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:

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!*

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

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!*

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 &?

(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:

In the correctly worked addition problem above, M, N, 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

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

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!*

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

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

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

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?

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

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:

**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.

**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.

**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.

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

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.

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

**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!

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

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.

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

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.

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