… s/v ……………..

(since m, p, s and v are all positive (not necessarily integers though) so m/p and s/v are to the right of 0)

Let’s look at statement II and III first since they look relatively easy.

II. ms/pv

Think of the case when m/p and s/v are both less than 1. When you multiply them, they will become even smaller. Say .2*.3 = .06. So the product ms/pv may not lie between m/p and s/v.

Tip: When working with number properties, you should imagine the number line split into four parts:

• less than -1
• between -1 and 0
• between 0 and 1
• greater than 1

Numbers lying in these different parts behave differently. You should have a good idea about how they behave.

III. s/v – m/p

Think of a case such as this:

…………. 0 ………………………… m/p … s/v ……….

s/v – m/p will be much smaller than both m/p and s/v and will lie somewhere “here”:

…………. 0 ……… here ………………… m/p … s/v ……….

So the difference between them needn’t actually lie between them on the number line.

Hence s/v – m/p may not be between m/p and s/v.

I. (m+s)/(p+v)

This is a little tricky. Think of the four numbers as N1, N2, D1, D2 for ease and given fractions as N1/D1 and N2/D2.

(m+s)/(p+v)

= [(m+s)/2]/[(p+v)/2]

= (Average of N1 and N2)/(Average of D1 and D2)

Now average of the numerators will lie between N1 and N2 and average of the denominators will lie between D1 and D2. So (Average of N1 and N2)/(Average of D1 and D2) will lie between N1/D1 and N2/D2. Try to think this through.

We will try to explain this but you must take some examples to ensure that you understand it fully. When is one fraction smaller than another fraction?

When N1/D1 < N2/D2, one of these five cases will hold:

• N1 < N2 and D1 = D2 . For example: 2/9 and 4/9

Average of numerators/Average of denominators = 3/9 (between N1/D1 and N2/D2)

• N1 < N2 and D1 > D2. For example: 2/11 and 4/9

Average of numerators/Average of denominators = 3/10 (between N1/D1 and N2/D2)

• N1 << N2 and D1 < D2. For example: 2/9 and 20/19 i.e. N1 is much smaller than N2 as compared with D1 to D2.

Average of numerators/Average of denominators = 11/14 (between N1/D1 and N2/D2)

• N1 = N2 but D1 > D2. For example: 2/9 and 2/7

Average of numerators/Average of denominators = 2/8 (between N1/D1 and N2/D2)

• N1 > N2 but D1 >> D2. For example: 4/9 and 2/1

Average of numerators/Average of denominators = 3/5 (between N1/D1 and N2/D2)

In each of these cases, (average of N1 and N2)/(average of D1 and D2) will be greater than N1/D1 but smaller than N2/D2. Take some more numbers to understand why this makes sense. Note that you are not expected to conduct this analysis during the test. The following should be your takeaway from this question:

Takeaway: (Average of N1 and N2)/(Average of D1 and D2) will lie somewhere in between N1/D1 and N2/D2 (provided N1. N2, D1 and D2 are positive)

(m+s)/(p+v) must lie between m/p and s/v.

Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on 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!

# Solving Inference-Based 700+ Level Official GMAT Questions

Sometimes, to solve some tough questions, we need to make inferences. Those inferences may not be apparent at first but once you practice, they do become intuitive. Today we will discuss one such inference based high level question of an official GMAT practice test.

Question: In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

(A) 65

(B) 55

(C) 45

(D) 35

(E) 25

Solution: We need to find the value of x – y

What is x? It is the greatest possible number of households that have all three devices

What is y? It is the lowest possible number of households that have all three devices

Say there are 100 households and we have three sets:

Set DVD including 75 households

Set Cell including 80 households

Set MP3 including 55 households

We need to find the values of x and y to get x – y.

We need to maximise the overlap of all three sets to get the value of x and we need to minimise the overlap of all three sets to get the value of y.

Maximum number of households that have all three devices:

We want to bring the circles to overlap as much as possible.

The smallest set is the MP3 set which has 55 households. Let’s make it overlap with both DVD set and Cell set. These 55 households are the maximum that can have all 3 things. The rest of the 45 households will definitely not have an MP3 player. Hence the value of x must be 55.

Note here that the number of households having no device may or may not be 0 (it doesn’t concern us anyway but confuses people sometimes). There are 75 – 55 = 20 households that have DVD but no MP3 player. There are 80 – 55 = 25 households that have Cell phone but no MP3 player. So they could make up the rest of the 45 households (20 + 25) such that these 45 households have exactly one device or there could be an overlap in them and hence there may be some households with no device. In the figure we show the case where none = 0.

Now, let’s focus on the value of y i.e. minimum number of households with all three devices:

How will we do that? Before we delve into it, let us consider a simpler example:

Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do?

Will you leave out one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them (one each). This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them.

This is the same concept.

When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all three. So you give at least one to all of them. Here there will be no household which has no device. Every household will have at least one device.

So you have 80 households which have cell phone. The rest of the 20 households say, have a DVD player so the leftover 55 households (75 – 20) with DVD player will have both a cell phone and a DVD player. There are 55 households who already have two devices and 45 households with just one device.

Now how will you distribute the MP3 players such that the overlap between all three is minimum? Give the MP3 players to the households which have just one device so 45 MP3 player households are accounted for. But we still need to distribute 10 more MP3 players. These 10 will fall on the 55 overlap of the previous two sets. Hence there are a minimum of 10 households which will have all three devices. This means y = 10

x – y = 55 – 10 = 45

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!

# When to Make Assumptions on GMAT Problem Solving Questions

Today we will discuss the flip side of “do not assume anything in Data Sufficiency” i.e. we will discuss “go ahead and assume in Problem Solving!”

Problem solving questions have five definite options, that is, “cannot be determined” and “data not sufficient” are not given as options. So this means that in all cases, data is sufficient for us to answer the question. So as long as the data we assume conforms to all the data given in the question, we are free to assume and make the problem simpler for ourselves. The concept is not new – you have been already doing it all along – every time you assume the total to be 100 in percentage questions or the value of n to be 0 or 1, you are assuming that as long as your assumed data conforms to the data given, the relation should hold for every value of the unknown. So the relation should be the same when n is 0 and also the same when n is 1.

Now all you have to do is go a step further and, using the same concept, assume that the given figure is more symmetrical than may seem. The reason is that say, you want to find the value of x. Since in problem solving questions, you are required to find a single unique value of x, the value will stay the same even if you make the figure more symmetrical – provided it conforms to the given data.

Let us give an example from Official Guide 13th edition to show you what we mean:

Question: In the figure shown, what is the value of v+x+y+z+w?

(A) 45

(B) 90

(C) 180

(D) 270

(E) 360

We see that the leg with the angle w seems a bit narrower – i.e. the star does not look symmetrical. But the good news is that we can assume it to be symmetrical because we are not given that angle w is smaller than the other angles.  We can do this because the value of v+x+y+z+w would be unique. So whether w is much smaller than the other angles or almost the same, it doesn’t matter to us. The total sum will remain the same. Whatever is the total sum when w is very close to the other angles, will also be the sum when w is much smaller. So for our convenience, we can assume that all the angles are the same.

Now it is very simple to solve. Imagine that the star is inscribed in a circle.

Now, arc MN subtends the angle w at the circumference of the circle; this angle w will be half of the central angle subtended by MN (by the central angle theorem discussed in your book).

Arc NP subtends angle v at the circumference of the circle; this angle v will be half of the central angle subtended by NP and so on for all the arcs which form the full circle i.e. PQ, QR and RM.

All the central angles combined measure 360 degrees so all the subtended angles w + v + x + y + z will add up to half of it i.e. 360/2 = 180.

There are many other ways of solving this question including long winded algebraic methods but this is the best method, in my opinion.

This was possible because we assumed that the figure is symmetrical, which we can in problem solving questions!

But beware of question prompts which look like this:

– Which of the following cannot be the value of x?

– Which of the following must be true?

You cannot assume anything here since we are not looking for a unique value that exists. If a bunch of values are possible for x, then x will take different values in different circumstances.

If we know that the unknown has a unique value, then we are free to assume as long as we are working under the constraints of the question. Finally, we would like to mention here that this is a relatively advanced technique. Use it only if you understand fully when and what you can assume.

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

# Advanced Applications of Common Factors on the GMAT – Part II

There is something about factors and divisibility that people find hard to wrap their heads around. Every advanced application of a basic concept knocks people out of their seats! Needless to say, that the topic is quite important so we are trying to cover the ground for you. Here is another post on the topic discussing another important concept.

In a previous post, we saw that

“Two consecutive integers can have only 1 common factor and that is 1.”

This implies that N and N+1 have no common factor other than 1. (N is an integer)

Similarly,

N + 5 and N + 6 have no common factor other than 1. (N is an integer)

N – 3 and N – 2 have no common factor other than 1. (N is an integer)

2N and 2N + 1 have no common factor other than 1. (N is an integer)

We are sure you have no problem up until now.

N and 2N+1 have no common factor other than 1. (N is an integer)

It is a simple application of the same concept but makes for a 700 level question!

2N and 2N+1 have no common factor other than 1 – we know

The factors of N will be a subset of the factors of 2N. It will not have any factors which are not there in the list of factors of 2N. So if 2N and another number have no common factors other than 1, N and the same other number can certainly not have any common factor other than 1.

Taking an example, say N = 6

Factors of 2N (which is 12) are 1, 2, 3, 4, 6, 12.

Factors of 2N + 1 (which is 13) are 1, 13.

2N and 2N + 1 can have no common factors.

Now think, what are the factors of N? They are 1, 2, 3, 6 (a subset of the factors of 2N)

They will obviously not have any factor in common with 2N+1 (except 1) since these are the same factors as those of 2N except that these are fewer.

So we can deduce the following (N and M are integers):

M and NM +1 will have no common factor other than 1.

8 and 8M + 1 will have no common factor other than 1.

M and NM – 1 will have no common factor other than 1.

and so on…

Here is the 700 level official question of this concept:

Question: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

Statement 1: x = 12u, where u is an integer.

Statement 2: y = 12z, where z is an integer.

Solution:

x = 8y + 12

We need to find the greatest common divisor of x and y. We have 8y in the equation. A couple of immediate deductions:

The factors of y will be a subset of the factors of 8y.

The difference between x and 8y is 12 so the greatest common divisor of x and 8y will be a factor of 12 (discussed in this post a few weeks back).

This implies that the greatest factor that x and y can have must be a factor of 12.

Looking at the statements now:

Statement 1: x = 12u, where u is an integer.

Now we know that x has 12 as a factor. The problem is that we don’t know whether y has 12 as a factor.

y could be 3 —> x = 8*3 + 12 = 36 (a multiple of 12). Here greatest common divisor of x and y will be 3.

or y could be 12 —> x = 8*12 + 12 = 108 (a multiple 12). Here greatest common divisor of x and y will be 12.

So this statement alone is not sufficient.

Statement 2: y = 12z, where z is an integer.

This statement tells us that y also has 12 as a factor. So now do we just mark (C) as the answer and move on? Well no! It seems like an easy (C) now, doesn’t it? We must analyse this statement alone.

Substituting y = 12z in the given equation:

x = 8*12z + 12

x = 12*(8z + 1)

So this already gives us that x has 12 as a factor. We don’t really need statement 1.

Since both x and y have 12 as a factor and the highest common factor they can have is 12, greatest common divisor of x and y must be 12.

This statement alone is sufficient to find the greatest common divisor of x and y.

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!

# Important Caveat on Joint Variation GMAT Questions

Before we start today’s discussion, recall a previous post on joint variation. A question arose some days back on the applicability of this concept. This official question was the case in point:

Question: In a certain business, production index p is directly proportional to efficiency index e, which is in turn directly proportional to investment i. What is p if i = 70?

Statement I: e = 0.5 whenever i = 60

Statement II: p = 2.0 whenever i = 50

This was the issue that was raised:

If one were to follow the method given in the post on joint variation, one would arrive at this solution:

p/e = k (a constant)

e/i = m (another constant)

Hence, p*i/e = n is the joint variation expression

(where k, m and n are constants)

So we get that p is inversely proportional to i, that is, p*i = Constant

Statement II gives us the values of p and i which can help us get the value of the Constant.

2*50 = Constant

The question asks us the value of p given the value of i = 70. If Constant = 100,

p = 100/70.

But actually, this is wrong and the value that you get for p in this question is different.

The question is “why is it wrong?”

Valid question, right? It certainly seems like a joint variation scenario – relation between three variables. Then why does’t it work in this case?

The takeaway from this question is very important and before you proceed, we would like you to think about it on your own for a while and then proceed to the the rest of the discussion.

Here is how this question is actually done:

Taking one statement at a time:

“production index p is directly proportional to efficiency index e,”

implies p = ke (k is the constant of proportionality)

“e is in turn directly proportional to investment i”

implies e = mi (m is the constant of proportionality. Note here that we haven’t taken the constant of proportionality as k since the constant above and this constant could be different)

Then, p = kmi (km is the constant of proportionality here. It doesn’t matter that we depict it using two variables. It is still just a number)

Here, p seems to be directly proportional to i!

So if you have i and need p, you either need this constant directly (as you can find from statement II) or you need both k and m (statement I only gives you m).

So the issue now is that is p inversely proportional to i or is it directly proportional to i?

Review the joint variation post – In it we discussed that joint variation gives you the relation between 2 quantities keeping the third (or more) constant.

p will vary inversely with i if and only if e is kept constant.

Think of it this way: if p increases, e increases. But we need to keep e constant, we will have to decrease i to decrease e back to original value. So an increase in p leads to a decrease in i to keep e constant.

But if we don’t have to keep e constant, an increase in p will lead to an increase in e which will increase i.

It is all about the sequence of increases/decreases

Here, we are not given that e needs to be kept constant. So we will not use the joint variation approach.

Note how the independent question is framed in the joint variation post:

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?

You need relation between N and M when reaction rate is constant.

You are given no such constraint here. So an increase in p leads to an increase in e which in turn, increases i.

So let’s complete the solution to our original question:

p = ke

e = mi

p = kmi

Statement I: e = 0.5 whenever i = 60

0.5 = m * 60

m = 0.5/60

We do not know k so we cannot find p given i and m.

This statement alone is not sufficient.

Statement II: p = 2.0 whenever i = 50

2 = km * 50

km = 1/25

If i = 70, p = (1/25)*70 = 14/5

This statement alone is sufficient.

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!

# When Not to Use Parallelism on the GMAT

We know that we are often tested on parallelism on the GMAT. The logically parallel entities should be grammatically parallel. But today, we need to talk about circumstances where you might be tempted to employ parallelism but it would be incorrect to do so.

For example, look at this sentence:

A New York City ordinance of 1897 regulated the use of bicycles, mandated a maximum speed of eight miles an hour, required cyclists to keep feet on pedals and hands on handlebars at all times, and granted pedestrians right-of-way.

Is everything ok here? Well, it certainly seems so. We have four elements in parallel:

regulated …

mandated …

required …

granted …

But actually, there is a problem in this sentence:

‘regulated…’ will not be parallel to the rest of the three elements. The rest of the three elements will be in parallel.

Before we explain why, let’s take a simpler example:

The girl sitting next to me wears blue everyday, eats only waffles, and listens to music in office.

The sentence will not be ‘The girl sits next to me…’ because ‘sit’ is not parallel to other verbs. “sit” modifies the girl and is not used as a verb here. It is a present participle modifier modifying ‘girl’. It specifies the girl about whom we are talking.

Similarly, in the original sentence, ‘regulate’ is modifying ‘ordinance of 1897’. It is telling you which ordinance of 1897.

The other verbs ‘mandated’, ‘required’ and ‘granted’ are used as verbs and are parallel. They are assimilated under ‘regulate’. They tell you how the ordinance regulated.

How did it regulate?

mandated …

required …

granted …

Hence, you cannot use ‘regulated’ here. You must use ‘regulating’  – the present participle modifier to modify the ordinance. So you have to think logically – are the items in the given list actually parallel? Are they equal elements? If yes, then they need to be grammatically parallel too; else not.

Here is the complete official question:

Question: A New York City ordinance of 1897 regulated the use of bicycles, mandated a maximum speed of eight miles an hour, required of cyclists to keep feet on pedals and hands on handlebars at all times, and it granted pedestrians right-of-way.

(A) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required of cyclists to keep feet on pedals and hands on handlebars at all

times, and it granted

(B) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, granting

(C) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists that they keep feet on pedals and hands on handlebars

at all times, and it granted

(D) regulating the use of bicycles, mandating a maximum speed of eight miles an

hour, requiring of cyclists that they keep feet on pedals and hands on

handlebars at all times, and granted

(E) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, and granted

Solution:

From our above discussion, we know that we have choose one of (C), and (E).

(A), (B) and (D) put regulate parallel to the other verbs.

Still, let’s point out all the errors of these options:

(A) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required of cyclists to keep feet on pedals and hands on handlebars at all

times, and it granted

Parallelism problem – regulated cannot be parallel to mandated and other verbs. Also, ‘mandated’ is not parallel to ‘it granted’. Besides, ‘required of X to do Y’ is unidiomatic.

(B) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, granting

Parallelism problem – ‘regulated’ is parallel to ‘mandated’ though it should not be.

‘granting’ is not parallel to ‘mandated’ and ‘required’ though it needs to be parallel.

You also need an ‘and’ before the last element of the list ‘and granted …’

(D) regulating the use of bicycles, mandating a maximum speed of eight miles an

hour, requiring of cyclists that they keep feet on pedals and hands on

handlebars at all times, and granted

This is not a valid sentence because the main clause does not have a verb. ‘regulating…’, ‘mandating…’ and ‘requiring…’ are the present participle modifiers.

‘granted…’ is not parallel to the other elements. Besides, ‘requiring of X that they do Y’ is unidiomatic.

Now let’s look at the leftover options:

(C) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists that they keep feet on pedals and hands on handlebars

at all times, and it granted

‘it granted’ is not parallel to the other verbs. Besides, ‘required X that they do Y’ is unidiomatic.

(E) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, and granted

Perfect! All issues sorted out!

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!

# 2 Possible Ways to Solve this GMAT Quant Question

Process of elimination is only next to number plugging in popularity as a strategy for solving Quant questions on the GMAT. I am not a fan of either method. Yes, they are useful sometimes, and even necessary in some questions but for most questions, I like to use logic/reasoning.

That said, there is a set of questions in which we should think of these strategies. Number plugging is very useful when you have one or two variables in the options. Algebra can be time consuming in these cases because of equation manipulation required.

Similarly, some questions beg you to use the process of elimination. Their question stem goes something like ”which of the following options can be the value of x?”, “which of  the following options cannot be the sum of a and b?” etc. These questions are framed like this because often they have multiple solutions. x could possibly take many different values but the options would have only one of them. So it makes sense to check which values x can take from the options. Let’s look at one such instance of a tricky question where process of elimination can be very useful.

Question:

A list of numbers has six positive integers. Three of those integers are known – 4, 5 and 24 and three of those are unknown – x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Solution: The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.

Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.

Now we know that the mean = 10 and median = 7.5

Method 1: Algebra/Logic

Let’s try to solve the question algebraically/logically first.

There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60

4 + 5 + 24 + x + y + z = 60

x + y + z = 27

Median of the list = 7.5

So sum of third and fourth elements must be 7.5 * 2 = 15

There are two cases possible:

Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.

The sum of the third and fourth elements of the list is 15 so

15 + z = 27

z = 12

So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be

5 and 10

6 and 9

7 and 8

x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.

Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.

If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.

The sum of the other two unknown numbers would be 27 – 10 = 17.

One of them must be less than 5 and the other greater than 10. So possible options are

4 and 13

3 and 14

2 and 15

1 and 16

x, y and z could take various values but none of them could be 11

Method 2: Process of Elimination

Let’s now try to look at the process of elimination here and see if we can find an easier way.

The three unknowns need to add up to 10*6 – 4 – 5 – 24 = 27.

Two of the given options are 5 and 10. They have a median of 7.5 so lets assume that two of the unknown numbers are 5 and 10 (5 can be one of the unknowns since we are not given that all six integers need to be distinct). If two unknowns make up third and fourth numbers in the list and have a median of 7.5, their sum would be 15 and the third unknown will be 12 (to get the mean of 10). This case (5, 10, 12) satisfies all conditions so options (B), (D) and (E) are out of play.

Now we are left with two options 13 and 11. Check any one of them and you will know which one is not possible. Let’s check 13.

From the given options, any number greater than 7.5 must be either the fourth number or the fifth number. 13 cannot be the fourth number since the third number would need to be 2 in that case to get median 7.5. But we have 4 and 5 more than 2 so it cannot be the third number. So 13 must be the fifth number of the list. We saw in the case above that if two unknowns are third and fourth numbers then the fifth number HAS TO BE 12. So the already present 5 must be the third number and the fourth number must be 10. In that case, the leftover unknown would be 4 (to get a sum of 27). So the three unknowns would be 4, 10 and 13. This satisfies all conditions and is possible. Hence answer must be (C). 11 will not be possible.

Let’s see what would have happened had you picked 11 to try out. If 11 were the fourth number, to get a median of 7.5, we would need 4 as the third number. That is not possible since we already have a 5 given. So 11 must have been the fifth number. This would mean that the already present 5 and one unknown 10 would make the median of 7.5. So the third unknown in this case would be 6 (to get a sum of 27). But 6 would be the third number and the median in this case would be (6 + 10)/2 = 8. So one of the numbers cannot be 11.

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 Symmetry Puzzle on the GMAT

A few days back, a student of ours asked me this question – in which cases is symmetry useful to us? Honestly, I don’t think I can create an exhaustive list of the topics where it could be useful. The first thing that comes to mind is of course, Geometry. Circles/equilateral triangles/squares/cubes are symmetrical figures. Symmetry helps us simplify questions which are based on these figures. We have also seen the uses of symmetry in dice throwing. In arrangements too, symmetry helped decrease our work substantially.

Today, let’s look at a puzzle where symmetry helps.

Question: A spider is sitting on one corner of a cube. It wants to get to the most distant corner but it can crawl only along the edges of the cube and cannot revisit a place where it has already been. In how many different ways can the spider go to the most distant corner?

(A) 6

(B) 12

(C) 18

(D) 24

(E) 30

Solution: The question is a puzzle type combinatorics questions. It seems like we will have to painstakingly calculate the various paths that the spider can take. But notice that the figure we have is a cube – a symmetrical figure. Let’s draw the figure to see what the question is asking.

Now, assume that the spider is at A. In that case, he has to go to F – the farthest vertex from A. Every vertex has only one vertex farthest to it. C, E and G are equidistant from A but they are in the same plane. F is further off than C, E and G. So it needs to go from A to F:

Step 1:

It can crawl only along the edges so from A, it can take three different paths – AB or AH or AD. As far as F is concerned, all the points D, H and B  are similar.

Step 2:

Now, from each of these 3 points, the spider has two path options. If it is at D, it can crawl on DE or DC. The third path from D leads back to A but the spider is not allowed to revisit a place. So there are only two forward options for it – DE or DC.

Similarly, if it is at H, it can crawl on HE or on HG. If it is at B, it can crawl on BG or BC. So the total number of paths that we have found till now are  3*2.

Till now, we hope you did not face any problems.

Step 3:

Now comes the tricky part. The spider is at one of three vertices – E, C or G. Assume it came the AD – DE route and is now at E. There are multiple ways in which it can reach F. The obvious one is directly from E to F. But it can also go to F via H because it has not visited a number of other vertices (H, G, B, C)

There are three ways in which it can reach F now:

• directly E to F
• a three path EH – HG – GF
• a five path EH – HG – GB – BC – CF

This takes care of all the ways in which it can reach F from E.

Since we found 3 different paths from E, it is obvious that we will find 3 different paths from C and from G too. It is a symmetrical figure and hence we don’t need to calculate the number of paths from each point. In any case, we have 3 ways to reach F now.

So total different paths to reach the farthest vertex = 3*2*3 = 18

Hope you see how symmetry helps us reduce our work substantially.

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!

# Easy Logic to a Difficult Combinatorics GMAT Question!

Sometimes, you come across some seriously interesting questions in Combinatorics. For example, this question I came across seemed like any other Combinatorics question, though it was a little cumbersome. But when I saw the answer, it got me thinking – it couldn’t have been a coincidence. There had to be a simpler logic to it and there was! I just wish I had thought of it before going the long route. So I must share it with you; you never know what might come in handy on test day!

But before I tell you what that question was, let’s solve a couple of questions which are similar to some questions you might have seen before  (for the sake of brevity,  let’s ignore the options):

Question 1: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get either three paintings or five paintings? (All paintings should be given away).

Solution 1:

There are two ways of distributing the paintings in this case:

Dave gets 3 paintings and Mona gets the rest: You select 3 of the 10 paintings and give them to Dave. This can be done in 10C3 = 120 ways

Dave gets 5 paintings and Mona gets the rest: You select 5 of the 10 paintings and give them to Dave. This can be done in 10C5 = 252 ways

Total number of ways in which you can distribute the paintings = 120 + 252 = 372 ways

Simple enough, right? Let’s take a  look at another simple similar question.

Question 2: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get at least two paintings? (All paintings should be given away.)

Solution 2:

Dave should get at least two paintings so it means he can get 2 or 3 or 4 or more up to 10 paintings. Calculating all those cases would be tedious so this is a perfect opportunity to use ‘Total – Opposite’ method.

Total ways in which you can distribute 10 paintings between two people without any constraints: Each painting can be given away in two ways – either to Dave or to Mona. So the paintings can be distributed in 2*2*2*…*2 = 2^10 = 1024 ways

Number of ways in which Dave gets 0 paintings or 1 painting: 1 + 10C1 = 11 ways

So number of ways in which Dave gets 2 or 3 or 4 … upto 10 (i.e. at least 2 paintings) = 1024 – 11 = 1013 ways

Another ‘seen before, know how to solve it’ kind of question. Now let’s come to the question of the day which doesn’t look much different but actually is.

Question 3: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if both collectors should get an even number of paintings? (All paintings should be given away.)

Solution 3:

Paintings can be distributed in the following ways:

0, 10 – One person gets 0 paintings and the other gets 10

2, 8 – One person gets 2 paintings and the other gets 8

4, 6 – One person gets 4 paintings and the other gets 6

You will need to calculate each one of these ways and then add them. Note that the ‘Total – Opposite’ method does not work here because finding the number of ways in which each person gets odd number of paintings is equally daunting.

Case 1: 0, 10

One person gets 0 paintings and the other gets 10. This can be done in 2 ways – either Dave gets all the paintings or Mona gets them.

Case 2: 2, 8

One person gets 2 paintings and the other gets 8. Select 2 paintings out of 10 for Dave in 10C2 = 45 ways. Mona could also get the 2 selected paintings so total number of ways = 45*2 = 90 ways

Case 3: 4, 6

One person gets 4 paintings and the other gets 6. Select 4 paintings out of 10 for Dave in 10C4 = 210 ways. Mona could also get the 4 selected paintings so total number of ways = 210*2 = 420

Total number of ways such that each person gets even number of paintings = 2 + 90 + 420 = 512 ways

But 512 is 2^9 – in form, suspiciously close to 2^10 we used in question 2 above. Is there some logic which leads to the answer 2^(n-1)? There is!

You have 10 different paintings. Each painting can be given to one of the 2 people in 2 ways. You do that with 9 paintings in 2*2*2…  = 2^9 ways. When you distribute 9 paintings, one person will have odd number of paintings and one will have even number of paintings (0 + 9 or 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5).

The tenth painting needs to be given to the person who has the odd number of paintings so you give the tenth painting in only one way. This accounts for all cases in which both get even number of paintings.

Total ways = 2^9 * 1 = 512

Question 4: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if both collectors should get an odd number of paintings? (All paintings should be given away.)

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

# Advanced Applications of Common Factors on the GMAT

Today we will discuss the logic behind common factors (other than 1) of two numbers.

Without actually finding all the factors of two numbers, how do we know whether they have any common factors (ignoring 1)?

Let’s take some examples:

• If the integers are even, we know that they must have at least one common factor – 2. Let’s say we have two numbers 476 and 478. How many common factors can they have? We know that 2 is a factor common to them. Can they have any other common factor? Note that the difference between them is 2. So if 4 were a factor of 476, could it be a factor of 478? No. If 4 were a factor of 476, it would be a factor of 480 next (4 away from 476). Similarly, if 7 were a factor of 476, it would not be a factor of 478, definitely. It would be a factor of 483 (7 away from 476). In fact, since the difference between the two numbers is 2, the only factor they can have in common is 2.
• Now consider that the two numbers are 476 and 484. They have a difference of 8 between them. The common factors they can have are 2, 4 and 8 (the factors of 8). If any of these factors is a factor of 476, it will be a factor of 484 too. Obviously, 476 and 484 will have many other factors but they will not have any other common factor. 7 is a factor of 476. The next multiple of 7 will be 483 and the next will be 490. 7 cannot be a factor of 484.
• What happens when both integers are odd? Say 525 and 531. The difference between them is 6. The factors of 6 are 2 and 3. Both 525 and 531 are odd numbers so 2 cannot be their factor. If 3 is a factor of 525, it will be a factor of 531 too else it will not be a factor of both the numbers. Any other number can be a factor of one of them, but not both.

This is what we can deduce:

The only factors that CAN be common (it’s not necessary that they will be common) between two numbers are the factors of the difference between them.

If any factor of the difference between them is a factor of one of the numbers, it will be a factor of the other number too. If it is not a factor of one number, it will not be a factor of the other number.

Take a look at a question based on these concepts:

Question: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

Statement 1: m^2 – 10m + 16 = 0

Statement 2: x + 26 is a prime number.

Solution:

The two given positive integers are (x + m) and (x – m). x is a positive integer so m must be an integer too. Whether m is positive or negative, we don’t know.

To know the GCD of two numbers, we need to know their common divisors. As of now, we have no idea about their common divisors, but we know that the difference between the two numbers is 2m. Their common factors must be factors of 2m.

Let’s look at the two statements:

Statement 1: m^2 – 10m + 16 = 0

We know that the quadratic will give us two values for m so we will not be able to find a unique value for m. But let’s solve it in case we get some other clues from it.

m^2 – 10m +16 = 0

m^2 – 2m – 8m + 16 = 0

m (m – 2) – 8 (m – 2) = 0

(m – 2)*(m – 8) = 0

m is either 2 or 8. So 2m is either 4 or 16.

The factors of 2m will be 1, 2 and 4 and additionally, 8 and 16 (if 2m is 16). We have no idea whether x+m and x-m will have these factors so this statement alone is not sufficient.

Statement 2: x + 26 is a prime number.

What does it tell us about x? Other than 2, all prime numbers are odd numbers. Since x is a positive integer,  x+26 cannot be 2. It must be a prime number greater than 2 and hence, must be odd. But 26 is even. So x must be an odd integer (Odd + Even = Odd). But we have no information about m so this statement alone is not sufficient.

Using both statements together, since x is an odd integer and m is definitely even (either 2 or 8), both the numbers (x + m) and (x – m) are odd integers. Odd integers will not have any of these factors: 2, 4, 8, 16.

So (x + m) and (x – m) must have 1 as the only common factor. Hence their greatest common divisor must be 1.

Together, the two statements are sufficient to answer the question.

To recap: Any common factor of two numbers has to be a factor of the difference between them. This also implies that the GCD of two numbers has to be a factor of the difference between them.

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

# Is this GMAT Question an Alphametic or Simple Number Properties Question?

As noticed in the first post of Alphametics, a data sufficiency alphametic is far more complicated than a problem solving alphametic. An alphametic can have multiple solutions and establishing that it does not, is time consuming. Hence, it is less likely that you will see a DS alphametic in the actual exam.

In fact, what may look like an alphametic problem, might actually be a number properties problem only.

We will look at an example below:

Question:

In the correctly-worked multiplication problem above, each symbol represents a different nonzero digit. What is the value of C?

Statement 1: D is prime.

Statement 2: B is not prime.

Solution: We multiply two two-digit integers and get 1995. The good thing is that we know the result of the multiplication will be 1995. Usually, multiplication alphametics are harder since they involve multiple levels, but here the multiplication is actually a blessing. There are many many ways in which you can ADD two integers to give 1995 but there are only a few ways in which you can multiply two integers to give you 1995.

Let’s prime factorize 1995:

1995 = 3*5*7*19

We can probably count on our fingers the number of ways in which we can select AB and CD.

19 needs to be multiplied with one other factor to give us a two digit number since 5*3*7 = 105 (a three digit number) so AB and CD cannot be 19 and 105.

19*3 = 57, 5*7 = 35 – This is not possible since two of the four digits are same here – 5.

19*5 = 95, 3*7 = 21 – This is one option for AB and CD.

19*7 = 133 – Three digit number not possible.

Hence AB and CD can only take values out of 21 and 95.

As of now, C can be 2 or 9. We need to find whether the given statements give us a unique value of C.

Statement 1: D is prime

D is the units digit of CD. So D can be 1 or 5.

1 is not prime so CD cannot be 21. Hence, CD must be 95 and AB must be 21.

Hence, C must be 9.

This statement alone is sufficient.

Statement 2: B is not prime

If B is not prime then AB cannot be 95. Hence AB must be 21.

This means CD will be 95 and C will be 9.

This statement alone is sufficient.