The Holistic Approach to Mods on the GMAT – Solutions

Quarter Wit, Quarter WisdomFirst, we would like to refer you back to a post we put up quite a while ago: The Holistic Approach to Mods

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?

Solution 1

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

Solution 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!

A 750+ Level Question on SD

Quarter Wit, Quarter Wisdom A couple of weeks back, we looked at a 750+ level question on mean, median and range concepts of Statistics. This week, we have a 750+ level question on standard deviation concept of Statistics. We do hope you enjoy checking it out.

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!

Busting Some GMAT SC Myths

Quarter Wit, Quarter WisdomToday we will bust some SC myths using a question. The following are the myths:

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!

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 750 Level GMAT Question on Statistics!

Quarter Wit, Quarter WisdomToday, we have a very interesting statistics question for you. We have already discussed statistics concepts such as mean, median, range etc in our QWQW series. Check them out here if you haven’t already done so:

The Meaning of Arithmetic Mean

Can You Solve these Mean GMAT Questions?

Finding Arithmetic Mean Using Deviations

Application of Arithmetic Means

Mean Questions on Median

A Range of Questions

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!

2 Sentence Correction GMAT Questions Involving Participle Modifiers

Quarter Wit, Quarter WisdomToday, as promised last week, we will look at a couple of questions involving participle modifiers. We will take one question in which you should use the participle and another in which you should not.

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!

Understanding Participles on the GMAT

Quarter Wit, Quarter WisdomThere is a lot of confusion surrounding the topic of Participles so let’s take a look at it today.

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!

First Do What You Know on GMAT Questions

Quarter Wit, Quarter WisdomWe have read a lot about one way of handling complex questions – simplify them to a question you know how to solve. Here is another way – first do what you do know, and then figure out the rest!

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!

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!

How to Expect the Unexpected on the GMAT

Quarter Wit, Quarter WisdomMost of us know that GMAT is a shrew, (euphemism for a more choice adjective that comes to mind!) and is very hard to tame. It is well established that it is able to give a pretty accurate estimate of aptitude with just a few questions, and that the only way to “deceive” it is by actually improving your aptitude! It has numerous tricks up its sleeves to uncloak a rather basic player.

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.

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!

95% of Students Find This GMAT Quant Question Difficult

Quarter Wit, Quarter WisdomToday we continue to look at ways to achieve that much desired score of 51 in Quant. Obviously, we don’t need Sheldon Cooper’s smarts to realize that for that revered high score, we must do well on the high level questions but the actual question is – how to do well on the high level questions?

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 1/3rd 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!!!

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 Absolute Phrases on the GMAT

Quarter Wit, Quarter WisdomRead the following sentences:

  1. About 70 percent of the tomatoes grown in the United States come from seeds that have been engineered in a laboratory, their DNA modified with genetic material not naturally found in tomato species.
  2. The defense lawyer and witnesses portrayed the accused as a victim of circumstance, his life uprooted by the media pressure to punish someone in the case.
  3. Researchers in Germany have unearthed 400,000-year-old wooden spears from what appears to be an ancient lakeshore hunting ground, stunning evidence that human ancestors systematically hunted big game much earlier than believed.

Which grammatical construct is represented by the underlined portions of these sentences?

These are called absolute phrases. They often confuse people but once you understand properly what they are and what they do, they will not be intimidating.

What is an Absolute Phrase?

An absolute phrase is a type of modifier that modifies an independent clause as a whole.

Structure of an Absolute Phrase

Often (but not always), this is the structure of an absolute phrase:

noun + participle (could be -ing or -ed) + optional modifier or object

Usage of an Absolute Phrase

It is often useful in describing one part of the whole person/place/thing or in explaining a cause or condition etc.

For example:

There was no one in sight and Sanders, his hands still jammed in his pockets, scowled down the empty street. (The underlined absolute phrase describes just the hands of Sanders)

We devoured the yummy pastries, our fingers scraping the leftover frosting off the plates. (The underlined absolute phrase describes just our fingers)

The underlined absolute phrase in sentence 1 above describes the DNA of the seeds.

The underlined absolute phrases in sentences 2 and 3 above describe conditions.

Some Alternative Structures of Absolute Phrases

Some absolute phrases have a different structure.

  1. The participle being is often omitted in an absolute phrase, leaving only a noun and a modifier:

The boys set off for school, faces glum, to begin the winter term.

  1. Also, an absolute phrase may contain a pronoun instead of a noun, or an infinitive (to + a verb) instead of a participle:

The customers filed out, some to return home, others to gather at the piazza.
[pronoun ‘some’ + infinitive ‘to return’ ; pronoun ‘others’ + infinitive ‘to gather’]

Now let’s look at the sentence correction question which uses statement 2.

Question: The defense lawyer and witnesses portrayed the accused as a victim of circumstance, his life uprooted by the media pressure to punish someone in the case.

(A) circumstance, his life
(B) circumstance, and his life
(C) circumstance, and his life being
(D) circumstance; his life
(E) circumstance: his life being

Solution:

“his life uprooted by the media pressure to punish someone in the case.” and “his life being uprooted by the media pressure to punish someone in the case.” are not independent clauses because they have no finite verbs in them.

With the coordinating conjunction (‘and’) and semi colon, you need an independent clause.

Accuracy wise, the use of ‘being’ is still suspect. ‘Being’ is not used to describe a state; it is used to describe an ongoing action such as ‘the tree is being uprooted’.

Colon is used if you need to give a list and hence, is not suitable here. Hence, options (B), (C), (D) and (E) are wrong.

Only option (A) describes circumstances suitably using the absolute phrase: his life uprooted by the media pressure to punish someone in the case.

Answer (A)

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 700+ GMAT Quant Question on Races

Quarter Wit, Quarter WisdomThis week we will look at the question on races that we gave you last week.

Question 3: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately?

(A)   8, 10

(B)    4, 5

(C)   5, 9

(D)   6, 9

(E)    7, 10

Solution: Now this question is a little tougher than the previous ones we saw last week.

There are two scenarios given:

1 – A gives B a head start of 200 m and beats him by 30 seconds.

2 – A gives B a head start of 3 mins and is beaten by 1000m.

Let’s study both of them and see what we can derive from them.

Scenario 1: A gives B a start of 200m and beats him by 30 seconds.

As we suggested before, we will start by making a diagram.

A runs from the Start line till the finish line i.e. a total distance of 2000 m.

A gives B a head start of 200 m so B starts, not from the starting point, but from 200 m ahead. A still beats him by 30 sec which means that A completes the race while B takes another 30 sec to complete it. So obviously A is much faster than B.
In this race, A covers 2000m. In the same time, B covers the distance shown by the red line. Since B needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance. The green line distance is given by (1/2)*s where s is the speed of B in meters per minute. The distance B has actually covered in the same time as A is the distance shown by the red line. This distance will be (1/2)*s less than  1800 i.e. it will be [1800 – (1/2)*s].

Scenario 2: A gives B a head start of 3mins and is beaten by 1000m.

A gives B a head start of 3 mins means B starts running first while A sits at the starting point. After 3 mins, B covers the distance shown by the red line which we do not know yet. Now, A starts running too. B beats A by 1000 m which means that B reaches the end point while A is still 1000 m away from the end i.e. at the mid point of the 2000 m track.

In this race, A covers a distance of 1000 m only. In that time, B covers the distance shown by the green line. The distance shown by the red line was covered by B in his first 3 mins i.e. this distance is 3*s. This distance shown by the green line is given by (2000 – 3s).

Now you see that in the first race, A covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, B would also have covered half the previous distance.

Distance covered by B in first race = 2*Distance covered by B in second race

1800 – (1/2)*s = 2*(2000 – 3s) (where s is the speed of B in meters/min)
s = 400 meters/min

Time taken by B to run a 2000 m race = Distance/Speed = 2000/400 = 5 min

Only one option has time taken by B as 5 mins and that must be the answer.

If required, you can easily calculate the time required by A too.

Distance covered by B in scenario 1 = 1800 – (1/2)*s = 1600 m

In the same time, A covers 2000 m which is a ratio of A:B = 5:4. Hence time taken by A:B will be 4:5.

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!

2 Simple GMAT Quant Questions That Will Help You Score Higher

Quarter Wit, Quarter WisdomLet’s discuss races today. It is a very simple concept but questions on it tend to be tricky. But if you understand how to handle them, most questions can be done easily.

A few points to remember in races:

1. Make a diagram. Draw a straight line to show the track and assume all racers are at start at 12:00. Then according to headstart, place the participants.

2. There are two types of head starts: Time and distance

Say there is a 1000 feet race between A and B which starts at 12:00.

Time – A gives B a headstart of 1 min means B starts running at 12:00 and A waits at the start point. Then A starts running from the start point at 12:01.

Distance – A gives B a headstart of 10 feet means A starts from the start point but B starts from the point 10 feet ahead (and hence runs only 990 feet to complete the race)

3. A dead heat is a race in which both the participants finish exactly at the same time. Most races in race questions end in a dead heat!

4. There are two ways in which a participant can beat another: Time and distance

Say A beats B in the 1000 feet race in which both start from the start point at 12:00.

Distance – If A beats B by 20 feet,  it means A finishes the race (full 1000 feet) and at that time, B is 20 feet away from the finish line.

Time – If A beats B by 2 mins, it means that if A finished at 12:10, B is still 2 mins away from the finish line i.e. at his/her speed, B takes 2 mins to reach the finish line.

That is all! Now let’s look at some questions:

Question 1: A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?

(A)   1/17

(B)   3/17

(C)   1/10

(D)   3/20

(E)    3/10

Solution: We have the ratio of A’s speed and B’s speed. This means, we know how much distance A covers compared with B in the same time.

This is what the beginning of the race will look like:

(Start) A_________B______________________________

If A covers 20 meters, B covers 17 meters in that time. So if the race is 20 meters long, when A reaches the finish line, B would be 3 meters behind him. If we want the race to end in a dead heat, we want B to be at the finish line too at the same time. This means B should get a head start of 3 meters so that he doesn’t need to cover that. In that case, the time required by A (to cover 20 meters) would be the same as the time required by B (to cover 17 meters) to reach the finish line.

So B should get a head start of 3/20th of the race.

Answer (D)

This question was relatively very straight forward and we gave it only to help you apply the concepts discussed above. Let’s make it slightly tricky now.

Question 2: A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that B beats A by 20% of the length of the race?

(A)   15%

(B)   20%

(C)   28%

(D)   32%

(E)    35%

SolutionAgain, we have ratio of A’s speed and B’s speed given as 20:17. If A covers 20 meters, B covers 17 meters in that time. This time, let’s assume that the length of the race is 25 meters.

At the beginning, this is what the 25 meter track will look like with a head start to B:

(Start ) A_________B_______________________________

Since A will give B a head start so A must start from the start line while B will start from ahead.

Since A should cover only 80% of the length of the race, when B reaches the finish line, A should still have 20% of the track leftover.

20% of the track will be (20/100)*25 = 5 meters. So A should be at 20 meters when B is at the finish line.

So this is what the finish of the race will look like:

____________20_________________A____5_____B (Finish)

A will cover a total of 20 meters when B should be at the finish line. In this time, B will cover only 17 meters. But the total track is of 25 meters. So the rest of the 25-17 = 8 meters, B should get as a head start.
Head start will be 8/25 *100 = 32% of the race.

Answer (D)

If you found it tricky, we would suggest you to practice some more races questions. It is usually easy to “figure out” the answer logically and the calculations required are minimum.

Now try this official question. We will solve it for you next week.

Question 3: A and B run a race of 2000 m. First, A gives B a start of 200m and beats him by 30 seconds. Next, A gives B a start of 3mins and is beaten by 1000m. Find the time in minutes in which A and B can run the race separately.

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 Go from a 48 to 51 in GMAT Quant – Part IV

Quarter Wit, Quarter WisdomTo take a look at the previous posts of this thread, check: Part I, Part II and Part III.

Another point to keep in mind while targeting Q50+ in GMAT: don’t buy complex official solutions. Most GMAT questions can be solved in a few steps. The point is that sometimes it is hard to identify those “few steps” and we keep going round and round in circles for a while till we arrive at the answer. The way to hit 51 is to look for simple solutions for difficult questions. The best example of this would be question number 148 of Official Guide 12. The question is tough, no doubt about it but just because it is tough, don’t think that the solution needs to be tough too – you don’t have to live with the solution provided.

If, even after reading the solution a couple of times, you know that if you try the question again in a week, you won’t be able to solve it on your own, this means you need to review either the concept or the solution. If the given solution is too complex and you almost have to learn it up step by step, it means you need a better solution. The next step of the solution should be apparent to you – you should be able to solve it on your own within two minutes.

Also, even if one method looks good, try to find other ways of solving the question. Often, there are multiple good ways of solving a particular question.

Here is a question similar to question number 148 of OG12. Let me give a few good methods of solving it:

Question: If x, y, and k are positive numbers such that {x/(x+y)}*20 + {y/(x+y)}*40 = k, and if x < y, which of the following could be the value of k?

(A) 15
(B) 20
(C) 25
(D) 35
(E) 40

Solution: One solution you have in the OG. Three more are provided here:

Method 2: Algebra

Note the “could be” in the question. This means that k can take multiple values and one of them is provided here.

20*x/(x+y) + 40*y/(x+y) = k

20(x+2y)/(x+y) = k

20*{(x + y)/(x + y) + y/(x + y)} = k

20*{ 1 + y/(x + y)} = k

Now since y is greater than x,  y/x+y  will be more than 1/2 but definitely less than 1 (x and y are positive numbers).

So the value of k will lie in the range 20*{1 + 1/2} < k < 20*{1 + 1}

i.e. 30 < k < 40

Only option (D) falls in this range.

Answer (D)

Method 3: Weighted Average

Does this equation remind you of something: 20*x/(x+y) + 40*y/(x+y) = k?

If you are a weighted average fan like me, you will notice that this is just the weighted average formula applied:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

Where Cavg = k

C1 = 20

C2 = 40

w1 = x

w2 = y

k = (20*x + 40*y)/(x + y)

It might be hard to see this on your own but the point is that if you do see it, the return is very high.

We know that the average of two quantities will lie in between them. So k must lie between 20 and 40. Also, we are given that x is less than y i.e. weight given to 20 is less than the weight given to 40. So the weighted average will lie toward 40. Between 30 and 40, there is only option (D)

Hence, answer (D).

Method 4: Plugging Numbers

Now, what if neither of the above given methods work for you during the test and your mind goes blank? Then you can pick some numbers to get an idea of the kind of values you will get. This is absolute brute force and may not always work out but it will give you a fighting chance of getting the correct answer.

20*x/(x+y) + 40*y/(x+y) = k

– Say, x = 1, y = 3 (x and y are positive numbers and x < y)

Then 20*1/(1+3) + 40*3/(1+3) = k = 35

– Say, x = 2, y = 3 (when you assume numbers, assume those which make the denominator a factor of 20 and 40 for ease of calculations. So assume numbers such that x+y is 4 or 5 or 10 etc)

Then 20*2/(2+3) + 40*3/(2+3) = k = 32

– Say, x = 1, y = 4

Then 20*1/(1+4) + 40*4/(1+4) = k = 36

Even if you do not get 35, note that the other values of k lie in the 30s. So your best bet would be to mark answer as (D).

Hope you see that there are many different ways of solving a given question, so you don’t usually require complex solutions. Practice on!

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 Number Properties on the GMAT – Part IV

Quarter Wit, Quarter WisdomAs pointed out by a reader, we need to complete the discussion on a question discussed in our previous ‘Advanced Number Properties’ posts so let’s do that today. Note that the discussion that follows doesn’t fall in the purview of GMAT and you needn’t know it. You will be able to solve any question without taking this post into account but that has never stopped us from letting loose our curiosity so here goes…

Question 1: Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 58
(D) 79
(E) 88

Solution: We discussed in that post that the sum of two prime numbers is usually even because prime numbers are usually odd. We also discussed that if the sum of two prime numbers is odd, it means one of the prime numbers is certainly 2 – the only even prime number.

For example:

2 + 3 = 5

2 + 7 = 9

2 + 17 = 19

Then it makes perfect sense to first look at the options which are odd. To be sum of two prime numbers, the sum must be of the form 2 + Another Prime Number.

We saw that (D) 79 = 2 + 77 (77 is not prime.) and hence we got (D) as our answer.

Now the question we raised there was: What happens if instead of 79, we had 81?

81 = 2 + 79

Then all three odd options would have been sum of two prime numbers and we would have needed to check the even options too. How do you figure whether an even number can be written as the sum of two prime numbers?

This is where Goldbach’s Conjecture comes into play (you don’t really need to know it. We are doing it for intellectual purposes. GMAC will never put you in this fix).

It says “Every even integer greater than 2 can be expressed as the sum of two primes.”

Mind you, it’s a conjecture i.e. it hasn’t been proven for all even numbers (only for even numbers till 4 * 10^{18}) but it does seem to hold.

For example:

4 = 2 + 2

6 = 3 + 3

8 = 3 + 5

10 = 3 + 7 = 5 + 5

12 = 5 + 7

and so on…

So given any even sum greater than 2, you can say that it CAN be written as sum of two prime numbers, for all practical purposes.

In fact, and here we are going into really geeky territory, we expect that every large even integer has not just one representation as the sum of two primes, but in fact has very many such representations. For all we know, 6 may be the only even number greater than 2 which cannot be written as the sum of two distinct prime numbers.

Coming back to our original question, we will actually check only odd numbers to see whether they can be written as sum of two primes. One of them has to be such that it cannot be written as sum of two primes and finding that is very simple! (as discussed in the previous post)

So all in all, the question that seemed very tedious turned out to be very simple!

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 Go From a 48 to 51 in GMAT Quant – Part III

Quarter Wit, Quarter WisdomLet’s get back to strategies that will help us reach the coveted 51 in Quant. First, take a look at Part I and Part II of this blog series. Since the Quant section is not a Math test, you need conceptual understanding and then some ingenuity for the hard questions (since they look unique). Today we look at a Quant problem which is very easy if the method “strikes”. Else, it can be a little daunting. What we will do is look at a “brute force” method for times when the textbook method is not easily identifiable.

Question: What is 0.99999999/1.0001 – 0.99999991/1.0003?

(A)   10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

Solution: Usually, when we have decimals such as .99999999 or 1.0001, we round them off to 1 without thinking twice. The issue here is that all numbers are very close to 1 so if we round them off to 1, we will get 1/1 – 1/1 = 0. This, obviously, doesn’t work and we need to work with the complicated numbers only.

Now here is the official method, something a Math professor will give us:

Method 1:

For simplification, you will need to use a^2 – b^2  = (a – b)(a + b)

Note that 0.99999999 is .00000001 less than 1 and 1.0001 is .0001 more than 1.

.00000001 is the square of .0001.

0.99999999/1.0001 – 0.99999991/1.0003

{1 – .00000001}/{1 + .0001} – {1 – .00000009}/{1 + .0003}

{1^2 – .0001^2}/{1 + .0001} – {1^2 – 0.0003^2}/{1 + .0003}

{(1 – .0001)(1 + .0001)}/{1 + .0001} – {(1 – .0003)(1 + .0003)}/{1 + .0003}

(1 – .0001) – (1 – .0003)

.0002 = 2*10^{-4}

All in all, the question only required us to recall something we learnt in 7th standard: a^2 – b^2  = (a – b)(a + b)

Does it mean it is a very simple question? Not really. The problem is that it is hard to identify that all you need is this formula and that you need to bring the terms in this format.

So here is a “brute force” method that people came up with and that we can use when Math fails us:

Method 2:

The fractions are quite complicated but the options are not fractions. This means that we are able to get rid of the denominator somehow. This brings an idea to mind: 0.99999999 might be a multiple of 1.0001. But how do we find ‘which multiple’?

For that, we need to use some pattern recognition.

9*1.0001 = 9.0009

99*1.0001 = 99.0099

and so on till 9999*1.0001 = 9999.9999 (to get eight 9s)

Now since the decimal is 4 digits to the left i.e. the number is actually 0.99999999,

0.9999 * 1.0001 = 0.99999999

This means 0.99999999/1.0001 = 0.9999

On the same lines, we might guess that 0.99999991 is a multiple of 1.0003. To find ‘which multiple’, we might need to think even harder now.  Note that something needs to multiply 1.0003 to give something ending in 1. Perhaps this multiple ends with a 7 because 3*7 ends in 1.

And sure, 9997*1.0003 = 9999.9991 which gives us 0.9997 * 1.0003 = 0.99999991

This gives us 0.99999991/1.0003 = 0.9997

Thus, the problem boils down to

0.9999 – 0.9997 = .0002

We encourage you to look for some more brute force methods.

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!

Determine Which Type of GMAT Question This Is: Assumption or Inference?

Quarter Wit, Quarter WisdomWe will continue our Quant 48 to 51 journey in the coming weeks but today, we need to discuss an important distinction between assumptions and inferences. Most of you will be able to explain the difference between an assumption and an inference but some questions will still surprise you. After all, both assumptions and inferences deal with the same elements in the argument. The way they are worded makes all the difference.

In simple words, when you have enough data given and you can infer something from it without doubt, it is called an inference/conclusion.
When you have author’s opinion (conclusion of the argument) and you need something to be true for the opinion to hold, that is an assumption.

Let us explain with a simple example.

All A are B.
All B are C.

You can conclude that: All A are C. This must be true. It is a conclusion.

If you conclude that ‘All C are A’ (your opinion, not necessarily a fact), you are assuming that A, B and C overlap i.e. they all have exactly the same elements.

Look again:

Argument 1:

Premises:

All A are B.
All B are C.

Conclusion: All A are C.

Argument 2:

Premises:

All A are B.
All B are C.

Conclusion: All C are A.

Assumption: A, B and C overlap.

The conclusion of argument 2 will not hold if the assumption is negated.

If you are wondering why we are emphasizing it again and again even though it looks really simple, here is an official question that might help you understand the reasons for our misgivings. We will give you the argument but not the question stem. We will also give you the correct answer. You will need to decide whether the question stem asks for a conclusion or an assumption.

Question: Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published. The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book’s publisher.

(A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book.

(B) Because the financial advantage of excerpting a new book in a magazine usually accrues to the book’s publisher, magazine editors are unwilling to publish excerpts from new books.

(C) In calculating the total number of copies that a book has sold, publishers include sales of copies of magazines that featured an excerpt of the book.

(D) The effectiveness of having excerpts of a book published in a magazine, measured in terms of increased sales of a book, is proportional to the circulation of the magazine in which the excerpts are published.

(E) Books that are suitable for excerpting in high-circulation magazines sell more copies than book that are not suitable for excerpting.

The correct answer is (A). What is the question?

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 Go From a 48 to 51 in GMAT Quant – Part II

Quarter Wit, Quarter WisdomThis post is continuation of last week’s post which you can check here.

Another method of saving time on simple questions – use data given in one statement to examine the other!

Now you might think we have lost it! After all, you know very well that in Data Sufficiency questions of GMAT, you must examine each statement independently. You CANNOT use data from one to analyze the other – absolutely correct. So you should ignore the other statement completely while examining one – hmm, not necessarily!

Sometimes, one statement could give us ideas about the next one such that we could save time while examining it. Needless to say, we need to be very careful but it certainly is a useful strategy. Also, it could help us verify that our calculations are correct. Here is why…

When we say DS question, think of a puzzle. The question stem gives you the statement of a puzzle ending with something like “What is the value of x?” or “Is x 7?” etc. You have to answer the question asked in the puzzle. Think of the two statements that come with the question as clues to the puzzle. So the puzzlemaster gives you the first clue (statement 1) and asks you: can you answer the question now? If you are able to, your answer is either (A) or (D).

Then he tells you to ignore the first clue and gives you another clue (statement 2). Again he asks you: can you answer the question now? Again, you may or may not able to. If you are able to, your answer will be (B) or (D) depending on how you fared in statement 1. If you are unable to answer the question, he tells you to consider both statements together and then try to answer. If you are able to, your answer is (C).

The point to note here is that both clues lead you to answer the same puzzle. Say if the puzzle is: What is x? If clue 1 tells you that x is 6, clue 2 cannot tell you that x is 9. They both must lead you to the same value of x. Clue 1 could tell you that x is either 6 or 8 and clue 2 could tell you that x is either 8 or 9. In this case, when we use both clues together, we find that x must be 8 to satisfy both. Hence the statements never contradict each other. This means, if we get possible values of x from statement 1, we know that statement 2 will also give us at least one of those values.

This is how one statement could give us a starting point for the next one. Now that you understand the “why”, let’s go on to “how”, using a question.

Question: If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

Statement 1: N is divisible by 3

Statement 2: N is divisible by 7

Solution:

Given: N = 4321 + K

1 <= K < 10

So N could range from 4322 (when K = 1) to 4330 (when K = 9). To find the value of K, we need to find the unique value of N.

Statement 1 tells us that N is divisible by 3.

4321 is not divisible by 3 since the sum of its digits is 4+3+2+1 = 10. It is 1 more than a multiple of 3. So the next multiple of 3 will be 4323. Hence N could be 4323. But there are some other multiples of 3 which could be the value of N. After 4323, 4326 and 4329 could also be the values of N since they are multiples of 3 too. We know this because if A is a multiple of 3, A+3, A+6, A+9, A-3, A-6 etc are also multiples of 3. So since 4323 is a multiple of 3, 4326 and 4329 will also be multiples of 3. We did not get a unique value for N so statement 1 alone is not sufficient.

Now let’s go on to statement 2. This tells us that N must be a multiple of 7. In 10 consecutive numbers, there will be either one multiple of 7 or two multiples of 7. If there is only one multiple of 7 in the range 4322 to 4331, statement 2 alone will be sufficient to give us the value of N. If there are two multiples of 7 in this range, then statement 2 alone will not be sufficient.

Recall that from statement 1, we already know that N will take one of three values: 4323, 4326 or 4329.

Let’s check for 4326 because it is in the middle. If 4326 is divisible by 7, there will be no other multiple of 7 in the range 4322 to 4331 because the closest multiples of 7 to 4326 will be 4326 – 7 and 4326 + 7. When we divide 4326 by 7, we find that it is divisible. This means that statement 2 gives us a single value of N. Hence statement 2 alone is sufficient.

Hypothetically, what if we had found that 4326 is not divisible by 7? Then we would have known that either 4323 or 4329 must be a multiple of 7. In both cases, statement 2 would have given us 2 multiples of 7 because both 4330 (7 more than 4323) and 4322 (7 less than 4329) are in the possible range. Then we would have known that the answer will be (C) i.e. we will need both statements to answer the question since the possible values from the two statements will have only one overlap in either case.

Note that what we gleaned from statement 1 helped us quickly examine statement 2 and get to the answer right away. But this is an advanced technique and you should use it only if you understand it very well. Else, it is best to stick to completely ignoring one statement while working on the other.

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 Go From a 48 to 51 in GMAT Quant – Part I

Quarter Wit, Quarter WisdomPeople often ask – how do we go from 48 to 51 in Quant? This question is very hard to answer since we don’t have a step by step plan – do theory from here – do questions from there – take a test from here – read posts from there etc. Today and in the next few weeks, we will discuss how to go from 48 to 51 in Quant.

Above Q48, the waters are pretty choppy! Questions are hard less because of the content and more because they look so unique – even though they’re testing the same concepts.  Training yourself to see familiarity in the obscure is difficult, and that happens from seeing a lot of problems. There is barely any scope for making silly mistakes – you must run through all simple questions quickly and neatly, leaving you plenty of time to think through the tougher ones. It’s important to have enough time and keep your cool, which is easier to do if you have more time.

The question for today is: how do you handle simple questions quickly? We have mentioned many times that most GMAT Quant questions do not need Algebra. We can easily solve them by just analyzing while reading the question stem!

Here is how we can do that:

Question: School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

Solution:  This is a pretty simple non-tricky PS question. To solve it, most people use an algebraic method which looks something like this.

Girls in school A : Girls in school B = 4 : 3
So number of girls in school A = 4n and number of girls in school B = 3n
Since in school A, 40% students are girls and 60% are boys, number of boys is 6n.
Since in school B, 60% students are girls and 40% students are boys, number of boys is 2n.

If we transfer 20 boys from school A, we are left with 6n – 20 and when 20 boys are added to school B, we get 2n + 20 boys in school B.
(6n – 20)/(2n + 20) = 5/3
You get n = 20

Boys at school A after transfer = 6*20 – 20 = 100
Boys at school B after transfer = 2*20 + 20 = 60
Difference = 40

Answer (B)

This method gives you the correct answer, obviously, but it does take quite a bit of time. On the other hand, this is what should go through your mind while reading the question if you are focused on using logic:

“School A is 40% girls and school B is 60% girls.”

School A – 40% girls
School B – 60% girls

“The ratio of the number of girls at school A to the number of girls at school B is 4:3”
When we read this line, we should take a step back to the previous line with the % figures. We see that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100 (use easy numbers). So school A has 80 girls while school B has 60 girls. This gives us a ratio of 4:3. (If you do not get 4:3 on your first try, you should tweak the assumed numbers a bit but you should stick to simple numbers.) Then verify the rest of the data against these numbers and get your answer.

School A has 120 boys and school B has 40 boys. Transfer 20 boys from school A to school B to get 100 boys in school A and 60 boys in school B giving us a difference of 40 boys.

This takes lesser time but requires some ingenuity. That could be the difference between Q48 and Q51.

Hope this gave you some ideas. Try the reasoning approach on other simple questions. With practice, you can save a ton of time!

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!

Find the Correct Answer for Diagonals of a Polygon in This GMAT Question

Quarter Wit, Quarter WisdomIn today’s post, we will give you a question with two solutions and two different answers. You have to find out the correct answer and explain why the other is wrong. But before we do that, let’s give you some background.

Given an n sided polygon, how many diagonals will it have?

An n sided polygon has n vertices. If you join every distinct pair of vertices you will get nC2 lines. These nC2 lines account for the n sides of the polygon as well as for the diagonals.

So the number of diagonals is given by nC2 – n.

nC2 – n = n(n-1)/2 – n = n(n – 3)/2

Taking quick examples:

Example 1: How many diagonals does a polygon with 25 sides have?

No. of diagonals = n(n – 3)/2 = 25*(25 – 3)/2 = 275

Example 2: How many diagonals does a polygon with 20 sides have, if one of its vertices does not send any diagonal?

The number of diagonals of a 20 sided figure = 20*(20 – 3)/2 = 170

But one vertex does not send any diagonals. Each vertex makes a diagonal with (n-3) other vertices – it makes no diagonal with 3 vertices: itself, the vertex immediately to its left, and the vertex immediately to its right. With all other vertices, it makes a diagonal. So we need to remove 20 – 3 = 17 diagonals from the total.

Total number of diagonals if one vertex does not make any diagonals = 170 – 17 = 153 diagonals.

We hope everything done till now makes sense. Now let’s go on to the part which seems to make no sense at all!

Question: How many diagonals does a polygon with 18 sides have if three of its vertices, which are adjacent to each other, do not send any diagonals?

Answer: We will use two different methods to solve this question:

Method 1: Using the formula discussed above

Number of diagonals in a polygon of 18 sides = 18*(18 – 3)/2 = 135 diagonals

Each vertex makes a diagonal with n-3 other vertices – as discussed before.

So each vertex will make 15 diagonals.

Total number of diagonals if 3 vertices do not send any diagonals = 135 – 15*3 = 90 diagonals.

Method 2:

The polygon has a total of 18 vertices. 3 vertices do not participate so we need to make all diagonals that we can with 15 vertices.

Number of lines you can make with 15 vertices = 15C2 = 15*14/2 = 105

But this 105 includes the sides as well. A polygon with 18 vertices has 18 sides. Since 3 adjacent vertices do not participate, 4 sides will not be formed. 15 vertices will have 14 sides which will be a part of the 105 we calculated before.

Total number of diagonals if 3 vertices do not send any diagonals = 105 – 14 = 91

Note that the two answers do not match. Method 1 gives us 90 and method 2 gives us 91. Both methods look correct but only one is actually correct. Your job is to tell us which method is correct and why the other method is incorrect.

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 Reason Behind Absolute Value Questions on the GMAT

Quarter Wit, Quarter WisdomEven after working extensively on absolute value questions, sometimes students come up with “why?” i.e. why do we have to take positive and negative values? Why do we have to consider ranges etc. They know the process but they do not understand the reason they need to follow the process. So here today, in this post, we will try to explain the reason.

You know how to solve an equation such as x + 2x = 4. Simple enough, right? Just add x with 2x to get 3x and separate out the x on one side. But what do you do when you have an equation with absolute values? How will you handle that equation? Say, you have |x| + 2x = 4. Is this your regular equation? No! You CANNOT say that x + 2x = 4 => 3x = 4 => x = 4/3. You have an absolute value and that complicates matters. You need to get rid of it to get a solution for x. How do you get rid of absolute values? The definition of absolute value helps us here:

|x| = x if x >= 0

|x| = -x if x < 0

So you can substitute x for |x| to make it a regular equation but only if x is non negative. If x is negative, then you put -x instead of |x| to convert it into a simple equation. And that is the reason you need to take positive and negative values of what is inside the absolute value sign.

Similarly,

|x-5| = (x-5) if (x-5) >= 0 i.e. if x >= 5

|x-5| = -(x-5) if (x-5) < 0 i.e. if x < 5

Let’s go back to the previous example and see how we can get rid of the absolute value to make it a regular equation:

Question 1: What is the value of x given |x| + 2x = 4?

We don’t know whether x is positive or negative so we will look at what happens in both cases:

Case 1: x is positive or 0

If x >= 0 then equation becomes x + 2x = 4 => x = 4/3

Our initial condition is that x is non negative. We get a positive solution on solving it and hence 4/3 is a valid solution.

Case 2: x is negative

If x < 0 then equation becomes -x + 2x = 4 => x = 4

Our initial condition is that x is negative. We get a positive solution on solving it and hence x = 4 is not a valid solution. Had we obtained a negative solution, it would have been valid.

So there is only one solution x = 4/3.

We hope the entire process makes more sense now. Let’s follow it up with a complex question from our algebra book.

Question 2:  If x and y are integers and y = |x+3| + |4-x|, does y equal 7?

Statement 1: x < 4

Statement 2: x > -3

Solution: Now what do you do when you have y = |x+3| + |4-x|? How do you convert this into a regular equation? You don’t know whether whatever is in the absolute value sign is positive or negative. How will you get rid of the sign then? You will work on all the cases (messy algebra coming up!).

Now, we see the same logic in this question:

y = |x+3| + |4-x|

|x+3| = (x+3) if (x+3) >= 0. In other words, if x >= -3

|x+3| = -(x+3) if (x+3) < 0. In other words, if x < -3

|4-x| = (4-x) if (4-x) >= 0. In other words, if x <= 4

|4-x| = -(4-x) if (4-x) < 0. In other words, if x > 4

So our absolute values behave differently when x < -3, between -3 and 4 and when x > 4. We say that -3 and 4 are our transition points.

Case 1:

When x < -3, |x+3| = -(x+3) and |4-x| = (4-x).

So the equation becomes y = -(x+3) + (4-x)

y = 1 – 2x

For different values of x, y will take different values. Recall that x must be less than -3. Say x = -4, then y = 9. If x = -5, y = 11.

Case 2:

When -3 <= x <= 4, |x+3| = (x+3) and |4-x| = (4-x).

So the equation becomes y = (x+3) + (4-x)

y = 7

In this range, y will always be 7.

Case 3:

When x > 4, |x+3| = (x+3) and |4-x| = -(4-x)

So the equation becomes y = (x+3) – (4-x)

y = 2x – 1

For different values of x, y will take different values. Recall that x must be more than 4. Say x = 5, then y is 9. If x = 6, then y is 11.

Note that y equals 7 only when x is between -3 and 4. Both statements together tell us that x is between -3 and 4. No statement alone gives us this information. Hence, using both statements, we know that y must be 7.

Answer (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!

Easy (A)/(B) Trap in Data Sufficiency Questions on the GMAT

Quarter Wit, Quarter WisdomWe know that ‘Easy C’ is a common trap of DS questions – have you wondered whether there could be trap called ‘Easy A/B’ such that the answer would actually be (C)? Such questions also exist! The point is that whenever you feel that the question was way too simple, you might want to take a step back and review. GMAT will try every trick in the trade to delineate you. Let us show you a question which looks like an easy (A) but isn’t:

Question: 25 integers are written on a board. Are there at least two consecutive integers among them?

Statement 1: For every value in the list, if the value is increased by 1, the number of distinct values in the list does not change.

Statement 2: At least one value occurs more than once in the list.

Solution: Let’s first review the information given to us here:

25 integers are written on the board – we don’t know whether they are all distinct. We want to know if there is any pair of consecutive integers among them.

Let’s look at the statements:

Statement 1: For every value in the list, if the value is increased by 1, the number of distinct values in the list does not change.

It is easy to fall for statement 1 and think that it is sufficient alone. Say, if any single value is increased by 1 and it doesn’t match any other value already there in the list, it means that there are no consecutive integers, doesn’t it? Well, no! But we will talk about that in a minute. Let’s first look at why we might think that statement 1 is sufficient.

Say, the numbers are: 1, 5, 8, 10, 35, 76 …

If you increase 1 by 1, you get 2 and the list looks like this:

Now the numbers are 2, 5, 8, 10, 35, 76 …

Note that the number of distinct integers is the same.

Had there been two consecutive integers such as 1, 2, 8, 10, 35, 76 …

If we increase 1 by 1, the list would have become 2, 2, 8, 10, 35, 76 … – this would have decreased the number of distinct integers.

You might be tempted to say here that statement 1 alone is sufficient. What you might forget is that when you increase a number by 1, one distinct integer could be getting wiped out and another taking its place! It may not occur to you that the case might be different when one value occurs more than once, but statement 2 should give you a hint. Obviously, statement 2 alone is not sufficient but let’s analyze what happens when we take both statements together.

Since statement 1 doesn’t tell you that all values are distinct, statement 2 should make you think that you need to consider the case where one value occurs more than once in the list. In that case, is it possible that number of different values in the list does not change even though there is a pair of consecutive integers?

Say the numbers are 1, 1, 2, 8, 10, 35, 76 …
Now if you increase 1 by 1, the list would look like 1, 2, 2, 8, 10, 35, 76 …

Here, the number of distinct integers stays the same even when you increase a number by 1 and you have consecutive integers! In this case, if there were no consecutive integers, the number of distinct integers would have increased. Hence if the numbers are not all distinct and the number of distinct numbers needs to stay the same, there must be a pair of consecutive integers.

This tells you that statement 1 is not sufficient alone but both statements together answer the question with a ‘Yes’.

Answer (C)

Takeaway – Just as when you get an easy (C), you must check whether the answer could be (A) or (B), when you feel that the answer is an easy (A) or (B), you might want to check whether the other statement gives some relevant data and is necessary.

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!

Understanding Conjunctions on the GMAT

Quarter Wit, Quarter WisdomWe would like to discuss a bit about conjunctions today – just whatever is relevant for GMAT. We will start by defining the kinds of conjunctions, then move on to the different ways in which they are used, and finally, we will see how they can be tested in a question.

Conjunction is a word that connects or joins together words, phrases, clauses, or sentences. There are two kinds of conjunctions:

1. Coordinating conjunctions – Connect two equal parts of a sentence

Further, coordinating conjunctions are of two types:

Pure Conjunctions – and, but, or, for, nor, yet, so (the first letters of these make the acronym FANBOYS) – try to keep these in mind.

Conjunctive Adverbs – These words sometimes act as conjunctions and at other times, as adverbs – accordingly, in fact, again, instead, also, likewise, besides, moreover, consequently, namely, finally, nevertheless, for example, otherwise, further, still, furthermore, that is, hence, then, however, therefore, indeed, thus

2. Subordinating conjunctions – Connect two unequal parts of a sentence e.g. independent and dependent clauses – after, since, when, although, so that, whenever, as, supposing, where, because, than, whereas, before, that, wherever, but that, though, whether, if, though, which, in order that, till, while, lest, unless, who, no matter, until, why, how, what, even though

Things to note about conjunctions:

1. Two independent clauses can be joined by a comma and a pure conjunction. However, a comma by itself will not work to join together two sentences and will create a comma splice!

Examples:

The rain slashed the town, and the people scurried for shelter.

The policeman dodged the bullets, but a bystander was shot.

If you omit the conjunctions ’or’ and ‘but’ above, you will create a comma splice.

2. When two independent clauses are joined by a conjunctive adverb we need to insert a semicolon between the two clauses. Note that conjunctive adverbs are not really full conjunctions, and they can’t do that job by themselves. It is the semicolon that does the real job of joining the two independent clauses.

Examples:

The rain slashed the town; furthermore, the people scurried for shelter.

The policeman dodged the bullets; however, a bystander was shot.

Note that if we use a comma instead of a semicolon in the examples above, we will create a comma splice.

3. A dependent clause at the beginning of a sentence is introductory, and it is usually followed by a comma.

Examples:

While the rain slashed the town, the people scurried for shelter.

Although the policeman dodged the bullets, a bystander was shot.

On the other hand, no punctuation is necessary for the dependent clause following the main clause.

Let’s take one of our own questions to understand the application of these concepts:

Question: Unlike the previous year’s bidding, the contract was awarded not simply to the firm offering to complete the work on time for the least cost; the thoroughness of the design submission was also factored into the decision.

(A) Unlike the previous year’s bidding, the contract this year was awarded not simply to the firm offering to complete the work on time for the least cost;

(B)   This year, unlike last year, the contract was awarded not simply to the firm offering to complete the work on time for the least cost;

(C)   Unlike the previous year’s bidding, this year the contract was awarded not simply to the firm offering to complete the work on time for the least cost;

(D)   Unlike the previous year’s bidding, the bidding for the contract this year was awarded not simply to the firm offering to complete the work on time for the least cost, instead

(E)    Unlike the previous year’s bidding, the contract’s bidding this year were awarded not simply to the firm offering to complete the work on time for the least cost;

Solution: Other than the comparison errors contained in (A) – compares bidding with contract – and (C) – compares bidding with year – we have sentence structure errors.

There are two independent clauses here:

–          the contract was awarded not simply to the firm offering to complete the work on time for the least cost.

–          the thoroughness of the design submission was also factored into the decision.

There are two ways to join them – we can use a conjunction or a semi colon. Options (A), (B), (C) and (E) use a semi colon.

Option (D) tries to use a conjunction with a comma but note that “instead” is a conjunctive adverb. It needs a semi colon before it. The use of instead with a comma has created a comma splice. Options (D) and (E) also have meaning errors since they award ‘bidding’ to the firm instead of awarding the ‘contract’ to the firm. (E) is also incorrect in its use of ‘were awarded’. The contract is singular and hence, ‘was awarded’ should be used.

Option (B) rectifies all these errors and is the answer!

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!

Rounding Up Some Official GMAT Questions!

Quarter Wit, Quarter WisdomLast week we looked at some rounding rules. Today, let’s go over some official questions on rounding. They are quite simple and if we just keep the “Slip to the side and look for a 5” rule in mind, they can be easily solved.

Question 1: If n = 2.0453 and n* is the decimal obtained by rounding n to the nearest hundredth, what is the value of n* – n?

(A) -0.0053
(B) -0.0003
(C) 0.0007
(D) 0.0047
(E) 0.0153

Solution: A quick note on place value nomenclature:

Given a decimal 345.789, we know that 5 represents the units digit, 4 the tens digit and 3 the hundreds digit. Also, 7 represents the tenths digit, 8 the hundredths digit and 9 the thousandths digit and so on…

Now let’s go back to this question:

n = 2.0453

We need to round n to the nearest hundredth which means we will retain 2 digits after the decimal. The third digit after the decimal is 5 so 2.0453 rounded to the nearest hundredth is 2.05.

Thus n* – n = 2.05 – 2.0453 = 0.0047

Answer (D)

Question 2: If digit h is the hundredths digit in the decimal n = 0.2h6, what is the value of n, rounded to the nearest tenth?

Statement 1: n < 1/4

Statement 2: h < 5

Solution: Given that n = 0.2h6

We need to find the value of n rounded to the nearest tenth i.e. we need to keep only one digit after the decimal.

Statement 1: n < 1/4

In decimal form, it means n < 0.25

If h were 5 or greater, n would become 0.256 or 0.266 or higher. All these values would be more than 0.25 so h must be less than 5 such as 0.246 or 0.236 etc. In all such cases, n would be rounded to 0.2

This statement alone is sufficient.

Statement 2: h < 5

This is even simpler. Since we have been given that h is less than 5, when we round n to the tenths digit, we will get 0.2

This statement alone is also sufficient.

Answer (D)

Question 3: If d denotes a decimal number, is d >= 0.5?

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

Statement 2: When d is rounded to the nearest integer, the result is 1.

Solution: Again, a simple question!

We need to find whether d is greater than or equal to 0.5 or not.

Statement 1: When d is rounded to the nearest tenth, the result is 0.5.

This means that whatever d is, when we round it to the nearest tenth, we get 0.5. What are the possible values of d? If d is anywhere from 0.450 to 0.5499999…, it will be rounded to 0.5

Some of these numbers are less than 0.5 and others are greater than 0.5 so this statement alone is not sufficient.

Statement 2: When d is rounded to the nearest integer, the result is 1.

In this case d must be at least 0.5; only then can it be rounded to 1.

d can be anything from 0.50 to 1.499999… In any case, d will be greater than or equal to 0.5.

This statement alone is sufficient to answer the question.

Answer (B)

We hope you see that if we just remember the rules, we can solve most rounding questions very quickly and efficiently.

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!

Rounding Rules on the GMAT: Slip to the Side and Look for a Five!

Quarter Wit, Quarter WisdomThe famous rounding song by Joe Crone is pretty much all you need to solve the trickiest of rounding questions on GMAT:

You just slip to the side, and you look for a five.

Well if the number that you see is a five or more, 

You gotta round up now, that’s for sure.

If the number that you see is a four or less, 

   You gotta round down to avoid a mess.

To put it in our own words, when we round a decimal, we drop the extra decimal places and apply certain rules:

–          If the first dropped digit is 5 or greater, we round up the last digit that we keep.

–          If the first dropped digit is 4 or smaller, we keep the last digit that we keep, the same.

For Example, we need to round the following decimals to two digits after decimal:

(a) 3.857

We drop 7. Since 7 is ‘5 or greater’, we are left with 3.86

(b) 12.983

We drop 3. Since 3 is ‘4 or smaller’, we are left with 12.98

(c) 26.75463

We drop 463. Since 4 is ‘4 or smaller’, we are left with 26.75

(d) 8.9675

We drop 75. Since 7 is ‘5 or greater’, we are left with 8.97

Note example (c) carefully:

When we round 26.75463 to two decimal places, we do not start rounding from the rightmost digit i.e. this is incorrect: 26.75463 becomes 26.7546 which becomes 26.755 which further becomes 26.76 – this is not correct. .00463 is less than .005 and hence should be ignored. You only need to worry about the digit right next to the digit you are keeping. Just slip to the side, and look for a five!

A logical question arises: what happens when we have, say, 2.5 and we need to round it to the nearest integer? 2.5 is midway between 2 and 3. In that case, why do we round the number up, as the rule suggests? Note that a 2.5 is a tie and we have many tie breaking rules that can be used. They are ‘Round half to odd’, ‘Round half to even’, ‘Round up’, ‘Round down’, ‘Round towards 0’, ‘Round away from 0’ etc. We don’t need to worry about all these since GMAT uses only Round up i.e. 2.5 will be rounded up to 3.

Let’s take a look at a question now which uses these fundamentals.

Question: The exact cost price to make each unit of a widget is $7.6xy7, where x and y represent single digits. What is the value of y?

Statement 1: When the cost is rounded to the nearest cent, it becomes $7.65.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes $7.65.

Solution: The question is based on rounding. We need to figure out the value of y given some rounding scenarios. Let’s look at them one by one.

Statement 1: When the cost is rounded to the nearest cent, it becomes $7.65.

When rounded to the nearest cent, the cost becomes 7 dollars and 65 cents. 6xy7 cents got rounded to 65 cents. When will .6xy7 get rounded to .65? When .6xy7 lies anywhere in the range .6457 to .6547. Note that in all these cases, when you round the number to 2 digits, it will become .65.

Say price is 7.6468. We need to drop 68 but since 6 is ‘5 or greater’, 4 gets rounded up to 5.

Similarly, say the price is 7.6543. We need to drop 43. Since 4 is ‘4 or smaller’, 5 stays as it is.

So x and y can take various different values. This statement alone is not sufficient.

Statement 2: When the cost is rounded to the nearest tenth of a cent, it becomes $7.65

Now the cost is rounded to the tenth of a cent which means 3 places after the decimal. But the cost is given to us as $7.65. Since we need 3 places, the cost must be $7.650 (which will be written as $7.65)

When will 7.6xy7 get rounded to 7.650? Now this is the tricky part of the question – from 7.6xy7, you need to drop the 7 and round up y. When you do that, you get 7.650. This means 7.6xy7 must have been 7.6497. Only in this case, when we drop the 7, we round up the 9 to make 10, carry the 1 over to 4 and make it 5. This is the only way to get 7.650 on rounding 7.6xy7 to the tenth of a cent. Hence x must be 4 and y must be 9. This statement alone is sufficient to answer the question.

Answer (B)

Hope you see that a few simple rules can make rounding questions quite easy.

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 Permutations & Combinations and Data Sufficiency Come Together on the GMAT!

Quarter Wit, Quarter WisdomWhile discussing Permutations and Combinations many months back, we worked through several examples of arranging people in seats. Today we bring you an interesting question based on those concepts. It brings to the fore the tricky nature of both Data Sufficiency and Combinatorics – so much so that when the two get together, it is unlimited fun!

In some circumstances, we suggest you to travel the whole nine yards – i.e. solve for the answer in Data Sufficiency questions too even if you feel that sufficiency has already been established. This is especially true for quadratic equations which we assume will give us two values of x but might actually give just a single unique value (such that both roots are the same). In Combinatorics too, sometimes what may look like two distinct cases could actually give the same answer. Let’s jump on to the question.

Question 1: There are x children and y chairs in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12

Statement 2: There are more chairs than children.

Solution:

There are x children and y chairs.

x and y are prime numbers.

Statement 1: x + y = 12

Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases:

Case 1: x=5 and y=7

There are 5 children and 7 chairs.

Case 2: x=7 and y=5

There are 7 children and 5 chairs

At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient  alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements.

Case 1: x=5 and y=7

If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.

Total number of arrangements would be 7C5 * 5!

Case 2: x = 7 and y = 5

If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways.

Total number of arrangements would be 7C5 * 5!

Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways.

So actually this statement alone is sufficient! Most people would not have seen that coming!

Statement 2: There are more chairs than people.

We don’t know how many children or chairs there are. This statement alone is not sufficient.

Answer: A

We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something!

Now, what if we alter the question slightly and make it:

Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

Statement 1: x + y = 12

Statement 2: There are more chairs than children.

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!

Medians, Altitudes and Angle Bisectors in Special Triangles on the GMAT

Quarter Wit, Quarter WisdomWe are assuming you know the terms median, angle bisector and altitude but still, just to be sure, we will start our discussion today by defining them:

Median – A line segment joining a vertex of a triangle with the mid-point of the opposite side.

Angle Bisector – A line segment joining a vertex of a triangle with the opposite side such that the angle at the vertex is split into two equal parts.

Altitude – A line segment joining a vertex of a triangle with the opposite side such that the segment is perpendicular to the opposite side.

Usually, medians, angle bisectors and altitudes drawn from the same vertex of a triangle are different line segments. But in special triangles such as isosceles and equilateral, they can overlap. We will now give you some properties which can be very useful.

I.

In an isosceles triangle (where base is the side which is not equal to any other side):

– the altitude drawn to the base is the median and the angle bisector;

– the median drawn to the base is the altitude and the angle bisector;

– the bisector of the angle opposite to the base is the altitude and the median.

II.

The reverse is also true. Consider a triangle ABC:

– If angle bisector of vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this angle bisector is also the altitude.

– If altitude drawn from vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this altitude is also the angle bisector.

– If median drawn from vertex A is also the angle bisector, the triangle is isosceles such that AB = AC and BC is the base. Hence this median is also the altitude.

and so on…

III.

In an equilateral triangle, each altitude, median and angle bisector drawn from the same vertex, overlap.

Try to prove all these properties on your own. That way, you will not forget them.

A few things this implies:

–          Should an angle bisector in a triangle which is also a median be perpendicular to the opposite side? Yes.

–          Can we have an angle bisector which is also a median which is not perpendicular? No. Angle bisector which is also a median implies isosceles triangle which implies it is also the altitude.

–          Can we have a median from vertex A which is perpendicular to BC but does not bisect the angle A? No. A median which is an altitude implies the triangle is isosceles which implies it is also the angle bisector.

and so on…

Let’s take a quick question on these concepts:

Question: What is ?A in triangle ABC?

Statement 1: The bisector of ?A is a median in triangle ABC.
Statement 2: The altitude of B to AC is a median in triangle ABC.

Solution: We are given a triangle ABC but we don’t know what kind of a triangle it is.

Jump on to the statements directly.

Statement 1: The bisector of ?A is a median in triangle ABC.

The angle bisector is also a median. This means triangle ABC must be an isosceles triangle such that AB = AC. But we have no idea about the measure of angle A. This statement alone is not sufficient.

Statement 2: The altitude of ?B to AC is a median in triangle ABC.

The altitude is also a median. This means triangle ABC must be an isosceles triangle such that AB = BC (Note that the altitude is drawn from vertex B here). But we have no idea about the measure of angle A. This statement alone is not sufficient.

Using both statements together, we see that AB = AC = BC. So the triangle is equilateral! So angle A must be 60 degrees. Sufficient!

Answer (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!

A Remainders Shortcut for the GMAT

Quarter Wit, Quarter WisdomWe firmly believe that teaching someone is a most productive learning for oneself and every now and then, something happens that strengthens this belief of ours. It’s the questions people ask – knowingly or unknowingly – that connect strings in our mind such that we feel we have gained more from the discussion than even our students!

The other day, we came across this common GMAT question on remainders and many people had solved it the way we would expect them to solve. One person, perhaps erroneously, used a shortcut which upon reflection made perfect sense. Let me give you that question and the shortcut and the problem with the shortcut. We would like you to reflect on why the shortcut actually does make sense and is worth noting down in your log book.

Question: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

Solution: We are assuming you know how people do the question usually:

The logic it uses is discussed here and the solution is given below as Method I.

Method I:

Positive integer n leaves a remainder of 4 after division by 6. So n = 6a + 4 

n can take various values depending on the values of a (which can be any non negative integer).

Some values n can take are: 4, 10, 16, 22, 28, …

Positive integer n leaves a remainder of 3 after division by 5. So n = 5b + 3

n can take various values depending on the values of a (which can be any non negative integer).

Some values n can take are: 3, 8, 13, 18, 23, 28, …

The first common value is 28. So n = 30k + 28

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

Perfect! But one fine gentleman came up with the following solution wondering whether he had made a mistake since it seemed to be “super simple Math”.

Method II:

Given in question: “n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5.”

Divide the options by 6 and 5. The one that gives a remainder of 4 and 3 respectively will be correct.

(A) 3 / 6 gives Remainder = 3 -> INCORRECT
(B) 12 / 6 gives Remainder = 0 -> INCORRECT
(C) 18 / 6 gives Remainder = 0 -> INCORRECT
(D) 22 / 6 gives Remainder = 4 but 22 / 5 gives  Remainder = 2 -> INCORRECT
(E) 28 / 6 gives Remainder = 4 and 28 / 5 gives Remainder = 3 ->  CORRECT

Now let us point out that the options are not the values of n; they are the values of remainder that is leftover after you divide n by 30. The question says that n must give a remainder of 4 upon division by 6 and a remainder of 3 upon division by 5. This solution divided the options (which are not the values of n) by 6 and 5 and got the remainder as 4 and 3 respectively. So the premise that when you divide the correct option by 6 and 5, you should get a remainder of 4 and 3 respectively is faulty, right?

This is where we want you to take a moment and think: Is this premise actually faulty?

The fun part is that method II is perfectly correct too. Method I seems a little complicated when compared with Method II, doesn’t it? Let us give you the logic of why method II is correct:

Recall that division is nothing but grouping. When you divide n by 30, you make complete groups of 30 each. The number of groups you get is called the quotient (not relevant here) and the leftover is called the remainder. If this is not clear, check this post first.

When n is divided by 30, groups of 30 are made. Whatever is leftover is given in the options. 30 is completely divisible by 6 and by 5 hence the groups of 30 can be evenly divided into groups of 6 as well as groups of 5. Now, whatever is leftover (given in the options) after division by 30, we need to split that into further groups of 6 and 5. When we split it into groups of 6 (i.e. divide the option by 6), we must have remainder 4 since n leaves remainder 4. When we split it into groups of 5 (i.e. divide the option by 5), we must have remainder 3 since n leaves remainder 3. And, that is the reason we can divide the options by 6 and 5, check their remainders and get the answer!

Now, isn’t that neat!

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 Take on GMAT Takeaways

Quarter Wit, Quarter WisdomOnce you have covered your fundamentals, we suggest you to practice advanced questions and jot down your takeaways from them. Sometimes students wonder how to find that all important “takeaway”. Today, let’s discuss how to elicit a takeaway from a question which seems to have none.

What is a takeaway? It is a small note to yourself which you would do well to remember while going for the exam. Even if you don’t remember the exact property you jotted down, knowing that such a property exists is enough. You can always try it on a couple of numbers in the test to recall the exact content.

The question we will discuss today serves another purpose – it discusses properties of squares of odd and even integers so in a sense is a continuation of our advanced number properties discussion.

Question: Given x and y are positive integers such that y is odd, is x divisible by 4?

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.
Statement 2: x – y = 3

Solution: As of now, we don’t know any specific properties of squares of odd and even integers. However, we do have a good (presumably!) understanding of divisibility. To recap quickly, divisibility is nothing but grouping. To take an example, if we divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is the quotient and 1 is the remainder. For more on these concepts, check out our previous posts on divisibility.

Coming back to our question,

First thing that comes to mind is that if y is odd, y = (2k + 1).

We have no information on x so let’s proceed to the two statements.

Statement 1: When (x^2 + y^2) is divided by 8, the remainder is 5.

The statement tell us something about y^2 so let’s get that.

If y = (2k + 1)

y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1

Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), k(k+1) will be even. So 4k(k+1) will be divisible by 4*2 i.e. by 8. So when y^2 is divided by 8, it will leave a remainder 1.

When y^2 is divided by 8, remainder is 1. To get a remainder of 5 when x^2 + y^2 is divided by 8, we should get a remainder of 4 when x^2 is divided by 8. So x must be even. If x were odd, the remainder when x^2 were divided by 8 would have been 1. So we know that x is divisible by 2 but we don’t know whether it is divisible by 4 yet.

x^2 = 8a + 4 (when x^2 is divided by 8, it leaves remainder 4)

x^2 = 4(2a + 1)

So x = 2*?Odd Number

Square root of an odd number will be an odd number so we can see that x is even but not divisible by 4. This statement alone is sufficient to say that x is NOT divisible by 4.

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (Since Even – Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say. This statement alone is not sufficient.

Answer (A)

So could you point out the takeaway from this question?

Note that when we were analyzing y, we used no information other than that it is odd. We found out that the square of any odd number when divided by 8 will always yield a remainder of 1.

Now what can you say about the square of an even number? Say you have an even number x.

x = 2a

x^2 = 4a^2

This tells us that x^2 will be divisible by 4 i.e. we can make groups of 4 with nothing leftover. What happens when we try to make groups of 8? We join two groups of 4 each to make groups of 8. If the number of groups of 4 is even, we will have no remainder leftover. If the number of groups of 4 is odd, we will have 1 group leftover i.e. 4 leftover. So when the square of an even number is divided by 8, the remainder is either 0 or 4.

Looking at it in another way, we can say that if a is odd, x^2 will be divisible by 4 and will leave a remainder of 4 when divided by 8. If a is even, x^2 will be divisible by 16 and will leave a remainder of 0 when divided by 8.

Takeaways

– The square of any odd number when divided by 8 will always yield a remainder of 1.

– The square of any even number will be either divisible by 4 but not by 8 or it will be divisible by 16 (obvious from the fact that squares have even powers of prime factors so 2 will have a power of 2 or 4 or 6 etc). In the first case, the remainder when it is divided by 8 will be 4; in the second case the remainder will be 0.

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 Number Properties on the GMAT – Part III

Quarter Wit, Quarter WisdomContinuing our discussion on number properties, today we will discuss how factorials affect the behavior of odd and even integers. Since we are going to deal with factorials, positive integers will be our concern. Using a question, we will see how factorials are divided.

Question: If x and y are positive integers, is y odd?

Statement 1: (y+2)!/x! = odd
Statement 2: (y+2)!/x! is greater than 2

Solution: The question stem doesn’t give us much information – just that x and y are positive integers.

Question: Is y odd?

Statement 1: (y+2)!/x! = odd

Note that odd and even are identified only for integers. Since (y+2)!/x! is odd, it must be a positive integer. This means that x! must be equal to or less than (y+2)!

Now think, how are y and y+2 related? If y+2 is odd, y+1 is even and hence y is odd. If y+2 is even, by the same logic, y is even.

(y+2)! = 1*2*3*4*…*y*(y+1)*(y+2)

x! = 1*2*3*4*…*x

Note that (y+2)! and x! have common factors starting from 1. Since x! is less than or equal to (y+2)!, x will be less than or equal to (y+2). So all factors in the denominator, from 1 to x will be there in the numerator too and will get canceled leaving us with the last few factors of (y+2)!

To explain this, let us take a few examples:

Example 1: Say, y+2 = 6, x = 6

(y+2)!/x! = 6!/6! = 1

Example 2: Say, y+2 = 7, x = 6

(y+2)!/x! = 7!/6! = (1*2*3*4*5*6*7)/(1*2*3*4*5*6) = 7 (only one leftover factor)

Example 3: Say, y+2 = 6, x = 4

(y+2)!/x! = 6!/4! = (1*2*3*4*5*6)/(1*2*3*4) = 5*6 (two leftover factors)

If the division of two factorials is an integer, the factorial in the numerator must be larger than or equal to the factorial in the denominator.

So what does (y+2)!/x! is odd imply? It means that the leftover factors must be all odd. But the leftover factors will be consecutive integers. So after one odd factor, there will be an even factor. If we want (y+2)!/x! to be odd, we must have either no leftover factors (such that (y+2)!/x! = 1) or only one leftover factor and that too odd.

If we have no leftover factor, it doesn’t matter what y+2 is as long as it is equal to x. It could be odd or even. If there is one leftover factor, then y+2 must be odd and hence y must be odd. Hence y could be odd or even. This statement alone is not sufficient.

Statement 2: (y+2)!/x! is greater than 2

This tells us that y+2 is not equal to x since (y+2)!/x! is not 1. But all we know is that it is greater than 2. It could be anything as seen in examples 2 and 3 above. This statement alone is not sufficient.

Both statements together tell us that y+2 is greater than x such that (y+2)!/x! is odd. So there must be only one leftover factor and it must be odd. The leftover factor will be the last factor i.e. y+2. This tells us that y+2 must be odd. Hence y must be odd too.

Answer (C)

Takeaways: Assuming a and b are positive integers,

–          a!/b! will be an integer only if a >= b

–          a!/b! will be an odd integer whenever a = b or a is odd and a = b+1

–          a!/b! will be an even integer whenever a is even and a = b+1 or a > b+1

Think about this: what happens when we put 0 in the mix?

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 Number Properties on the GMAT – Part II

Quarter Wit, Quarter WisdomBefore we get started, be sure to take a look at Part I of this article. Number properties concepts come across as pretty easy, theoretically, but they have some of the toughest questions. Today let’s take a look at some properties of prime numbers and their sum. Note that don’t try to “learn” all the takeaways you come across for number properties – it will be very stressful. Instead, try to understand why the properties are such so that if you get a question related to some such properties, you can replicate the results effortlessly.

To start off, we would like to take up a simple question and then using the takeaway derived from it, we will look at a harder problem.

Question 1: Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Solution: What do we know about sum of two prime numbers?

e.g. 3 + 5 = 8

5 + 11 = 16

5 + 17 = 22

23 + 41 = 64

Do you notice something? The sum is even in all these cases. Why? Because most prime numbers are odd. When we add two odd numbers, we get an even sum.

We have only 1 even prime number and that is 2. Hence to obtain an odd sum, one number must be 2 and the other must be odd.

2 + 3 = 5

2 + 7 = 9

2 + 17 = 19

Look at the options given in the question. Three of them are odd which means they must be of the form 2 + Another Prime Number.

Let’s check the odd options first:

(A)   19 = 2 + 17 (Both Prime. Can be written as sum of two prime numbers.)
(B) 45 = 2 + 43 (Both Prime. Can be written as sum of two prime numbers.)
(D) 79 = 2 + 77 (77 is not prime.)

79 cannot be written as sum of two prime numbers. Note that 79 cannot be written as sum of two primes in any other way. One prime number has to be 2 to get a sum of 79. Hence there is no way in which we can obtain 79 by adding two prime numbers.

(D) is the answer.

Now think what happens if instead of 79, we had 81?

81 = 2 + 79

Both numbers are prime hence all three odd options can be written as sum of two prime numbers. Then we would have had to check the even options too (at least one of which would be different from the given even options). Think, how would we find which even numbers can be written as sum of two primes? We will give the solution of that next week. So the takeaway here is that if you get an odd sum on adding two prime numbers, one of the numbers must be 2.

Question 2: If m, n and p are positive integers such that m < n < p, is m a factor of the odd integer p?

Statement 1: m and n are prime numbers such that (m + n) is a factor of 119.

Statement 2: p is a factor of 119.

Solution: First of all, we are dealing with positive integers here – good. No negative numbers, 0 or fraction complications. Let’s move on.

The question stem tells us that p is an odd integer. Also, m < n < p.

Question: Is m a factor of p?

There isn’t much information in the question stem for us to process so let’s jump on to the statements directly.

Statement 1: m and n are prime numbers such that (m + n) is a factor of 119.

Write down the factors of 119 first to get the feasible range of values.

119 = 1, 7, 17, 119

All factors of 119 are odd numbers. So (m + n), a sum of two primes must be odd. This means one of m and n is 2. There are many possible values of m and n e.g. 2 and 5 (to give sum 7) or 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119).

Note that we also have m < n. This means that in each case, m must be 2 and n must be the other number of the pair.

So now we know that m is 2. We also know that p is an odd integer. Is m a factor of p? No. Odd integers are those which do not have 2 as a factor. Since m is 2, p does not have m as a factor.

This statement alone is sufficient to answer the question!

Statement 2: p is a factor of 119

This tells us that p is one of 7, 17 and 119. p cannot be 1 because m < n < p where all are positive integers.

But it tells us nothing about m. All we know is that it is less than p. For example, if p is 7, m could be 1 and hence a factor of p or it could be 5 and not a factor of p. Hence this statement alone is not sufficient.

Answer (A)

Something to think about: In this question, if you are given that m is not 1, does it change our answer?

Key Takeaways:

– When two distinct prime numbers are added, their sum is usually even. If their sum is odd, one of the numbers must be 2.

– Think what happens in case you add three distinct prime numbers. The sum will be usually odd. In case the sum is even, one number must be 2.

– Remember the special position 2 occupies among prime numbers – it is the only even prime 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!

Advanced Number Properties on the GMAT – Part I

Don’t worry, we are not going to discuss (Even + Even = Even) and (Odd + Odd = Even) type of basic number properties in this post. What we have in mind for today is something based on this but far more advanced. Often, people complain that they thoroughly understand the theory but have difficulties applying it and hence are stuck at a score of 600. They look for practice questions and tend to ignore concepts since they already “know” them. We often ask them to go back to concepts since we believe that a strong foundation of concepts is necessary for ‘score increase’. Mind you, when we do that, we don’t mean to ask them to review the basic concepts again, we mean to ask them to deduce and work on advanced concepts. Let’s show you with the help of a question.

Question: If two integers are chosen at random out of first 5 positive integers, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/5
B. 3/5
C. 7/10
D. 4/5
E. 9/10

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

– Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.
If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.

If the product is odd, it can always be written as product of two odd numbers.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Answer (C)

Takeaway:

When can we write a number as difference of squares?

– When the number is odd

or

– When the number has 4 as a factor

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 GMAT Formula to Remember: Profit on One, Loss on Another

Quarter Wit, Quarter WisdomI am no fan of formulas, especially the un-intuitive ones but the one we are going to discuss today has proved quite useful. It is for a concept tested on GMAT Prep so it might be worth your while to remember this little formula.

When two items are sold at the same selling price, one at a profit of x% and the other at a loss of x%, there is an overall loss. The loss% = (x^2/100)%

We will see how this formula is derived but the algebra involved is tedious. You can skip it if you wish.

Say two items are sold at $S each. On one, a profit of x% is made and on the other a loss of x% is made.

Say, cost price of the article on which profit was made = Ct

Ct (1 + x/100) = S

Ct = S/(1 + x/100)

Cost Price of the article on which loss was made = Cs

Cs (1 – x/100) = S

Cs = S/(1 – x/100)

Total Cost Price of both articles together = Ct + Cs = S/(1 + x/100) + S/(1 – x/100)

Ct + Cs = S[1/(1 + x/100) + 1/(1 – x/100)]

Ct + Cs = 2S/(1 – (x/100)^2)

Total Selling Price of both articles together = 2S

Overall Profit/Loss = 2S – (Ct + Cs)

Overall Profit/Loss % = [2S – (Ct + Cs)]/[Ct + Cs] * 100

= [2S/(Ct + Cs) – 1] * 100

= [2S/[2S/(1 – (x/100)^2)] – 1] * 100

= (x/100)^2 * 100

= x^2/100

Overall there is a loss of (x^2/100)%.

Let’ see how this formula works on a GMAT Prep question.

Question: John bought 2 shares and sold them for $96 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had

(A) a profit of $10
(B) a profit of $8
(C) a loss of $8
(D) a loss of $10
(E) neither a profit nor a loss

Solution:

Note that the question would have been straight forward had the COST price been the same, say $100. A 20% profit would mean a gain of $20 and a 20% loss would mean a loss of $20. Overall, there would have been no profit no loss.

Here the two shares are sold at the same SALE price. One at a profit of 20% on cost price which must be lower than the sale price (to get a profit) and the other at a loss of 20% on cost price which must be higher than the sale price (to get a loss). 20% of a lower amount will be less in dollar terms and hence overall, there will be a loss.

The loss % = (20)^2/100 % = 4%.
But we need the amount of loss, not the percentage of loss.

Total Sale price of the two shares = 2*96 = $192
Since there is a loss of 4%, the 96% of the total cost price must be the total sale price

(96/100)*Cost Price = Sale Price
Cost Price = $200
Loss = $200 – $192 = $8

Answer (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!

Determining the Area of Similar Triangles on the GMAT

Quarter Wit, Quarter WisdomRecall the important property that we discussed about the relation between the areas of the two similar triangles last week – if the ratio of their sides is ‘k’, the ratio of their areas will be k^2. As mentioned last week, it’s an important property and helps you easily solve otherwise difficult questions. The question I have in mind today also brings in focus the Pythagorean triplets.

There are some triplets that you should know out cold: (3, 4, 5), (5, 12, 13) and (8, 15, 17). Usually you will find one of these three or their multiples on GMAT. Given a right triangle and two sides, say the two legs, of length 20 and 48, we need to immediately bring them down to the lowest form 20:48 = 5:12. So we know that we are talking about the 5, 12, 13 triplet and the hypotenuse will be 13*4 = 52. These little things help us save a lot of time. Why is it that some people get done with the Quant section in less than an hour while others fall short on time? It is these little things that an adept test taker has mastered which make all the difference.

Anyway, let us go on to the question we have in mind.

Question: In the figure given below, the length of PQ is 12 and the length of PR is 15. The area of right triangle STU is equal to the area of the shaded region. If the ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR, what is the length of TU?

(A) (9?2)/4

(B) 9/2

(C) (9?2)/2

(D) 6?2

(E) 12

Solution: The information given in the question seems to overwhelm us but let’s take it a bit at a time.

“length of PQ is 12 and the length of PR is 15”

PQR is a right triangle such that PQ = 12 and PR = 15. So PQ:PR = 4:5. Recall the 3-4-5 triplet. A multiple triplet of 3-4-5 is 9-12-15. This means QR = 9.

“ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR”
ST/TU = PQ/QR

The ratio of two sides of PQR is equal to the ratio of two sides of STU and the included angle between the sides is same ( = 90). Using SAS, triangles PQR and STU are similar.

“The area of right triangle STU is equal to the area of the shaded region”

Area of triangle PQR = Area of triangle STU + Area of Shaded Region

Since area of triangle STU = area of shaded region, (area of triangle PQR) = 2*(area of triangle STU)

In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^2. If the ratio of the areas is given as 2 (i.e. k^2 is 2), the sides must be in the ratio ?2 (i.e. k must be ?2).

Since QR = 9, TU must be 9/?2. But there is no 9/?2 in the options – in the options the denominators are rationalized. TU = 9/?2 = (9*?2)/(?2*?2) = (9*?2)/2.

Answer (C)

The question could take a long time if we do not remember the Pythagorean triplets and the area of similar triangles property.

Takeaways:

  1. Pythagorean triplets you should know: (3, 4, 5), (5, 12, 13) and (8, 15, 17) and their multiples.
  2. In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^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!

Looking for Similar Triangles on the GMAT

Quarter Wit, Quarter WisdomOur Geometry book discusses the various rules we use to recognize similar triangles such as SSS, AA, SAS and RHS so we are assuming that we needn’t take those up here.

We are also assuming that you are comfortable with the figures that beg you to think about similar triangles such as

 

Try to figure out the similar triangles and the reason they are similar in each one of these cases. (Angles that look 90 are 90). Most of the figures have right angles/parallel lines.

This topic was also discussed by David Newland in a rather engaging post last week. You must check it out for its content as well as its context!

What we would like to discuss today are situations where most people do not think about similar triangles but if they do, it would make the question very easy for them. But before we do that, we would like to discuss a concept related to similar triangles which is very useful but not discussed often.

We already know that sides of similar triangles are in the same ratio. Say two triangles have sides a, b, c and A, B, C respectively. Then, a/A = b/B = c/C = k

Note that the altitudes of the two triangles will also be in the same ratio, ‘k’, since all lengths have the ratio ‘k’.

Then what is the relation between the areas of the two triangles? Since the ratio of the bases is k and the ratio of the altitudes is also k, the ratio of the areas will be k*k = k^2.

So if there are two similar triangles such that their sides are in the ratio 1:2, their areas will be in the ratio 1:4.

Now we are all ready to tackle the question we have in mind.

Question: In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area?

(A)   3:8

(B)   3:5

(C)   5:8

(D)   8:5

(E)    5:3

Solution: There are many ways to do this question but we will look at the method using similar triangles (obviously!).

Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below)

Now look at the original figure.

HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4.

Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P.

The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P.

Hence, area of shaded region : Area of unshaded region = 5:3

Answer (E)

Try to think of other ways in which you can solve this question.

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 Tricky Question on Negative Remainders

Quarter Wit, Quarter WisdomToday, we will discuss the question we left you with last week. It involves a lot of different concepts – remainder on division by 5, cyclicity and negative remainders. Since we did not get any replies with the solution, we are assuming that it turned out to be a little hard.

It actually is a little harder than your standard GMAT questions but the point is that it can be easily solved using all concepts relevant to GMAT. Hence it certainly makes sense to understand how to solve it.

 

Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Solution: As we said last week, this question can easily be solved using cyclicity and negative remainders. What is the remainder when a number is divided by 5? Say, what is the remainder when 2387646 is divided by 5? Are you going to do this division to find the remainder? No! Note that every number ending in 5 or 0 is divisible by 5.

2387646 = 2387645 + 1

i.e. the given number is 1 more than a multiple of 5. Obviously then, when the number is divided by 5, the remainder will be 1. Hence the last digit of a number decides what the remainder is when the number is divided by 5.

On the same lines,

What is the remainder when 36793 is divided by 5? It is 3 (since it is 3 more than 36790 – a multiple of 5).

What is the remainder when 46^8 is divided by 5? It is 1. Why? Because 46 to any power will always end with 6 so it will always be 1 more than a multiple of 5.

On the same lines, if we can find the last digit of 3^(7^11), we will be able to find the remainder when it is divided by 5.

Recall from the discussion in your books, 3 has a cyclicity of 4 i.e. the last digit of 3 to any power takes one of 4 values in succession.

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

and so on… The last digits of powers of 3 are 3, 9, 7, 1, 3, 9, 7, 1 … Every time the power is a multiple of 4, the last digit is 1. If it is 1 more than a multiple of 4, the last digit is 3. If it is 2 more than a multiple of 4, the last digit is 9 and if it 3 more than a multiple of 4, the last digit is 7.

What about the power here 7^(11)? Is it a multiple of 4, 1 more than a multiple of 4, 2 more than a multiple of 4 or 3 more than a multiple of 4? We need to find the remainder when 7^(11) is divided by 4 to know that.

Do you remember the binomial theorem concept we discussed many weeks back? If no, check it out here.

7^(11) = (8 – 1)^(11)

When this is divided by 4, the remainder will be the last term of this expansion which will be (-1)^11. A remainder of -1 means a positive remainder of 3 (if you are not sure why this is so, check last week’s post here). Mind you, you are not to mark the answer as (D) here and move on! The solution is not complete yet. 3 is just the remainder when 7^(11) is divided by 4.

So 7^(11) is 3 more than a multiple of 4.

Review what we just discussed above: If the power of 3 is 3 more than a multiple of 4, the last digit of 3^(power) will be 7.

So the last digit of 3^(7^11) is 7.

If the last digit of a number is 7, when it is divided by 5, the remainder will be 2. Now we got the answer!

Answer (C)

Interesting question, isn’t 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!

All About Negative Remainders on the GMAT

Quarter Wit, Quarter WisdomI could have sworn that I had discussed negative remainders on my blog but the other day I was looking for a post discussing it and much as I would try, I could not find one. I am a little surprised since this concept is quite useful and I should have taken it in detail while discussing divisibility. Though we did have a fleeting discussion of it here.

Since we did miss it, we will discuss it in detail today but you must review the link given above before we proceed.

Consider this: When n is divided by 3, it leaves a remainder 1.

This means that when we divide n balls in groups of 3 balls each, we are left with 1 ball.

This also means that n is 1 MORE than a multiple of 3. Or, it also means that n is 2 less than the next multiple of 3, doesn’t it?

Say, n is 16. When you split 16 balls into groups of 3 balls each, you get 5 groups of 3 balls each and there is one ball leftover. n is 1 more than a multiple of 3 (the multiple being 15). But we can also say that it is 2 LESS than the next multiple of 3 (which is 18). Hence, the negative remainder in this case is -2 which is equivalent to a positive remainder of 1.

Generally speaking, if n is divided by m and it leaves a remainder r, the negative remainder in this case is -(m – r).

When n is divided by 7, it leaves a remainder of 4. This is equivalent to a remainder of -3.

n is 3 more than a multiple of m. It is also 2 less than the next multiple of m. This means m is 5.

This concept is very useful to us sometimes, especially when the divisor and the remainder are big numbers.

Let’s take a question to see how.

Question 1: What is the remainder when 1555 * 1557 * 1559 is divided by 13?

(A)   0

(B)   2

(C)   4

(D)   9

(E)    11

Solution: Since it is a GMAT question (a question for which we will have no calculator), multiplying the 3 numbers and then dividing by 13 is absolutely out of question! There has to be another method.

Say n = 1555 * 1557 * 1559

When we divide 1555 by 13, we get a quotient of 119 (irrelevant to our question) and remainder of 8. So the remainder when we divide 1557 by 13 will be 8+2 = 10 (since 1557 is 2 more than 1555) and when we divide 1559 by 13, the remainder will be 10+2 = 12 (since 1559 is 2 more than 1557).

So n = (13*119 + 8)*(13*119 + 10)*(13*119 + 12) (you can choose to ignore the quotient and just write it as ‘a’ since it is irrelevant to our discussion)

So we need to find the remainder when n is divided by 13.

Note that when we multiply these factors, all terms we obtain will have 13 in them except the last term which is obtained by multiplying the constants together i.e. 8*10*12.

Since all other terms are multiples of 13, we can say that n is 8*10*12 (= 960) more than a multiple of 13. There are many more groups of 13 balls that we can form out of 960.

960 divided by 13 gives a remainder of 11.

Hence n is actually 11 more than a multiple of 13.

We did not use the negative remainders concept here. Let’s see how using negative remainders makes our calculations easier here. The remainder of 8, 10 and 12 imply that the negative remainders are -5, -3 and -1 respectively.

Now n = (13a – 5) * (13a – 3) * (13a – 1)

The last term in this case is -5*-3*-1 = -15

This means that n is 15 less than a multiple of 13 i.e. actually 2 less than a multiple of 13 because when you go back 13 steps, you get another multiple of 13. This gives us a negative remainder of -2 which means the positive remainder in this case will be 11.

Here we avoided some big calculations.

I will leave you now with a question which you should try to solve using negative remainders.

Question 2: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Hint: I solved this question orally in a few secs using cyclicity and negative remainders. Don’t get lost in calculations!

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 Suspect?

Quarter Wit, Quarter WisdomI came across a discussion on one of our questions where the respondents were split over whether it is a strengthen question or weaken! Mind you, both sides had many supporters but the majority was with the incorrect side. You must have read the write up on ‘support’ in your Veritas Prep CR book where we discuss how question stems having the word ‘support’ could indicate either strengthen or inference questions. I realized that we need a write up on the word ‘suspect’ too so here goes.

First let me give you the question stem of that question:

Which of the following, if true, supplies the best reason to suspect that the proposed new course will increase interest in the metropolitan cooking academy?

So do we have to find a reason which indicates that the new course WILL increase interest in the academy or do we have to find a reason that indicates that the new course WILL NOT increase interest in the academy? That is, is this a type of ‘strengthen’ question or the opposite – a ‘weaken’ question?

I would have expected most people to tag it as a strengthen question i.e. we are looking for a reason which indicates that the new course WILL increase interest in the academy but that is not the case. Many people get waylaid by the word ‘suspect’ and incorrectly tag it as a weaken question. Yes, suspect could mean ‘doubtful’ but just because the question stem has it, it doesn’t mean you have a weaken question at hand. Similar to the situation where the word ‘support’ in the question stem doesn’t necessarily imply that you have been given a strengthen question to deal with.

Let’s discuss various meanings of the word ‘suspect’ and how they are used (using merriam-webster.com):

Suspect can be used as a verb, noun or adjective. In our question stem, it is used as a verb and that’s what we will focus most on but let’s take up the other two briefly first.

SUSPECT (- NOUN) – one that is suspected; especially a person suspected of a crime

  1. One suspect has been arrested.
  2. She is a possible suspect in connection with the kidnapping.

SUSPECT (- ADJECTIVE) regarded or deserving to be regarded with suspicion

doubtful, questionable

  1. The room had a suspect odor.
  2. Since she was carrying no cash or credit cards, her claim to the store’s detectives that she had intended to pay for the items was suspect.

SUSPECT ( – VERB) – The verb suspect can be used in 3 different ways:

  1. to imagine (one) to be guilty or culpable on slight evidence or without proof

For Example: He’s suspected in four burglaries.

2.  to have doubts of: distrust

For Example: The fire chief suspects arson. I suspect his intentions.

3. to imagine to exist or be true, likely, or probable

For Example: I suspect it will rain.

Given a construction “I suspect A will happen”, which meaning will it have? In this case, it has the meaning of ‘imagine to be true’ or ‘think to be true’. There is absolutely no ambiguity here. When I ask “Which option supplies the best reason to suspect that the new course will increase interest in the academy?” you are definitely looking for the option that indicates that the new course WILL increase interest. Let me give you the whole question now. I am sure you will be able to solve it easily.

Question: The metropolitan cooking academy surveyed prospective students and found that students wanted a curriculum that focused on today’s healthy dining trends. In order to reverse the trend of declining interest in the school’s programs, administrators propose a series of new courses focused on cooking exotic species of fish, alternative grains such as quinoa, and organically produced vegetables.

Which of the following, if true, supplies the best reason to suspect that the proposed new course will increase interest in the metropolitan cooking academy?

(A) Cooking fish, grains, and vegetables relies on same culinary fundamentals as does the preparation of other ingredients.
(B) In the food and beverage industry, many employers no longer have time to train apprentices and therefore demand basic culinary skills from their new hires.
(C) Local producers in the area near the Metropolitan cooking academy are excellent sources of exotic fish and organic vegetables.
(D) Many other cooking schools have found a decline in the level of interest in their program.
(E) Many advocates of healthy dining stress the importance of including fish, grains and organically produced vegetables in one’s diet.

Solution: Let’s break down the argument:

Premises:

A survey found that students wanted a curriculum that focused on today’s healthy dining trends.

Administrators propose a series of new courses focused on cooking fish, alternative grains and organically produced vegetables

What will indicate that the new course will increase interest in the academy? If fish, grains and organic vegetables are considered ‘today’s healthy dining trends’, then probably the course will become popular. That is what option (E) says. Hence the answer is (E).

I suspect that the word suspect will not waylay any reader of this article anymore.

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!

An Official Question on Absolute Values

Quarter Wit, Quarter WisdomNow that we have discussed some important absolute value properties, let’s look at how they can help us in solving official questions.

Knowing these basic properties can help us quickly analyze the question and arrive at the answer without getting stuck in analyzing different ranges, a cumbersome procedure.

First we will look at a GMAT Prep question.

 

Question 1: Is |m – n| > |m| – |n|?

Statement 1: n < m
Statement 2: mn < 0

Solution 1:

Recall the property number 2

Property  2: For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y| when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

So if m and n have the same sign with |m| >= |n|, equality will hold.

Also, if n is 0, equality will hold.

If we can prove that both these conditions are not met, then we can say that |m – n| is definitely greater than |m| – |n|.

Statement 1: n < m

We have no idea about the signs of m and n. Are they same? Are they opposite? We don’t know. Also n may or may not be 0. Hence we don’t know whether the equality will hold or the inequality. Statement 1 alone is not sufficient to answer the question.

Statement 2: mn < 0

Since mn is negative, it means one of m and n is positive and the other is negative. This also implies that n is definitely not 0. So we know that m and n do not have the same sign and n is not 0. So under no condition will the equality hold.  Hence |m – n| is definitely greater than |m| – |n|. Statement 2 alone is sufficient to answer the question.

Answer (B)

Let’s look at one more question now.

Question 2: If xyz ? 0, is x(y + z) >= 0?

Statement 1: |y + z| = |y| + |z|
Statement 2: |x + y| = |x| + |y|

Solution 2:  xyz ? 0 implies that all, x, y and z, are non zero numbers.

Question: Is x(y + z) >= 0?

If we can prove that x(y + z) is not negative that is x and (y+z) do not have opposite signs, we can say that x(y + z) is positive or 0.

Looking at the statements given, let’s review our property number 1:

Property 1: For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

The two statements give us equalities which means that the relevant part of the property is this:
|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

We are also given in the question stem that x, y and z are not 0. Hence, given |x + y| = |x| + |y|, we can infer that x and y have the same sign.

Statement 1: |y + z| = |y| + |z|

This implies that y and z have the same signs. But we have no information about the sign of x hence this statement alone is not sufficient.

Statement 2: |x + y| = |x| + |y|

This implies that x and y have the same signs. But we have no information about the sign of z hence this statement alone is not sufficient.

Using both statements together, we know that x, y and z have the same sign. Whatever is the sign of y and z, the same will be the sign of (y+z). Hence x and (y+z) have the same sign. This implies that x(y + z) cannot be negative.

Hence we can answer our question with a definite ‘yes’.

Answer (C).

Mind you, both these questions can get time consuming (even though they aren’t really tough) if you don’t understand these properties well. You can certainly start your thinking from the scratch, arrive at the properties and then proceed or resort to more desperate measures such as number plugging but that is best avoided in DS questions.

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!

Properties of Absolute Values on the GMAT – Part II

Quarter Wit, Quarter WisdomWe pick up this post from where we left the post of last week in which we looked at a few properties of absolute values in two variables. There is one more property that we would like to talk about today. Thereafter, we will look at a question based on some of these properties.

(III)     |x – y| = 0 implies x = y

x and y could be positive/negative integer/fraction; if the absolute value of their difference is 0, it means x = y. They cannot have opposite signs while having the same absolute value. They must be equal. This also means that if and only if x = y, the absolute value of their difference will be 0.

Mind you, this is different from ‘difference of their absolute values’

|x| – |y| = 0 implies that the absolute value of x is equal to the absolute value of y. So x and y could be equal or they could have opposite signs while having the same absolute value.

Let’s now take up the question we were talking about.

Question: Is |x + y| < |x| + |y|?

Statement 1: | x | is not equal to | y |
Statement 2: | x – y | > | x + y |

Solution: One of the properties we discussed last week was

“For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs”

We discussed in detail the reason absolute values behave this way.

So our question “Is |x + y| < |x| + |y|?” now becomes:

Question: Do x and y have opposite signs?

We do not care which one is greater – the one with the positive sign or the one with the negative sign. All we want to know is whether they have opposite signs (opposite sign also implies that neither one of x and y can be 0)? If we can answer this question definitively with a ‘Yes’ or a ‘No’, the statement will be sufficient to answer the question. Let’s go on to the statements now.

Statement 1: | x | is not equal to | y |
This statement tells us that absolute value of x is not equal to absolute value of y. It doesn’t tell us anything about the signs of x and y and whether they are same or opposite. So this statement alone is not sufficient.

Statement 2:| x – y | > | x + y |
Let’s think along the same lines as last week – when will | x – y | be greater than | x + y |? When will the absolute value of subtraction of two numbers be greater than the absolute value of their addition? This will happen only when x and y have opposite signs. In that case, while subtracting, we would actually be adding the absolute values of the two and while adding, we would actually be subtracting the absolute values of the two. That is when the absolute value of the subtraction will be more than the absolute value of the addition.

For Example: x = 3, y = -2

| x – y | = |3 – (-2)| = 5

| x + y | = |3 – 2| = 1

or

x = -3, y = 2

| x – y | = |-3 – 2| = 5

| x + y | = |-3 + 2| = 1

If instead, x and y have the same sign, | x + y | will be greater than| x – y |.

If at least one of x and y is 0, | x + y | will be equal to| x – y |.

Since this statement tells us that | x – y | > | x + y |, it implies that x and y have opposite signs. So this statement alone is sufficient to answer the question with a ‘Yes’.

Answer (B)

Takeaway from this question:

If x and y have the same signs, | x + y | >| x – y |.

If x and y have opposite signs, | x + y | <| x – y |.

If at least one of x and y is 0, | x + y | =| x – y |.

You don’t need to ‘learn this up’. Understand the logic here. You can easily recreate it in the exam if need be.

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!

 

Properties of Absolute Values on the GMAT

Quarter Wit, Quarter WisdomWe have talked about quite a few concepts involving absolute value of x in our previous posts. But some absolute value questions involve two variables. Then do we need to consider the positive and negative values of both x and y? Certainly! But there are some properties of absolute value that could come in handy in such questions. Let’s take a look at them:

(I)    For all real x and y, |x + y| <= |x| + |y|

(II)   For all real x and y, |x – y| >= |x| – |y|

We don’t need to learn them of course and there is no need to look at how to prove them either. All we need to do is understand them – why do they hold, when is the equality sign applicable and when can they be useful. Let’s look at both the properties one by one.

(I)     For all real x and y, |x + y| <= |x| + |y|

The result of both the left hand side and the right hand side will be positive or zero. On the right hand side, the absolute values of x and y will always get added irrespective of the signs of x and y. On the left hand side, the absolute values of x and y might get added or subtracted depending on whether they have the same sign or different signs. Hence the result of the left hand side might be smaller than or equal to that of the right hand side.

For which values of x and y will the equality hold and for which values will the inequality hold? Let’s think logically about it.

The absolute values of x and y get added on the right hand side. We want the absolute values of x and y to get added on the left hand side too for the equality to hold. This will happen when x and y have the same sign. So the equality should hold when they have the same signs.

For example, x = 4, y = 8:

|4 + 8| = |4| + |8| = 12

OR x = -3, y = -4:

|-3 -4| = |-3| + |-4| = 7

Also, when at least one of x and y is 0, the equality will hold.

For example, x = 0, y = 8:

|0 + 8| = |0| + |8| = 8

OR x = -3, y = 0:

|-3 + 0| = |-3| + |0| = 3

What happens when x and y have opposite signs? On the left hand side, the absolute values of x and y get subtracted hence the left hand side will be smaller than the right hand side (where they still get added). That is when the inequality holds i.e. |x + y| < |x| + |y|

For example, x = -4, y = 8:

|-4 + 8| < |-4| + |8|

4 < 12

OR x = 3, y = -4:

|3 -4| < |3| + |-4|

1 < 7

Let’s look at our second property now:

(II) For all real x and y, |x – y| >= |x| – |y|

Thinking on similar lines as above, we see that the right hand side of the inequality will always lead to subtraction of the absolute values of x and y whereas the left hand side could lead to addition or subtraction depending on the signs of x and y. The left hand side will always be positive whereas the right hand side could be negative too. So in any case, the left hand side will be either greater than or equal to the right hand side.

When will the equality hold?

When x and y have the same sign and x has greater (or equal) absolute value than y, both sides will yield a positive result which will be the difference between their absolute values

For example, x = 9, y = 2;

|9 – 2| = |9| – |2| = 7

OR x = -7, y = -3

|-7 – (-3)| = |-7| – |-3| = 4

Also when y is 0, the equality will hold.

For example, x = 8, y = 0:

|8 – 0| = |8| – |0| = 8

OR x = -3, y = 0:

|-3 – 0| = |-3| – |0| = 3

What happens when x and y have the same sign but absolute value of y is greater than that of x?

It is easy to see that in that case both sides have the same absolute value but the right hand side becomes negative.

For example, x = -4, y = -9

|x – y| = |-4 – (-9)| = 5

|x| – |y| = |-4| – |-9| = -5

So even though the absolute values will be the same since we will get the difference of the absolute values of x and y on both sides, the right hand side will be negative. If we were to take further absolute value of the right hand side, the two will become equal i.e. the right hand side will become |(|x| – |y|)| = |-5| = 5 in our example above. In that case, the equality will hold again.

Similarly, what happens when only x = 0? The right hand side becomes negative again so taking further absolute value will make both sides equal.

For example, x = 0, y = -5

|x – y| = |0 – (-5)| = 5

|x| – |y| = |0| – |5| = -5

Taking further absolute value, |(|x| – |y|)| = |-5| = 5

So when we take further absolute value of the right hand side, this property becomes similar to property 1 above: |x – y| = |(|x| – |y|)| when x and y have the same sign or at least one of x and y is 0.

Now let’s look at the inequality part of property 2.

Whenever x and y have opposite signs, |x – y| > |x| – |y|

On the left hand side, the absolute values will get added while on the right hand side, the absolute values will get subtracted. So the absolute value of the left hand side will always be greater than the absolute value of the right hand side. The left hand side will always be positive while the right hand side could be negative too. Hence even if we take the further absolute value of the right hand side, the inequality will hold: |x – y| > |(|x| – |y|)| when x and y have opposite signs

For example, x = -4, y = 8:

|-4 – 8| > |-4| – |8|

12 > -4

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-4| = 4

Still, 12 > 4 i.e. |x – y| > |(|x| – |y|)|

OR x = 3, y = -4:

|3 –(-4)| > |3| – |-4|

7 > -1

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-1| = 1

Still, 7 > 1 i.e. |x – y| > |(|x| – |y|)|

Note that the inequality of the original property 2 also holds when x and y have the same sign but absolute value of y is greater than the absolute value of x since the right hand side becomes negative. It also holds when x is 0 but y is not.

To sum it all neatly,

(I) For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

(II) For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y|when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

(III) For all real x and y, |x – y| >= |(|x| – |y|)|

|x – y| = |(|x| – |y|)| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x – y| > |(|x| – |y|)| when (1) x and y have opposite signs

Note that property (III) matches property (I).

There is another property we would like to discuss but let’s take it up next week along with some GMAT questions where we put these properties to use.

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!