Quarter Wit, Quarter Wisdom: Stuck in Assumptions Again

Quarter Wit, Quarter WisdomThere is a particular issue in assumption questions that I would like to discuss today. We discussed in our previous posts that assumptions are ‘necessary missing premises’. Many students get stuck between two options in assumption questions. The correct option is the necessary premise. The incorrect one is often a sufficient premise. Due to the sufficiency, they believe that that particular option is a stronger assumption. But the point to remember is that an assumption is only necessary for the conclusion to be true. It may not actually lead to the conclusion beyond a reasonable doubt. You only have to answer what has been asked (which is an assumption), not what you think is better to make the conclusion true.

Let me explain this with an example:

Question: Exports of United States cotton will rise considerably during this year. The reason for the rise is that the falling value of the dollar will make it cheaper for cloth manufacturers in Japan and Western Europe to buy American cotton than to get it from any other source.

Which of the following is an assumption made in drawing the conclusion above?

(A) Factory output of cloth products in Japan and Western Europe will increase sharply during this year.
(B) The quality of the cotton produced in the United States would be adequate for the purposes of Japanese and Western European cloth manufacturers.
(C) Cloth manufacturers in Japan and Western Europe would prefer to use cotton produced in the United States if cost were not a factor.
(D) Demand for cloth products made in Japan and Western Europe will not increase sharply during this year.
(E) Production of cotton by United States companies will not increase sharply during this year.

Solution:

First, let’s analyze the given argument:

Premises:
– Dollar is falling.
– It will be cheaper for cloth manufacturers in Japan and Western Europe to buy American cotton than to get it from any other source.

Conclusion:
– Exports of United States cotton will rise considerably during this year.

The conclusion links ‘sale of cotton’ to ‘cost of cotton’. It says that since the cost of American cotton will be lower than the cost of cotton from any other source, American cotton will sell. We are assuming here that the American cotton is adequate in all other qualities that the cloth manufacturers look for while buying cotton or that lower cost is all that matters. We are assuming that lowest cost will automatically lead to sale.

Let’s look at each of the options now:

(A)   Factory output of cloth products in Japan and Western Europe will increase sharply during this year.

Notice that we don’t NEED the factory output to increase. Even if it stays the same or in fact, even if it falls, as long as the manufacturers find American cotton suitable, the cotton exports of US could rise. Hence this is not the assumption.

(B)   The quality of the cotton produced in the United States would be adequate for the purposes of Japanese and Western European cloth manufacturers.

This option says that the quality is adequate and hence this is an assumption. Notice that it is necessary for our conclusion. If the quality is not adequate, no matter what the cost, US cotton sale may not increase. Hence, answer is (B).

(C)   Cloth manufacturers in Japan and Western Europe would prefer to use cotton produced in the United States if cost were not a factor.

This is the tricky non-correct option! Many people will swing between (B) and (C) for a while and then choose (C).  This option says that Japanese and Europeans prefer to use US cotton if cost does not matter. Do we NEED this to be true? No. It is good if it is true because it means that if cost of US cotton goes down, US cotton will sell more (hence, it is sufficient for the conclusion to be true – assuming all else stays constant). But do we NEED them to prefer US cotton? No. It is not necessary for our conclusion to be true. Even if the manufacturers don’t particularly prefer US cotton, US cotton exports could still increase if the price is the lowest.

Beware of this difference between ‘necessary’ and ‘sufficient’ conditions. Remember that assumptions are NECESSARY conditions; they don’t need to be sufficient. We end up incorrectly choosing sufficient conditions because they seem to be all encompassing and hence more attractive for our conclusion. If the sufficient condition is satisfied, then the conclusion has to be true. But mind you, that is not what the question is asking you. The question is looking for only a necessary condition, not a sufficient one. Also notice that sufficient conditions may not be necessary.

(D)   Demand for cloth made in Japan and Western Europe will not increase sharply during this year.

This is incorrect. We are not assuming that the demand for their cloth will not increase.

(E)    Production of cotton by United States companies will not increase sharply during this year.

It doesn’t matter what happens to the production of cotton in US. All we care about is that exports should rise.

The correct answer is (B). 

Hope you will be careful next time when you come across such a 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!

Quarter Wit, Quarter Wisdom: And Now, Evading Formulas!

Quarter Wit, Quarter WisdomToday, we again pay homage to the lazy bum within each one of us in our QWQW series. If you are wondering what we mean by ‘again’, check out our last two posts of the QWQW series. We have been discussing how to avoid calculations. Today let’s learn why it is advisable to avoid learning formulas too!

You really don’t need to know many formulas for GMAT – just the basic ones e.g. Distance = Speed*Time, Work = Rate*Time (which are actually the same if you look at them closely) etc. If a Time-Distance-Speed question pertains to GMAT, rest assured it can be solved using just the formula given above and that too, within 1-2 mins. Then, do you need to learn the many formulas that people claim speed up question solving? No! In fact, the more specific the formula, the more constraints it has. It can be used in only particular circumstances and hence when the situation differs even a little bit from the ideal, you could end up using the formula incorrectly. Therefore, we recommend our students to stay away from the umpteen, less generic formulas until and unless they have already used them extensively. Let’s discuss this point with an example:

Question: A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?

(A) 1 hour
(B) 1 hour 10 minutes
(C) 2 hours 30 minutes
(D) 1 hour 40 minutes
(E) 2 hours 10 minutes

Solution: People often like to use a formula for this situation. Let’s quickly discuss that first.

If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:

Sa/Sb = sqrt(b/a)

i.e. Ratio of speeds is given by the square root of the inverse ratio of time taken.

Sa/Sb = sqrt(90/40) = 3/2

This gives us that the ratio of the speed of A : speed of B as 3:2. We know that time taken is inversely proportional to speed. If ratio of speed of A and B is 3:2, the time taken to travel the same distance will be in the ratio 2:3. Therefore, since B takes 90 mins to travel from the meeting point to Opladen, A must have taken 60 (= 90*2/3) mins to travel from Opladen to the meeting point

So time taken by A to travel from Opladen to Cologne must be 60 + 40 mins = 1 hr 40 mins

Now let’s see how we can solve the question without using the formula.

Think of the point in time when they meet:

A starts from Opladen and B from Cologne simultaneously. After some time, say t mins of travel, they meet. Since A covers the entire distance of Opladen to Cologne in (t + 40) mins and B covers it in (t + 90) mins, A is certainly faster than B and hence the Meeting point is closer to Cologne.

Now think, what information do we have? We know the time taken by A and B to reach their respective destinations from the meeting point. We also know that they both traveled the same distance i.e. the distance between Opladen and Cologne. So let’s try to link distance with time taken. We know that ‘Distance’ varies directly with ‘Time taken’. (Check out this post if you don’t know what we are talking about here.)

Distance between Opladen and Meeting point /Distance between Meeting point and Cologne = Time taken to go from Opladen to Meeting point/Time taken to go from Meeting point to Cologne = t/40 (in case of A) = 90/t (in case of B)
t = 60 mins

So A takes 60 mins + 40 mins = 1 hr 40 mins to cover the entire distance.

Answer (D)

We could easily solve the question without using any specific formula. So stick to your basics and kick those little grey cells to get to 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!

Quarter Wit, Quarter Wisdom: Evading Calculations Part II

Quarter Wit, Quarter WisdomLast week we discussed how to solve equations with the variable in the denominator. We also said that the technique generally works for PS questions but you need to be careful while working on DS questions. Today, let’s look at the reason behind the caveat.

Say, the question stem of a DS question asks you to find the value of n, the number of people in the room. Statement 1 of the question gives you the following equation:

60/(n – 5) – 60/n = 2

We can easily figure out that a value of n that satisfies this equation is 15. Now, is that enough to say that statement 1 is sufficient alone? No! It could be a trap! The equation, when manipulated, gives us a quadratic. It is important to find out whether the second solution of the quadratic works for us. When n is the number of people, it must be positive. So one extra step that we should take is re-arrange the equation to get the quadratic. If the constant term i.e. the product of the roots is negative, it means one root is positive and one is negative. Since we have already found the positive root, it is the only answer and hence we can say that the statement 1 is sufficient alone.

60/(n – 5) – 60/n = 2

60*n – 60*(n – 5) = 2*n*(n – 5)

n^2 – 5n – 150 = 0

The constant term, -150, is negative so the product of the roots must be negative. This means one root must be negative and the other must be positive. Since we have already found the positive root i.e. the number of people in the room, we can say that statement 1 is sufficient alone.

Let’s look at an example where we could fall in the trap.

Say statement 1 gives us an equation which looks like this:

60/(n +5) – 10/(n – 5) = 2

As discussed last week, we will easily see that n = 10 satisfies this equation. So should we move on now and say that statement 1 is sufficient alone? No, not so fast! Let’s try to manipulate the equation to get the quadratic.

60/(n +5) – 10/(n – 5) = 2

60*(n – 5) – 10*(n + 5) = 2*(n – 5)(n + 5)

n^2 – 25n + 150 = 0

n = 10 or 15

So actually, there are two values of n that satisfy this equation. In PS questions, since we have a single answer, there would be only one solution so once you get one, you are done. In DS questions, you need to be certain that only one value satisfies. There is a possibility that both values satisfy your constraints in which case your answer would change.

Therefore, it may not be necessary to solve the equation for the PS question, but it is certainly necessary to solve it for DS. That’s counter intuitive, isn’t it? We hope you understand the reason.

Another related trap in DS questions: Statement 1 gives you a quadratic and asks you for the value of x (no constraints that x must be an integer or positive number etc). You know that it is a quadratic and it will give you two values of x so you say that statement 1 is not sufficient alone and move on. But hold it! What if both the roots of the equation are same? It may not apparent to you when you look at the equation. When you solve it, you realize that the roots are the same. Hence, ensure that you solve the equation in DS questions before you decide on the sufficiency.

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!

Quarter Wit, Quarter Wisdom: Evading Calculations!

Quarter Wit, Quarter WisdomWe have discussed before how GMAT is not a calculation intensive exam. Whenever you land on an equation which looks something like this: 60/(n – 5) – 60/n = 2, you probably think that we don’t know what we are talking about! You obviously need to cross multiply, make a quadratic and finally, solve the quadratic to get the value of n. Actually, you usually don’t need to do any of that for GMAT questions. You have an important leverage – the options. Even if the options don’t directly give you the values of n or n-5, you can use the knowledge that every GMAT question is do-able in 2 mins and that the numbers fit in beautifully well.

Let’ see whether we can get a value of n which satisfies this equation without going the whole nine yards. We will not use any options and will try to rely on our knowledge that GMAT questions don’t take much time.

60/(n – 5) – 60/n = 2

So, the difference between the two terms of the left hand side is 2. Try to look for values of n which give us simple numbers i.e. try to plug in values which are factors of the numerator.
Say, if n = 10, you get 60/5 – 60/10 = 12 – 6 = 6. The difference between them is much more than 2. 60/n and 60/(n – 5) need to be much closer to each other so that the difference between them is 2. The two terms should be smaller to bring them closer together. So increase the value of n.

Put n = 15 since it is the next number such that (15 – 5 =) 10 as well as 15 divide 60 completely. You get 60/10 – 60/15 = 6 – 4 = 2. It satisfies and you know that a value that n can take is 15. Usually, you will get a solution within 2-3 iterations. This is enough for a PS question. Notice that this equation gives us a quadratic so be careful while working on DS questions. You might need to manipulate the equation a little to figure out whether the other root is a possible solution as well. Anyway, today we will focus on the application of such equations in PS questions only. Let’s take a question now to understand the concept properly:

Question: Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Solution: We need to find the time taken by machine A to make 40 widgets. It will be best to take the time taken by machine A to make 40 widgets as the variable x. Then, when we get the value of x, we will not need to perform any other calculations on it and hence the scope of making an error will reduce. Also, value of x will be one of the options and hence plugging in to check will be easy.

Machine A takes x hrs to make 40 widgets.

Rate of work done by machine A = Work done/Time taken = 40/x

Machine B take 2 hrs less than machine A to make 20 widgets hence it will take 4 hrs less than machine B to make 40 widgets. Think of it this way: Break down the 40 widgets job into two 20 widget jobs. For each job, machine B will take 2 hrs less than machine A so it will take 4 hrs less than machine A for both the jobs together.

Time taken by machine B to make 40 widgets = x – 4

Rate of work done by machine B = Work done/Time taken = 40/(x – 4).

We know the combined rate of the machines is 25/3

So here is the equation:

40/x + 40/(x – 4) = 25/3

The steps till here are not complicated. Getting the value of x poses a bit of a problem.

Notice here that that the right hand side is not an integer. This will make the question a little harder for us, right? Wrong! Everything has its pros and cons. The 3 of the denominator gives us ideas for the values of x (as do the options).  To get a 3 in the denominator, we need a 3 in the denominator on the left hand side too.

x cannot be 3 but it can be 6. If x = 6, 40/(6 – 4) = 20 i.e. the sum will certainly not be 20 or more since we have 25/3 = 8.33 on the right hand side.

The only other option that makes sense is x = 12 since it has 3 in it.

40/12 + 40/(12 – 4) = 10/3 + 5 = 25/3

Answer (E)

If we did not have the options, we might have tried x = 9 too before landing on x = 12. Nevertheless, these calculations are not time consuming at all since you can get rid of the incorrect numbers orally. Making a quadratic and solving it is certainly much more time consuming.

Another method could be to bring 3 to the left hand side to get the following equation:

120/x + 120/(x – 4) = 25

This step doesn’t change anything but it helps if you face a mental block while working with fractions. Try to practice such questions using these techniques – they will save you a lot 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!

Quarter Wit, Quarter Wisdom: An Official Assumption Question

Quarter Wit, Quarter WisdomToday we will look at an OG question of critical reasoning (as promised last week). We will use the concept discussed last week – remember what an assumption is. An assumption is a missing necessary premise. It will bring in new information essential to the conclusion.

Now let’s jump on to the OG question.

Question: A recent report determined that although only three percent of drivers on Maryland highways equipped their vehicles with radar detectors, thirty-three percent of all vehicles ticketed for exceeding the speed limit were equipped with them. Clearly, drivers who equip their vehicles with radar detectors are more likely to exceed the speed limit regularly than are drivers who do not.

The conclusion drawn above depends on which of the following assumptions?

(A) Drivers who equip their vehicles with radar detectors are less likely to be ticketed for exceeding the speed limit than are drivers who do not.

(B) Drivers who are ticketed for exceeding the speed limit are more likely to exceed the speed limit regularly than are drivers who are not ticketed.

(C) The number of vehicles that were ticketed for exceeding the speed limit was greater than the number of vehicles that were equipped with radar detectors.

(D) Many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.

(E) Drivers on Maryland highways exceeded the speed limit more often than did drivers on other state highways not covered in the report.

Solution:

Let’s look at the question stem first. We need to find an assumption. An assumption is a missing necessary premise. Something that will not only strengthen the conclusion but also be essential to the argument.
An assumption is a statement that needs to be added to the premises for the conclusion to be true. Let’s first find the premises and the conclusion of this argument.

Premises:

– Only 3% of drivers on Maryland highways had radar detectors.

– 33% of vehicles that got speeding tickets had radar detectors.

Conclusion:

Drivers with radar detectors are more likely to exceed the speed limit regularly than other drivers.

There must be some disconnect between the premises and conclusion since there is an assumption in the argument. Look carefully. Premises give you the connection between ‘vehicles that have radar detectors’ and ‘vehicles that get speeding tickets’. The conclusion, on the other hand, concludes a relation between ‘vehicles that have radar detectors’ and ‘vehicles that exceed the speed limit’. The assumption must then give a connection between ‘vehicles that get speeding tickets’ and ‘vehicles that exceed speed limit’.

To clarify it further,

A – vehicles that have radar detectors

B – vehicles that get speeding tickets/vehicles that were ticketed for speeding

C – vehicles that exceed the speed limit

Premises:

– Only 3% of all vehicles are A

– 33% of B are A

Conclusion:

– A are  more likely to be C

The assumption needs to be something that links B to C i.e. that links ‘vehicles that get speeding tickets’

to ‘vehicles that exceed the speed limit’. Option (B) gives us that relation. It says ‘B are more likely to be C’.

Lets add it to premises and see if the conclusion makes more sense now:

– Only 3% of drivers on Maryland highways had radar detectors.

– 33% of vehicles that got speeding tickets had radar detectors.

– Drivers who get speeding tickets are more likely to exceed the speed limit regularly than others.

Conclusion: Drivers with radar detectors are more likely to exceed the speed limit regularly than other drivers.

Now it makes sense!

Let’s take a quick look at the other options and see why they don’t work. We will retain the A, B, C structure given above.

Option (A) says ‘A are less likely to be B’ – Cannot be our assumption

Option (C) only tells us that number of B are greater than number of A.

Option (D) tells us that many vehicles were ticketed multiple times.

Option (E) compares drivers on Maryland highways with drivers on other state highways. This is out of scope.

It is clear that option (B) is the outright winner. This question is one of the tougher questions. You can easily handle it using this technique. We hope you will be able to put this technique to good 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!

Quarter Wit, Quarter Wisdom: Making Sense of Assumptions

Quarter Wit, Quarter WisdomToday we would like to discuss a technique which is very useful in solving assumption questions. No, I am not talking about the ‘Assumption Negation Technique’ (ANT), which, by the way, is extremely useful, no doubt. The point is that ANT is explained beautifully and in detail in your book so there is no point of re-doing it here. You already know how to use it.

What we are going to discuss today is not so much a technique as a revelation of how you can identify the assumption in a particular argument by just fully understanding this – ‘what is an assumption?’. Actually, this is also already discussed in your CR book but I would like to draw your attention to it. We usually end up ignoring the finer points here and hence get stuck on something that is supposed to be quite obvious.

Ok so, what is an assumption?

An assumption is a missing necessary premise. (Doesn’t seem like much a revelation, right? You already knew that! Right! Focus on every word now.)

An assumption is a premise – it gives you some new fact/information.

It is also necessary – necessary for the conclusion to be true. The conclusion cannot be true if the assumption doesn’t hold. Our ANT is based on this premise.

To add, it is also missing – it is not something already mentioned in the argument.

Let’s take a very simplistic example to understand the implication of a missing necessary premise.

Argument: A implies B. B implies C. Hence, A implies D.

Premises given in the argument:
– A implies B
– B implies C

Conclusion given in the argument:
– A implies D

Is it apparent that something is missing here? Sure! The premises give us the relations between A, B and C. They do not mention D. But while drawing the conclusion, we are concluding about the relation between A and D. We can’t do that. We must know something about D too to be able to conclude a relation between A and D. Hence, there is a necessary premise that is missing here. What we are looking for is something that says ‘C implies D’.

When we add this to our premises, our argument makes sense.

Argument: A implies B. B implies C. C implies D. Hence, A implies D.

This little point will help us in solving the trickiest of questions. We get so lost in the n number of things mentioned in the argument that we forget to consider this aspect.

We will discuss an LSAT question today because it seems to be created just to exemplify this concept! Many people falter on this question. After going through it with us here, you will wonder why.

Question:

Therapist: The ability to trust other people is essential to happiness, for without trust there can be no meaningful emotional connection to another human being, and without meaningful emotional connections to others we feel isolated.
Which one of the following, if assumed, allows the conclusion of the therapist’s argument to be properly inferred?
(A) No one who is feeling isolated can feel happy.
(B) Anyone who has a meaningful emotional connection to another human being can be happy.
(C) To avoid feeling isolated, it is essential to trust other people.
(D) At least some people who do not feel isolated are happy.
(E) Anyone who is able to trust other people has a meaningful emotional connection to at least one other human being.

Solution:

First, we break down the argument into premises and conclusion.

Premises:

– Without trust there can be no meaningful emotional connection.

– Without meaningful emotional connections, we feel isolated.

Conclusion:

Ability to trust is essential to happiness.

Do you see something missing here? We are concluding about trust and happiness but in the premises, the link between ‘feeling isolated’ and ‘happiness’ is missing. The premises do not talk about happiness at all. So we need a premise which says, ‘feeling isolated’ means ‘not happy’ for the conclusion to make sense.
Look at the premises now:

Premises:

– Without trust there can be no meaningful emotional connection.

– Without meaningful emotional connections, we feel isolated.

– When we feel isolated, we cannot be happy. (The assumption)

Conclusion:

Ability to trust is essential to happiness.

Now it all makes sense, doesn’t it?

Look at the options now.

Option (A) says – ‘No one who is feeling isolated can feel happy.’ – exactly what we needed.

Hence (A) will be your answer.

Hope this makes sense to you. Next week, we will see how you can easily solve OG questions using this concept.

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

Quarter Wit, Quarter Wisdom: The Efficiency of Using Variation

Quarter Wit, Quarter WisdomToday, we would like to discuss one of our own work questions. The intent is to show you how simple your calculations can get when you use the methods we discussed in the last few weeks. I couldn’t say it enough – develop a love for ratios. You will save a huge amount of time in lots of questions. If you haven’t been following the last few weeks’ posts, take a look at this link before checking out the question. Otherwise the method may not make sense to you.

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

(1) Mules work more slowly than horses.
(2) 48 mules can haul the same load of lumber in 16 minutes.

Solution: First do this question on your own and see the calculations involved. Thereafter, check out the solution given below to know how we can solve the question using our joint variation method.

We are given that 16 horses can complete the work in 24 mins. Let’s find out how much work is done by 12 horses in 14 mins (before the mules join in)

16 horses ……… 24 mins ………. 1 work
12 horses ……… 14 mins ………. ?? work

Work done = 1*(14/24)*(12/16) = 7/16 work (if you don’t know how we arrived at this, seriously, check out last week’s post first)

So in 14 mins, the 12 horses can complete 7/16 of the work i.e. they do 1/16 of the work every 2 mins.

How much work is leftover for the mules and horses to do together? 1 – 7/16 = 9/16

Leftover work = 9/16

This makes us think that 12 horses alone will take 9*2 = 18 mins to finish the work. When 12 mules join in, depending on the rate of work of mules, time taken to complete this work could be less than or more than 15 mins.

Statement 1: Mules work more slowly than horses.

This statement doesn’t give us enough information. It just tells us that mules work slower than horses. Say if they work very slowly so that, effectively, they are not adding much to the work done, the work will get done in approximately 18 mins. If they work faster, time taken will keep decreasing. If they work as fast as the horses, the rate at which the work will be done will double (because we already have 12 horses and we will add 12 mules which will be equivalent to 12 horses) and time taken will become half i.e. it will be 9 mins. So the time taken will vary in the range 9 mins to 18 mins depending on the rate of work of mules. This statement alone is not sufficient.

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

We now know the rate of work of mules. The point is that now we can easily calculate the exact time taken by 12 horses and 12 mules to complete 9/16 of the work. Once we calculate the exact time, we will be able to say whether the time taken will be less than or more than 15 mins. Hence this statement alone is sufficient to answer the question. We don’t really need to find out exactly how much is taken by the 24 animals together since it is a DS question. Ideally, we should mark the answer as (B) and move on.

Nevertheless, let’s do the calculations if only to practice application of work concepts.

Let’s try to find the equivalence of mules and horses (the way we did with cars in the previous post)

We know that 16 horses can haul a load of lumber in 24 minutes. Let’s find out the number of mules that are needed to complete the work in 24 mins.

48 mules …….16 mins.

?? mules ……..24 mins

No. of mules required = 48*16/24 = 32 mules

So, 32 mules do the same work in the same time as done by 16 horses. Or we can say that 2 mules are equivalent to 1 horse. Hence, 12 mules are equivalent to 6 horses. When 12 mules join the 12 horses, equivalently we get 12+6 = 18 horses.

16 horses ……… 24 mins ………. 1 work

18 horses ……….. ?? mins ……….. 9/16 work

Time taken by 18 horses (i.e. 12 horses and 12 mules) = 24*(9/16)*(16/18) = 12 mins

Yes, the horses and mules together will take less than a quarter-hour to finish hauling the load.

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!

Quarter Wit, Quarter Wisdom: Work-Rate Using Joint Variation

Quarter Wit, Quarter WisdomThis week, let’s look at some work-rate questions which use joint variation. Check out the last three posts of QWQW series if you are not comfortable with joint variation.

Question 1: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?

(A) 60

(B) 70

(C) 75

(D) 80

(E) 100

Solution: Can we say that 10 people can finish the work in 100 days? No. If that were the case, after 20 days, only 1/5th of the work would have been over. But actually 1/4th of the work is over. This means that ‘10 people can complete the work in 100 days’ was just the contractor’s estimate (which turned out to be incorrect). Actually 10 people can do 1/4th of the work in 20 days. The contractor fires 2 people. So the question is how many days are needed to complete 3/4th of the work if 8 people are working?

We need to find the number of days. How is ‘no. of days’ related to ‘no. of people’ and ‘work done’?

If we have more ‘no. of days’ available, we need fewer people. So ‘no. of days’ varies inversely with ‘no. of people’.

If we have more ‘no. of days’ available, ‘work done’ will be more too. So ‘no. of days’ varies directly with ‘work done’.

Therefore,

‘no. of days’ * ‘no. of people’/’work done’ = constant

20*10/(1/4) = ‘no. of days’*8/(3/4)

No. of days = 75

So, the work will get done in 75 days if 8 people are working.

We can also do this question using simpler logic. The concept used is joint variation only. Just the thought process is simpler.

10 people can do 1/4th of the work in 20 days.

8 people can do 3/4th of the work in x days.

Start with the no. of days since you want to find the no of days:

x = 20*(10/8)*(3/1) = 75

From where do we get 10/8? No. of people decreases from 10 to 8. If no. of people is lower, the no of days taken to do the work will be more. So 20 (the initial no. of days) is multiplied by 10/8, a number greater than 1, to increase the number of days.

From where do we get (3/1)? Amount of work increases from 1/4 to 3/4. If more work has to be done, no. of days required will be more. So we further multiply by (3/4)/(1/4) i.e. 3/1, a number greater than 1 to further increase the number of days.

This gives us the expression 20*(10/8)*(3/1)

We get that the work will be complete in another 75 days.

Answer (C)

Let’s take another question to ensure we understand the logic.

Question 2: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?

(A) 1200 lt

(B) 1555 lt

(C) 1664 lt

(D) 1728 lt

(E) 4800 lt

Solution: First let’s try to figure out what is meant by ‘consume 20% less fuel than the company’s cars’. It means that if company’s each car consumes 1 lt per hour, the hired cars consume only 4/5 lt per hour. So renting 5 more cars is equivalent to renting 4 cars which are same as the company’s cars. Hence, the total number of cars that will be run for the next 6 days is 8 company-equivalent cars.

4 cars running 10 hrs for 10 days consume 1200 lt of fuel

8 cars running 12 hrs for 6 days consume x lt of fuel

x = 1200*(8/4)*(12/10)*(6/10) = 1728 lt

We multiply by 8/4 because more cars implies more fuel so we multiply by a number greater than 1.

We multiply by 12/10 because more hours implies more fuel so we multiply by a number greater than 1.

We multiply by 6/10 because fewer days implies less fuel so we multiply by a number smaller than 1.

Answer (D)

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

Quarter Wit, Quarter Wisdom: Varying Jointly

Quarter Wit, Quarter WisdomNow that we have discussed direct and inverse variation, joint variation will be quite intuitive. We use joint variation when a variable varies with (is proportional to) two or more variables.

Say, x varies directly with y and inversely with z. If y doubles and z becomes half, what happens to x?

“x varies directly with y” implies x/y = k (keeping z constant)

If y doubles, x doubles too.

“x varies inversely with z” implies xz = k (now keeping y constant)

If z becomes half, x doubles.

So the overall effect is that x becomes four times of its initial value.

The joint variation expression in this case will be xz/y = k. Notice that when z is constant, x/y = k and when y is constant, xz = k; hence both conditions are being met. Once you get the expression, it’s very simple to solve for any given conditions.

x1*z1/y1 = x2*z2/y2 = k (In any two instances, xz/y must remain the same)

x1*z1/y1 = x2*(1/2)z1/2*y1

x2 = 4*x1

Let’s look at some more examples. How will you write the joint variation expression in the following cases?

1. x varies directly with y and directly with z.

2. x varies directly with y and y varies inversely with z.

3. x varies inversely with y^2 and inversely with z^3.

4. x varies directly with y^2 and y varies directly with z.

5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.

Solution: Note that the expression has to satisfy all the conditions.

1. x varies directly with y and directly with z.

x/y = k

x/z = k

Joint variation: x/yz = k

2. x varies directly with y and y varies inversely with z.

x/y = k

yz = k

Joint variation: x/yz = k

3. x varies inversely with y^2 and inversely with z^3.

x*y^2 = k

x*z^3 = k

Joint variation: x*y^2*z^3 = k

4. x varies directly with y^2 and y varies directly with z.

x/y^2 = k

y/z = k which implies that y^2/z^2 = k

Joint variation: x*z^2/y^2 = k

5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.

x/y^2 = k

yz = k which implies y^2*z^2 = k

z/p^3 = k which implies z^2/p^6 = k

Joint variation: (x*p^6)/(y^2*z^2) = k

Let’s take a GMAT prep question now to see these concepts in action:

Question 1: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?

(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase

Solution:

Rate/M^2 = k

Rate*N = k

Rate*N/M^2 = k

If Rate has to remain constant, N/M^2 must remain the same too.

If N is doubled, M^2 must be doubled too i.e. M must become ?2 times. Since ?2 = 1.4 (approximately),

M must increase by 40%.

Answer (D)

Simple enough?

Quarter Wit, Quarter Wisdom: Varying Inversely

Quarter Wit, Quarter WisdomAs promised, we will discuss inverse variation today. The concept of inverse variation is very simple – two quantities x and y vary inversely if increasing one decreases the other proportionally.

If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1*y1 = x2*y2 = x3*y3 = some constant value

This means that if x doubles, y becomes half; if x becomes 1/3, y becomes 3 times etc. In other words, product of x and y stays the same in different instances.

Notice that x1/x2 = y2/y1; The ratio of x is inverse of the ratio of y.

The concept will become clearer after working on a few examples.

Question 1: The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with 0.02% impurities is $2500. What is the cost of a diamond with 0.05% impurities (keeping everything else constant)?

(A)   $400

(B)   $500

(C)   $1000

(D)   $4000

(E)    $8000

Solution:

Price1*(% of Impurities1)^2 = Price2*(% of Impurities2)^2

2500*(.02)^2 = Price2*(.05)^2

Price = $400

Answer (A)

The answer is quite intuitive in the sense that if % of impurities in the diamond increases, the price of the diamond decreases.

There is an important question type related to inverse variation. It often uses the formula:

Total Price = Number of units*Price per unit

If, due to budgetary constraints, we need to keep the total money spent on a commodity constant, number of units consumed varies inversely with price per unit. If price per unit increases, we need to reduce the consumption proportionally.

Question 2: The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A)   5%

(B)   9%

(C)   10%

(D)   11%

(E)    20%

Solution: Do you think the answer is 10%? Think again.

Total Cost = Number of units*Price per unit

If the price per unit increases by 10%, it becomes 11/10 of its original value. To keep the total cost same, you need to multiply number of units by 10/11. i.e. you need to decrease the number of units by 1/11 i.e. 9.09%. In that case,

New Total Cost = (10/11)*Number of units*(11/10)*Price per unit

This new total cost will be the same as the previous total cost.

Answer (B)

Let’s look at one more example of the same concept but this one is a little trickier.

Question 3: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy a new car with better mileage as compared to his current car. By what % should the new mileage be more than the previous mileage to ensure that Mr X’s total fuel cost stays the same for the month? (assuming the distance traveled every month stays the same)

(A)   10%

(B)   17%

(C)   20%

(D)   21%

(E)    25%

Solution:  The problem here is ‘how is mileage related to fuel price?’

Total fuel cost = Fuel price * Quantity of fuel used

Since the ‘total fuel cost’ needs to stay the same, ‘fuel price’ varies inversely with ‘quantity of fuel used’.

Quantity of fuel used = Distance traveled/Mileage

Distance traveled = Quantity of fuel used*Mileage

Since the same distance needs to be traveled, ‘quantity of fuel used’ varies inversely with the ‘mileage’.

We see that ‘fuel price’ varies inversely with ‘quantity of fuel used’ and ‘quantity of fuel used’ varies inversely with ‘mileage’. So, if fuel price increases, quantity of fuel used decreases proportionally and if quantity of fuel used decreases, mileage increases proportionally. Hence, if fuel price increases, mileage increases proportionally or we can say that fuel price varies directly with mileage.

If fuel price becomes 6/5 (20% increase) of previous fuel price, we need the mileage to become 6/5 of the previous mileage too i.e. mileage should increase by 20% too.

Another method is that you can directly plug in the expression for ‘Quantity of fuel used’ in the original equation.

Total fuel cost = Fuel price * Distance traveled/Mileage

Since ‘total fuel cost’ and ‘distance traveled’ need to stay the same, ‘fuel price’ is directly proportional to ‘mileage’.

Answer (C)

We hope you are comfortable with fundamentals of direct and inverse variation now. More next week!

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

 

Quarter Wit, Quarter Wisdom: Varying Directly

Quarter Wit, Quarter WisdomWe can keep working on ‘pattern recognition’ questions for a long time and not run out of questions of different types on which it can be used. We hope you have understood the basic concepts involved. So let’s move on to another topic now: Variation.

Basically, variation describes the relation between two or more quantities. e.g. workers and work done, children and noise, entrepreneurs and start ups. More workers means more work done; more children means more noise; more entrepreneurs means more start ups and so on… These are examples of direct variation i.e. if one quantity increases, the other increases proportionally. Then there are quantities that have inverse variation between them e.g. workers and time taken. If there are more workers, time taken to complete a work will be less.

Let’s discuss direct variation today.

Formally, let’s say x varies directly with y. If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1/y1 = x2/y2 = x3/y3 = Some constant value

In other words, ratio of x and y stays the same in different instances.

(Notice that this is the same as x1/x2 = y1/y2)

It might seem a little cumbersome when put this way but the truth is that direct variation is quite intuitive. A couple of questions will make it clear.

Question 1: 20 workmen can make 35 widgets in 5 days. How many workmen are needed to make 105 widgets in 5 days?

(A)   7

(B)   20

(C)   25

(D)   30

(E)    60

Solution: Notice that the number of days stays the same so we can ignore it. Now think, how are workmen and widgets related? If the number of workmen increases, the number of widget made also increases proportionally. You need to find the new number of workmen required. The number of widgets has become thrice (105/35 = 3) so number of workmen needed will become thrice as well (remember, the number of workmen will increase in the same proportion).

We need 20*3 = 60 workmen

Answer (E)

The concept of variation is very intuitive. If the number of widgets required doubles, the number of workmen required to make them in the same amount of time will double too. If the number of widgets required becomes one fourth, the number of workmen required to make them in the same amount of time will become one fourth too.

A quantity can directly vary with some power of another quantity. Let’s take an example of this scenario too.

Question 2: If the ratio of the volumes of two right circular cylinders is given by 64/9, what is the ratio of their radii? (Both the cylinders have the same height)

(A)   4/3

(B)   8/3

(C)   16/9

(D)   4/1

(E)    16/3

Solution: This question involves a little bit of geometry too. The volume of a right circular cylinder is given by Area of base * height i.e.

Volume of a right circular cylinder = pi*radius^2 * height

So volume varies directly with the square of radius.

Va/Vb = 64/9 = Ra^2/Rb^2

Ra/Rb = 8/3

Answer (B)

We hope this little concept is not hard to understand. We will work on inverse proportion next week and then work on problems involving both (that’s where the good questions are!).

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

Quarter Wit, Quarter Wisdom: Pattern Recognition or Number Properties?

Quarter Wit, Quarter WisdomContinuing our quest to master ‘pattern recognition’, let’s discuss a tricky little question today. It is best done using divisibility and remainders logic we discussed in some previous posts. We suggest you check out these divisibility posts if you haven’t yet.

We are first going to see how to solve the question conceptually. The interesting thing is – what do you do if you are under immense pressure during the test and are unable to remember anything you ever read on divisibility? It is a fairly common phenomenon – students have reported that they had blanked out during the test and couldn’t think of an appropriate approach. Our suggestion is that in that case, you should lean on trying to figure out the pattern. Try out a couple of values and see what you get. You may not understand why you are getting what you are getting but that doesn’t stop you from getting the correct answer. Let’s jump on to the question – we will first discuss the ideal approach and then go on to what happens if you don’t have a clue of what to do in the question.

Question: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

Statement 1: When p is divided by 8, the remainder is 5.
Statement 2: m – n = 3

Solution:  Given that p, m and n are positive integers. If n is odd, n^2 must also be odd. How do you represent an odd integer? As (2k + 1)

n^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 – discussed in detail here), 4k(k+1) will be divisible by 8. Therefore, when n^2 is divided by 8, it will leave a remainder of 1.

Statement 1: When p is divided by 8, the remainder is 5.

When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. When n^2 is divided by 8, the remainder will be 1. To get a remainder of 5, when m^2 is divided by 8, we should get a remainder of 4.
m^2 = 8a + 4 (i.e. we can make ‘a’ groups of 8 and 4 will be leftover)
m^2 = 4(2a + 1)

This implies m = 2*?(Odd Number) because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that m is not divisible by 4.

This statement alone is sufficient.

Statement 2: m – n = 3

The difference between m and n is 3 i.e. an odd number. Since n is odd, we can say that m will be even (Even – Odd = Odd). But whether m is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p.

This statement alone is not sufficient.

Answer (A)

In this question, analyzing the question stem and statement 1 is a little complicated. Say we don’t analyze the question stem and jump to statement 1. Let’s see how we can use pattern recognition to make it easier.

Question: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

Statement 1: When p is divided by 8, the remainder is 5.
Statement 2: m – n = 3

Solution:

Statement 1: When p is divided by 8, the remainder is 5.
When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5.  We need a remainder of 5 when m^2 + n^2 is divided by 8. Let’s try to find the remainders when m^2 and n^2 are divided by 8.

We are given that n is odd. Let’s try to figure out what this implies.

n = 1; When n^2 ( = 1) is divided by 8, the remainder is 1.
n = 3; When n^2 ( = 9) is divided by 8, the remainder is 1.
n = 5; When n^2 ( = 25) is divided by 8, the remainder is 1.
n = 7; When n^2 (= 49) is divided by 8, the remainder is 1.

There is a pattern here! Whenever you divide the square of an odd number by 8, you get the remainder 1. (We have discussed ‘why’ above in the logical explanation of statement 1)

This implies that when we divide m^2 by 8, we get 4 as remainder. If m^2 gives 4 as remainder, it means it is of the form m^2 = 8a + 4 = 4(2a + 1). So m must be of the form 2?(Odd Number). Hence m is not divisible by 4.

This statement is sufficient alone.

Through this example, you can see that pattern recognition is a very important tool (in easy as well as difficult questios)

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

Quarter Wit, Quarter Wisdom: Pattern Recognition – Part II

Quarter Wit, Quarter Wisdom
Today’s post comes from Karishma, a Veritas Prep GMAT instructor. Before reading, be sure to check out Part I from last week!

Last week we saw how to use pattern recognition. Today, let’s take up another question in which this concept will help us. Mind you, there are various ways of solving a question. Most questions we solve using pattern recognition can be solved using another method. But pattern recognition is a method we can use in various cases. It is something that comes to our aid when we forget everything else. If you don’t know from where to start on a question, try to give some values to the variables. You might see a pattern. You may not ‘know’ something. Even then, you can ‘figure out’ the answer because GMAT is not a test of your knowledge; it is a test of your wits. It is a test of whether you can keep your cool when faced with the unknown and use whatever you know to solve the question.

Let’s look at a question now.
Continue reading “Quarter Wit, Quarter Wisdom: Pattern Recognition – Part II”

Quarter Wit, Quarter Wisdom: Pattern Recognition

Quarter Wit, Quarter WisdomIf you are hoping for a 700+ in GMAT, you need to develop the ability to recognize patterns. GMAT does not test advanced concepts but you can certainly get advanced questions on simple concepts. For such questions, the ability to quickly observe patterns can come in quite handy. We will discuss a complicated question today which can be easily solved by observing the pattern.

Question: If a and b are distinct integers and a^b = b^a, how many solutions does the ordered pair (a, b) have? Continue reading “Quarter Wit, Quarter Wisdom: Pattern Recognition”

Quarter Wit Quarter Wisdom: GCF and LCM of Fractions

Quarter Wit, Quarter WisdomLast week we discussed some concepts of GCF. Today we will talk about GCF and LCM of fractions.

LCM of two or more fractions is given by: LCM of numerators/GCF of denominators

GCF of two or more fractions is given by: GCF of numerators/LCM of denominators

Why do we calculate LCM and GCF of fractions in this way? Let’s look at the algebraic explanation first. Then we will look at a more intuitive reason.

Algebraic Approach:

Consider 2 fractions a/b and c/d in their lowest form, their LCM is Ln/Ld and GCF is Gn/Gd, also in their lowest forms.
Let’s work on figuring out the LCM first.

LCM is a multiple of both the numbers so Ln/Ld must be divisible by a/b. This implies (Ln/Ld)/(a/b) is an integer. We can re-write this as:

Ln*b/Ld*a is an integer.

Since a/b and Ln/Ld are in their lowest forms, Ln must be divisible by a; also, b must be divisible by Ld (because a and b have no common factors and Ln and Ld have no common factors).

Using the same logic, Ln must be divisible by c; also d must be divisible by Ld.

Ln, the numerator of LCM, must be divisible by both a and c and hence should be the LCM of a and c, the numerators. Ln cannot be just any multiple of a and c; it must be the lowest common multiple so that Ln/Ld is the lowest multiple of the two fractions.

b and d both must be divisible by Ld, the denominator of LCM, and hence Ld must be their highest common factor. Mind you, it cannot be just any common factor; it needs to be the highest common factor so that Ln/Ld is the lowest multiple possible.

This is why LCM of two or more fractions is given by: LCM of numerators/GCF of denominators.

Using similar reasoning, you can figure out why we find GCF of fractions the way we do.

Now let me give you some feelers. They are more important than the algebraic explanation above. They build intuition.

Intuitive Approach:

Let me remind you first that LCM is the lowest common multiple. It is that smallest number which is a multiple of both the given numbers.

Say, we have two fractions: 1/4 and 1/2. What is their LCM? It’s 1/2 because 1/2 is the smallest fraction which is a multiple of both 1/2 and 1/4. It will be easier to understand in this way:

1/2 = 2/4. (Fractions with the same denominator are comparable.)

LCM of 2/4 and 1/4 will obviously be 2/4.

If this is still tricky to see, think about their equivalents in decimal form:

1/2 = 0.50 and 1/4 = 0.25. You can see that 0.50 is the lowest common multiple they have.

Let’s look at GCF now.

What is GCF of two fractions? It is that greatest factor which is common between the two fractions. Again, let’s take 1/2 and 1/4. What is the greatest common factor between them?

Think of the numbers as 2/4 and 1/4. The greatest common factor between them is 1/4.

(Note that 1/2 and 1/4 are both divisible by other factors too e.g. 1/8, 1/24 etc but 1/4 is the greatest such common factor)

Now think, what will be the LCM of 2/3 and 1/8?

We know that 2/3 = 16/24 and 1/8 = 3/24.
LCM = 16*3/24 = 48/24 = 2

LCM is a fraction greater than both the fractions or equal to one or both of them (when both fractions are equal). When you take the LCM of the numerator and GCF of the denominator, you are making a fraction greater than (or equal to) the numbers.

Also, what will be the GCF of 2/3 and 1/8?

We know that 2/3 = 16/24 and 1/8 = 3/24.

GCF = 1/24

GCF is a fraction smaller than both the fractions or equal to one or both of them (when both fractions are equal). When you take the GCF of the numerator and LCM of the denominator, you are making a fraction smaller than (or equal to) the numbers.

We hope the concept of GCF and LCM of fractions makes sense to you 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!

Quarter Wit, Quarter Wisdom: Some GCF Concepts

Quarter Wit, Quarter WisdomSometimes students come up looking for explanations of concepts they come across in books. Actually, in Quant, you can establish innumerable inferences from the theory of any topic. The point is that you should be comfortable with the theory. You should be able to deduce your own inferences from your understanding of the topic. If you come across some so-called rules, you should be able to say why they hold. Let’s discuss a couple of such rules from number properties regarding GCF (greatest common factor). Many of you might read them for the first time. Stop and think why they must hold.

Rule 1: Consecutive multiples of ‘x’ have a GCF of ‘x’

Explanation: What do we mean by consecutive multiples of x? They are the consecutive terms in the multiplication table of x. For example, 4x and 5x are consecutive multiples of x. So are 18x and 19x…

What will be the greatest common factor of 18x and 19x? We know that x is their common factor. Do 18 and 19 have any common factors (except 1)? No. So greatest common factor will be x. Take any two consecutive numbers. They will have no common factors except 1. Hence, if we have two consecutive factors of x, their GCF will always be x.

For more on common factors of consecutive numbers, check:
http://www.veritasprep.com/blog/2011/09 … c-or-math/
http://www.veritasprep.com/blog/2011/09 … h-part-ii/

Can you derive some of your own ‘rules’ based on this now? Let’s give you some ideas:

Two consecutive integers have GCF of 1.

Two consecutive odd multiples of x have GCF of x.

Rule 2: The G.C.F of two distinct numbers cannot be larger than the difference between the two numbers.

Explanation: GCF is a factor of both the numbers. Say, the GCF of two distinct numbers is x. This means the two numbers are mx and nx where m and n have no common factor. What can be the smallest difference between m and n? m and n cannot be equal since the numbers are distinct. The smallest difference between them can be 1 i.e. they can be consecutive numbers. In that case, the difference between mx and nx will be x which is equal to the GCF. If m and n are not consecutive integers, the difference between them will be much larger than x. The difference between mx and nx cannot be less than x.
Say, GCF of two numbers is 6. The numbers can be 12 and 18 (GCF = 6) or 12 and 30 (GCF = 6) etc but they cannot be 12 and 16 since both numbers must have 6 as a factor. So after a multiple of 6, the other multiple of 6 must be at least 6 away.

Let’s look at a third party question based on these concepts now.

Question 1: What is the greatest common factor of x and y?

Statement 1: Both x and y are divisible by 4.
Statement 2: x – y = 4

Solution:

Statement 1: Both x and y are divisible by 4

We know that 4 is a factor of both x and y. But is it the highest common factor? We do not know. There could be another factor common between x and y and hence highest common factor could be greater than 4. e.g. 4 and 16 have 4 as the highest common factor but 12 and 36 have 12 as the highest common factor though both pairs have 4 as a common factor.

Statement 2: x – y = 4

We know that x and y differ by 4. So their GCF cannot be greater than 4 (as discussed above). The GCF could be any of 1/2/4 e.g. 7 and 11 have GCF of 1 while 2 and 6 have GCF of 2.

Taking both statements together: From statement 1, we know that x and y have 4 as a common factor. From statement 2, we know that x and y have one of 1/2/4 as highest common factor. Hence 4 is the highest common factor.

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!

Quarter Wit, Quarter Wisdom: Have a Game Plan

Quarter Wit, Quarter WisdomFor the past few weeks, we have been discussing conditional statements. Let’s switch back to Quant today. I have been meaning to discuss a question for a while. We can easily solve it by plugging in the right values. The only issue is in figuring out the right values quickly. The point we are going to discuss is that there has to be a plan.

Question 1: Six countries in a certain region sent a total of 75 representatives to an international congress. No two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?

Statement 1: One of the six countries sent 41 representatives to the congress.
Statement 2: Country A sent fewer than 12 representatives to the congress. Continue reading “Quarter Wit, Quarter Wisdom: Have a Game Plan”

Quarter Wit, Quarter Wisdom: Necessary Conditions

Quarter Wit, Quarter WisdomLast week we discussed a very tricky CR question based on conditional statements. This week, we would like to discuss another CR question based on necessary conditions. Note that you don’t need to be given ‘only if’ or ‘only when’ to mark a necessary condition. The wording of the statement could imply it. You need to keep a keen eye to figure out necessary and sufficient conditions.

Question: Successfully launching a new product for supermarket sale requires either that supermarkets give prominent shelf space to the product or that plenty of consumers who have not tried it seek it out. Consumers will seek out a new product only if it is extensively advertised, either on television or in the press. One way for a manufacturer to obtain prominent shelf space for a new product is to promote it in trade journals.
Continue reading “Quarter Wit, Quarter Wisdom: Necessary Conditions”

Quarter Wit, Quarter Wisdom: A Question on Conditional Statements

Quarter Wit, Quarter WisdomWe hope that you have understood conditional sentences we discussed in the last post. The concept is very important and you will come across questions using this concept often. Now, let’s discuss the GMAT question we gave you last week.

Question: A newborn kangaroo, or joey, is born after a short gestation period of only 39 days. At this stage, the joey’s hind limbs are not well developed, but its forelimbs are well developed, so that it can climb from the cloaca into its mother’s pouch for further development. The recent discovery that ancient marsupial lions were also born with only their forelimbs developed supports the hypothesis that newborn marsupial lions must also have needed to climb into their mothers’ pouches.
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Quarter Wit, Quarter Wisdom: Conditional Statements

Quarter Wit, Quarter WisdomLast week we discussed a Critical Reasoning question in detail. Today, I want to discuss a very important concept of CR — analyzing a conditional statement. You will often encounter these even though they may not be in the exact same format as the one we will discuss below. We will discuss the basic framework and then we will look at questions where this concept will be very helpful. Mind you, without this framework, it can get a little tricky to wrap your head around these questions.

Statement 1:

If you trouble your teacher, you will be punished.

What does this imply? It implies that ‘troubling your teacher’ is a sufficient condition to get punished. If you trouble, you will get punished.
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Quarter Wit, Quarter Wisdom: Evaluate the Conclusion

Quarter Wit, Quarter WisdomAt the risk of generalizing from a relatively small sample, let me say that people who are good at Quant, tend to be good at Critical Reasoning in Verbal. I certainly cannot comment about their SC and RC prowess but they are either good at CR right from the start or improve dramatically after just a couple of our sessions. The reason for this is very simple – CR is more like Quant than like Verbal. CR is very mathematical. You need to keep in mind some basic rules and based on those, you can easily crack the most difficult of problems. There is only one catch – don’t get distracted by options put there to distract you!

Today I would like to discuss a great CR question from our book. It upsets a lot of students even though it is simple – just like a GMAT question is supposed to be. Here is the question:
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Quarter Wit, Quarter Wisdom: Rates Revisited

Quarter Wit, Quarter WisdomPeople often complain about getting stuck in work-rate problems. Hence, I would like to take some 700+ level questions on rate today. I have discussed the basic concepts of work-rate (using ratios) in a previous post:

Cracking the Work Rate Problems

You  might want to go through that post before you set out to work on these problems. Ensure that you are very comfortable with the relation: Work = Rate*Time and its implications: If rate doubles, work done doubles too if the time remains constant; if one work is done, rate = 1/time etc. Thorough understanding of these implications is fundamental to ‘reasoning out’ the answer.
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Quarter Wit, Quarter Wisdom: Questions on Factorials

Quarter Wit, Quarter WisdomLast week we discussed factorials – how we can take something common when we have factorials in some equations. Today let’s discuss a couple of questions based on factorials. They look intimidating but they are pretty simple. Factorial is all about multiplication and hence there is a high probability that you will be able to take something common and cancel something. These techniques reduce our work significantly. Hence, seeing a factorial in a question should bring a smile to your face!

Question 1: Given that x, y and z are positive integers, is y!/x! an integer?

Statement 1: (x + y)(x – y) = z! + 1

Statement 2: x + y = 121
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Quarter Wit, Quarter Wisdom: Managing Factorials in Equations

Quarter Wit, Quarter WisdomA concept we have not yet covered in this series is factorials (though we used some factorials in the post Power in Factorials). Let’s first discuss the basics of factorials. Once we do, we will see that most factorial expressions can be easily solved using a single method: taking common!

First of all, what is (n!)?

n! = 1*2*3*4*5*6*…*(n – 2)*(n -1)*n

Let’s take some examples:
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Quarter Wit, Quarter Wisdom: Successive Division

Quarter Wit, Quarter WisdomWe discussed divisibility and remainders many weeks ago. Today, we will use those concepts and discuss another type of question – successive division. But before we do, you need to go through the previous related posts on division if you haven’t read them already:

Divisibility Unraveled

Divisibility Applied on the GMAT

Divisibility Applied to Remainders
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Quarter Wit, Quarter Wisdom: Working on Getting the Full Picture Again!

Quarter Wit, Quarter WisdomToday, we will continue our discussion on why it is important to understand the workings behind seemingly miraculous shortcuts. We will use another example from probability.

Question 1: A bag contains 4 white balls, 2 black balls & 3 red balls. One by one three balls are drawn out with replacement (i.e. a ball is drawn and then put back. Thereafter, another ball is drawn). What is the probability that the third ball is red?

Solution:

The question is simple, isn’t it? When you draw balls with replacement, the probability stays the same at the beginning of every cycle. On first draw, the probability of drawing a red ball is 3/9 (since there are 3 red balls and 9 balls in all). When you draw the ball and put in back, the probability of drawing a red ball again stays the same i.e. 3/9 (since again there are 3 red balls and 9 balls in all). The situation at the beginning of every draw is the same.
Continue reading “Quarter Wit, Quarter Wisdom: Working on Getting the Full Picture Again!”

Quarter Wit, Quarter Wisdom: Get the Full Picture – Part II

Quarter Wit, Quarter WisdomLet’s refer back to last week’s post. We discussed why it is important to fully understand what you are doing and why you are doing it, especially while using an innovative method. We talked about it using a combinatorics example.

Let’s revisit it here:

Question 1: What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

We discussed the ways of obtaining various sums. The regular way of obtaining a sum of 8 is enumerating all the possibilities. An innovative way was using 7C2 (as discussed last week).
Continue reading “Quarter Wit, Quarter Wisdom: Get the Full Picture – Part II”

Quarter Wit, Quarter Wisdom: Get the Full Picture

Quarter Wit, Quarter WisdomToday we will discuss why ‘understanding’ rather than just ‘learning’ a concept is important. Most questions can be solved using different methods. Sometimes, a particular method seems really easy and quick and we tend to ‘learn’ it without actually knowing why we are doing what we are doing. We need to understand the strengths and the weaknesses of the method before we use it. Let me elaborate with an example.

Question: What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

When you throw three dice simultaneously, you can obtain a total sum ranging from 3 (1 on each die) to 18 (6 on each die). Let’s consider all the possible sums that we can obtain:

Sum of 3: This can happen in only 1 way. Each die shows 1 in this case

Sum of 4: This can happen in 3 ways. One die shows 2 and the other two show 1 each. Any of the three dice could show 2 so there are a total of 3 ways of getting a sum of 4.

Sum of 5: There are different ways of obtaining a sum of 5:

1, 1, 3 – Any of the three dice could show 3 so there are a total of 3 ways of obtaining 5 in this way.

1, 2, 2 – Any of the three dice could show 1 so there are a total of 3 ways of obtaining 5 in this way.

Total number of ways of obtaining a sum of 5 = 3 + 3 = 6

Sum of 6: There are different ways of obtaining a sum of 6:

1, 1, 4 – Any of the three dice could show 4 so there are a total of 3 ways of obtaining 6 in this way.

1, 2, 3 – These 3 numbers could be arranged among the 3 dice in 3! ways (using basic counting principle). There are a total of 3! = 6 ways of obtaining 6 in this way.

2, 2, 2 – This can happen in only one way. All dice show 2.

Total number of ways of obtaining a sum of 6 = 3 + 6 + 1 = 10

Similarly, we can find the number of ways in which all other sums can be obtained. Below I will list the number of ways in each case.

Sum of 3: 1 way                          Sum of 18: 1 way

Sum of 4: 3 ways                        Sum of 17: 3 ways

Sum of 5: 6 ways                        Sum of 16: 6 ways

Sum of 6: 10 ways                     Sum of 15: 10 ways

Sum of 7: 15 ways                    Sum of 14: 15 ways

Sum of 8: 21 ways                    Sum of 13: 21 ways

Sum of 9: 25 ways                    Sum of 12: 25 ways

Sum of 10: 27 ways                 Sum of 11: 27 ways

Notice the symmetry here. There is only 1 way of obtaining 3 and only one way of obtaining 18. Similarly, there are 3 ways in which you can obtain 4 and 3 ways in which you can obtain 17. Is it a co-incidence? No. The second half of the column can be obtained by replacing 1 by 6, 2 by 5, 3 by 4, 4 by 3, 5 by 2 and 6 by 1. The number of ways of obtaining a sum is symmetrical about the center.

We have simplified one aspect of this problem. If we need to find the number of ways of obtaining 15, we can instead find the number of ways of obtaining a sum of 6 (which is psychologically easier to handle). Now the problem is whether there is an easier way of obtaining the number of ways in which you can get a sum of 6 or do we have to enumerate it in every case.

You might have come across something like this:

Sum of 3: 2C2 = 1 way

Sum of 4: 3C2 = 3 ways

Sum of 5: 4C2 = 6 ways

Sum of 6: 5C2 = 10 ways

Sum of 7: 6C2 = 15 ways

Sum of 8: 7C2 = 21 ways

Perfect till now! Matches the numbers we obtained above. It seems like a good method to use instead of writing down all the cases, doesn’t it? But what happens further on?

Sum of 9: 8C2 = 28 ways

Sum of 10: 9C2 = 36 ways

These don’t match! The key to understanding this is to understand why it works in the above given cases. First review this post.

Focus on method II of question 2. Notice how you divide n identical objects among m distinct groups. Let’s take the example of a sum of 7. You have to divide 7 among 3 dice such that each die must have at least 1 (no die face can show 0). First step is to take 3 out of the 7 and give one each to the three dice. Now you have 4 left to distribute among 3 distinct groups such that it is possible that some groups may get none of the four. Think of partitioning 4 in 3 groups. This can be done in (4+2)!/4!*2! = 6C2 ways (check out the given link if you do not understand this)

This is how you obtain 6C2 for the sum of 7.

The concept works perfectly till the sum of 8. Thereafter it fails. Think why. I will explain it next week.

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

Quarter Wit, Quarter Wisdom: Three Overlapping Sets

Today, let’s take a look at how to use Venn diagrams to solve questions involving three overlapping sets.

First, let me show you what the three overlapping sets diagram looks like.

Notice that the total comprises of the elements that do not fall in any of the three sets and the elements that are a part of at least one of the three sets.

The elements falling in the red, yellow or green region (region a, b or c) fall in only one set. The elements falling in region d, e or f fall in exactly two sets and the elements falling in region g fall in all three sets.

Now some quick questions to get a clear picture:

Question 1: Which regions represent the elements that belong to at least 2 sets?

Answer 1: d + e + f + g

Question 2: Which regions represent the elements that belong to at least 1 set?

Answer 2: a + b + c + d + e + f + g = Total – None

Question 3: Which regions represent the elements that belong to at most 2 sets?

Answer 3: None + a + b + c + d + e + f = Total – g

Hope there are no doubts up till now. Let’s look at a question to see how to apply these concepts.

Question: Three table runners have a combined area of 200 square inches. By overlapping the runners to cover 80% of a table of area 175 square inches, the area that is covered by exactly two layers of runner is 24 square inches. What is the area of the table that is covered with three layers of runner?

(A) 18 square inches
(B) 20 square inches
(C) 24 square inches
(D) 28 square inches
(E) 30 square inches

Solution: Let’s first try to understand what exactly is given to us. The area of all the runners is equal to 200 square inches.

Runner 1 + Runner 2 + Runner 3 = 200

In our diagram, this area is represented by
(a + d + g + e) + (b + d + g + f) + (c + e + g + f) = 200

(We need to find the value of g i.e. the area of the table that is covered with three layers of runner.)

Area of table covered is only 80% of 175 i.e. only 140 square inches. This means that if each section is counted only once, the total area covered is 140 square inches.
a + b + c + d + e + f + g = 140
So the overlapping regions are obtained by subtracting second equation from the first. We get d + e + f + 2g = 60
But d + e + f (area with exactly two layers of runner) = 24
So 2g = 60 – 24 = 36

g = 18 square inches

Note that you don’t need to make all these equations and can directly jump to d + e + f + 2g = 60. We wrote these equations down only for clarity. It is a matter of thinking vs solving. If we think more, we have to solve less. Let’s see how.

Combined area of runners is 200 square inches while area of table they cover is only 140 square inches. So what does the extra 60 square inches of runner do? It covers another runner!
Wherever there are two runners overlapping, one runner is not covering the table but just another runner. Wherever there are three runners overlapping, two runners are not covering the table but just the third runner at the bottom.
So can we say that (d + e + f) represents the area where one runner is covering another runner and g is the area where two runners are covering another runner?
Put another way, can we say d + e + f + 2g = 60?
We know that d + e + f = 24 giving us g = 18 square inches
This entire ‘thinking process’ takes ten seconds once you are comfortable with it and your answer would be out in about 30 sec!

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

Quarter Wit, Quarter Wisdom: A Sets Question that Upsets Many

Quarter Wit, Quarter WisdomWe hope last week’s discussion improved your understanding of sets and showed you how you can come across some tricky sets questions even though the concepts seem very simple. Today, let’s further build up on what we learned last week with the help of an example.

Example: A group of people were given 2 puzzles. 79% people solved puzzle X and 89% people solved puzzle Y. What is the maximum  and minimum percentage of people who could have solved both the puzzles?
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Quarter Wit, Quarter Wisdom: Nuances of Sets

Quarter Wit, Quarter WisdomWe will start with sets today. Your Veritas Prep GMAT book explains you the basics of sets very well so I am not going to get into those. If you have gone through the concepts, you know that we can use Venn diagrams to solve the sets questions.

First, let’s look at why we should focus on terminology in sets question. Thereafter, we will put up a very nice question from our very own book which is simple but takes down many people (just like a typical GMAT question):

Say, there are a total of 100 people in a housing society. There are two clubs close to the society – A and B. You are given that of the 100 people of the housing society, 60 people are members of club A and 50 people are members of club B.
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Quarter Wit, Quarter Wisdom: Working like Clockwork

Quarter Wit, Quarter WisdomIn the last few weeks, we have gone through the concepts of relative speed. You might be surprised to know that we can use the same concept to solve some clock problems too. The reason some clock problem can be tricky is that the hour hand and the minute hand move simultaneously so handling them separately is not easy. In such questions you can easily use relative speed i.e. speed of the minute hand relative to the hour hand. Let’s try to understand this with the help of an example.

Example: In a circular clock, the minute hand is the radius of the circle. At what time is the smaller angle between the minute hand and the hour hand of the clock not divisible by 10?
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Quarter Wit, Quarter Wisdom: Taking the Official GMAT Prep Test

Quarter Wit, Quarter WisdomToday’s post of this series is not based on Quant section of the GMAT. Instead, we will try to answer the often asked question: When should I take the two GMAT prep tests available on mba.com?

Let’s first discuss a hypothetical situation: Say, your goal is to beat a particular opponent called Mr. Dude in a game of chess in a competition being held three months from now. Over three months, you can play four friendly matches against him on a day of your choice. The results of the friendly matches are immaterial and will not be made known to any third party. What will be your strategy?
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Quarter Wit, Quarter Wisdom: Moving in Circles

Quarter Wit, Quarter WisdomHopefully, you are a little comfortable with the relative speed concept now. The concept can come in handy in some circular motion questions too. Today, we will show you how you can use the fundamentals we learned in the last few weeks to solve questions involving moving in a circle. Let’s try a couple of GMAT questions to put what we learned to use:

Question 1: Two cars run in opposite directions on a circular path. Car A travels at a rate of 4? miles per hour and car B runs at a rate of 6? miles per hour. If the path has a radius of 8 miles and both the cars start from point S at the same time, how long, in hours, after the cars depart will they meet again for the first time after leaving?
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Quarter Wit, Quarter Wisdom: Some Tricky Relative Speed Concepts

Quarter Wit, Quarter WisdomAs promised, we tackle some 700+ level questions today. Mind you, these questions are not your typical GMAT type questions. The reason we are discussing them is that they look mind boggling but are easily workable when the concepts of relative speed are used. They give insights that help you understand relative speed. Once you are good with the concepts, you can solve most of the relative speed based questions easily.

Question 1: Cities A and B are 20 miles apart. From both of these cities, simultaneously, two people start walking toward each other at a constant speed of 2 miles/hr. At the same time, a dog leaves city B and runs at a constant speed of 5 miles/hr toward city A. When it reaches the person from city A, it immediately turns around and runs back to the person from city B. When it reaches the person from city B, it turns around and runs back to the person from city A. It keeps doing so until the two people meet. How many miles did the dog run?

(A)   15 miles

(B)   20 miles

(C)   25 miles

(D)   30 miles

(E)    35  miles

Solution: Looks really tricky, doesn’t it? Actually it is really easy once you look at it from the right perspective. It would take you less than a minute to solve almost any 700+ level GMAT question if you can figure out the most optimum method to solve it. The point is – how long will it take you to figure out the most optimum method? Take a minute to think what you would do in this question.

We know the dog’s speed. We need to know the distance it has run. Directly calculating distance is a little complicated since we need to consider the distance covered by the two men too. Instead, if we can figure out the time for which the dog ran, we can easily calculate the distance. For how long did the dog run? He started when the two men started from their respective cities and stopped when the two men met. Therefore, if we can find the time taken by the men to cover the 20 miles, we will get the time for which the dog ran.

This is where the relative speed concept comes in handy.

Total distance between the two men = 20 miles

Relative speed of the men with respect to each other = 2 + 2 mph (since they are travelling in opposite directions)

Time taken by the men to meet = 20/4 = 5 hrs

Therefore, the dog also ran for 5 hrs.

In 5 hrs, the dog must have run 5*5 = 25 miles

Answer (C)

I hope you see that the right perspective simplifies the question immensely. Let’s look at another 700+ level question.

Question 2: A man cycling along the road at a constant speed noticed that every 12 minutes a bus overtakes him and every 6 minutes he meets an oncoming bus. If all buses move at the same constant speed and leave the bus station at fixed time intervals, what is the time interval between consecutive buses?

(A) 5 minutes

(B) 6 minutes

(C) 8 minutes

(D) 9 minutes

(E) 10 minutes

Solution: Again, the question is a little tricky but once you understand how to tackle it, it takes less than a minute.

Buses are coming from opposite directions. A bus overtakes the man every 12 minutes i.e. a bus moving in the same direction as the man overtakes him every 12 mins. To clarify: say, at constant intervals, buses leave a bus station located from where the man left and travel on the same road as the man. Since they are faster, they overtake the man. The man noticed that a bus overtakes him every 12 mins. Obviously then, they must be leaving at constant intervals. Also, he meets a bus coming from the opposite direction every 6 mins. So buses must be leaving from a bus station located at the opposite end of the road at constant intervals.

I hope the problem is clear to you. Now let’s try to work out the solution.

Say the cyclist is stationary at a point. Buses are coming from opposite directions (same speed, same time interval). A bus will meet the cyclist every t minutes from either direction. This ‘t minutes’ must be the time interval between consecutive buses. Let’s say, a bus from each direction just met him. After t minutes, 2 more buses from opposite directions will meet him and so on… We need to find ‘t’.

Now the cyclist starts moving at speed c. His speed relative to the bus going in the same direction becomes b – c. His speed relative to the bus from the opposite direction becomes b + c. This is the reason that the time interval between two buses is different for the opposite directions. Time interval is in the ratio 12:6. Then, the ratio of the relative speed in the two cases must be inverse i.e. 6:12

(b-c):(b+c) = 6:12 which gives you c = (1/3)b

This means that the bus travelling at a relative speed which is 2/3rd of its usual speed (b-c = 2b/3) takes 12 minutes to meet the man. If it were travelling at its usual speed, it would have taken 12*(2/3) = 8 mins to meet the man. This 8 mins is the value of ‘t’ i.e. the time interval between buses.

Answer (C)

You might need to go through the question a few times before you fully understand it. It will also be helpful to draw a diagram and see what the situation looks like.

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

Quarter Wit, Quarter Wisdom: Questions on Speeding

Quarter Wit, Quarter WisdomWe discussed the concepts of relative speed in GMAT questions last week. This week, we will work on using those concepts to solve questions. The questions we take today will be 600-700 level. I intend to take the 700+ level questions next week (don’t want to scare you away just yet!). Let’s get going now.

Question 1: A man walking at a constant rate of 4 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 20 miles per hour. The woman stops to wait for the man 5 minutes after passing him, while the man continues to walk at his constant rate. How many minutes must the woman wait until the man catches up?
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Quarter Wit, Quarter Wisdom: Speeding, Relatively

Quarter Wit, Quarter WisdomLet’s look at the concept of relative speed today. A good understanding of relative speed can be very useful in some questions. If you don’t use relative speed in these GMAT questions, you can still solve them but they would be rather painful to work through (they might need multiple variables and you know my policy on variables – use one, if you must)

So the first question is – what is relative speed? To understand this, let’s first consider ‘speed’. When your car’s speedometer shows 60 mph, what does it mean? It means that you are traveling at a speed of 60 mph relative to Earth. For an observer in space, your speed could be different. It would be the resultant of your speed and Earth’s speed. If this is hard to imagine, think of a plane flying in the sky. Say, it is flying at a speed of 500 mph. What happens if a strong wind starts blowing in the same direction as the plane? The wind’s speed gets added to the plane’s speed and the plane travels even faster. So anyway, the speed we usually talk about is relative to Earth.

It is sometimes useful to consider the speed of an object relative to another object. Say, two people are sitting next to each other in the same train which is moving at a speed of 100 mph. Are they moving relative to each other? No. Their speed relative to each other is 0. Ofcourse, relative to the Earth, they are moving at the same speed as the train i.e. 100 mph. When working on relative speed questions, just focus on the two objects we are talking about and how fast they are moving relative to each other. Their actual speed may not be of much relevance to us.

Let’s consider a few cases related to relative speed. Say, there are two people, A and B.

Case 1: A is standing still and B is moving at a speed of 5 mph.

What is B’s speed relative to A? B is moving away from A at a rate of 5 miles every hour. So B’s speed relative to A is 5 mph. What is A’s speed relative to B? Is it 0? No! A’s speed relative to Earth is 0. A’s speed relative to B is 5 mph since distance between A and B is increasing at a rate of 5 mph. Confused? When we say ‘relative to B’, we assume that B is stationary. Since distance between the two is increasing at a rate of 5 mph, we say A’s speed relative to B is 5 mph.

Case 2: A is walking due east at a speed of 5 mph and B is walking due east at a speed of 2 mph.

What is A’s speed relative to B? Distance between A and B is increasing by 3 miles every hour. So A’s speed relative to B is only 3 mph, not 5 mph. Think of it this way – A is moving fast but not as fast when compared with B since B is also moving. Similarly, B’s speed relative to A is also 3 mph. Notice that when they move in the same direction, their relative speed is the difference of their speeds.

Case 3: A is walking due east at a speed of 5 mph and B is walking due west, away from A, at a speed of 2 mph.

What is A’s speed relative to B? Distance between A and B is increasing by 7 miles every hour. So A’s speed relative to B is 7 mph, not 5 mph. Think of it this way – A is moving fast and even faster as compared with B since B is moving in the opposite direction. Similarly, B’s speed relative to A is also 7 mph. Notice that when they move in the opposite directions, their relative speed is the sum of their speeds. It doesn’t matter whether they are moving toward each other or away from each other.

Let’s look at some easy questions on relative speed to cement the concepts. We will look at some tough nuts next week.

Question 1: Train A starts from station A traveling at 30 miles per hour toward station B. At the same time, on a parallel track, train ? leaves station B at 40 miles per hour toward station A. When the two trains meet, how far is train A from station B if the distance between stations A and B is 700 miles?
(A) 140
(B) 240
(C) 300
(D) 340
(E) 400

Solution: Both the trains start at the same time from stations that are 700 miles away from each other. This means that in the beginning, distance between them is 700 miles. When they meet, they have together covered the entire 700 miles. Relative to each other, their speed is 30 + 40 = 70 mph.

Time for which they travel till they meet = 700/70 = 10 hrs

Train A covered 30*10 = 300 miles. This means, it is 400 miles away from station B. Answer (E)

Meanwhile, train B covered 40*10 = 400 miles.

Let’s look at a variation of this question now.

Question 2: Train A leaves the station and travels at 30 miles per hour. Three hours later, train ? leaves the same station traveling in the same direction at 40 miles per hour on a parallel track. How far from the station was train A overtaken by train B?
(A) 90
(B) 180
(C) 200
(D) 300
(E) 360

Solution: Train A travels at 30 mph for 3 hrs and covers 30*3 = 90 miles in this time. This is when train B leaves the station. Now both the trains are running in the same direction and their relative speed is 40 – 30 = 10 mph. This means that train B covers an extra 10 miles every hour. Since the initial distance between the two trains is 90 miles, it takes train B ( 90/10 = ) 9 hrs to catch up with train A. In 9 hrs, train B must have traveled 40*9 = 360 miles. Hence, train A must also be 360 miles away from the station.

Answer (E)

You can easily do these questions using variables. The relative speed concept is not a must. But, you definitely end up saving some time if you use relative speed. You don’t take any variables and don’t make any equations. Next week, we will look at some tougher relative speed 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 GMAT prep for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Some Inequalities, Mods and Sets

Quarter Wit, Quarter WisdomToday, let’s look at a question that involves inequalities and modulus and is best understood using the concept of sets. It is not a difficult question but it is still very tricky. You could easily get it right the first time around but if you get it wrong, it could take someone many trials before he/she is able to convince you of the right answer. Even after I write a whole post on it, I wouldn’t be surprised if I see “but I still don’t get it” in the comments below!

Anyway, enough of introduction! Let’s get to the question now.

Question: If x/|x| < x, which of the following must be true about x?

(A)   x > 1

(B)   x > -1

(C)   |x|< 1

(D)   |x| = 1

(E)    |x|^2 > 1

Solution:

First thing we do is tackle the mod. We know that |x| is just the absolute value of x.

So, x/|x| can take only 2 values: 1 or -1
If x is positive, x/|x| = 1 e.g. if x = 4, then 4/|4| = 1
If x is negative, x/|x| = -1 e.g. if x = -4, then -4/|-4| = -4/4 = -1
x cannot be 0 because we cannot have 0 in the denominator of an expression.

Now let’s work on the inequality.

x > x/|x| implies x > 1 if x is positive or x > -1 if x is negative.
Hence, for this inequality to hold, either x > 1 (when x is positive) or -1 < x< 0 (when x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc.

Now think – which of the following MUST BE TRUE about every value that x can take?

(A)   x > 1

or

(B)   x > -1

I hope that you agree that x > 1 doesn’t hold for every possible value of x whereas x > -1 holds for every possible value of x. Mind you, every value greater than -1 need not be a possible value of x.

This concept might need some more work. Let me explain with another example.

Forget this question for a minute. Say instead you have this question:

Example 1: x > 2 and x < 7. What integral values can x take?
I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.

The top arrow shows x > 2 and the bottom arrow shows x < 7. You see that the overlapping area includes 3, 4, 5 and 6. That is the region that satisfies both the inequalities.

Now consider this:

Example 2: x > 2 or x > 5. What integral values can x take?
Let’s draw that number line again.

So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. the overlapping part) as was the case in example 1. OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.

x/|x| is either 1 or -1.
So x > 1 or x > -1
So which values can x take? All values included in at least one of the sets. Therefore, x > -1.

Note that the confusion lies only between the first two options. The other three options are rejected outright.

(C)   |x|< 1 implies -1 < x < 1. Definitely doesn’t hold.

(D)   |x| = 1 implies x = 1 or -1. Definitely doesn’t hold.

(E)    |x|^2 > 1 implies either x < -1 or x > 1. Definitely doesn’t hold.

So what do you say? Are you convinced that the answer is (B)?

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

Quarter Wit, Quarter Wisdom: Questions on Inequalities

Quarter Wit, Quarter WisdomNow that we have covered some variations that arise in inequalities in GMAT problems, let’s look at some questions to consolidate the learning.

We will first take up a relatively easy OG question and then a relatively tougher question which looks harder than it is because of the use of mods in the options (even though, we don’t really need to deal with the mods at all).

Question 1: Is n between 0 and 1?
Statement 1: n^2 is less than n
Statement 2: n^3 is greater than 0
Continue reading “Quarter Wit, Quarter Wisdom: Questions on Inequalities”

Quarter Wit, Quarter Wisdom: Inequalities with Complications – Part II

Quarter Wit, Quarter WisdomLast week, we discussed a couple of variations of inequality questions with many factors. Let’s now look at some more complications that we should know how to handle.

Complication No 3: Even powers

(x  – a)^2(x – b)(x – c)(x – d) > 0

How would you handle even powers of factors? Say, if the question has a factor (x – a) which has an even power, would you still plot ‘a’ on the number line? No, you will just ignore that factor while making the number line! Why? This is so because this factor will never be negative. It will be either 0 (in case the inequality includes the equality sign e.g. (x  – a)^2(x – b)(x – c)(x – d) ? 0) or it will be positive. Therefore, a factor with an even power acts just like a positive constant. Then, does it mean factors with even powers have no role to play at all? No, it doesn’t. While writing out the range, they could impact the final answer. We will discuss this in more detail using an example later on.

e.g. (x – 4)(x – 2)(x + 8)^4 < 0

You don’t need to plot -8 here. Since (x + 8)^4 can never be negative, it doesn’t change the sign of the expression.

The expression will be negative in the range 2 < x < 4.

Complication No 4: Odd powers

(x  – a)^3(x – b)(x – c)(x – d) > 0

What will you do in case of odd powers? Notice that in the last two posts, we have handled questions where the power of all the factors is 1. 1 is an odd power. So when you have any other odd power, you will handle it the same way. You can assume that the odd power is equal to 1 and proceed as usual. This is so because the sign of (x – a) will be the same as the sign of (x – a)^3.

e.g. (x – 4)(x – 2)(x + 1)^3 < 0

The expression will be negative in the range x < -1 or 2 < x < 4.

Now let’s look at a question involving both these complications.

Question: Find the range of x for which the given inequality holds.

[(2x – 5) * (6 – x)^3]/x^2 ? 0

Solution:

I hope you agree that it doesn’t matter whether the factors are multiplied or divided. We are only concerned with the sign of the factors.

[2(x – 5/2) * (-1)(x – 6)^3]/x^2 ? 0 (take 2 common)

[2(x – 5/2) * (x – 6)^3]/(x – 0)^2 ? 0 (multiply both sides by -1)

Now let’s draw the number line. We don’t need to plot 0 since (x – 0) has an even power.

We want to find the range where the expression is positive. The required range is x ? 5/2 or x ? 6. But we are missing something here. x ? 5/2 implies that all values less than 5/2 are acceptable but note that x cannot be 0 since x^2 is in the denominator. Hence the acceptable range is x ? 5/2 or x ? 6 but x ? 0.

When you have the equal to sign, you have to be careful about the way you choose your range.

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!