Today, we bring another tip for you to help get that dream score of 51 – if you must write down the data given, write down all of it! Let us explain.

If you think that you will need to jot down the data given in the question and then solve it on your scratch pad (instead of in your mind), you must jot down every single detail. It is easy to overlook small things which are difficult to express algebraically such as ‘x is an integer’. These details are often critical and could make all the difference between an ‘unsolvable’ question and a ‘solvable within 2 minutes’ one. Once you start solving the question on your scratch pad, you will not refer back to the original question again and again and hence might forget these details. Have them along with the rest of the data. Read every word of the question carefully, and ensure that it is consolidated on your scratch pad. For example, look at this question:

A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?

It is a difficult question because it incorporates statistics as well as max-min – both tricky topics. On top of it, people often overlook the ‘are equal’ part of the question here. The reason for that is that they are actively looking for implications of the sentences and the moment they read “The rest three numbers lie between these two numbers”, they go back to the previous sentence which tells us “A particular number among the five exceeds another by 100”. They then make a note of the fact that 100 is the range of the five positive integers. In all this excitement, they miss the three critical words “and are equal”. Ensure that when you go to the sentence above, you pick the next sentence from the point where you left it. Another thing to note here is that all numbers are positive integers. This information will be critical to us.

Let’s demonstrate how you will solve this question after incorporating all the information given.

**Question**: A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?

(A) 18

(B) 19

(C) 21

(D) 42

(E) 59

Solution:

Let’s assume that the 5 natural numbers in increasing order are: a, b, b, b, a+100

We are given that a < b < a+100.

Also, we are given that a and b are positive integers. This information is critical – we will see later why.

The average of the 5 numbers is (a+b+b+b+a+100)/5 = 150

(a+b+b+b+a+100) = 5*150

2a+3b = 650

We need to find the number of distinct values that a can take because a+100 will also take the same number of distinct values.

Now there are two methods to proceed. Let’s discuss both of them.

**Method 1: Pure Algebra** – Write b in terms of a and plug it in the inequality

b = (650 – 2a)/3

a < (650 – 2a)/3 < a+100

3a < 650 – 2a < 3a + 300

Now split it into two inequalities: 3a < 650 – 2a and 650 – 2a < 3a + 300

Inequality 1: 3a < 650 – 2a

5a < 650

a < 130

Inequality 2: 650 – 2a < 3a + 300

5a > 350

a > 70

So we get that 70 < a < 130. Since a is an integer, can we say that a can take all values from 71 to 129? No. What we are forgetting is that b is also an integer. We know that

b = (650 – 2a)/3

For which values will be get b as an integer? Note that 650 is not divisible by 3. You need to add 1 to it or subtract 2 out of it to make it divisible by 3. So a should be of the form 3x+1.

b = (650 – 2*(3x+1))/3 = (648 – 6x)/3 = 216 – 2x

Here, for any positive integer x, b will be an integer.

From 71 to 129, we have the following numbers which are of the form 3x+1:

73, 76, 79, 82, 85, … 127

This is an Arithmetic Progression. How many terms are there here?

Last term = First term + (n – 1)*Common Difference

127 = 73 + (n – 1)*3

n = 19

a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values.

**Method 2: Using Transition Points**

Note that a < b < a+100

Since a < b, let’s find the point where a = b, i.e. the transition point

2a + 3a = 650

a = 130 = b

But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But decreasing a by 3 decreases the largest number i.e. a+100 by 3 too; so we need to increase b by another 1.

We get a = 127 and b = 132. This give us the numbers as 127, 132, 132, 132, 227. Here the average is 150

Since b < a+100, let’s find the point where b = a+100

2a + 3(a+100) = 650

a = 70, b = 170

But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But increasing a by 3 increases the largest number, i.e. a+100 by 3 too, so we need to decrease b by another 1.

We get a = 73 and b = 168. This gives us the numbers as 73, 168, 168, 168, 173. Here the average is 150

Values of a will be: 73, 76, 79, ….127 (Difference of 3 to make b an integer)

This is an Arithmetic Progression.

Last term = First term + (n – 1)*Common difference

127 = 73 + (n – 1)*3

n = 19

a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values.

Answer (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!*

In this post, we discussed how to use graphing techniques to easily solve very high level questions on nested absolute values. We don’t think you will see such high level questions on actual GMAT. The aim of putting up the post was to illustrate the use of graphing technique and how it can be used to solve simple as well as complicated questions with equal ease. It was aimed at encouraging you to equip yourself with more visual approaches.

We gave you two questions at the end of that post to try on your own. We have seen quite a bit of interest in them and hence will discuss their solutions today.

The solutions involve a number of graphs and hence we have made pdf files for them.

Question 1: Given that y = |||x – 5| – 10| -5|, for how many values of x is y = 2?

Question 2: Given that y = |||x| – 3| – x|, for what range of x is y = 3?

*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!*

Before you begin, you might want to review the post that discusses standard deviation: Dealing With Standard Deviation

So here goes the question.

Question: Given that set S has four odd integers and their range is 4, how** **many distinct values can the standard deviation of S take**?**

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

Solution: Recall what standard deviation is. It measures the dispersion of all the elements from the mean. It doesn’t matter what the actual elements are and what the arithmetic mean is – the standard deviation of set {1, 3, 5} will be the same as the standard deviation of set {6, 8, 10} since in each set there are 3 elements such that one is at mean, one is 2 below the mean and one is 2 above the mean. So when we calculate the standard deviation, it will give us exactly the same value for both sets. Similarly, standard deviation of set {1, 3, 3, 5, 6} will be the same as standard deviation of {10, 12, 12, 14, 15} and so on. But note that the standard deviation of set {25, 27, 29, 29, 30} will be different because it represents a different arrangement on the number line.

Let’s look at the given question now.

Set S has four odd integers such that their range is 4. So it could look something like this {1, x, y, 5} when the elements are arranged in ascending order. Note that we have taken just one example of what set S could look like. There are innumerable other ways of representing it such as {3, x, y, 7} or {11, x, y, 15} etc.

Now in our example, x and y can take 3 different values: 1, 3 or 5

x and y could be same or different but x would always be smaller than or equal to y.

- If x and y were same, we could select the values of x and y in 3 different ways: both could be 1; both could be 3; both could be 5

- If x and y were different, we could select the values of x and y in 3C2 ways: x could be 1 and y could be 3; x could be 1 and y could be 5; x could be 3 and y could be 5.

For clarification, let’s enumerate the different ways in which we can write set S:

{1, 1, 1, 5}, {1, 3, 3, 5}, {1, 5, 5, 5}, {1, 1, 3, 5}, {1, 1, 5, 5}, {1, 3, 5, 5}

These are the 6 ways in which we can choose the numbers in our example.

Will all of them have unique standard deviations? Do all of them represent different distributions on the number line? Actually, no!

Standard deviations of {1, 1, 1, 5} and {1, 5, 5, 5} are the same. Why?

Standard deviation measures distance from mean. It has nothing to do with the actual value of mean and actual value of numbers. Note that the distribution of numbers on the number line is the same in both cases. The two sets are just mirror images of each other.

For the set {1, 1, 1, 5}, mean is 2. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

For the set {1, 5, 5, 5}, mean is 4. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean.

The deviations in both cases are the same -> 1, 1, 1 and 3. So when we square the deviations, add them up, divide by 4 and then find the square root, the figure we will get will be the same.

Similarly, {1, 1, 3, 5} and {1, 3, 5, 5} will have the same SD. Again, they are mirror images of each other on the number line.

The rest of the two sets: {1, 3, 3, 5} and {1, 1, 5, 5} will have distinct standard deviations since their distributions on the number line are unique.

In all, there are 4 different values that standard deviation can take in such a case.

Answer (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!*

Myth 1: Passive voice is always wrong.

Active voice is preferred over passive voice but that doesn’t make passive voice wrong.

Myth 2: The same pronoun cannot refer to two different antecedents in a sentence.

A pronoun, say ‘it’, can refer to two different objects in a single sentence, but it should not refer to two different objects in the same clause since that creates ambiguity.

We will explain these two points to you using a sentence correction question:

**Question**: Once the computer generates the financial reports, they are then used to program a company-wide balance sheet, so named because it demonstrates that every department’s accounting elements are in balance.

(A) Once the computer generates the financial reports, they are then used to program a company-wide balance sheet, so named because it demonstrates that every department’s accounting elements balance.

(B) Once the computer generates the financial reports, it is then used to program a company-wide balance sheet, named such because it demonstrated the balance of every department’s accounting elements.

(C) Once the computer generates the financial reports they are then used to program a company-wide balance sheet, which demonstrates the balance of every department’s accounting elements.

(D) Once the financial reports are generated by the computer, it is then used to program a company-wide balance sheet, so named because it demonstrates the balance of every department’s accounting elements.

(E) Once the financial reports are generated by the computer, they are then used to program a company-wide balance sheet, named such because it demonstrates that every department’s accounting elements are in balance.

**Solution**: Let’s split the sentence into clauses:

- Once the computer generates the financial reports,

- they are then used to program a company-wide balance sheet,

- so named because it demonstrates that every department’s accounting elements are in balance.

Find the decision points. The first clause is in active voice in the first three options and in passive in the other two. Both are correct.

The first decision point is they vs it. Should we use they or should we use it? The first clause talks about two things – computer (singular) and financial reports (plural). What do we want to refer to in the second clause? What do we use to program a balance sheet? A computer is used to program something. Reports cannot program anything. They can be used while programming but they cannot program. Hence, the use of ‘it’ would be correct here.

Only in options (B) and (D) do we use ‘it’ (singular) which refers back to the computer (singular). It cannot refer back to financial reports (plural). So eliminate options (A), (C) and (E).

Now comes our next decision point – we have to choose one of ‘named such’ and ‘so named’. ‘named such’ which is used in option (B) is awkward. Also, we use the past tense of the verb ‘demonstrate’ in option (B). This is not correct since a balance sheet is so called because is always demonstrates the balance of every element. It did not demonstrate it only in the past. Hence we need to use simple present tense.

This leads us to option (D). Everything is taken care of here.

Here are a couple of points about option (D):

(D) Once the financial reports are generated by the computer, it is then used to program a company-wide balance sheet, so named because it demonstrates the balance of every department’s accounting elements.

Sometimes, people eliminate it because it uses passive voice “the financial reports are generated by the computer”. Be aware that passive is not wrong. You have learned active passive in school. Passive is just a bit weaker form of writing than active and hence, given a choice, active is preferred but not at the expense of grammatical correctness! Using passive is not incorrect.

At other times, people have problems with the use of the pronoun ‘it’ for two different antecedents

- **it** (the computer) is then used to program a company-wide balance sheet,

- so named because **it** (the balance sheet) demonstrates the balance of every department’s accounting elements.

An OG problem has been pointed out here:

Starfish, with anywhere from five to eight arms, have a strong regenerative ability, and if one arm is lost **it** [animal] quickly replaces **it** [arm], sometimes by the animal overcompensating and growing an extra one or two.

The above answer is incorrect since the pronoun ‘it’ refers to two different antecedents in a single clause. Note that the pronoun ‘it’ refers to two different antecedents in the same clause. It is hard to understand what ‘it’ refers to.

But that is not the case in our option (D).

The first ‘it’ clearly refers to the computer since there is only one singular antecedent before it.

The second ‘it’ in the third clause clearly refers to the balance sheet because the clause talks about the balance sheet: … company wide balance sheet, so named because it …

There is no ambiguity of pronoun reference here.

We can’t re-iterate it enough – don’t try to learn up ‘rules’ for sentence correction. Every so called “rule” is not applicable in every situation. Use logic!

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

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The Meaning of Arithmetic Mean

Can You Solve these Mean GMAT Questions?

Finding Arithmetic Mean Using Deviations

Application of Arithmetic Means

This question needs you to apply all these concepts but can still be easily done in under two minutes. Now, without further ado, let’s go on to the question – there is a lot to discuss there.

**Question**: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?

(A)10

(B)12

(C)14

(D)15

(E)20

**Solution**: Let’s first assimilate the information we have. We need to find the minimum number of experts that must be there. Why should there be a minimum number of people satisfying these statistics? Let’s try to understand that with some numbers.

Say, N cannot be 1 i.e. there cannot be a single expert in the unit because then you cannot have the range of $10,000. You need at least two people to have a range – the difference of their salaries would be the range in that case.

So there are at least 2 people – say one with salary 0 and the other with 10,000. No salary will lie outside this range.

Median is $5000 – i.e. when all salaries are listed in increasing order, the middle salary (or average of middle two) is $5000. With 2 people, one at 0 and the other at 10,000, the median will be the average of the two i.e. (0 + 10,000)/2 = $5000. Since there are at least 10 people, there is probably someone earning $5000. Let’s put in 5000 there for reference.

0 … 5000 … 10,000

Arithmetic mean of all the salaries is $7000. Now, mean of 0, 5000 and 10,000 is $5000, not $7000 so this means that we need to add some more people. We need to add them more toward 10,000 than toward 0 to get a higher mean. So we will try to get a mean of $7000.

Let’s use deviations from the mean method to find where we need to add more people.

0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right.

Now we have

0 … 5000 … 10000, 10000, 10000

Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000.

5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 – this is the most important step. Ensure that you have understood this before moving ahead.

When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left.

Let’s try one such set of addition:

0 … 5000, *5000, 5000, 5000* … 10000, 10000, 10000, *10000, 10000*

The median is not $5000 yet. Let’s try one more set of addition.

0 … 5000, 5000, 5000, 5000, *5000, 5000, 5000* … 10000, 10000, 10000, 10000, 10000, *10000, 10000*

The median now is $5000 and we have maintained the mean at $7000.

This gives us a total of 15 people.

Answer (D)

Granted, the question is tough but note that it uses very basic concepts and that is the hallmark of a good GMAT question!

Try to come up with some other methods of solving this.

*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 in Detroit, Michigan, and regularly participates in content development projects such as this blog!

Let’s see how we decide that.

**Question 1**: In the wake of the global housing crisis, and amid dramatically changing demographics, it is likely that a widespread shift in thinking is ahead, which will reduce demand for large suburban homes, thus increasing demand for smaller urban apartments.

(A) it is likely that a widespread shift in thinking is ahead, which will reduce demand for large suburban homes, thus increasing demand for smaller urban apartments.

(B) it is likely that a widespread shift in thinking is ahead, which will reduce demand for large suburban homes, and thus increase demand for smaller urban apartments.

(C) it is not unlikely that a widespread shift in thinking is ahead, reducing demand for large suburban homes, thus creating an increase in demand for smaller urban apartments.

(D) it is not unlikely that a widespread shift in thinking is ahead, reducing demand for large suburban homes and increasing demand for smaller urban apartments.

(E) it is not unlikely that a widespread shift in thinking is ahead, reducing demand for large suburban homes, increasing demand for smaller urban apartments.

**Solution**: Let’s start looking for decision points – the first decision point is ‘it is likely’ vs ‘it is not unlikely’ – both have similar meanings and are grammatically correct so we cannot eliminate any option based on this right now. The next decision point is the beginning of the modifier. Options (A) and (B) use ‘which clauses’. Options (C), (D) and (E) use present participle modifiers.

‘which’ is a relative pronoun but there is no noun before it which can act as an antecedent. Hence, the use of which is incorrect here. On the other hand, the use of participle modifier is acceptable here. Last week, we discussed that present participle modifier after a comma will modify the preceding clause. It provides additional information about the preceding clause. ‘reducing …’ tells us more about ‘widespread shift in thinking‘. Hence, let’s focus on options (C), (D) and (E).

In (C), the “thus” used to introduce the second participle is incorrect: the two participles should be linked with a coordinating conjunction without a comma. One is not really leading to the other – they are both byproducts of the change in thinking – reducing demand for large homes and increasing demand for urban apartments. Lastly, in option (C), the “creating an…” is unnecessary and redundant – you just need “increasing demand.”

For option (E), you need something to link the two participle phrases together – without it, there is a comma splice error. Hence we eliminate (E) as well.

Option (D) gets the structure and meaning correct – “the shift in thinking is reducing … and increasing …”

Answer is (D).

Now, let’s look at an official GMAT question.

**Question 2**: In 1984, medical researchers at Harvard and Stanford universities concluded that sedentary life-styles lead to heart and lung diseases that shorten lives, strongly recommending middle-aged people to undertake some form of regular exercise.

(A) strongly recommending middle-aged people to

(B) strongly recommending that middle-aged people should

(C) and strongly recommended for middle-aged people to

(D) and their strong recommendation was for middle-aged people to

(E) and they strongly recommended that middle-aged people

**Solution**: The given sentence has two clauses:

Main clause – medical researchers at Harvard and Stanford universities concluded

That clause – that sedentary life-styles lead to heart and lung diseases that shorten lives

If we use a comma and the present participle ‘recommending’ here, it will modify the ‘that clause’. So ‘recommending’ will be done by ‘sedentary life-styles’. Obviously, this is incorrect since the researchers are the ones who recommend exercise. So we cannot use the participle here. Hence we eliminate options (A) and (B).

Options (C), (D) and (E) use ‘recommend’ in verb form.

Options (C) and (D) are unidiomatic in their usage of the verb recommend.

You recommend X for Y (say a person X for position Y)

or

You recommend that X do Y (say a person X do Y)

Option (C) says ‘recommended for X to do Y’ and option (D) says ‘recommendation was for X to do Y’ – both are incorrect.

Option (E) uses recommend properly – ‘recommended that X do Y’. Also, ‘… researchers concluded that … and recommended that …’ have parallel structure. Hence, option (E) is correct.

Answer (E)

Hope you now understand how participle phrases are used.

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

Quite simply, participles are words formed from verbs which can be used as describing words (on the other hand, gerunds are verbs used as nouns, but that is a topic for another day!).

There are two types of participles:

1. The Past Participle – usually ends in -ed, -d, -t, -en, or –n

For Example: chosen, danced, known, sung etc

2. The Present Participle – ends in –ing

For Example: choosing, dancing, knowing, singing etc

These participles often start the participle phrases used to describe nouns/noun phrases/entire sentences. The participial phrases are underlined in the examples given below.

Examples:

I want to stand next to the girl wearing the yellow dress.

Standing next to the tall gentleman, she looked petite.

Battered by hail, the car collapsed.

The most important crop of this region is rice, sown in the month of June and harvested in October.

Here is how participle phrases are usually used:

**Present Participle Phrases** (the underlined parts of the sentences are participial phrases):

1. At the beginning of a sentence followed by a comma and then a clause (present participle phrase + comma + clause) – In this case, the participle phrase could modify the subject of the clause or the entire clause.

Examples:

Wagging its tail, my dog ran up to me. (modifies ‘my dog’)

Silencing the students, the principal stepped on to the podium. (modifies the entire clause because the principal silenced the students by stepping on to the podium)

2. At the end of a sentence separated from the clause using a comma (clause + comma + present participle phrase) – In this case, the participle phrase modifies the entire preceding sentence.

Example: The principal stepped on to the podium, silencing the students. (modifies the entire preceding clause)

3. Following a noun without a comma – In this case, the participle phrase modifies the noun.

Example: I want to stand next to the girl wearing the yellow dress. (modifies ‘the girl’)

**Past Participle Phrases **(the underlined parts of the sentences are participial phrases):

1. Following a noun separated by a comma (noun + comma + past participle phrase) – In this case, the participle phrase modifies the noun.

Example: The most important crop of this region is rice, sown in the month of June and harvested in October . (modifies ‘rice’)

2. At the beginning of a sentence followed by a comma and then a clause (past participle phrase + comma + clause) – In this case, the participle phrase modifies the subject of the clause.

Example: Battered by hail, the car collapsed. (modifies ‘the car’)

Next week, we will take some questions to show the classic usage of participle modifiers in GMAT. But today we need to move on and discuss an important point regarding the rules discussed.

Important Note: In regular English grammar, a past participle phrase following a clause and separated by a comma (clause + comma + past participle phrase) could modify the entire preceding clause. But GMAT is not very keen on this usage; so avoid it. That said, remember that studying grammar rules in isolation is worthless. If the sentence demands such a construction, then it is correct to use it. We cannot explain this point without a question so let’s take one from our own collection.

**Question**: Due to the slow-moving nature of tectonic plate movement, the oldest ocean crust is thought to date from the Jurassic period, formed from huge fragments of the Earth’s lithosphere and lasted 200 million years.

(A) formed from huge fragments of the Earth’s lithosphere and lasted 200 million years.

(B) forming from huge fragments of the Earth’s lithosphere and lasting 200 million years.

(C) forming from huge fragments of the Earth’s lithosphere and lasted 200 million years.

(D) formed from huge fragments of the Earth’s lithosphere and lasting 200 million years.

(E) formed from huge fragments of the Earth’s lithosphere and has been lasting 200 million years.

Here is our official solution:

*The correct response is (D).*

*The meaning of the sentence is that the “oldest ocean crust” was “formed” in the past during the Jurassic period and is currently still “lasting” (since if it’s the “oldest” it must still be around!). We need the past tense/participle verbs to be used correctly.*

*If you chose (A), the ocean crust was “formed” in the past” but if “lasted” is past tense then the oldest ocean crust is no longer around, which would mean it couldn’t be the “oldest.”*

*If you chose (B) or (C), “forming” implies the crust is still being formed. While it’s true the Earth’s crust is constantly in flux, we’re concerned with the “oldest ocean crust” – that part that is no longer continuing to form, but was formed at some point during the Jurassic period.*

*If you chose (E), you correctly used “formed,” however the present perfect “has been lasting” is unnecessarily wordy. The simple participle verb form will suffice.*

Does logic dictate that (D) is the correct answer? Yes. Will you ignore it because it uses past participle form modifying the previous subject/clause instead of ‘Jurassic Period’? No. Note that it is correct grammatically and you should know it. Whatever we can infer about the preferences of GMAT is from the questions it gives. GMAT doesn’t clarify its stand on every grammatical issue and the stand is probably flexible depending on the sentence under examination. So you need to be flexible in your understanding of what is and is not acceptable in GMAT. Use logic – remember, GMAT is a test of your reasoning skills. Get to the best answer under given circumstances.

*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 in Detroit, Michigan, and regularly participates in content development projects such as this blog!

We know that basic concepts are twisted to make advanced questions. Our aim is to break down the question into two parts – ‘the basic concept’ and ‘the complexity’. You can either deal with the complexity first and then glide through the basic concept or you can glide through the basic concept first and then face the complexity. The method you use will depend on the question. If the question seems too complex at the outset, it means you will have to deal with the complexity first. If the question seems familiar but has some extra not-so-familiar elements, it means you should get the familiar out of the way first. Let’s take a question today to see how to do that.

Question: During a sale of 20% on everything in a store, a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies. The store still makes a profit of 12% on this sale. What is the mark up percentage on each candy?

(A) 100%

(B) 80%

(C) 75%

(D) 66+2/3%

(E) 55%

Solution:

This question can get very messy if you let it! We have seen people working on this question with multiple variables: C for cost price, S for sale price, M for marked price etc. That can get very confusing because there are two types of mark up – the actual mark up (the store marks up the price of every candy by this percentage and lists it on the candy) and the effective mark up (because the kid takes 6 extra candies, this is the effective mark up). So let’s not go the algebra way.

Instead, let’s focus on what we can do without much effort. As a first step, let’s do what we know already (and hope that the rest will work out!).

We already know the relation between mark-up, discount and profit. The problem is that this question has another aspect – the kid takes 20 candies but pays the price of only 14 candies (which is the price obtained by reducing the marked price by 20% of discount). But let’s worry about it later.

Let’s first deal with the mark-up, discount and profit aspect of the question.

We know that (1 + m/100)(1 – d/100) = (1 + p/100) (already discussed in detail in this post)

Since p is the effective profit that the store got, m must be the effective mark up here.

(1 + m/100)(1 – 20/100) = (1 + 12/100)

(1 + m/100) = (5/4)*(28/25)

(1 + m/100) = 7/5

m = 40

So effective mark up was 40% – i.e. 40% was the mark up in a situation where 14 articles were sold and charged for. This tells us this – effective mark up turned out to be 40% though his actual mark up must have been higher since he gave away 20 articles for the cost of 14.

Now what we already know is done. We get to the really tricky part – the thing that makes this question different – how do we find the actual mark up?

Let’s say the cost price of each of the 20 candies was $1. Then total cost price for the 20 candies was $20. This is the cost of the candies to the store. The effective markup was 40% i.e. the articles were effectively marked at 20 + (40/100)*20 = $28. The store gave a discount of 20% on this amount and made a profit of 12%. But this amount of $28 actually represents the mark up on 14 candies only. The cost price of 14 candies is $14 to the store. So the actual mark up percentage on the 14 candies is (28 – 14)/14 * 100 = 100%

Answer (A)

Obviously, there are many other ways of solving this question. See if you can figure out another one on your own!

Let’s discuss one such trick today – a trick in which you need to realize that the situation calls for a complete U-turn of the usual.

Let’s take an example:

**Question**: Two cars run in opposite directions on a circular track. Car A travels at a rate of 6? miles per hour and Car B runs at a rate of 8? miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

(A) 6/7 hrs

(B) 12/7 hrs

(C) 4 hrs

(D) 6 hrs

(E) 12 hrs

**Solution**: What would we usually do in such a question? Two cars start from the same point and run in opposite directions – their speeds are given. This would remind us of relative speed. When two objects move in opposite directions, their relative speed is the sum of their speeds. So we might be tempted to do something like this:

Perimeter of the circle = 2?r = 2?*6 = 12? miles

Time taken to meet = Distance/Relative Speed = 12?/(6? + 8?) = 6/7 hrs

But take a step back and think – what does 6/7 hrs give us? It gives us the time taken by the two of them to complete one circle together. In this much time, they will meet somewhere on the circle but not at the starting point. So this is definitely not our answer.

The actual time taken to meet at point S will be given by 12?/(8? – 6?) = 6 hrs

This is what we mean by unexpected! The relative speed should be the sum of their speeds. Why did we divide the distance by the difference of their speeds? Here is why:

For the two objects to meet again at the starting point, obviously they both must be at the starting point. So the faster object must complete at least one full round more than the slower object. In every hour, car B – the one that runs at a speed of 8? mph covers 2? miles more compared with the distance covered by car A in that time (which runs at a speed of 6? mph). We want car B to complete one full circle more than car A. In how much time will car B cover 12? miles (a full circle) more than car A? In 12?/2? hrs = 6 hrs.

Now we will keep the question the same but will change the figures a bit:

**Modified Question**: Two cars run in opposite directions on a circular track. Car A travels at a rate of 3? miles per hour and Car B runs at a rate of 5? miles per hour. If the track has a radius of 7.5 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

So following the same logic as above,

Perimeter of the circle = 2?r = 2?*7.5 = 15? miles

The time taken to meet at point S will be given by 15?/(5? – 3?) = 7.5 hrs

But note that the two cars will not even be at the starting point, S, in 7.5 hrs. So this answer is wrong. Why? It has something to do with the word “at least” used in the explanation above i.e. “So the faster object must complete at least one full round more than the slower object. “

Try to put it all together.

Meanwhile, let’s give you **another method**. This will not fail you no matter what the figures.

**Using the original question**:

Time taken by car A to complete one full circle = 12?/6? = 2 hrs

Time taken by car B to complete one full circle 12?/8? = 1.5 hrs

So every 2 hrs car A is at S and every 1.5 hrs, car B is at S. When will they both be together at S?

Car A at S -> 2 hrs, 4 hrs, 6 hrs, 8 hrs …

Car B at S -> 1.5 hrs, 3 hrs, 4.5 hrs, 6 hrs …

In 6 hrs – the first common time, both cars will be at the point S together. So answer is 6 hours.

**Using the same method on the Modified Question**,

Time taken by car A to complete one full circle = 15?/3? = 5 hrs

Time taken by car B to complete one full circle = 15?/5? = 3 hrs

So every 5 hrs, car A is at S and every 3 hrs, car B is at S. When will they both be together at S?

Car A at S -> 5 hrs, 10 hrs, 15 hrs, 20 hrs

Car B at S -> 3 hrs, 6 hrs, 9 hrs, 12 hrs, 15 hrs

In 15 hrs – the first common time (LCM of 3 and 5), both cars will be at the point S together.

This all makes sense now.

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

We will illustrate that with the help of a supremely beguiling official question today. We are sure you wouldn’t call an academician’s work exactly thrilling but questions like these do add a decent bit of joie de vivre to our lives. It’s hard to explain the gratification we get when it all falls into place in your mind and you light up with – “shoot, so simple, and yet, it seemed like a monster a few minutes back!” – we basically live for those moments!

Let us first give you some stats which indicate the difficulty level of this question:

95% of people find this question hard. Only 1/3^{rd} of respondents answer it correctly (which includes the ton of people who had tried it before and hence knew the correct answer).

Let us give you the question now:

**Question**: Each piglet in a litter is fed exactly one-half pound of a mixture of oats and barley. The ratio of the amount of barley to that of oats varies from piglet to piglet, but each piglet is fed some of both grains. How many piglets are there in the litter?

**Statement 1**: Piglet A was fed exactly 1/4 of the oats today.

**Statement 2**: Piglet A was fed exactly 1/6 of the barley today.

First think, what concept does it test? Fractions? Ratios? Or is it just a word problem requiring algebraic manipulation?

Actually, none of these. We can look at the question and say straight away that the answer is (C). It needs no manipulation and no calculation. Of course, what it does need is a solid understanding of the weighted averages principle!

For now, forget the data given in the question.

Consider this:

Say, 10% of total Oats and 20% of total Barley was fed to a piglet.

The question now is – Of the total food (Oats + Barley) what percentage was fed to this piglet?

We hope you agree that it will depend on the ratio of Oats and Barley. If the mixture was only oats, the piglet was fed 10% of the total food. If the mixture was only Barley, the piglet was fed 20% of the total mixture. If the mixture was half oats and half barley, the piglet was fed 15% of the total mixture. If the mixture was 1 part Oats for every 4 parts of Barley, the piglet was fed 18% of the mixture (it is just weighted average with weights being the amount of initial quantity of Oats and Barley). Whatever the case, the piglet was fed more than 10% of total food and less than 20% of total food if the mixture consisted of both Oats and Barley.

If this is not clear, look at this example:

Say a meal consists of a sandwich and a milkshake. You eat 1/2 of the sandwich and drink 1/2 of the milkshake. Can we say that you have had 1/2 of the meal? Sure.

If you eat only 1/4 of the sandwich and drink 1/4 of the milkshake, then you would have had only 1/4 of the meal.

What happens in case you eat 1/2 of the sandwich but drink only 1/4 of the milkshake? In that case, you have had less than 1/2 of the meal but certainly more than 1/4 of the meal, right?

Go through this again till you are satisfied with this logic.

If this sounds good, consider data given in the question – piglet A was fed 25% Oats (1/4 Oats) and 16.66% Barley (1/6 Barley). So definitely, the piglet was fed more than 16.66% (which is 1/6) of the total mixture and less than 25% (which is 1/4) of the total mixture (as reasoned above). Stay with this idea.

Another piece of information from the question stem: the total food mixture was split equally among all the piglets. Since all piglets got the same quantity of food, we can say that all piglets were fed more than 1/6 of the total mixture but less than 1/4 of the total mixture. Number of piglets has to be an integer, say n. Then, each piglet gets the same amount of food i.e. 1/n of the total mixture. This 1/n must lie between 1/4 and 1/6. Note that the number of pigs i.e. n, must be a positive integer. What integer value can n take? Can it be 7? Will 1/7 lie between 1/6 and 1/4? No. 1/7 will be less than 1/6. Can n be 3? Will 1/3 lie between 1/4 and 1/6? No, because 1/3 will be greater than 1/4. n cannot be greater than 6 or less than 4 because it goes out of range. Only 1/5 lies between 1/4 and 1/6 (such that n is a positive integer). Hence n must be 5.

Notice that we did not need to do any calculations – just looking at the two statements, we can say that 1/n must lie between 1/4 and 1/6 and hence n must be 5.

Questions such as this one set GMAT apart from other tests. It tests you on basic concepts but how!!!