Advanced Applications of Common Factors on the GMAT

Quarter Wit, Quarter WisdomToday we will discuss the logic behind common factors (other than 1) of two numbers.

Without actually finding all the factors of two numbers, how do we know whether they have any common factors (ignoring 1)?

Let’s take some examples:

  • If the integers are even, we know that they must have at least one common factor – 2. Let’s say we have two numbers 476 and 478. How many common factors can they have? We know that 2 is a factor common to them. Can they have any other common factor? Note that the difference between them is 2. So if 4 were a factor of 476, could it be a factor of 478? No. If 4 were a factor of 476, it would be a factor of 480 next (4 away from 476). Similarly, if 7 were a factor of 476, it would not be a factor of 478, definitely. It would be a factor of 483 (7 away from 476). In fact, since the difference between the two numbers is 2, the only factor they can have in common is 2.
  • Now consider that the two numbers are 476 and 484. They have a difference of 8 between them. The common factors they can have are 2, 4 and 8 (the factors of 8). If any of these factors is a factor of 476, it will be a factor of 484 too. Obviously, 476 and 484 will have many other factors but they will not have any other common factor. 7 is a factor of 476. The next multiple of 7 will be 483 and the next will be 490. 7 cannot be a factor of 484.
  • What happens when both integers are odd? Say 525 and 531. The difference between them is 6. The factors of 6 are 2 and 3. Both 525 and 531 are odd numbers so 2 cannot be their factor. If 3 is a factor of 525, it will be a factor of 531 too else it will not be a factor of both the numbers. Any other number can be a factor of one of them, but not both.

This is what we can deduce:

The only factors that CAN be common (it’s not necessary that they will be common) between two numbers are the factors of the difference between them.

If any factor of the difference between them is a factor of one of the numbers, it will be a factor of the other number too. If it is not a factor of one number, it will not be a factor of the other number.

Take a look at a question based on these concepts:

Question: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

Statement 1: m^2 – 10m + 16 = 0

Statement 2: x + 26 is a prime number.

Solution:

The two given positive integers are (x + m) and (x – m). x is a positive integer so m must be an integer too. Whether m is positive or negative, we don’t know.

To know the GCD of two numbers, we need to know their common divisors. As of now, we have no idea about their common divisors, but we know that the difference between the two numbers is 2m. Their common factors must be factors of 2m.

Let’s look at the two statements:

Statement 1: m^2 – 10m + 16 = 0

We know that the quadratic will give us two values for m so we will not be able to find a unique value for m. But let’s solve it in case we get some other clues from it.

m^2 – 10m +16 = 0

m^2 – 2m – 8m + 16 = 0

m (m – 2) – 8 (m – 2) = 0

(m – 2)*(m – 8) = 0

m is either 2 or 8. So 2m is either 4 or 16.

The factors of 2m will be 1, 2 and 4 and additionally, 8 and 16 (if 2m is 16). We have no idea whether x+m and x-m will have these factors so this statement alone is not sufficient.

Statement 2: x + 26 is a prime number.

What does it tell us about x? Other than 2, all prime numbers are odd numbers. Since x is a positive integer,  x+26 cannot be 2. It must be a prime number greater than 2 and hence, must be odd. But 26 is even. So x must be an odd integer (Odd + Even = Odd). But we have no information about m so this statement alone is not sufficient.

Using both statements together, since x is an odd integer and m is definitely even (either 2 or 8), both the numbers (x + m) and (x – m) are odd integers. Odd integers will not have any of these factors: 2, 4, 8, 16.

So (x + m) and (x – m) must have 1 as the only common factor. Hence their greatest common divisor must be 1.

Together, the two statements are sufficient to answer the question.

Answer (C)

To recap: Any common factor of two numbers has to be a factor of the difference between them. This also implies that the GCD of two numbers has to be a factor of the difference between them.

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

Is this GMAT Question an Alphametic or Simple Number Properties Question?

Quarter Wit, Quarter WisdomAs noticed in the first post of Alphametics, a data sufficiency alphametic is far more complicated than a problem solving alphametic. An alphametic can have multiple solutions and establishing that it does not, is time consuming. Hence, it is less likely that you will see a DS alphametic in the actual exam.

In fact, what may look like an alphametic problem, might actually be a number properties problem only.

We will look at an example below:

Question:

AlphameticPost3

In the correctly-worked multiplication problem above, each symbol represents a different nonzero digit. What is the value of C?

Statement 1: D is prime.

Statement 2: B is not prime.

Solution: We multiply two two-digit integers and get 1995. The good thing is that we know the result of the multiplication will be 1995. Usually, multiplication alphametics are harder since they involve multiple levels, but here the multiplication is actually a blessing. There are many many ways in which you can ADD two integers to give 1995 but there are only a few ways in which you can multiply two integers to give you 1995.

Let’s prime factorize 1995:

1995 = 3*5*7*19

We can probably count on our fingers the number of ways in which we can select AB and CD.

19 needs to be multiplied with one other factor to give us a two digit number since 5*3*7 = 105 (a three digit number) so AB and CD cannot be 19 and 105.

19*3 = 57, 5*7 = 35 – This is not possible since two of the four digits are same here – 5.

19*5 = 95, 3*7 = 21 – This is one option for AB and CD.

19*7 = 133 – Three digit number not possible.

Hence AB and CD can only take values out of 21 and 95.

As of now, C can be 2 or 9. We need to find whether the given statements give us a unique value of C.

Statement 1: D is prime

D is the units digit of CD. So D can be 1 or 5.

1 is not prime so CD cannot be 21. Hence, CD must be 95 and AB must be 21.

Hence, C must be 9.

This statement alone is sufficient.

Statement 2: B is not prime

If B is not prime then AB cannot be 95. Hence AB must be 21.

This means CD will be 95 and C will be 9.

This statement alone is sufficient.

Answer (D)

Note that the entire question was just about number properties – prime factors, prime numbers etc. Actually it required no iterative steps and no hit and trial. Rest assured that if it is a GMAT question, it will be reasoning based and will not require painful calculations.

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

How to Solve Alphametic Multiplication Questions on the GMAT

Quarter Wit, Quarter WisdomLast week, we looked at alphametics involving addition and subtraction. The logic becomes a little more involved when the alphametic involves multiplication. When a two digit number is multiplied by another two digit number, the process of finding the result is composed of multiple levels. Today, let’s see how to handle those multiple levels.  The question involves quite a few steps and observations using number properties. Hence, you are unlikely to see such a question in actual GMAT but you might see a simpler version so it’s good to be prepared.

Question: The following alphametic shows multiplication of two numbers, IF and DR. The product you obtain is DORF.

AlphameticPost2Fig1

What is the value of D + O + R + F?
(A) 17
(B) 20
(C) 22
(D) 23
(E) 30

Solution: The good thing is that we know D + O + R + F has a single value. This means there will be a logic to obtain the value of each of D, O, R and F.

As discussed last week, we first focus on the big picture, but we will have to go one level at a time.

(i) IF * R = OFF

(ii) IF * D = IF

(iii) OF + IF = DOR

A few interesting points to note from the above:

– From (ii), IF * D = IF

When you multiply IF by D, you get IF itself. This means that D must be 1. D can take no other value.

D = 1

– From (iii), F + F has unit’s digit of R.

Also O + I gives O as unit’s digit and 1 as tens digit (D of DORF obtained from above). How can this happen? Say, if O = 4, 4 + I = 14. This is possible only when I = 9 and there is a 1 carry over from the previous addition of F + F. This means that F must be 5 or greater to have a carryover of 1. It cannot be 5 because 5+5 will give you 10 making R = 0. This would mean that F*R would end in R (0). But in (i), F * R has unit’s digit of F, not R. So F cannot be 5.

D = 1, I  = 9

– Another interesting point: From (i), F * R has unit’s digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so)

Since F has to be greater than 5 (as seen above), R must be 6.
If R = 6, then from (iii), F + F has unit’s digit of 6 and a carryover of 1 so F = 8. When you add 8 + 8, you will get 16 (units digit of 6 and a carryover)

D = 1, I = 9, R = 6, F = 8

– From (i), when we multiply IF by R, we get OFF. That is, when we multiply 98 by 6, we get 588. So O must be 5.

This gives us: D + O + R + F = 1 + 5 + 6 + 8 = 20

Answer (B)

This question uses your understanding of numbers and how they are added and multiplied. It certainly takes time to get to the answer. Such questions can help you get a feel for numbers and their behavior.

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

An Introduction to Solving Alphametic Questions on the GMAT

Quarter Wit, Quarter WisdomToday, let’s learn how to solve alphametics. An alphametic is a mathematical puzzle where every letter stands for a digit from 0 – 9. The mapping of letters to numbers is one-to-one; that is, the same letter always stands for the same digit, and the same digit is always represented by the same letter.

First focus on the big picture of the alphametic – such as, a two number is added to another two digit number to give a three digit number etc. Then look at the nitty gritty – for which digit can each letter stand?

Question 1: With # and & each representing different digits in the problem below, the difference between #&& and ## is 667. What is the value of &?

Alphametic0

(A) 3
(B) 4
(C) 5
(D) 8
(E) 9

Solution: The big picture: A two digit number is subtracted from a three digit number to give 667. So the three digit number must be a bit larger than 667. This means that the hundreds digit of #&& must be either 6 or 7. It cannot be 8 because you cannot obtain 800+ by adding a two digit number to 667.

Let’s look at both cases:

# is 6: If you subtract 66 from 6&&, you will not get 667 – the largest value you can get is 699 – 66 = 633. So # cannot be 6.

# must be 7.

Now the question is very simple
7&& – 77 = 667
7&& = 667 + 77 = 744

Answer (B)

There are many other ways in which you can solve this question including plugging in the answer choices. We should now take a look at a DS question on alphametics.

Question 2:

Alphametic

In the correctly worked addition problem above, M, N, P, R, S, T and V are distinct digits. Is R > 3?

Statement 1: M, N and P are positive even integers.

Statement 2: S = 2

Solution: This is certainly harder than the PS question but our process will remain the same.

First, let’s see what information we are given in the question – the units digits of all three numbers are the same. The two-digit numbers add up to give a three digit number. The hundreds digit, S, is either 1 or 2. Three two-digit numbers cannot add up to give a number 300 or more since 99 + 99 + 99 = 297. We have no information on what the value of R can be. All we know is that R cannot be 0 because 0+0+0 = 0 but V needs to be different from R.

Let’s look at the statements now.

Statement 1: M, N and P are positive even integers.

At first, it may seem that this has nothing to do with the value of R but we must analyze what is given to be sure.

M, N and P must take distinct values out of 2, 4, 6 and 8 and add up to give the units digit of T (again, distinct)

Every time you add three even numbers, you will get an even number. Let’s see which combinations we can get:

2 + 4 + 6 = 12

2 + 4 + 8 = 14

2 + 6 + 8 = 16

4 + 6 + 8 = 18

Note that in all four cases, the units digit is one of the numbers but T must be distinct. This means that there must have been a carryover from the previous addition. So when we added the three Rs, we must have got a carryover. Had R been 3 or less, we would not have got a carryover since 1+1+1 = 3, 2+2+2 = 6 and 3+3+3 = 9. So R must be greater than 3.

One such case would be

Alphametic1

This statement alone is sufficient.

Statement 2: S = 2

The result of addition gives us a number which is more than 200. In statement 1 we saw a case in which S is 2 and R is greater than 3. Now all we have to do is find a case in which S is 2 and R is less than 3. One of these cases is

Alphametic2

So this statement alone is not sufficient.

Answer (A)

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

How to Solve Advanced Compound Interest Questions on the GMAT

Quarter Wit, Quarter WisdomWe have discussed simple and compound interest in a previous post.

We saw that simple and compound interest (compounded annually) in the first year is the same. In the second year, the only difference is that in compound interest, you earn interest on previous year’s interest too. Hence, the total two year interest in compound interest exceeds the two year interest in case of simple interest by an amount which is interest on year 1 interest.

So a question such as this one is very simple to solve:

Question 1: Bob invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 2 years at the same rate of interest and received $605 as interest. What was the annual rate of interest?

(A) 5%
(B) 10%
(C) 12%
(D) 15%
(E) 20%

Solution:

Simple Interest for two years = $550

So simple interest per year = 550/2 = 275

But in case of compound interest, you earn an extra 605 – 550 = $55

This $55 is interest earned on year 1 interest i.e. if rate of interest is R, it is

55 = R% of 275

R = 20

Answer (E)

The question is – what happens in case you have 3 years here, instead of 2? How do you solve it then? Here is a small table of the difference between simple and compound interest to help you.

Say the Principal is P and the rate of interest if R

Compound Interest

It gets a bit more complicated though not very hard to solve. All you need to do is solve a quadratic, which, if the values are well thought out, is fairly simple to solve. Let’s look at the same question adjusted for three years.

Question 2: Bob invested one half of his savings in a bond that paid simple interest for 3 years and received $825 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 3 years at the same rate of interest and received $1001 as interest. What was the annual rate of interest?

(A) 5%
(B) 10%
(C) 12%
(D) 15%
(E) 20%

Simple Interest for three years = $825

So simple interest per year = 825/3 = $275

But in case of compound interest, you earn an extra $1001 – $825 = $176

What all is included in this extra $176? This is the extra interest earned by compounding.

This is R% of interest of Year1 + R% of total interest accumulated in Year2

This is R% of 275 + R% of (275 + 275 + R% of 275) = 176

(R/100) *[825 + (R/100)*275] = 176

Assuming R/100 = x to make the equation easier,

275x^2 + 825x – 176 = 0

25x^2 + 75x – 16 = 0

25x^2 + 80x – 5x – 16 = 0

5x(5x + 16) – 1(5x + 16) = 0

x = 1/5 or -16/5

Ignore the negative value to get R/100 = 1/5 or R = 20

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!

Taking the Best Approach When Solving Data Sufficiency GMAT Questions

Quarter Wit, Quarter WisdomSometimes, we get questions which we cannot neatly bracket as Arithmetic/Algebra/Geometry etc. In fact, the higher level questions usually focus on more than one subject area.  The trick in these questions is to assimilate all your knowledge from various areas and then think how best to solve.

Today we have such a question for you – you could get really lost in it or could solve it in a few seconds if you take the right track. The trick is starting on the right track and that is why you have 2 mins per question available to you else 40 secs per question would have been sufficient!

Question: This game season, five divisions are going to play. Out of all the teams in each division, 6, 9, 12, 13 and 14 teams have qualified from the respective divisions. Each division will hold its own tournament – where a team is eliminated from the tournament upon losing two games – in order to determine its champion. The five division champions will then play in a knock-off tournament – a team is eliminated as soon as it loses a game – in order to determine the overall champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?

(A) 89

(B) 100

(C) 102

(D) 107

(E) 112

Solution: Is it a max-min problem? Perhaps, but which guiding principle of max-min will we use to solve this problem? First think on your own how you will solve this problem.

Will you focus on the method or the result i.e. will you worry about who plays against whom or just focus on each result which gives one loss and one win? If you don’t worry about the method and just focus on the result, you can use a concept of mixtures here. In mixture questions, we focus on one component and how it changes. Here, we need to keep track of losses. Let’s focus on those and forget about the wins. As given, there were no ties so every loss has a win on the other side.

Every time a game is played, someone loses. We can give at most 2 losses to a team since after that it is out of the tournament. Don’t worry about against who it plays those two games. Whenever a team loses 2 games, it is out. The team could have won many games but we are not counting the wins and hence, are not concerned about its wins. As discussed, we are counting the losses so each win of that team will be counted on the other side i.e. as a loss for the other team.

Consider the division which has 6 teams – what happens when 12 games are played? There are 12 losses and each team gets 2 losses (we can’t give more than 2 to a team since the team gets kicked out after 2 losses), so all are out of the tournament. But we need a winner so we play only 11 games so that the winning team gets only 1 loss. We want to maximize the losses (and hence the number of games), therefore the winning team must be given a loss too.

So maximum number of games that can be played by the district in its own tournament = 2*6 – 1 = 11

Similarly, the division with 9 teams can play at most 2*9 – 1 = 17 games.

The division with 12 teams can play at most 2*12 – 1 = 23 games.

The division with 13 teams can play at most 2*13 – 1 = 25 games.

The division with 14 teams can play at most 2*14 – 1 = 27 games.

This totals up to 11 + 17 + 23 + 25 + 27 = 103 games

Now we come to the games between the district champions.

We have 5 teams. 1 loss gets a team kicked out. If the teams play 4 games, there are 4 losses and 4 teams get kicked out. We have a final winner!
Hence the total number of games = 103 + 4 = 107

There are a lot of variations you can consider for this question.

Say, if we need to minimize the number of games, how many total games would have been played?

Notice that the only games you can avoid are the ones in which the 5 district champions lost. You do still need 2 losses for each team to get the district champion and one loss each for four district champions to get the winner. Hence, at least 107 – 5 = 102 games need to be played.

Look at it in another way:

To kick out a team, it needs to have 2 losses so if the district had 6 teams, there would be 5*2 = 10 games played.

Similarly, the division with 9 teams will play at least 2*8 = 16 games.

The division with 12 teams will play at least 2*11 = 22 games.

The division with 13 teams will play at least 2*12 = 24 games.

The division with 14 teams will play at least 2*13 = 26 games.

This totals up to 10 + 16 + 22 + 24 + 26 = 98 games

Now we have 5 champions and they will need to play at least 4 games to pick a winner.

Therefore, at least 98 + 4 = 102 games need to be played.

You can try other similar variations – what happens when a team is kicked out after it loses 3 games instead of 2? What happens if you don’t have the knock-off tournament and instead need each district champion to lose 2 games to get knocked out?

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

Identifying and Correcting Run-On Sentences on GMAT Verbal Questions

Quarter Wit, Quarter WisdomOn the GMAT, most sentence correction questions involve compound/complex sentences with multiple phrases, clauses and modifiers. Hence it is very likely that you will see some run-on sentences on your test. In the complicated sentences that we get on the GMAT, it is very easy to overlook that we are dealing with run-on sentences.

A run-on sentence has at least two independent clauses which are not connected properly. There are various ways in which a sentence may be run-on. Here are some of the most common circumstances:

  1. When an independent clause gives a suggestion/advice/command based on what was said in the prior independent clause:

GMAT is a very tricky test, you should work hard.

Here, we should either split the two clauses into two sentences by putting in a full stop or we should put a semi colon between the two clauses.

  1. When two independent clauses are connected by a conjunctive adverb such as however, moreover, nevertheless.

My grandmother is supposed to travel tomorrow, however, she is not feeling well.

Here, we should either split the two clauses into two sentences by putting in a full stop or we should put a semi colon between the two clauses

To read more on conjunctive adverbs, check out this post.

  1. When the second of two independent clauses contains a pronoun that connects it to the first independent clause.

Marcy is thrilled, she got permission to go to the school dance.

Although these two clauses are quite brief, and the ideas are closely related, this is a run-on sentence. We need to put a full stop or a semi colon in place of the comma.

Now that we have an idea of what run-on sentences are, let’s look at a GMAT Prep question where this concept is tested extensively.

Question: The Anasazi settlements at Chaco Canyon were built on a spectacular scale with more than 75 carefully engineered structures, of up to 600 rooms each, were connected by a complex regional system of roads.

(A) with more than 75 carefully engineered structures, of up to 600 rooms each, were

(B) with more than 75 carefully engineered structures, of up to 600 rooms each,

(C) of more than 75 carefully engineered structures of up to 600 rooms, each that had been

(D) of more than 75 carefully engineered structures of up to 600 rooms and with each

(E) of more than 75 carefully engineered structures of up to 600 rooms each had been

Solution:

Consider option (A): Remove all the unnecessary elements and get the skeleton of the sentence (primarily the subject and the verbs):

The settlements were built on a spectacular scale…, were connected …

The action verb “were connected” has no subject here. If it were to have the same subject as the first clause “the Anasazi settlements”, then there should have been a conjunction joining the two clauses together. This is a run-on sentence.

Consider option (B): The problem of run-on sentence has been rectified here by using past participle instead.
The settlements were built on a spectacular scale with more than 75 carefully engineered structures, of up to 600 rooms each, connected by a complex regional system of roads.

Remove the non essential modifier “of up to 600 rooms each” and you see that the 75 carefully engineered structures were the ones connected by a complex system of roads. Now it all makes sense.

To read more about participles, check this post.

Let’s look at the other options too.

Option (C): You cannot say “built on a spectacular scale of more than 75 structures”. You need “with” instead of “of”. The same problem exists with options (D) and (E) too.

Also, in option (C), the use of past perfect “had been” is not justified.

Option (D): As discussed above, “scale of” is incorrect in option (D).
It is also illogical “and with each connected” doesn’t clarify what of each is connected by roads.

Option (E): As discussed above, “scale of” is incorrect in option (E).

Also, it is a run-on sentence.

The settlements were built on a spectacular scale of more than 75 carefully engineered structures of up to 600 rooms each had been connected by a complex regional system of roads.

The two different clauses do not even have a comma in between here. Also, the use of past perfect “had been” is not justified.

Hope you understand run-on sentences a little better 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!

Playing the Devil’s Advocate on the GMAT

Quarter Wit, Quarter WisdomConfess it – while watching Harvey Specter and Mike Ross on ‘Suits’, many of you have wondered how ‘cool’ it would be to be a lawyer. It’s surprising how they question every assumption, every reason and come up with an innovative solution which looks as if the magician just pulled a rabbit out of a hat.

Well, in high level GMAT questions, you have a chance to play the Devil’s Advocate. If your best thought out logic says that answer has to be 2, still think why it cannot be 1. The higher level questions are quite tricky and if you play at 700+ level, you will need to be extra careful – if it seems too easy, it probably is! To illustrate, we have quite a brilliant little question from GMAT Prep.

Question 1: If 5x^2 has two different prime factors, at most how many different prime factors does x have?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Solution: So here is the logic with which most of us would come up – 5x^2 has two different prime factors – one would be 5 since it is already there so x must have one more prime factor. x^2  has only those prime factors that x has so 5x^2 will have two prime factors – one 5 and the other from x. Sounds perfectly reasonable and the answer should be 1 – x has 1 prime factor. In fact, it must have at least one prime factor and it cannot have more than one prime factor.

But then, and here we have a hint in the question to play the devil’s advocate – why does the question ask “at most how many different prime factors”. If there were a single value for different prime factors of x, the question would have probably said “how many different prime factors…”. There would have been no need for those words ‘at most’.

Then look at the other options. Is it possible that x has 2 prime factors? It certainly cannot have more than 2 distinct prime factors since then, 5x^2 will have more than two distinct prime factors. Actually, x can have two prime factors! x can have 5 as a factor too. Sneaky – eh? We already have a 5 in 5x^2 but that doesn’t mean that we cannot have a 5 in x^2 too. It will still count as a single prime factor. x can have another prime factor such as 2 or 3 or 7 or 11 etc. In that case, x can have two distinct prime factors.

So x can be 15 (two different prime factors) and x can be 25 (one prime factor)

Answer (B)

Note that this question has no calculations and no time consuming equations but still, this little trick makes this question quite hard. If most people get it wrong because of missing this trick, the question will be termed hard.

Now here is a trickier version of this question:

Question 2: How many prime factors does positive integer n have?

Statement 1: n/7 has only one prime factor.
Statement 2: 3*n^2 has two different prime factors.

Solution: Let’s keep in mind our learning from above while trying to solve this question.

Statement 1: n/7 has only one prime factor.

n/7 has a factor so obviously, it is an integer. Hence n must have a 7 as a factor. So we might jump to the conclusion that n has two prime factors –7 and another one which is left when n is divided off by 7. So n would be something like 7*3 so that n/7 = 7*3/7 = 3 (only one prime factor).

But what we wouldn’t have considered in this case is that n may have multiple 7s so that when a 7 is cancelled in n/7, you would still be left with 7 i.e. if n is 7*7, then n/7 = 7*7/7 = 7. In this case, n has only one distinct prime factor.

So n can have either one or two prime factors. This statement alone is not sufficient.

Statement 2: 3*n^2 has two different prime factors.

This is the same as our previous question. 3n^2 has two different prime factors but n itself can have either one or two prime factors (one of which will be 3). For example, n can be 7 or n can be 3*7. This statement alone is not sufficient.

Using both statements, n could have one or two prime factors i.e. n could be 49 (only one prime factor – 7) or n could be 21 (two prime factors).

Hence, even using both the statements, we cannot say how many prime factors n has.

Answer (E)

Now this question might not have seemed very complicated since we discussed the logic in the first question above. Remember to play the devil’s advocate in high level questions.

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

A Closer Look at GMAT Function Questions

Quarter Wit, Quarter WisdomLast week, we looked at the basics of how to handle function questions. Today, let’s look at a couple of questions. We will start with an easier one and then go on to a slightly tougher one.

Question 1: If f(x) = 343/x^3, what is the value of f(7x)* f(x/7) in terms of f(x)?

(A)  f(x^2)

(B) (f(x))^2

(C) f(x^3)

(D) (f(x))^3

(E) f(343x)

Solution: We discussed that to get f(a) given f(x), all you need to do is substitute x with a.

f(x) = 343/x^3

f(7x) = 343/(7x)^3 = 1/x^3

f(x/7) = 343/(x/7)^3 = 343*343/x^3

So we get f(7x) * f(x/7) = (1/x^3) * (343*343/x^3) = (343/x^3)^2

But we know that 343/x^3 = f(x)

So, f(7x) * f(x/7) = (f(x))^2

Answer(B)

There are other ways of solving this too:

Say x = 1, then f(1) = 343

f(7x)* f(x/7) = f(7)*f(1/7) = (343/7^3) * (343/(1/7)^3 = (343)^2

So f(7x)* f(x/7) = (f(1))^2

We hope you see that the question was not difficult to solve. Once you get over your fear of symbols, it is quite straight forward.

Now, let’s take a question similar to an official question from the GMAT paper tests:

Question 2: The function f is defined for each positive three-digit integer n by f(n) = 2^x * 3^y * 5^z, where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m) = 25f(v), then m-v=?

(A) 2
(B) 9
(C) 18
(D) 20
(E) 100

Solution: The question may seem a bit difficult to understand first so let’s take one sentence at a time:

f is defined for each positive three-digit integer n by f(n) = 2^x * 3^y * 5^z, where x, y and z are the hundreds, tens, and units digits of n

Let’s take an example to see how to make sense of it: say 146 is a three digit positive integer. So f(146) = 2^1 * 3^4 * 5^6

In the same way, f(283) = 2^2 * 3^8 * 5^3

If m and v are three-digit positive integers such that f(m) = 25f(v)

So f(m) = 5^2 * f(v)

If m is represented as abc and v as def, then (2^a * 3^b * 5^c) = 5^2 * (2^d * 3^e * 5^f)

Note that for the left hand side to be equal to right hand side, a = d, b = e and c = 2 + f.

So the units digit of m is 2 more than the units digit of v but their tens and hundreds digits are the same.

So m – v = 2.

Answer (A)

If you are still not sure how we arrived at this, take an example.

f(m) = f(266) = 2^2 * 3^6 * 5^6

f(v) = f(264) = 2^2 * 3^6 * 5^4

The difference between f(m) and f(v) will be of 5^2 since their units digits are 2 away from each other.

Hope next time you see a functions question, you will not get spooked and will, instead, take it in your stride!

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!

Functions on GMAT

Quarter Wit, Quarter WisdomLet’s discuss how to handle functions today. People usually perceive functions as an advanced topic mainly because of the notation. But actually, the function questions are very simplistic and can be solved with a simple process. If we ask you the value of 5x^3 where x = 3, would you be worried about what to do? We assume you won’t be. Then there should be no problem with “given f(x) = 5x^3, what is the value of f(3)?”

Just keep a few simple things in mind:

– f(x) = …. will be followed by an expression in x. This is the core of your calculations. You can turn a blind eye to f(x) – just focus on the expression. For example: f(x) = (x^2+1)/5x. Keep your eye on (x^2+1)/5x.

– When faced with “what is f(a)?” all you have to do is recall the expression given and put x = a in that. It doesn’t matter what a is – wherever you have x, just put ‘a’ there. For example: if f(x) = (x^2+1)/5x, what is f(5x^3)? Don’t get confused. Here, a = 5x^3. Look for x in the expression and put 5x^3 in its place.

f(5x^3) = ((5x^3)^2+1)/5(5x^3)

If you seem to be getting lost in too many exponents, terms etc, in place of x, put 5x^3 and put brackets around it as done above. Then simplify by opening the brackets.

f(5x^3) = (25x^6+1)/25x^3

– When we are given that f(a) = B, put x = a in the expression and equate the whole expression to B. For example: f(x) = (x^2+1)/5x, given that f(a) = 2/5, what is the value of a?

We know how to find f(a). It is simply (a^2+1)/5a. We are given that this is 2/5.

(a^2+1)/5a = 2/5

a^2 + 1 = 2a

a^2 – 2a + 1 = 0

(a – 1)^2 = 0

a = 1

That is pretty much all you need. Let’s look at a GMAT Prep question on functions.

Question: For which of the following functions f is f(x) = f(1-x) for all x?

(A) f(x) = 1 – x

(B) f(x) = 1 – x^2

(C) f(x) = x^2 – (1 – x)^2

(D) f(x) = x^2*(1 – x)^2

(E) f(x) = x/(1 – x)

Solution: What does this mean: f(x) = f(1-x)? It means that given a certain expression in x called f(x), for which function will that be the same as f(1-x) i.e. when you substitute x by (1-x), which expression will stay the same? Let’s look at each option:

(A) f(x) = 1 – x

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = 1 – (1 – x)

f(1 – x) = x

f(x) is not the same as f(1-x) here. Ignore this option.

(B) f(x) = 1 – x^2

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = 1 – (1 – x)^2

f(1 – x) = 2x – x^2

f(x) is not the same as f(1-x) here. Ignore this option.

(C) f(x) = x^2 – (1 – x)^2

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = (1 – x)^2 – (1 – (1-x))^2

f(1 – x) = (1 – x)^2 – x^2

f(1 – x) = -x^2 + (1 – x)^2

f(x) is not the same as f(1-x) here. Ignore this option.

(D) f(x) = x^2*(1 – x)^2

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = (1 – x)^2 * (1 – (1 – x))^2

f(1 – x) = (1 – x)^2 * x^2

f(1 – x) = x^2 * (1 – x)^2

Note that here, f(x) = f(1 – x), so this must be our answer. Still, let’s take a look at (E) as well for practice.

(E) f(x) = x/(1 – x)

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = (1 – x)/(1 – (1-x))

f(1 – x) = (1 – x)/x

f(x) is not the same as f(1-x). Ignore this option.

Answer (D)

A cursory look back at the solution might make you feel that it involves some complicated manipulations but we hope you do see that it is anything but complicated. Now there are some other ways of handling this question too. If you are comfortable with the above, continue with the rest of the post.

Method 2: Number Plugging

We want the expression for which f(x) = f(1 – x) for ALL values of x. So no matter what value we give x, f(x) should be same as f(1 – x).

Say, if x = 0, for which function is f(x) = f(1 –x )? i.e. for which function is f(0) = f(1)

(A) f(x) = 1 – x

f(0) = 1 and f(1) = 0. Not equal.

(B) f(x) = 1 – x^2

f(0) = 1 and f(1) = 0. Not equal.

(C) f(x) = x^2 – (1 – x)^2

f(0) = -1 and f(1) = 1. Not equal.

(D) f(x) = x^2*(1 – x)^2

f(0) = 0 and f(1) = 0. Equal. But when using number plugging, you need to check all options because multiple options could give you equal values. In that case, you would need to try for another value of x.

(E) f(x) = x/(1 – x)

f(0) = 0 and f(1) is not defined. Just to be sure, say x = -1.

f(-1) = -1/2 and f(2) = -2. Not equal.

Answer (D)

Method 3: Intuitive Approach

Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. Intuitively, we should check (D) first since it involves multiplication of the terms.

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!

Past Perfect without Past Tense on GMAT Sentence Correction Questions

Quarter Wit, Quarter WisdomRecall the golden rule of past perfect tense –

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!

4 Average Speed Formulas You Need to Know for the GMAT

Quarter Wit, Quarter WisdomMany people have asked us to clear the confusion surrounding the various formulas of average speed. We will start with the bottom line – There is a single versatile formula for ALL average speed questions and that is

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!

Using Symmetry in Probability on the GMAT

Quarter Wit, Quarter WisdomWe know that Combinatorics and Probability are tricky topics. It is easy to misinterpret questions of these topics and get the incorrect answer – which, unfortunately, we often find in the options, giving us a false sense of accomplishment.

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!

Of Opinions and Facts in GMAT Critical Reasoning Questions

Quarter Wit, Quarter WisdomToday, we would like to discuss with you one of our most debated critical reasoning questions. It is an absolutely brilliant question – not just because the correct option fits in beautifully but because the other four options are also very well thought out. It is easy to write the incorrect four options such that the student community will be split between 2 options – the correct one and one of the four incorrect ones but when the jury is split between 4 or all 5 options, that’s when we know that we have come up with an absolute masterpiece. Of course, in such questions, a lot of effort is needed to convince everyone of the correct answer but it is well worth it.

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 21st 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)

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

Advanced Number Properties on the GMAT – Part V

Quarter Wit, Quarter WisdomToday, let’s look in detail at a relation between arithmetic mean and geometric mean of two numbers. It is one of those properties which make sense the moment someone explains to us but are very hard to arrive on our own.

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)

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!

Pre-thinking in Quant GMAT Questions

Quarter Wit, Quarter WisdomWe all know about the role of pre-thinking in Critical Reasoning and how anticipating the answer can be supremely beneficial in not just the physical aspect of saving time in analyzing options but also the psychological aspect of promoting our self-confidence – we were thinking that the answer should look like this and that is exactly what we found! Pre-thinking puts us in the driver’s seat and we feel energized without consuming any red bull!

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.

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!

Intelligent Guessing on GMAT

Quarter Wit, Quarter WisdomWe often tell you that if you are short on time, you can guess intelligently on a few questions and move on. Today we will discuss what we mean by “intelligent guessing”. There are many techniques – most of them involving your reasoning skills to eliminate some options and hence generating a higher probability of an accurate guess. Let’s look at one such method to get values in the ballpark.

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.

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

The Speed and Accuracy Trade Off on the GMAT

Quarter Wit, Quarter WisdomWe know that speed is important in GMAT. We have about 2 mins per question and we always have questions in which we get stuck, waste 3-4 mins and probably still answer incorrectly. So we are always trying to go faster, rush, complete the easy ones in less time! In our bid to save time, sometimes we sacrifice accuracy. We should know that accuracy is most important. No point running through questions and completing all of them before time if at the end of it all, most of our answers are incorrect – there are no bonus points for completing the test before time, after all!

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)

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!

Bringing Back the Lazy Genius to Solve GMAT Questions!

Quarter Wit, Quarter WisdomThose of you who have seen the previous version of our curriculum would know that we had tips and tricks under the heading of ‘Lazy Genius’. These used to discuss innovative shortcuts for various questions – the way very smart people would solve the question – without putting in too much effort!

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 1/4th of the tank and every hour drain Gamma empties 1/4th 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 (1/4th 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!

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

Finding the Last Two Digits on GMAT Quant Questions – Part III

Quarter Wit, Quarter WisdomAs promised last week, we will look at another question which involves finding the last two digits of the product of some random numbers. In this question, along with the concepts discussed last week, we will assimilate the concept of negative remainders too discussed some weeks ago.

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)

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

Finding the Last Two Digits on GMAT Quant Questions – Part II

Quarter Wit, Quarter WisdomLet’s continue the discussion of last two digits we started last week. We discussed the concept of pattern recognition and how it can help us determine the last two digits in case of numbers raised to some powers. Today we look at what happens when there is no pattern to determine! What if we are asked to determine the last two digits of the product of a bunch of numbers. We know that getting the last digit in this case is very easy – just multiply the last digits of the numbers together. But last TWO digits would seem much more complicated.

Actually, we can find the last two digits quite easily in most such cases by using the concepts of remainders.

There are two concepts you need to understand before we go on to see how to solve such questions:

I. When you divide a number by 100, the remainder is formed by the last two digits of the number.  Say, you divide 138 by 100, the remainder will be 38 (last two digits). Take another example – divide 1275 by 100, the remainder will be 75 and so on.

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. This should remind you of the binomial theorem we discussed many weeks ago.  When we multiply all these terms together (px + a), (qx + b) etc, each term obtained will have at least one x except the last term which is obtained by multiplying the remainders together. To get a better idea, let’s take some numbers:

Let’s say we need to find the remainder when we divide 12*23*52*81 by 10.

K = 12*23*52*81 = (10 + 2)*(20 + 3)*(50 +2)*(80 + 1)

When you multiply these four terms together, you will get many terms such as 10*20*50*80, 10*20*50*1, 10*20*2*80 etc. All these will have a multiple of 10 except the last one. The last one will be 2*3*2*1 = 12. That doesn’t have a multiple of 10. Now divide 12 by 10 to get the remainder 2. So when you divide K by 10, the remainder will be 2.

Now, let’s look at a question:

Question 1: What are the last two digits of 63*35*37*82*71*41?

(A) 10

(B) 30

(C) 40

(D) 70

(E) 80

Solution: Using concept 1, we know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

Note: 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.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)

Next week, we will see some more complicated questions using these and other fundamentals.

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

Finding the Last Two Digits on GMAT Quant Questions – Part I

Quarter Wit, Quarter WisdomWe all know how to find the last digit using cyclicity when we are given a number raised to a power. Last digit of a number depends only on the last digit of the base.  You must be quite familiar with something like this –

Last Digit of Base:

0 – Last digit of expression with any power will be 0.

1 – Last digit of expression with any power will be 1.

2 – 2, 4, 8, 6, 2, 4, 8, 6… Cyclicity is 4.

3 – 3, 9, 7, 1, 3, 9, 7, 1… Cyclicity is 4.

4 – 4, 6, 4, 6, 4, 6, 4, 6… Cyclicity is 2.

5 –  Last digit of expression with any power will be 5.

6 – Last digit of expression with any power will be 6.

7 – 7, 9, 3, 1, 7, 9, 3, 1… Cyclicity is 4.

8 – 8, 4, 2, 6, 8, 4, 2, 6… Cyclicity is 4.

9 – 9, 1, 9, 1, 9, 1, 9, 1… Cyclicity is 2.

Cyclicity is nothing but pattern recognition. You see that when you multiply 2 by  itself, there is a pattern of last digit which goes 2, 4, 8, 6, 2, 4, 8, 6 and so on. We can use the same principle for when a question asks us for the last two digits of the expression. Let me remind you first that here at QWQW, we sometimes flirt with the lines that define GMAT scope. Obviously, we do point out whenever we are indulging and that’s exactly what we are going to do in this post. We are carrying on for the love of Math and the Q51 score.

The last two digits of the base decide the last two digits of the expression. For example,

Example 1: Let’s look at powers of 11.

11^1 = 11

11^2 = 121

11^3 = 1331

11^4 = …41 (we should just multiply the last two digits together and ignore the rest)

11^5 = …51

11^6 = …61

11^7 = …71

Note that the last two digits are displaying a pattern depending on the power. So we expect the cyclicity here to be 10.

11^8 = …81

11^9 = …91

11^(10) = …01

11^(11) = …11

11^(12) = …21

and so on. So the last two digits should go from 11, 21 to 91, 01 and then go to 11 again. The cycle of 10 starts from power of 1, 11, 21 etc. This means that 11^(46) should have last two digits as 61, 11^(92) should have last two digits as 21 and 11^(168) should have last two digits as 81.

Let’s look at some other numbers now:

Example 2: Say, we need the last two digits of 6^{58}

6^1 = 6 (No second last digit)

6^2 = 36

6^3 = 216

6^4 = …96 (Just multiply the last two digits)

6^5 = …76

6^6 = …56

6^7 = …36

and hence starts the cycle again:

3, 1, 9, 7, 5, 3, 1, 9, 7, 5 and so on.

The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit.

Example 3: How about the last two digits of 7^102?

7^1 = 7 (No second last digit)

7^2 = 49

7^3 = 343

7^4 = …01

7^5 = …07

7^6 = …49

7^7 = …43

We see a cyclicity of 4 here: 49, 43, 01, 07, 49, 43, 01, 07 … and so on. The new cycle begins at 2, 6, 10, 14 i.e. even powers which are not multiples of 4. So a new cycle will begin at 102 too. So the last two digits of 7^(102) will be 49.

Now there can be many variations in the questions asking us to find the last two digits. We will use different concepts for different question types. Today we saw how to use pattern recognition. We will look at some other methods 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!

Figuring Out the Topic of Discussion on the GMAT

Quarter Wit, Quarter WisdomYou must have come across questions which you thought tested one concept but later found out could be easily dealt with using another concept.  Often, crafty little mixture problems belong to this category. For example:

Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark’s remaining chips are $20 chips, how much money did Mark bet?

You can view this as a word problem where you assume the number of chips and then go splitting them up or you can view this as a mixtures problem even though it doesn’t use words such as ‘mixture’, ‘solution’, ‘combined’ etc. As we have seen enough number of times, our mixture problems are solved in seconds using the weighted average concept.

The question discussed here also belongs to the same category – looks super tricky but can be easily solved with weighted averages formula. But we have seen plenty and more of such questions in our blog posts. Today we will take a look at a different type of sinister question and I suggest you to think about the concept being tested in that before trying to solve it.

Question: Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15). Watch3 loses 20 minutes every hour relative to watch2. Finally, watch4 gains 20 minutes every hour relative to watch3. If Mark resets all four watches to the correct time at 12 noon, what time will watch4 show at 12 midnight that day?

(A)10:00
(B)10:34
(C)11:02
(D)11:48
(E)12:20

Before we look at the solution, think about the concept being tested here – clocks? Circular motion?

Neither!

Solution: Note that when giving data about watch1, you are told how it varies with the actual time. Data about all other watches tells us about the time they show relative to the incorrect watches. The concept being tested here is Relative Speed!

What do we mean by “gains 15 mins” or “loses 20 mins” etc? When a watch gains 15 mins every hour, it means that even though it should show that one hour has passed, it shows that 1 hr 15 mins have passed. So the watch runs faster than it should. Hence the speed of the watch is more than the speed of a correct watch. Now the question is how much more? The minute hand of the correct watch travels one full circle in one hour. The minute hand of the incorrect watch travels one full circle and then a quarter circle in one hour (to show that 1 hour 15 mins have passed even when only an hour has passed). So it is 5/4 times the speed of a correct watch. On the same lines, let’s analyze each watch.

Say the speed of a correct watch is s.

– “Watch1 loses 15 minutes every hour. “

Watch1 covers only three quarters of the circle in an hour.

Speed of watch1 = (3/4)*s

– “Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15).”

Now we have the speed of watch2 relative to speed of watch1. Speed of watch2 is (5/4) times the speed of watch1.

Speed of watch2 = (5/4)*(3/4)s = (15/16)*s

– “Watch3 loses 20 minutes every hour relative to watch2.”

Watch3 loses 20 mins every hour means its speed is (2/3)rd the speed of watch2

Speed of watch3 = (2/3)*(15/16)*s = (5/8)*s

– “Finally, watch4 gains 20 minutes every hour relative to watch3.”

Speed of watch4 = (4/3)*Speed of watch3 = (4/3)*(5/8)*s = (5/6)*s

So the speed of watch4 is (5/6)th the speed of a correct watch. So if a correct watch shows that 6 hours have passed, watch4 will show that 5 hours have passed. If a correct watch shows that 12 hours have passed, watch4 will show that 10 hours have passed. From 12 noon to 12 midnight, a correct watch would have covered 12 hours. Watch4 will cover 10 hours and will show the time as 10:00.

Answer (A)

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

What are the Weights in Weighted Averages?

Quarter Wit, Quarter WisdomWe have discussed weighted averages in detail here but one thing we are yet to talk about is how you decide what the weights will be in weighted average problems. It is not always straight forward to identify the weights. For example, in a question such as this one,

While traveling from Detroit to Novi, a car averaged 10 miles per gallon, and while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

We have two figures for mileage given here – 10 miles per gallon and 18 miles per gallon. We need to find the average mileage. So we can use the weighted average formula but what will the weights be? Will they be 1:2 since the distance between the two cities is given to be in the ratio 1:2? If you think that taking the distance to be the weights in this problem is correct, then you fell for the trap in this question.

To explain the concept, let us use a simpler example first:

When talking about average speed, what are the weights? We know that the weight given to each speed is the time for which that speed was maintained, right? Yes! But why?

Let’s review our weighted average formula:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)
Average Speed = (Distance1 + Distance2)/(Time 1 + Time2)
Average Speed = Total Distance/Total Time

This is an accurate representation of average speed.

Now see what happens when you use distance as the weights.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Speed*Distance doesn’t represent any physical quantity. So this doesn’t make sense. The units of the quantities will help you see the relation clearly.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)

Average Speed = (miles/hour * hour + miles/hour * hour)/(hour + hour)

Average Speed = (miles + miles)/(hour + hour)

Average Speed = Total miles/Total hours

What happens when you take distance as the weights?

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Average Speed = (miles/hour * miles + miles/hour * miles)/(miles + miles)

miles^2/hour doesn’t represent a physical quantity and hence doesn’t make sense here. Therefore, whenever you are confused what the weights should be, look at the units.

Let’s go back to the original question now. Average required is miles per gallon. So you are trying to find the weighted average of two quantities whose units must be miles/gallon.

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

The unit of Cavg, C1 and C2 is miles/gallon so w1 and w2 should be in gallons to get

miles/gallon = (miles/gallon * gallon + miles/gallon * gallon)/(gallon + gallon)

miles/gallon = Total miles/Total gallons

So how will we actually solve this question?

Question: While traveling from Detroit to Novi, a car averaged 10 miles per gallon while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

Solution:

Let the distance between Detroit and Novi be D. So the distance between Novi and Lapeer must be 2D.

Amount of fuel used to cover distance D = D/10

Amount of fuel used to cover distance 2D = 2D/18 = D/9

So the two weights used must be D/10 and D/9

Average miles/gallon = (10*D/10 + 18*D/9)/(D/10 + D/9) = 3D*90/19D = 270/19 = 14.2 miles/gallon

Or simply, Average miles/gallon = Total miles/Total gallons = 3D/(D/10 + D/9) = 14.2 miles/gallon

Food for thought: Which one of the following can you solve?

– If a vendor sold 10 apples at a profit of 10% and 15 oranges at a profit of 20%, what was his overall profit%?

– If a vendor sold apples at a profit of 10% and oranges at a profit of 20%, what was his overall profit% if cost price of each apple was $0.20 and the cost price of each orange was $.06?

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

Busting Some GMAT Sentence Correction Myths – Part II

Quarter Wit, Quarter WisdomA few weeks back, we wrote a post busting some Sentence Correction myths. Let’s continue from where we left.  We discussed how we can have pronouns referring to different antecedents in different clauses of the same sentence. Let’s take another example illustrating that principle. Also, we learn how to use ‘being’ correctly in GMAT.

Myth 3: Use of ‘being’ is always wrong on GMAT!

Often, the way we use ‘being’ in our day-to-day communication, is incorrect. For example,

Being a doctor, he is very well respected.

But there are correct ways of using ‘being’. Since most students believe that ‘being’ is wrong, don’t trust the GMAC to not use this nugget of information to misdirect the test takers. The correct answers of questions at higher ability are worded in such a way that they make the test takers uncomfortable!

So how is ‘being’ used correctly?

‘Being’ is used to express a temporary state.

The little boy started screaming when he saw his dog being impounded.

‘Being impounded’ is a temporary state and would be over – unlike being a doctor. So the use of being is correct here.

Let’s look at one of our own sentence correction questions now:

Question: The data being collected in the current geological survey are providing a strong warning for engineers as they consider the new dam project, but their greatest importance might lie in how they influence the upcoming decision by those same engineers on whether to retrofit 75 bridges in the survey zone.

A. The data being collected in the current geological survey are providing a strong warning for engineers as they consider the new dam project, but their greatest importance

B. The data being collected in the current geological survey provide a strong warning for engineers as they consider the new dam project, but its greatest importance

C. The data collected in the current geological survey is providing a strong warning for engineers as they consider the new dam project, but their greatest importance

D. The data collected in the current geological survey provides a strong warning for engineers in consideration of the new dam project, but its greatest importance

E. The data collected in the current geological survey provide a strong warning for engineers in consideration for the new dam project, but the greatest importance

Solution:  Let’s find the decision points:

First decision point: being collected vs collected

‘The data being collected’ is a temporary state here. Data won’t always be collected, but are being collected for a short time right now, so ‘being’ is used properly here. And the sentence makes it very clear that this is temporary; look at the word ‘current’ before ‘geological survey.’ If you were skeptical of the word ‘being’ before, that is understandable, but the word ‘current’ should serve as a clear warning that this is a temporary, ongoing event. (And furthermore, the non-underlined portion talks about an upcoming decision, even further cementing the idea that this is a temporary survey with data ‘being’ collected for a short period of time).   Alas, even with that temporary state, there isn’t really anything wrong with ‘The data collected in …’ either so we retain all answer options.

It’s worth noting that they put the ‘being collected’ vs. ‘collected’ decision early in the sentence/answer-choices as a great place to put a ‘false decision point.’  The authors of these questions know that they want to reward emphases on logical meaning and core grammar rules, and they also know that students like to study quick tips and tricks, so they leave that bait there for the tips/tricks folks while they hold off the bigger reward for those willing to prioritize decision points from most important to least and not just from left to right in order of appearance.

Second decision point: are/is

Technically, data is plural of datum. In academic writing it is almost always treated as plural. It is treated as singular in informal writing but GMAT favors treating it as plural.
Even if you do not know this, the use of “they influence the upcoming” – in the portion of the sentence that is not underlined – should tell you that ‘data’ is used in plural form here.

Hence the use of ‘are’ is appropriate. Hence, options (C) and (D) are eliminated.

Third decision point: Pronouns

There are many pronouns used here. Antecedent of each pronoun is present in the sentence. The usage clarifies which pronoun refers to data and which refers to engineers.

Original Sentence: The data being collected in the current geological survey are providing a strong warning for engineers as they consider the new dam project, but their greatest importance might lie in how they influence the upcoming decision by those same engineers on whether to retrofit 75 bridges in the survey zone.

they – only engineers can consider the new dam project so ‘they’ refers to engineers
their/its  – greatest importance will be of data, which is plural, so ‘their’ would be the correct usage. Eliminate option (D)

There is no ambiguity in the use of pronouns. The nouns are present and the usage clarifies the antecedent.

Now we are left with options (A) and (E). In option (E), “in consideration for the new dam project” is bad diction.  Also, it doesn’t tell us ‘whose greatest importance?’.

Answer is (A)

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

How to Go from a 48 to 51 in GMAT Quant – Part V

Quarter Wit, Quarter WisdomFirst, let us give you the link to the last post of this series: Post IV. It contains links to previous parts too.

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

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

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

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

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

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

(A) 18
(B) 19
(C) 21
(D) 42
(E) 59

Solution:

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

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

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

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

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

2a+3b = 650

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

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

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

b = (650 – 2a)/3

a < (650 – 2a)/3 < a+100

3a < 650 – 2a < 3a + 300

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

Inequality 1: 3a < 650 – 2a

5a < 650

a < 130

Inequality 2: 650 – 2a < 3a + 300

5a > 350

a > 70

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

b = (650 – 2a)/3

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

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

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

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

73, 76, 79, 82, 85, … 127

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

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

127 = 73 + (n – 1)*3

n = 19

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

Method 2: Using Transition Points

Note that a < b < a+100

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

2a + 3a = 650

a = 130 = b

But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But decreasing a by 3 decreases the largest number i.e. a+100 by 3 too; so we need to increase b by another 1.
We get a = 127 and b = 132. This give us the numbers as 127, 132, 132, 132, 227. Here the average is 150

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

2a + 3(a+100) = 650

a = 70, b = 170

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

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

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

This is an Arithmetic Progression.

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

127 = 73 + (n – 1)*3

n = 19

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

Answer (B)

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

The Holistic Approach to Mods on the GMAT – Solutions

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

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

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

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

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

Solution 1

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

Solution 2

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

A 750+ Level Question on SD

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

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

So here goes the question.

Question: Given that set S has four odd integers and their range is 4, how many distinct values can the standard deviation of S take?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

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

Let’s look at the given question now.

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

Now in our example, x and y can take 3 different values: 1, 3 or 5
x and y could be same or different but x would always be smaller than or equal to y.

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

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

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

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

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

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

Standard deviations of {1, 1, 1, 5} and {1, 5, 5, 5} are the same. Why?
Standard deviation measures distance from mean. It has nothing to do with the actual value of mean and actual value of numbers. Note that the distribution of numbers on the number line is the same in both cases. The two sets are just mirror images of each other.

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

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

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

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

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

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

Answer (B)

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

Busting Some GMAT SC Myths

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

Myth 1: Passive voice is always wrong.

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

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

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

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

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

(A) Once the computer generates the financial reports, they are then used to program a company-wide balance sheet, so named because it demonstrates that every department’s accounting elements balance.
(B) Once the computer generates the financial reports, it is then used to program a company-wide balance sheet, named such because it demonstrated the balance of every department’s accounting elements.
(C) Once the computer generates the financial reports they are then used to program a company-wide balance sheet, which demonstrates the balance of every department’s accounting elements.
(D) Once the financial reports are generated by the computer, it is then used to program a company-wide balance sheet, so named because it demonstrates the balance of every department’s accounting elements.
(E) Once the financial reports are generated by the computer, they are then used to program a company-wide balance sheet, named such because it demonstrates that every department’s accounting elements are in balance.

Solution:  Let’s split the sentence into clauses:

–          Once the computer generates the financial reports,

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

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

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

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

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

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

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

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

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

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

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

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

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

An OG problem has been pointed out here:

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

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

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

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

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

There is no ambiguity of pronoun reference here.

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

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

 

A 750 Level GMAT Question on Statistics!

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

The Meaning of Arithmetic Mean

Can You Solve these Mean GMAT Questions?

Finding Arithmetic Mean Using Deviations

Application of Arithmetic Means

Mean Questions on Median

A Range of Questions

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

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

(A)10
(B)12
(C)14
(D)15
(E)20

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

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

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

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

0 … 5000 … 10,000

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

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

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

Now we have

0 … 5000 … 10000, 10000, 10000

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

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

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

Let’s try one such set of addition:

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

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

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

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

This gives us a total of 15 people.

Answer (D)

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

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

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

2 Sentence Correction GMAT Questions Involving Participle Modifiers

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

Let’s see how we decide that.

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

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

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

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

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

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

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

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

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

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

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

Answer is (D).

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

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

(A) strongly recommending middle-aged people to
(B) strongly recommending that middle-aged people should
(C) and strongly recommended for middle-aged people to
(D) and their strong recommendation was for middle-aged people to
(E) and they strongly recommended that middle-aged people

Solution: The given sentence has two clauses:

Main clause – medical researchers at Harvard and Stanford universities concluded

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

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

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

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

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

or

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

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

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

Answer (E)

Hope you now understand how participle phrases are used.

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

Understanding Participles on the GMAT

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

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

There are two types of participles:

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

For Example: chosen, danced, known, sung etc

2. The Present Participle – ends in –ing

For Example: choosing, dancing, knowing, singing etc

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

Examples:

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

Standing next to the tall gentleman, she looked petite.

Battered by hail, the car collapsed.

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

Here is how participle phrases are usually used:

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

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

Examples:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Here is our official solution:

The correct response is (D).

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

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

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

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

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

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

First Do What You Know on GMAT Questions

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

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

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

(A) 100%
(B) 80%
(C) 75%
(D) 66+2/3%
(E) 55%

Solution:

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

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

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

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

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

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

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

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

(1 + m/100) = 7/5
m = 40

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

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

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

Answer (A)

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

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

How to Expect the Unexpected on the GMAT

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

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

Let’s take an example:

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

(A) 6/7 hrs

(B) 12/7 hrs

(C) 4 hrs

(D) 6 hrs

(E) 12 hrs

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

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

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

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

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

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

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

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

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

So following the same logic as above,

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

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

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

Try to put it all together.

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

Using the original question:

Time taken by car A to complete one full circle = 12?/6? = 2 hrs
Time taken by car B to complete one full circle 12?/8? = 1.5 hrs

So every 2 hrs car A is at S and every 1.5 hrs, car B is at S. When will they both be together at S?
Car A at S -> 2 hrs, 4 hrs, 6 hrs, 8 hrs …
Car B at S -> 1.5 hrs, 3 hrs, 4.5 hrs, 6 hrs …

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

Using the same method on the Modified Question,

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

So every 5 hrs, car A is at S and every 3 hrs, car B is at S. When will they both be together at S?
Car A at S -> 5 hrs, 10 hrs, 15 hrs, 20 hrs
Car B at S -> 3 hrs, 6 hrs, 9 hrs, 12 hrs, 15 hrs

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

This all makes sense now.

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

95% of Students Find This GMAT Quant Question Difficult

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

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

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

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

Let us give you the question now:

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

Statement 1: Piglet A was fed exactly 1/4 of the oats today.
Statement 2: Piglet A was fed exactly 1/6 of the barley today.

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

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

For now, forget the data given in the question.

Consider this:

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

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

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

If this is not clear, look at this example:

Say a meal consists of a sandwich and a milkshake. You eat 1/2 of the sandwich and drink 1/2 of the milkshake. Can we say that you have had 1/2 of the meal? Sure.
If you eat only 1/4 of the sandwich and drink 1/4 of the milkshake, then you would have had only 1/4 of the meal.
What happens in case you eat 1/2 of the sandwich but drink only 1/4 of the milkshake? In that case, you have had less than 1/2 of the meal but certainly more than 1/4 of the meal, right?

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

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

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

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

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

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

A Closer Look at Absolute Phrases on the GMAT

Quarter Wit, Quarter WisdomRead the following sentences:

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

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

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

What is an Absolute Phrase?

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

Structure of an Absolute Phrase

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

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

Usage of an Absolute Phrase

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

For example:

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

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

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

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

Some Alternative Structures of Absolute Phrases

Some absolute phrases have a different structure.

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

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

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

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

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

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

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

Solution:

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

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

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

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

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

Answer (A)

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

A 700+ GMAT Quant Question on Races

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

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

(A)   8, 10

(B)    4, 5

(C)   5, 9

(D)   6, 9

(E)    7, 10

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

There are two scenarios given:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer (B)

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

2 Simple GMAT Quant Questions That Will Help You Score Higher

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

A few points to remember in races:

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

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

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

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

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

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

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

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

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

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

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

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

(A)   1/17

(B)   3/17

(C)   1/10

(D)   3/20

(E)    3/10

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

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

(Start) A_________B______________________________

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

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

Answer (D)

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

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

(A)   15%

(B)   20%

(C)   28%

(D)   32%

(E)    35%

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

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

(Start ) A_________B_______________________________

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

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

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

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

____________20_________________A____5_____B (Finish)

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

Answer (D)

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

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

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

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

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

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

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

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

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

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

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

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

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

Method 2: Algebra

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

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

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

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

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

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

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

i.e. 30 < k < 40

Only option (D) falls in this range.

Answer (D)

Method 3: Weighted Average

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

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

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

Where Cavg = k

C1 = 20

C2 = 40

w1 = x

w2 = y

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

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

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

Hence, answer (D).

Method 4: Plugging Numbers

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

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

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

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

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

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

– Say, x = 1, y = 4

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

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

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

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

Advanced Number Properties on the GMAT – Part IV

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

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

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

For example:

2 + 3 = 5

2 + 7 = 9

2 + 17 = 19

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

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

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

81 = 2 + 79

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

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

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

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

For example:

4 = 2 + 2

6 = 3 + 3

8 = 3 + 5

10 = 3 + 7 = 5 + 5

12 = 5 + 7

and so on…

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

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

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

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

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