The Past Perfect expresses the idea that something occurred before another action in the past. It can also show that something happened before a specific time in the past.

We often ignore the “something happened before a specific time in the past” part of the tense.

For example, look at this sentence: Robin had never cooked pasta before last night.

Here, we use past perfect “had cooked” without another verb in the past tense – why? Because we use past perfect to show that something happened before a specific time in the past i.e. before last night.

Similarly, sometimes in GMAT too, you may see past perfect where it seems reasonable but you may not find a verb in past tense. It could be because an action happened before a specific time in the past or there is an implied action in the past. There is a reason why we brought up this point – check out the sentence given below:

*According to some economists, the gains in the stock market reflect growing confidence that the economy will avoid the recession that many had feared earlier in the year and instead come in for a ‘soft landing’.*

The sentence is similar to a correct sentence given in Official Guide. Note the use of “had feared” – many people question the use of past perfect here. The reason past perfect is correct here is this:

“According to some economists” implies an action in the past – something similar to “Some economists said” or in other words, it implies a specific time in the past – the time when the economists expressed their opinion. In the sentence, “earlier in the year” is a time before the economists expressed their opinion and hence it makes sense to use past perfect. In such cases, our use of common sense is more important than the mere retention of grammar rules. Another thing that helps in such situations is that all other options would have a major fault.

Let’s show you the actual OG question:

Question: According to some analysts, the gains in the stock market reflect growing confidence that the economy will avoid the recession that many had feared earlier in the year and instead come in for a “soft landing,” followed by a gradual increase in business activity.

(A) that the economy will avoid the recession that many had feared earlier in the year and instead come

(B) in the economy to avoid the recession, what many feared earlier in the year, rather to come

(C) in the economy’s ability to avoid the recession, something earlier in the year many had feared,

and instead to come

(D) in the economy to avoid the recession many were fearing earlier in the year, and rather to come

(E) that the economy will avoid the recession that was feared earlier this year by many, with it instead coming

Let’s look at the errors in the other options:

(B) in the economy to avoid the recession, what many feared earlier in the year, rather to come

You cannot use “what” in place of “which”. Also, the use of “confidence in A to avoid” is not correct. It should be “confidence that A will avoid”.

(C) in the economy’s ability to avoid the recession, something earlier in the year many had feared,

and instead to come

The placement of “earlier in the year” is incorrect here. It should come after “had feared”.

(D) in the economy to avoid the recession many were fearing earlier in the year, and rather to come

Again, the use of “confidence in A to avoid” is not correct. It should be “confidence that A will avoid”.

(E) that the economy will avoid the recession that was feared earlier this year by many, with it instead coming

“With it instead coming” doesn’t make any sense so this option isn’t correct either.

So we see that all other options have fatal flaws. Hence, in this case, option (A) is our best bet even though the use of past perfect isn’t the way we usually see it.

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

**Average Speed = Total Distance/Total Time**

No matter which formula you choose to use, it will always boil down to this one. Keeping this in mind, let’s discuss the various formulas we come across:

1. **Average Speed = (a + b)/2**

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

2. **Average Speed = 2ab/(a + b)**

Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds. On similar lines, you can modify this formula for one-third distance.

3. **Average Speed = 3abc/(ab + bc + ca)**

Applicable when one travels at speed a for one-third of the distance, at speed b for another one-third of the distance and speed c for rest of the one-third of the distance.

Note that the generic Harmonic mean formula for n numbers is

Harmonic Mean = n/(1/a + 1/b + 1/c + …)

4. You can also use weighted averages. Note that in case of average speed, the weight is always ‘time’. So in case you are given the average speed, you can find the ratio of time as

**t1/t2 = (a – Avg)/(Avg – b)**

As you already know, this is just our weighted average formula.

Now, let’s look at some simple questions where you can use these formulas.

Question 1: Myra drove at an average speed of 30 miles per hour for T hours and then at an average speed of 60 miles/hr for the next T hours. If she made no stops during the trip and reached her destination in 2T hours, what was her average speed in miles per hour for the entire trip?

(A) 40

(B) 45

(C) 48

(D) 50

(E) 55

Solution: Here, time for which Myra traveled at the two speeds is same.

Average Speed = (a + b)/2 = (30 + 60)/2 = 45 miles per hour

Answer (B)

Question 2: Myra drove at an average speed of 30 miles per hour for the first 30 miles of a trip & then at an average speed of 60 miles/hr for the remaining 30 miles of the trip. If she made no stops during the trip what was her average speed in miles/hr for the entire trip?

(A) 35

(B) 40

(C) 45

(D) 50

(E) 55

Solution: Here, distance for which Myra traveled at the two speeds is same.

Average Speed = 2ab/(a+b) = 2*30*60/(30 + 60) = 40 mph

Answer (B)

Question 3: Myra drove at an average speed of 30 miles per hour for the first 30 miles of a trip, at an average speed of 60 miles per hour for the next 30 miles and at a average speed of 90 miles/hr for the remaining 30 miles of the trip. If she made no stops during the trip, Myra’s average speed in miles/hr for the entire trip was closest to

(A) 35

(B) 40

(C) 45

(D) 50

(E) 55

Solution: Here, Myra traveled at three speeds for one-third distance each.

Average Speed = 3abc/(ab + bc + ca) = 3*30*60*90/(30*60 + 60*90 + 30*90)

Average Speed = 3*2*90/(2 + 6 + 3) = 540/11

This is a bit less than 50 so answer (D).

Question 4: Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5

(B) 1/3

(C) 2/5

(D) 2/3

(E) 3/5

Solution: We know the average speed and must find the fraction of time taken at a particular speed.

t1/t2 = (A2 – Aavg)/(Aavg – A1)

t1/t2 = (60 – 50)/(50 – 30) = 1/2

So out of a total of 3 parts of the journey time, she drove at 30 mph for 1 part and at 60 mph for 2 parts of the time. Fraction of the total time for which she drove at 30 mph is 1/3.

Answer (B)

Hope this sorts out some of your average speed formula confusion.

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

In many questions, we need to account for different cases one by one but we don’t really see such questions on the GMAT since we have limited time. Also, we don’t tire of repeating this again and again – GMAT questions are more reasoning based than calculation intensive. Usually, there will be an intellectual method to solve every GMAT question – a method that will help you solve it in seconds.

We have discussed using symmetry in Combinatorics before. It can be used in many questions though most people don’t realize that. In our ongoing endeavor to expose you to intellectual methods, here we present how most people tackle a question and how you can tackle it instead to be in the top 1%ile.

**Question**: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?

(A) 5

(B) 6

(C) 11

(D) 16

(E) 19

Solution:

**Most Common Solution**:

What are the permutations of sequence S? They are the different ways in which we can arrange the elements of S. For example, 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc

In how many different ways can we make the sequence? The first element can be chosen in 4 ways – one of 3, 4, 5 and 6. (You are given that 2 cannot be the first element).

The second element can be chosen in 4 ways (2 and the leftover 3 numbers).

The third element can be chosen in 3 ways.

The fourth element can be chosen in 2 ways.

And finally there will be only 1 element left for the last spot.

Number of ways of making set S = 4*4*3*2*1 = 96

In how many of these sets will 5 be in the second spot?

If 5 is reserved for the second spot, there are only 3 ways of filling the first spot (3 or 4 or 6).

The second spot has to be taken by 5.

The third element will be chosen in 3 ways (ignoring 5 and the first spot)

The fourth element can be chosen in 2 ways.

And finally there will be only 1 element left for the last spot.

Number of favorable cases = 3*1*3*2*1 = 18

Required Probability = Favorable Cases/Total Cases = 18/96 = 3/16 = a/b

a+b = 3 + 16 = 19

Answer (E)

**Intellectual Approach:**

Use a bit of logic of symmetry to solve this question without any calculations.

Set S would include all such sequences as 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc – starting with 3, with 4, with 5 or with 6 with equal probability.

By symmetry, note that 1/4th of them will start with 5 – which we need to ignore – so we are left with the rest of the 3/4th sequences.

Now, in these 3/4th sequences which start with either 3 or 4 or 6, 5 could occupy any one of the 4 positions – second, third, fourth or fifth with equal probability. So we need 1/4th of these sequences i.e. only those sequences in which 5 is in the second spot.

Probability that 5 is the second element of the sequence = (3/4)*(1/4) = 3/16

Therefore, a+b = 3+16 = 19

Answer (E)

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

This question brings an important point to the fore – the correct option in strengthen/weaken question is the one that supplies new information but in most cases, the new information has to be a fact, not an opinion. Let’s explain this in detail with the help of this question.

**Question**: According to recent research, a blindfolded person whose nostrils have been pinched so that smelling is impossible will have great difficulty in differentiating a bite of an apple from a bite of a raw potato. This clearly demonstrates that taste buds are not the only sense organs involved in determining the taste of a piece of food.

Which of the following premises, is an assumption required by the argument?

(A) All people agree that an apple and a potato differ in taste.

(B) There are no other senses involved in tasting other than taste, smell, and sight.

(C) The word “taste” can be used to describe an experience that involves sight or smell or both.

(D) The research was based on experiments that were conducted on a broad spectrum of the general population.

(E) People who have been blindfolded and whose nostrils are pinched can differentiate a bite of an apple from a bite of an onion more easily than they can differentiate a bite of an apple from a bite of a raw potato.

**Solution**:

Argument:

– If you remove sight and smell, people will have great difficulty in differentiating a bite of an apple from a bite of a raw potato.

Conclusion: Taste buds are not the only sense organs involved in determining the taste of a piece of food.

We will look at the options one by one:

(A) All people agree that an apple and a potato differ in taste.

Note that usually, people’s opinion will not count for much. Facts are the ones which are important. The only opinion we care about is the author’s. We cannot strengthen/weaken the author’s opinion by giving similar/dissimilar opinions of other people.

Say, the conclusion of an argument is:

Daniel Day-Lewis is the greatest actor of the 21^{st} century.

The premises would perhaps list his great performances, talk about his acting prowess, his Oscars and so on.

Can you strengthen the conclusion by saying that “My friend also believes that he is the greatest actor.”? No. You cannot strengthen your opinion by giving the opinion of other people. You need to give facts to strengthen your view.

So this option is already suspect. It is giving you the opinion of people “All people agree that an apple and a potato differ in taste.” So it doesn’t seem to be the right choice.

Anyway, let’s try to negate (A) just to be sure since this is an assumption question.

Negation: Not all people agree that an apple and a potato differ in taste.

This means there is at least one person who does not agree that an apple and a potato differ in taste. Perhaps he feels that the experience of eating an apple – the smell, the look, the sweetness etc is the same as the experience of eating a potato. It is still possible that taste buds are not the only sense organs involved in determining the taste of a piece of food. Even after we negate (A), the conclusion is possible so (A) is not an assumption.

Think of it in another way: During the research, blindfolded people with pinched noses found it very hard to differentiate the taste. One person comes up and says that he himself cannot differentiate between the two while looking and smelling. Does it mean that senses other than taste buds are not involved? No. There could be many other people who feel that they can easily differentiate between an apple and a potato taste. So other senses could be involved and (A) is not your answer.

(B) There are no other senses involved in tasting other than taste, smell, and sight.

This is not an assumption. All we are saying is that taste buds are not the only sense organs involved in determining the taste of a piece of food. Any other organs could be involved including smell and sight.

(C) The word “taste” can be used to describe an experience that involves sight or smell or both.

This option highlights a very basic thing that needs to be true for our conclusion to hold. When we conclude: taste buds are not the only sense organs involved in determining the ** taste** of a piece of food, how do we define “taste”? Taste buds, we know, tell us whether the food is salty/sweet/sour etc. But how do we say that “taste” is not defined by only these features? We are assuming that taste is defined by not just how the food sits on our tongue but by other features such as sight/smell too. If this option were not true, then we would have needed only taste buds to find the taste of food. Hence, our conclusion would fall apart. Hence, (C) is the correct answer.

(D) The research was based on experiments that were conducted on a broad spectrum of the general population.

The conclusion does not say that for people of most classes/regions, taste buds are not the only sense organs involved in determining the taste of a piece of food. It is acceptable if the research was conducted on a few people and it was determined that other senses are involved. Even if some people found it difficult to differentiate between the two things, we can say that other senses are involved.

(E) People who have been blindfolded and whose nostrils are pinched can differentiate a bite of an apple from a bite of an onion more easily than they can differentiate a bite of an apple from a bite of a raw potato.

This option tells us that apples and onions are more different on the tongue than apples and potatoes. This is out of scope and is certainly not an assumption.

Answer (C)

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

When two positive numbers are equal, their Arithmetic Mean = Geometric Mean = The number itself

Say, the two numbers are m and n (and are equal). Their arithmetic mean = (m+n)/2 = 2m/2 = m

Their geometric mean = sqrt(m*n) = sqrt(m^2) = m (the numbers are positive so |m| = m)

We also know that Arithmetic Mean >= Geometric Mean

So when arithmetic mean is equal to geometric mean, it means the arithmetic mean is taking its minimum value. So when (m+n)/2 is minimum, it implies (m+n) is minimum. Therefore, sum of numbers takes its minimum value when the numbers are equal.

When geometric mean is equal to arithmetic mean, it means the geometric mean is taking its maximum value. So when sqrt(m*n) is maximum, it means m*n is maximum. Therefore, product of numbers takes its maximum value when the numbers are equal.

Let’s see how to solve a difficult question using this concept.

**Question**: If x and y are positive, is x^2 + y^2 > 100?

**Statement 1**: 2xy < 100

**Statement 2**: (x + y)^2 > 200

Solution:

We need to find whether x^2 + y^2 must be greater than 100.

Statement 1: 2xy < 100

Plug in some easy values to see that this is not sufficient alone.

If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100

If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100

So x^2 + y^2 may be less than or greater than 100.

Statement 2: (x + y)^2 > 200

There are two ways to deal with this statement. One is the algebra way which is easier to understand but far less intuitive. Another is using the concept we discussed above. Let’s look at both:

Algebra solution:

We know that (x – y)^2 >= 0 because a square is never negative.

So x^2 + y^2 – 2xy >= 0

x^2 + y^2 >= 2xy

This will be true for all values of x and y.

Now, statement 2 gives us x^2 + y^2 + 2xy > 200. The left hand side is greater than 200. If on the left we substitute 2xy with (x^2 + y^2), the left hand side will either become greater than or same as before. So in any case, the left hand side will remain greater than 200.

x^2 + y^2 + (x^2 + y^2) > 200

2(x^2 + y^2) > 200

x^2 + y^2 > 100

This statement alone is sufficient to say that x^2 + y^2 will be greater than 100. But, we agree that the first step where we start with (x – y)^2 is not intuitive. It may not hit you at all. Hence, here is another way to analyze this statement.

Logical solution:

Let’s try to find the minimum value of x^2 + y^2. It will take minimum value when x^2 = y^2 i.e. when x = y (x and y are both positive)

We are given that (x+y)^2 > 200

(x+x)^2 > 200

x > sqrt(50)

So x^2 + y^2 will take a value greater than [sqrt(50)]^2 + [sqrt(50)]^2 = 100.

So in any case, x^2 + y^2 will be greater than 100. This statement alone is sufficient to answer the question.

Answer (B)

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

The exciting thing is that pre-thinking is useful in Quant too. If you take a step back to review what the question asks and think about what you are going to do and what you expect to get, it is highly likely that you will not get distracted mid-way during your solution. Let’s show you with the help of an example:

**Question**: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

**Statement I**: Train B arrived at Newcastle before train A arrived at Birmingham.

**Statement II**: The distance between Newcastle and Birmingham is greater than 140 km.

Following are the things that would ideally constitute pre-thinking on this question:

– Quite a bit of data is given in the question stem with some speed and time taken.

– Distance traveled by both the trains is the same since they travel along the same route.

– We could possibly make an equation by equating the two distances and come up with multiple answers for the time at which train B arrived at Newcastle.

– The statements do not provide any concrete data. We cannot make any equation using them but they might help us choose one of the answers we get from the equation of the question stem.

Mind you, the thinking about the statements helping us to arrive at the answer is just speculation. The answer may well be (E). But all we wanted to do at this point was find a direction.

The diagram given above incorporates the data given in the question stem. Train A starts from Newcastle toward Birmingham at 3:00 and meets train B at 4:00. Train B starts from Birmingham toward Newcastle at 3:50 and meets train A at 4:00. Let x be the distance from Birmingham to the meeting point.

Speed of train A = 100 km/hr

Speed of train B = Distance/Time = x/(10 min) = x/(1/6) km/hr = 6x km/hr (converted min to hour)

If we get the value of x, we get the value of speed of train B and that tells us the time it takes to travel from the meeting point to Newcastle (a distance of 100 km). So all we need to figure out is whether the statements can give us a unique value of x.

By 4:00, train A has already travelled for 1 hour and train B has already travelled for 10 mins i.e. 1/6 hour. Total time taken by both is 2 hrs. The remaining (5/6) hrs is the time needed by both together to reach their respective destinations.

Time taken by train A to reach Birmingham + Time taken by train B to reach Newcastle = 5/6

Distance(x)/Speed of train A + 100/Speed of train B = 5/6

x/100 + 100/6x = 5/6

3x^2 – 250x + 5000 = 0

3x^2 – 150x – 100x + 5000 = 0

3x(x – 50) – 100(x – 50) = 0

(3x – 100)(x – 50) = 0

x = 100/3 or 50

So speed of train B = 6x = 200 km/hr or 300 km/hr

Statement 1: Train B arrived at Newcastle before Train A arrived at Birmingham.

If x = 50, time taken by train A to reach Birmingham = 50/100 = 1/2 hour and time taken by train B to reach Newcastle = 100/300 = 1/3 hour. Train B takes lesser time so it arrives first.

If x = 33.33, time taken by train A to reach Birmingham = (100/3)/100 = 1/3 hour and time taken by train B to reach Newcastle = 100/200 = 1/2 hour. Here, train A takes lesser time so it arrives first at its destination.

Since train B arrived first, x must be 50 and train B must have taken 1/3 hour i.e. 20 mins to arrive at Newcastle. So train B must have arrived at 4:20.

This statement is sufficient alone.

Statement 2: The distance between Newcastle and Birmingham is greater than 140 km.

Total distance between Newcastle and Birmingham = (100 + x) km. x must be 50 to make total distance more than 140.

Time taken by train B must be 1/3 hr (as calculated above) and it must have arrived at 4:20.

This statement is sufficient alone.

Answer (D)

So our speculation was right. Each of the statements provided us relevant information to choose one of the two values that the quadratic gave us.

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

A few months back, we had discussed a 700 level ‘Races’ question.

**Question 1**: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately.

(A) 8, 10

(B) 4, 5

(C) 5, 9

(D) 6, 9

(E) 7, 10

Check out its complete solution here.

Now, what if we had only 30 seconds to guess on it and move on? Then we could have easily guessed (B) here and moved on. Actually, the question implies that the only possible options are those in which the time taken by B is somewhere between 3 mins and 6 mins (excluding) – we would guess 4 mins or 5 mins. Since only option (B) has time taken by (B) as 5 mins, that must be the answer – no chances of error here – perfect! Had there been 2 options with 4 mins/5 mins, we would have increased the probability of getting the correct answer to 50% from a mere 20% within 30 seconds.

Now you are probably curious as to how we got the 3 min to 6 min range. Here is the logic:

Read one sentence of the question at a time –

“*A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds.*”

So first, A gives B a head start of 1/10th of the race but still beats him. This means B is certainly quite a bit slower than A. This should run through your mind on reading this sentence.

“*Next, A gives B a head start of 3 mins and is beaten by 1000 m.*”

Next, A gives B a head start of 3 mins and B beats him by 1000 m i.e. half of the race. What does this imply? It implies that B ran more than half the race in 3 mins. To understand this, say B covers x meters in 3 mins. Once A, who is faster, starts running, he starts reducing the distance between them since he is covering more distance than B every second. At the end, the distance between them is still 1000 m. This means the initial distance that B created between them by running for 3 mins was certainly more than 1000 m (This was intuitively shown in the diagram in this post). Since B covered more than 1000 m in 3 mins, he would have taken less than 6 mins to cover the length of the race i.e. 2000 m. A must be even faster and hence would take even lesser time.

Only option (B) has time taken by B as 5 mins (less than 6 mins) and hence satisfies our range! So the answer has to be (B).

Let’s try the same technique on another question.

**Question 2**: If 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days, how long will 7 men and 10 women take to do it?

(A) 4.2 days

(B) 6.8 days

(C) 8.3 days

(D) 9.8 days

(E) 10.2 days

Solution: If we try to use algebra here, the calculations involved will be quite complicated. The options are not very close together so we can try to get a ballpark value and move forward. Let’s take each sentence at a time:

“*If 12 men and 16 women can do a piece of work in 5 days*”

Say rate of work of each man is M and that of each woman is W. This statement gives us that

12M + 16W = 1/5 (Combined rate done per day)

In lowest terms, it is 3M + 4W = 1/20

“*13 men and 24 women can do it in 4 days*,”

This gives us 13M + 24W = 1/4

“*how long will 7 men and 10 women take to do it?”*

Required: 7M + 10W = ?

Solving the two equations above will be tedious so let’s estimate:

(3M + 4W = 1/20) * 6 gives 6M + 8W = 1/10

So 6 men and 8 women working together will take 10 days. Hence, 7 men and 10 women will certainly take fewer than 10 days.

(13M + 24W = 1/4) / 2 gives 6.5M + 12W = 1/8

So 7 men and 10 women might take about 8 or perhaps a little bit more than 8 days to complete the work. There is only 0.5 additional man (hypothetically) but 2 fewer women to complete it. So we would guess that the number of days would lie between 8 to 10 and closer to 8 days.

Answer (C) fits.

Note that it seems like there are many equations here but all you have actually done is made two equations. Once you write them down, you don’t even need to actually multiply them with some integer to get them close to the required equation. Just looking at the first one, you can say that 6 men and 8 women will take 10 days. It takes but a couple of seconds to arrive at these conclusions.

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

In your haste to complete the test on time, don’t overlook the important details. Getting too many easy questions wrong is certainly disastrous. Take a step back and ensure that what they asked is what you have found and that your logic is solid. To illustrate the problem, let’s give you a question – people gloss over it, consider it an easy remainders problem, answer it incorrectly and move on. But guess what, it isn’t as easy as it looks!

**Question**: If m and n are positive integers such that m > n, what is the remainder when m^2 – n^2 is divided by 21?

**Statement 1**: The remainder when (m + n) is divided by 7 is 1.

**Statement 2**: The remainder when (m – n) is divided by 3 is 1.

First let’s give you the **incorrect** solution provided by many.

Question: What is the remainder when (m^2 – n^2) is divided by 21?

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

Therefore, remainder of product (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) when it is divided by 21 is 1.

Answer (C)

This would have been correct had the statements been:

Statement 1: The remainder when (m + n) is divided by 21 is 1.

Statement 2: The remainder when (m – n) is divided by 21 is 1.

Statement 1: (m + n) = 21a + 1

Statement 2: (m – n) = 21b + 1

(m^2 – n^2) = (m + n)*(m – n) = (21a + 1)*(21b + 1) = 21*21ab + 21a + 21b + 1

Here, every term is divisible by 21 except the last term 1. So when we divide (m^2 – n^2) by 21, the remainder will be 1.

But let’s go back to our original question. If you solved it the way given above and got the answer as (C), you are not the only one who jumped the gun. Many people end up doing just that. But here is the correct solution:

The statements given are:

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

This gives us (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) = 21ab + 7a + 3b + 1

Here only the first term is divisible by 21. We have no clue about the other terms. We cannot say that 7a is divisible by 21. It may or may not be depending on the value of a. Similarly, 3b may or may not be divisible by 21 depending on the value of b. So how can we say here that the remainder must be 1? We cannot. We do not know what the remainder will be in this case even with both statements together.

Say, if a = 1 and b = 1,

m^2 – n^2 = 21*1*1 + 7*1 + 3*1 + 1 = 21 + 11

The remainder when you divide m^2 – n^2 by 21 will be 11.

Say, if a = 2 and b = 1,

m^2 – n^2 = 21*2*1 + 7*2 + 3*1 + 1 = 21*2 + 18

The remainder when you divide m^2 – n^2 by 21 will be 18.

Hence, both statements together are not sufficient to answer the question.

Answer (E)

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

Today, let’s bring back the beloved lazy genius through a question. Try to solve it lazily i.e. try to do minimum work on paper. This means making equations and solving them is a big no-no and doing too many calculations is cumbersome.

**Question**: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)

(B) yz/(yz + xz – xy)

(C) yz/(yz + xz + xy)

(D) xyz/(yz + xz – xy)

(E) (yz + xz – xy)/yz

Note that you have variables in the question and the options. Since we are looking for a lazy solution, making equations out of the variables is not acceptable. So then, should we plug in numbers? With three variables to take care of, that might involve a lot of calculations too. Then what else?

Here is our minimum-work-solution to this problem; try to think one of your own and don’t forget to share it with us.

Plugging in numbers for the variables can be troublesome but you can give some very convenient values to the variables so that the effect of a pump and a drain will cancel off.

There are no constraints on the values of x, y and z except z > x (drain Gamma empties slower than pipe Alpha fills)

Let’s say, x = 2 hrs, y = 4 hrs, z = 4 hrs

What did we do here? We made the rate of Beta same as the rate of Gamma i.e. 1/4 of the tank each. This means, whenever both of them are working together, drain Gamma cancels out the work of pump Beta. Every hour, pump Beta fills 1/4^{th} of the tank and every hour drain Gamma empties 1/4^{th} of the tank. So the entire tank will be filled by pump Alpha alone. Hence, if y = z, pump Alpha fills the entire tank i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1.

In the options, put y = z and see which option gives you 1. Note that you don’t have to put in the values of 2, 4 and 4. We gave those values only for illustration purpose.

If y = z, xy = xz.

So in option (B), xz cancels xy in the denominator giving yz/yz = 1

Again, in option (E), xz cancels xy in the numerator giving yz/yz = 1

The other options will not simplify to 1 even though when we put y = z, the answer should be 1 irrespective of the value of x, y and z. The other options will depend on the values of x and/or y. Hence the only possible options are (B) and (E). But we still need to pick one out of these two.

Now let’s say, x = 4, y = 2, z = 4.00001 ( z should be greater than x but let’s assume it is infinitesimally greater than x such that we can approximate it to 4 only)

Rate of work of Gamma (1/4^{th} of the tank per hour) is half the rate of work of Beta (1/2 the tank per hour). Rate of work of Gamma is same as rate of work of Alpha. Half the work done by pump Beta is removed by drain Gamma. So if pump Beta fills the tank, drain Gamma empties half of it in that time – the other half would be filled by pump Alpha i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1/2.

Put x = z in the options (B) and (E). The one that gives you 1/2 with these values should be the answer. Again, you don’t need to plug in the actual values till the end.

If x = z, yx = yz

(B) yz/(yz + xz – xy)

yz cancels xy in the denominator giving us yz/xz = y/x = 2/4 = 1/2

(E) (yz + xz – xy)/yz

yz cancels xy in the numerator giving us xz/yz = x/y = 4/2 = 2

Only option (B) gives 1/2. Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. If you do, you are on your way to becoming a lazy genius yourself!

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

Let’s recap the concepts before we see the question:

I. When you divide a number by 100, the remainder is formed by the last two digits of the number.

II. When you divide (px + a)(qx + b)*…*(tx + e) by x, the remainder will be the remainder obtained by dividing a*b*…*e by x.

III. We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

These three were discussed with examples last week.

IV. When m is divided by n and a negative remainder (–r) is obtained, we can find the actual remainder simply as (n – r).

This is discussed with examples in this post.

Now, let’s solve a question involving all these concepts.

**Question**: What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96

(B) 76

(C) 56

(D) 36

(E) 16

**Solution**: We need to find the last two digits of the product. It means we need to find the remainder when we divide the product by 100.

Find remainder of (301*402*503*604*646*547*448*349)^2/100

Note that 301 = 300 + 1 which gives us a small remainder 1 to work with but 349 = 300 + 49, a large remainder with which calculations will become cumbersome. But note that 349 is close to 350. All the numbers in the product are quite close to a multiple of 50, if not to a multiple of 100.

We need to find the remainder of:

(301*402*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/100

This implies we need to find the remainder of:

(301*201*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/50

We cancel off a 2 (of 402) from the numerator with a 2 of the denominator to make the divisor 50. We will multiply the remainder we obtain by 2 back at the end.

We need the remainder of:

(300 + 1)*(200+1)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)* (300+ 1)*(400+2)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)/50

Note that in this product, all terms obtained will have a multiple of 50 except the last term obtained by multiplying the remainders together:

We need:

the remainder of 1*1*3*4*(-4)*(-3)*(-2)*(-1)*1*2*3*4*(-4)*(-3)*(-2)*(-1) / 50

the remainder of (12)*(24)*(24)*(24) / 50

Notice now that the remainders are far too great but they are close to 25. So let’s bring the divisor down to 25 by canceling off another 2.

We need:

the remainder of (6)*(24)*(24)*(24) / 25

the remainder of (6)*(25-1)*(25-1)*(25-1) / 25

Again, the only product we need to worry about is the last one obtained by multiplying the remainders together:

the remainder of 6*(-1)*(-1)*(-1) / 25

The remainder is -6 which is negative. To get the positive remainder, 25 – 6 = 19. But remember that we had divided the divisor twice by 2 so to get the actual remainder, we must multiply the remainder obtained back by 4: the actual remainder is 4*19 = 76

Answer (B)

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