# Planning for the “Plan” Questions on the GMAT Critical Reasoning Section

At Veritas Prep, we are often asked to discuss how to handle the “plan” Critical Reasoning questions test takers are asked on the GMAT. Here is how these questions are different from your regular strengthen/weaken questions – instead of a conclusion, we are given situations and plans to remedy a particular problem. We are then asked to evaluate the success of the plan or identify a weakness in the plan or an assumption of the plan.

Note that a plan question is very similar to a strengthen/weaken/assumption question. The main difference between them is that instead of being given a conclusion, you are asked to strengthen/weaken the possibility of a plan working out or an assumption made in the plan (looking at a few example questions will make this clearer). Let’s look at some examples of each of the three types of “plan” questions you are likely to come across on the GMAT exam:

Example 1 (the most common one): Which of the following will help us in evaluating the success of the plan?

In the country of Bedenia, officials have recently implemented a new healthcare initiative to reduce dangerous wait times at emergency rooms in the country’s hospitals. This initiative increases the number of available emergency nurses and doctors in urban settings: scholarships and no-interest loans are being offered to prospective students in these fields if they work in major city hospitals, relocation packages to urban centers are being offered for current emergency practitioners, and immigration rules are being changed to enable foreign emergency doctors and nurses to more easily move to Bedenia’s major cities.

Which of the following would be most important to determine in assessing whether the initiative will be successful?

(A) What percentage of current nurses and doctors work in emergency medicine.
(B) Which hospitals in Bedenia have dangerous wait times in their emergency rooms.
(C) Whether a career in emergency medicine pays substantially less than other types of medicine.
(D) Whether wait times could be reduced by means other than increasing the number of available nurses and doctors.
(E) Whether many foreign doctors and nurses are currently not allowed to enter Bedenia.

Plan: Reduce the dangerous wait time by increasing the availability of emergency nurses and doctors in urban settings by providing scholarships, offering relocation packages and changing immigration rules.

We need to find out whether this given plan will actually reduce wait time. Note that we are not worried about what else could reduce the dangerous wait time or what else this plan could do. The only point of concern for us is whether this plan will reduce the wait time.

This plan intends to increase the availability of emergency nurses and doctors in urban settings, so ask yourself this question: is this actually what is required? Do the urban hospitals have dangerous wait times? What if only rural hospitals have wait times and that is where the impetus is required? Answer choice B addresses exactly this question and, hence, will allow us to determine whether or not the initiative will be successful. Therefore, the answer is B.

Now look at our second example:

Example 2: Which of the following provides an argument against the plan?

In the last two years alone, nearly a dozen of Central University’s most prominent professors have been lured away by the higher salaries offered by competing academic institutions. In order to protect the school’s ranking, Central University’s president has proposed increasing tuition by 10% and using the extra money to offer more attractive compensation packages to the most talented and well-known members of its faculty.

Which of the following provides the most persuasive argument against the university president’s proposed course of action?

(A) It is inevitable that at least some members of the faculty will ultimately take jobs at other universities, regardless of how much Central University offers to pay them.
(B) Other universities are also looking for ways to provide higher salaries to prominent members of the faculty.
(C) Central University slipped in the last year’s ranking of regional schools.
(D) The single most important factor in ranking a university is its racial and socioeconomic diversity.
(E) The president of Central University has only been in office for 18 months and has never managed such a large enterprise.

Plan: Protect the school’s ranking by retaining its most prominent members by increasing their compensation.

We need to find a persuasive argument against the given plan – something that leads us to believe the plan should not be implemented. Here, test takers often become confused between options B and D. Let’s break down each answer choice in detail to determine which one is correct:

(B) Other universities are also looking for ways to provide higher salaries to prominent members of the faculty.

This option supports the given plan. It is a reason to actually implement the plan since if more disparity gets created, more prominent professors will leave. Remember, we are looking for an option that is against the plan, so B cannot be our answer.

(D) The single most important factor in ranking a university is its racial and socioeconomic diversity.

This is an argument against the plan. It states that the single most important factor in ranking is “racial and socioeconomic diversity,” so trying to retain prominent professors is not likely to retain ranking. Hence, the correct answer would be D.

Now let’s look at our final example:

Example 3: Which of the following is an assumption of the plan?

The general availability of high-quality electronic scanners and color printers for computers has made the counterfeiting of checks much easier. In order to deter such counterfeiting, several banks plan to issue to their corporate customers checks that contain dots too small to be accurately duplicated by any electronic scanner currently available; when such checks are scanned and printed, the dots seem to blend together in such a way that the word “VOID” appears on the check.

A questionable assumption of the plan is that

(A) in the territory served by the banks the proportion of counterfeit checks that are made using electronic scanners has remained approximately constant over the past few years.
(B) most counterfeiters who use electronic scanners counterfeit checks only for relatively large amounts of money.
(C) the smallest dots on the proposed checks cannot be distinguished visually except under strong magnification.
(D) most corporations served by these banks will not have to pay more for the new checks than for traditional checks.
(E) the size of the smallest dots that generally available electronic scanners are able to reproduce accurately will not decrease significantly in the near future.

Plan: To deter counterfeiting, issue checks that contain dots too small to be accurately duplicated (which will form the word VOID) by any electronic scanner currently available.

We need to find an assumption that this given plan makes. Note that the plan is based on the capabilities of the currently available scanners and assumes that their capabilities will not improve in the near future. Hence, E is an assumption.

Some test takers get confused with  answer choice C:

(C) the smallest dots on the proposed checks cannot be distinguished visually except under strong magnification

This option is actually not an assumption. Even if the dots can be distinguished visually, they don’t form the word VOID. Only when current scanners scan the checks and then we print them do the dots merge to form the word. Thus, our answer is E.

We hope you have understood how to handle various “plan” questions on the GMAT. The most important aspect of such questions to remember is to first identify the plan and what one hopes to achieve through 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!

# Is It Incorrect to Use Multiple Verb Tenses in a Sentence?

Some GMAT test-takers wonder whether it is grammatically correct to use multiple tenses in a single sentence. Today we will discuss the cases in which mixing verb tenses is acceptable and those in which this is not. The bottom line is this: there is no restriction on what tenses we can use and mix within a sentence, as long as they are appropriate for the context.

Take a look at this example sentence featuring multiple different verb tenses:

I have heard that Mona left Manchester this morning, and has already arrived in London, where she will be for the next three weeks.

Here, we have present perfect tense, simple past tense and simple future tense all in the same sentence. We’re mixing different tenses, but they all make sense together to create a logical sequence of events.

The confusion over using multiple verb tenses in one sentence probably arises because we have heard that we need to maintain verb tense consistency. These two things are different.

Tense Consistency – We do not switch one tense to another unless the timing of the action demands that we do. We do not switch tenses when there is no time change for the actions.

Pay particular attention to the “unless” clause in that description – maintaining consistency of a single verb tense is not an absolute virtue! Verb tenses need to convey a logical timeline or sequence of events, so if there is a shift in the timeline of when events occurred, your job isn’t to preserve verb tense consistency at all costs, but rather to mix verb tenses as necessary to tell a logical story of how and when things happened.

Let’s take a look at some examples to understand this:

Example 1: During the match, my dad stood up and waved at me.

These two actions (“stood” and “waved”) happen at the same time and hence, need to have the same tense. This sentence could take place in the present or future tense too, but both verbs will still need to take on the same tense. For example:

Example 2: During my matches, my dad stands up and waves at me.
Example 3: During the match tomorrow, my dad will stand up and wave at me.

On the other hand, a sentence such as…

Example 4: During the match, my dad stood up and waves at me.

This sentence is grammatically incorrect. Since both actions (“stood” and “waves”) happen at the same time, we need them to be in the same tense, as shown in the variations of this sentence above. Consider this case, however:

Here, the two actions (“reached” and “eaten”) happen at different times in the past, so we use both the simple past and past perfect tenses. The shift in tense is correct in this context, and that mixing tenses is necessary to convey a logical sequence of events.

Takeaway: The tenses of verbs in a sentence must be consistent when the actions happen at the same time. When dealing with actions that occur at different points in time, however, we can – and probably should – use multiple tenses in the same sentence.

Let’s look at an official GMAT question now to see how multiple tenses can be a part of the same sentence:

For the farmer who takes care to keep them cool, providing them with high-energy feed, and milking them regularly, Holstein cows are producing an average of 2,275 gallons of milk each per year.

(A) providing them with high-energy feed, and milking them regularly, Holstein cows are producing
(B) providing them with high-energy feed, and milked regularly, the Holstein cow produces
(C) provided with high-energy feed, and milking them regularly, Holstein cows are producing
(D) provided with high-energy feed, and milked regularly, the Holstein cow produces
(E) provided with high-energy feed, and milked regularly, Holstein cows will produce

This is a very tricky question. Let’s first shortlist our options based on the obvious errors.

The non-underlined part of the sentence uses the pronoun “them” to refer to the cows, so using “the Holstein cow” (singular) as the antecedent will be incorrect. The antecedent must be “Holstein cows” (plural) – this means answer choices B and D are out.

Also, we know for sure that “provide” and “milk” are parallel elements in the sentence, so they should take the same verb tense. Hence, answer choice C is also out.

Let’s look at A now. If we assume this option is correct, “providing” and “milking” act as modifiers to “keep them cool”. That certainly does not make sense since “providing with high energy feed” and “milking regularly” are not ways of keeping cows cool.

This means the correct answer is E, but we need to see how.

For the farmer who takes care to keep them cool, provided with high-energy feed, and milked regularly, Holstein cows will produce an average of 2,275 gallons of milk each per year.

Let’s break down the sentence:

For the farmer who takes care to keep them…

• cool,
• provided with high-energy feed,
• milked regularly,

…Holstein cows will produce an average of 2,275 gallons of milk each per year.

Note that we’re mixing two different tenses here: “For the farmer who takes care…” and “cows will produce…”. The word “takes” is the present tense while “will produce” is the future, but that does not make this sentence incorrect. The context of the author could very well justify the use of the future tense. Perhaps the farmers have obtained Holstein cows recently, and hence, will see the produce of 2,275 gallons in the future, only.

A shift in the tense certainly doesn’t make the sentence incorrect. When you’re presented with multiple verbs in various tenses in a problem, check to determine whether the verbs convey a logical sequence of events. If a sentence covers a broad timeline, your goal isn’t to avoid mixing tenses, but instead to ensure that the tenses used properly reflect that timeline.

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! Note the proper use of mixed tenses there – Karishma’s many activities encompass the past, present, and future, so it takes multiple tenses to keep up with her.

# Dreaded Data Sufficiency Questions That Will Test Your Knowledge of Number Properties

Here is an often-repeated complaint we hear from test takers – Data Sufficiency questions that deal with number properties are very difficult to handle (even for people who find problem-solving number properties questions manageable)! They feel that such questions are time consuming and often involve too many cases.

Here is our advice – when solving number properties questions, imagine a number line. It reminds us that numbers behave differently “between 0 and 1”, “between -1 and 0”, “less than -1”, and “more than 1”, and that integers occur only at regular intervals and that there are infinite numbers in between them. The integers are, in turn, even and odd. Also, 0, 1 and -1 are special numbers, hence it is always a good idea to consider cases with them.

Let’s see how thinking along these lines can help us on a practice Data Sufficiency question:

If a and b are non-zero integers, is a^b an integer?

Statement 1: b^a is negative
Statement 2: a^b is negative

The answer to this problem does not lie in actually drawing a number line. The point is that we need to think along these lines: -1, 0, 1, ranges between them, integers, negatives-positives, even-odd, decimals and how each of these comes into play in this case.

What we know from the question stem is that a and b are non-zero integers, which means they occur at regular intervals on the number line. To answer the question, “Is a^b an integer?”, let’s first look at Statement 1:

Statement 1: b^a is negative

For a number to be negative, its base must be negative. But that is not enough – the exponent should not be an even integer. If the exponent is an even integer, the negative signs will cancel out. Since a and b are integers, if a is not an even integer, it must be an odd integer.

We know that the sign of the exponent is immaterial as far as the sign of the result is concerned (since a^(-n) is just 1/a^n). For b^a to be negative, then we know that b must be a negative integer and a must be an odd integer. Does this help us in deducing whether a^b is an integer? Not necessarily!

If b is negative, say -2, a^(-2) = 1/a^2. a could be 1, in which case 1/a^2 = 1 (an integer), or a could be 3, in which case 1/a^2 = 1/9 (not an integer). Because there are two possible answers, this statement alone is not sufficient.

Let’s look at Statement 2:

Statement 2: a^b is negative

Again, the logic remains the same – for a number to be negative, its base must also be negative and the exponent should not be an even integer. If the exponent is an even integer, the negative signs will cancel out. Since a and b are integers, if b is not an even integer, it must be an odd integer. Again, we know that the sign of the exponent is immaterial as far as the sign of the result is concerned (since a^(-n) is just 1/a^n).

For a^b to be negative, then we know that a must be a negative integer and b must be an odd integer. a could be -1/-2/-3/-4… etc, and b could be 1/3/5… or -1/-3/-5.

If a = -1 and b = 1, then a^b = -1 (an integer). If a = -2 and b = -3, then a^b = (-2)^(-3) = 1/(-2)^3 = -1/8 (not an integer). This statement alone is also not sufficient.

We hope you see how we are using values of 1 and -1 to enumerate our cases. Now, let’s consider using both statements together:

a is a negative, odd integer, so it can take values such as -1, -3, -5, -7, …
b is a negative, odd integer too, so it can also take values such as -1, -3, -5, -7, …

If a = -1 and b = -1, then a^b = -1 (an integer)
If a = -3 and b = -3, then a^b = (-3)^(-3) = -1/27 (not an integer)

Even using both statements together, we do not know whether a^b is an integer or not. therefore, our answer is E.

Thinking of a number line and knowing what it represents will help you tackle many Data Sufficiency questions that are about number properties.

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

# Quarter Wit, Quarter Wisdom: Using Ingenuity on GMAT Remainder Questions

We have looked at various types of GMAT remainder questions and discussed how to tackle them in a few previous posts. Specifically, we have examined the concepts of general divisibility, divisibility as applied to GMAT questions, and divisibility specifically applied to remainders. There is one concept, however, that we haven’t discussed yet, and that is using ingenuity on remainder questions.

Say “x” gives you a remainder of 2 when divided by 6. What will be the remainder when x + 1 is divided by 6?

Go back to the divisibility concepts discussed above. When x balls are split into groups of 6, we will have 2 balls leftover. If we are given 1 more ball, it will join the 2 balls and now we will have 3 balls leftover. The remainder will be 3.

What happens in the case of x + 6 – what will be the remainder when this is divided by 6? This additional 6 balls will just make an extra group of 6, so we will still have 2 balls leftover.

What about the case of x + 9? Now, of the extra 9 balls, we will make one group of 6 and will have 3 balls leftover. These 3 balls will join the 2 balls leftover from x, giving us a remainder of 5.

Now, what about the case of 2x? Recall that 2x = x + x. The number of groups will double and so will the remainder, so 2x will give us a remainder of 2*2 = 4.

On the other hand, if x gives us a remainder of 4 when divided by 6, then 2x divided by 6 will have a remainder of 2*4 = 8, which gives us a remainder of 2 (since another group of 6 will be formed from the 8 balls).

Let’s consider the tricky case of x^2 now. If x gives us a remainder of 2 when it is divided by 6, it means:

x = 6Q + 2
x^2 = (6Q + 2)*(6Q + 2) = 36Q^2 + 24Q + 4

Note here that the first and the second terms are divisible by 6. The remainder when you divide this by 6 will be 4.

We hope you understand how to deal with these various cases of remainders. Let’s take a look at a GMAT sample question now:

If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r?

Statement 1: When (z−3)^2 is divided by 8, the remainder is 4.
Statement 2: When 2z is divided by 8, the remainder is 2.

This is not our typical, “When z is divided by 8, r is the remainder” type of question. Instead, we are given a quadratic equation in the form of z that, when divided by 8, gives us a remainder of r. We need to find r. This question might feel complicated, but look at the statements – at least one of them gives us data on a quadratic! Looks promising!

Statement 1: When (z−3)^2 is divided by 8, the remainder is 4

(z – 3)^2 = z^2 – 6z + 9

We know that when z^2 – 6z + 9 is divided by 8, the remainder is 4. So no matter what z is, z^2 – 6z + 9 + 8z, when divided by 8, will only give us a remainder of 4 (8z is a multiple of 8, so will give remainder 0).

z^2 – 6z + 9 + 8z = z^2 + 2z + 9

z^2 + 2z + 9 when divided by 8, gives remainder 4. This means z^2 + 2z + 5 is divisible by 8 and would give remainder 0, further implying that z^2 + 2z + 4 would be 1 less than a multiple of 8, and hence, would give us a remainder of 7 when divided by 8. This statement alone is sufficient.

Let’s look at the second statement:

Statement 2: When 2z is divided by 8, the remainder is 2

2z = 8a + 2
z = 4a + 1
z^2 = (4a + 1)^2 = 16a^2 + 8a + 1

When z^2 is divided by 8, the remainder is 1. When 2z is divided by 8, the remainder is 2. So when z^2 + 2z is divided by 8 the remainder will be 1+2 = 3.

When z^2 + 2z + 4 is divided by 8, remainder will be 3 + 4 = 7. This statement alone is also sufficient. Because both statements alone are sufficient, our answer is D.

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

# Using Parallel Lines and Transversals to Your Advantage on the GMAT

Today, we will look at a Geometry concept involving parallel lines and transversals (a line that cuts through two parallel lines). This is the property:

The ratios of the intercepts of two transversals on parallel lines is the same.

Consider the diagram below:

Here, we can see that:

• “a” is the intercept of the first transversal between L1 and L2.
• “b” is the intercept of the first transversal between L2 and L3.
• “c” is the intercept of the second transversal between L1 and L2.
• “d” is the intercept of the second transversal between L2 and L3.

Therefore, the ratios of a/b = c/d. Let’s see how knowing this property could be useful to us on a GMAT question. Take a look at the following example problem:

In triangle ABC below, D is the mid-point of BC and E is the mid-point of AD. BF passes through E. What is the ratio of AF:FC ?

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

Here, the given triangle is neither a right triangle, nor is it an equilateral triangle. We don’t really know many properties of such triangles, so that will probably not help us. We do know, however, that AD is the median and E is its mid-point, but again, we don’t know any properties of mid-points of medians.

Instead, we need to think outside the box – parallel lines will come to our rescue. Let’s draw lines parallel to BF passing through the points A, D, and C, as shown in the diagram below:

Now we have four lines parallel to each other and two transversals, AD and AC, passing through them.

Consider the three parallel lines, “line passing through A”, “BF”, and “line passing through D”. The ratio of the intercepts of the two transversals on them will be the same.

AE/ED = AF/FP

We know that AE = ED since E is the mid point of AD. Hence, AE/ED = 1/1. This means we can say:

AE/ED = 1/1 = AF/FP
AF = FP

Now consider these three parallel lines: “BF”, “line passing through D”, and “line passing through C”. The ratio of the intercepts of the two transversals on them will also be the same.

BD/DC = FP/PC

We know that BD = DC since D is the mid point of BC. Hence, BD/DC = 1/1. This means we can also say:

BD/DC = 1/1 = FP/PC
FP = PC

From these two calculations, we will get AF = FP = PC, and hence, AF:FC = 1:(1+1) = 1:2.

Therefore, the answer is B. We hope you see that Geometry questions on the GMAT can be easily resolved once we bring in parallel lines.

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 Pythagorean Triples Properties You’ll See on the GMAT

Today, let’s discuss a few useful properties of primitive Pythagorean triples. A primitive Pythagorean triple is one in which a, b and c (the length of the two legs and the hypotenuse, respectively) are co-prime. So, for example, (3, 4, 5) is a primitive Pythagorean triple while its multiple, (6, 8, 10), is not.

Now, without further ado, here are the properties of primitive Pythagorean triples that you’ll probably encounter on the GMAT:

I. One of a and b is odd and the other is even.
II. From property I, we can then say that c is odd.
III. Exactly one of a, b is divisible by 3.
IV. Exactly one of a, b is divisible by 4.
V. Exactly one of a, b, c is divisible by 5.

If you keep in mind the first primitive Pythagorean triple that we used as an example (3, 4, 5), it is very easy to remember all these properties.

If we look at some other examples:

(3, 4, 5), (5, 12, 13), (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73), etc.

we will see that these properties hold for all primitive Pythagorean triples.

Now, let’s take a look at an example GMAT question which can be easily solved if we know these properties:

The three sides of a triangle have lengths p, q and r, each an integer. Is this triangle a right triangle?

Statement 1: The perimeter of the triangle is an odd integer.
Statement 2: If the triangle’s area is doubled, the result is not an integer.

We know that the three sides of the triangle are all integers. So if the triangle is a right triangle, the three sides will represent a Pythagorean triple. Given that p, q and r are all integers, let’s use the properties of primitive Pythagorean triples to break down each of the statements.

Statement 1: The perimeter of the triangle is an odd integer.

Looking at the properties above, we know that a primitive Pythagorean triple can be represented as:

(Odd, Even, Odd) (The first two are interchangeable.)

Non-primitive triples are made by multiplying each member of the primitive triple by an integer n greater than 1. Depending on whether n is odd or even, the three sides can be represented as:

(Odd*Odd, Even*Odd, Odd*Odd) = (Odd, Even, Odd)
or
(Odd*Even, Even*Even, Odd*Even) = (Even, Even, Even)

However, the perimeter of a right triangle can never be odd because:

Odd + Even + Odd = Even
Even + Even + Even = Even

Hence, the perimeter will be even in all cases. (If the perimeter of the given triangle is odd, we can say for sure that it is not a right triangle.) This statement alone is sufficient.

Statement 2: If the triangle’s area is doubled, the result is not an integer.

If p, q and r are the sides of a right triangle such that r is the hypotenuse (the hypotenuse could actually be either p, q, or r but for the sake of this example, let’s say it’s r), we can say that:

The area of this triangle = (1/2)*p*q
and
Double of area of this triangle = p*q

Double the area of the triangle has to be an integer because we are given that both p and q are integers, but this statement tells us that this is not an integer. In that case, this triangle cannot be a right triangle. If the triangle is not a right triangle, double the area would be the base * the altitude, and the altitude would not be an integer in this case.

This statement alone is sufficient, too. Therefore, our answer is D.

As you can see, understanding the special properties of primitive Pythagorean triples can come in handy on the GMAT – especially in tackling complicated geometry 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!

# How to Quickly Interpret Ranges of Variables in GMAT Questions

Sometimes a GMAT Quant question will give us multiple ranges of values that apply to a single variable, and when this happens it can really take us for a ride. Evaluating these ranges to arrive at deductions can extremely confusing, so today we will look at some strategies for how to deal with such problems.

To start off, let’s take a look at an example problem:

If it is true that z < 8 and 2z > -4, which of the following must be true?

(A) -8 < z < 4
(B) z > 2
(C) z > -8
(D) z < 4
(E) None of the above

Given that z < 8 and 2z > -4, we know that z > -2. This means -2 < z < 8. z must lie within that range, hence z can take values such as -1, 0, 5, 7.4, etc.

Now, which of the given answer choices would hold true for ALL such values? Let’s examine each option and see:

(A) -8 < z < 4
We know that z may be more than 4, so this range does not hold true for all possible values of z.

(B) z > 2
We know that z may be less than 2, so this also does not hold true for all possible values of z.

(C) z > -8
No matter what value z will take, it will always be more than -8. This range holds true for all values of z.

(D) z < 4
We know that z may be greater than 4, so this does not hold for all possible values of z.

To understand this concept more clearly, let’s use a real life example:

We know that Anna’s weight is more than 120 pounds but less than 130 pounds. Which of the following is definitely true about her weight?

(A) Her weight is 125 pounds.
(B) Her weight is more than 124 pounds.
(C) Her weight is less than 127 pounds.
(D) Her weight is more than 110 pounds.

Can we say that her weight is 125 pounds? No – we just know that it is more than 120 but less than 130. It could be anything in this range, such as 122, 125, 127.5, etc.

Can we say that her weight is more than 124 pounds? This may be true, but it might not be true. Knowing our given range, her weight could very well be 121 pounds, instead.

Can we say her weight is less than 127 pounds? Again, this might not necessarily be true. Her weight could be 128 pounds.

Now, can we say that her weight is more than 110 pounds? Yes – since we know Anna’s weight is between 120 and 130 pounds, it must be more than 110 pounds.

This question uses the same concept as the first question! If you look at that question again, it will hopefully make much more sense. Now try solving this example problem:

If 1/55 < x < 1/22 and 1/33 < x < 1/11, then which of the following could be the value of x?

(i) 1/54
(ii) 1/23
(iii) 1/12

(A) Only (i)
(B) Only (ii)
(C) (i) and (ii)
(D) (ii) and (iii)
(E) (i), (ii) and (iii)

In this problem, we are given two ranges of x. We know that 1/55 < x < 1/22 and 1/33 < x < 1/11, so x is greater than 1/55 AND it is greater than 1/33. Since 1/33 is greater than 1/55 (the smaller the denominator, the larger the number), we just need to know that x will be greater than 1/33.

We are also given that x is less than 1/22 AND it is less than 1/11. Since 1/22 is less than 1/11, we really just need to know that x is less than 1/22.

Hence, the range for x should be 1/33 < x < 1/22. x could take all values that lie within this range, such as 1/32, 1/31, 1/24, 1/23, etc.

Looking at the answer choices, we can see that 1/54 and 1/12 (i and iii) are both out of this range. Therefore, our answer is B.

If we go back to our real life example, this is what the question would look like now:

We know that Anna’s weight is more than 110 pounds but less than 130 pounds. We also know that her weight is more than 115 pounds but less than 140 pounds. Which of the following is definitely true about her weight?

(A) Her weight is 112 pounds.
(B) Her weight is 124 pounds.
(C) Her weight is 135 pounds.

We are given that Anna’s weight is more than 110 pounds and also more than 115 pounds. Since 115 is more than 110, we just need to know that her weight is more than 115 pounds. We are also given that Anna’s weight is less than 130 pounds and also less than 140 pounds. Since 130 is less than 140, we just need to know that her weight is less than 130 pounds.

Now we have the following range: 115 pounds < Anna’s weight < 130 pounds. Only answer choice B lies within this range, so that is our answer.

We hope you see that evaluating ranges of numbers on GMAT questions is not difficult when we consider them in terms of a real life example. The same logic that we use in the simple weight problem is also applicable when algebraic data is given.

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

# Quarter Wit, Quarter Wisdom: The 3-Step Method to Solving Complex GMAT Algebra Problems

If you have been practicing GMAT questions for a while, you will realize that not every question can be solved using pure algebra, especially at higher levels. There will be questions that will require logic and quite a bit of thinking on your part.  These questions tend to throw test-takers off – students often complain, “Where do I start from? Thinking through the question takes too much time!” Unfortunately, there is no getting away from such questions.

Today, let’s see how to handle such questions step-by-step by looking at an example problem:

N and M are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either N or M. What is the smallest possible positive difference between N and M?

(A) 29
(B) 49
(C) 58
(D) 113
(E) 131

This is not a simple algebra question, where we are asked to make equations and solve them.

We are given 6 digits: 1, 2, 3, 6, 7, 8. Each digit needs to be used to form two 3-digit numbers. This means that we will use each of the digits only once and in only one of the numbers.

We also need to minimize the difference between the two numbers so they are as close as possible to each other. Since the numbers cannot share any digits, they obviously cannot be equal, and hence, the smaller number needs to be as large as possible and the greater number needs to be as small as possible for the numbers to be close to each other.

Think of the numbers  of a number line. You need to reduce the difference between them. Then, under the given constraints, push the smaller number to the right on the number line and the greater number to the left to bring them as close as possible to each other.

STEP 1:
The first digit (hundreds digit) of both numbers should be consecutive integers – i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3** (the difference between the latter will certainly be more than 100).

We get lots of options for hundreds digits: (1** and 2**) or (2** and 3**) or (6** and 7**) or (7** and 8**). All of these options could satisfy our purpose.

STEP 2:
Now let’s think about what the next digit (the tens digit) should be. To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1, if possible) and the tens digit of the smaller number should be as large as possible (8, if possible). So let’s not use 1 or 8 in the hundreds places and reserve them for the tens places instead, since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?

Let’s try to make a pair of numbers in the form of 2** and 3**. We need to make the 2** number as large as possible and make the 3** number as small as possible. As discussed above, the tens digit of the smaller number should be 8 and the tens digit of the greater number should be 1. We now have 28* and 31*.

STEP 3:
Now let’s use the same logic for the units digit – make the units digit of the smaller number as large as possible and the units digit of the greater number as small as possible. We have only two digits left over – 6 and 7.

The two numbers could be 287 and 316 – the difference between them is 29.

Let’s try the same logic on another pair of hundreds digits, and make the pair of numbers in the form of 6** and 7**. We need the 6** number to be as large as possible and the 7** number to be as small as possible. Using the same logic as above, we’ll get 683 and 712. The difference between these two is also 29.

The smallest of the given answer choices is 29, so we need to think no more. The answer must be A.

Note that even if you try to express the numbers algebraically as:

N = 100a + 10b + c
M = 100d + 10e + f

a lot of thought will still be needed to find the answer, and there is no real process that can be followed.

Assuming N is the greater number, we need to minimize N – M.

N – M = 100 (a – d) + 10( b – e) + (c – f)

Since a and d cannot be the same, the minimum value a – d can take is 1. (a – d) also cannot be negative because we have assumed that N is greater than M. With this in mind, a and d must be consecutive (2 and 1, or 3 and 2, or 7 and 6, etc). This is another way of completing STEP 1 above.

Next, we need to minimize the value of (b – e). From the available digits, 1 and 8 are the farthest from each other and can give us a difference of -7. So b = 1 and e = 8. This leaves the consecutive pairs of 2, 3 and 6, 7 for hundreds digits. This takes care of our STEP 2 above.

(c – f) should also have a minimum value. We have only one pair of digits left over and they are consecutive, so the minimum value of (c – f) is -1. If the hundreds digits are 3 and 2, then c = 6 and f = 7. This is our STEP 3.

So, the pair of numbers could be 316 and 287 – the difference between them is 29. The pair of numbers could also be 712 and 683 – the difference between them is also 29.

In either case, note that you do not have a process-oriented approach to solving this problem. A bit of higher-order thinking is needed to find the correct answer.

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

# Quarter Wit, Quarter Wisdom: When Can You Divide by a Variable?

We have often come across test takers confused about division by a variable. When is it allowed, when is it not allowed? Why is it allowed in some cases and not in others? What are the constraints we need to look out for?

For example:

Is division by x allowed here: x^2 = 10x?
Is division by x allowed here: y = 4x?
Is division by x allowed here: x^2 < 4x?

Let’s take a detailed look at all these questions today.

The basic guidelines:

1. Division by 0 is not allowed, hence you cannot divide by a variable until and unless we know that it cannot be 0.
2. In the case of an inequality, when you divide by a negative number, the sign of the inequality flips. So we cannot divide by a variable until and unless we know that it cannot be 0 AND whether it is positive or negative.

Let’s look at the three questions given above and try to solve them using these guidelines:

Is division by x allowed here: x^2 = 10x?

The first thing to find out here is whether or not x can equal 0.

Case 1: If no other information has been given, then x can be 0 and we cannot divide by it. This is how we proceed in that case:

x^2 – 10x = 0
x(x – 10) = 0
x = 0 or 10

Case 2: If the question stem tells us that x is not 0, then we can divide by x.

x^2/x = 10x/x
x = 10

Obviously, we don’t get the second solution (x = 0) in this case, as we already know that x cannot be 0. Now let’s look at the second problem:

Is division by x allowed here: y = 4x?

Again, this is an equation and we need to know whether or not x can equal 0.

Case 1: If x can be 0, you cannot divide by it. In this case, x = 0 and y = 0 is one of the infinite possible solutions.

Case 2: If the question stem states that x cannot be 0, then we can do the following:

y/x = 4

Now let’s look at the final question:

Is division by x allowed here: x^2 > -4x?

Here, we have an inequality. Before deciding whether we can divide by x or not, we need to know not only whether x can be equal to 0, but also whether x is positive or negative.

Case 1: If we know nothing about the possible values that x can take, then this is how we proceed:

x^2 + 4x > 0
x(x + 4) > 0

Now we can use the method discussed in the first problem to arrive at the range of x.

x > 0 or x < -4

Case 2: If we know that x is positive, then we can proceed like this:

x^2/x > -4x/x
x > -4

Since we are given that x is positive, we know that that x > 0 (looking at the two options above).

Case 3: If we know that x is negative, then this is how we will proceed:

x^2/x < -4x/x (we flip the sign of the inequality because we divide by x, which is negative)
x < -4

The results obtained are logical, right? When x can be anywhere on the number line, we get the range as x > 0 or x < -4.

If x has to be positive, the range is x > 0.
If x has to be negative, the range is x < -4.

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!

# GMAT Tip of the Week: Evolving Your GMAT Quant Score with Help from The Evolution Of Rap

If it’s March, it must be Hip Hop Month at the GMAT Tip of the Week space, where this year we’ve been transfixed by Vox’s video on the evolution of rhyme schemes in the rap world.

And if you don’t have the study break time right now, we’ll summarize. While a standard rhyme might have a one-syllable rhyme at the end of each bar (do you like green eggs and HAM, yes I like them Sam I AM), rappers have continued to evolve to the point where nowadays each bar can contain multiple rhyme schemes. Consider Eminem’s “Lose Yourself”:

Snap back to reality, oh there goes gravity
Oh there goes Rabbit he choked, he’s so mad but he won’t
Give up that easy, nope, he won’t have it he knows
His whole back’s to these ropes, it don’t matter he’s dope
He knows that but he’s broke, he’s so stagnant he knows…

Where “gravity,” “Rabbit, he,” “mad but he,” “that easy,” “have it he,” “back’s to these,” “matter he’s,” “that but he’s,” and “stagnant, he” all rhyme with one another, the list of goes/goes/choked/so/won’t/knows/whole/ropes/don’t/dope… keeps that hard “O” sound rhyming consistently throughout, too. And that was 15 years ago…since them, Eminem, Kendrick, and others have continued to build elaborate rhyme schemes that reward those listeners who don’t just listen for the simple rhyme at the end of each bar, but pick up the subtle rhyme flows that sometimes don’t come back until a few lines later.

So what does this have to do with your GMAT score?

One of the most common study mistakes that test-takers make is that they study skills as individual, standalone entities, and don’t look for the subtle ways that the GMAT testmaker can layer in those sophisticated Andre-3000-style combinations. Consider an example of an important GMAT skill, the “Difference of Squares” rule that (x + y)(x – y) = x^2 – y^2. A standard (think early 1980s Sugarhill Gang or Grandmaster Flash) GMAT question might test it in a relatively “obvious” way:

What is the value of (x + y)?

(1) x^2 – y^2 = 0
(2) x does not equal y

Here if you factor Statement 1 you’ll get (x + y)(x – y) = 0, and then Statement 2 tells you that it’s not (x – y) that equals zero, so it must be x + y. This Data Sufficiency answer is C, and the test is essentially just rewarding you for knowing the Difference of Squares.

The GMAT it cares
’bout the Difference of Squares
When there’s squares and subtraction
Put this rule into action

A slightly more sophisticated question (think late 1980s/early 1990s Rob Bass or Run DMC) won’t so obviously show you the Difference of Squares. It might “hide” that behind a square that few people tend to see as a square, the number 1:

If y = 2^(16) – 1, the greatest prime factor of y is:

(A) Less than 6
(B) Between 6 and 10
(C) Between 10 and 14
(D) Between 14 and 18
(E) Greater than 18

Here, many people don’t recognize 1 as a perfect square, so they don’t see that the setup is 2^(16) – 1^(2), which can be factored as:

(2^8 + 1)(2^8 – 1)

And that 2^8 – 1 can be factored again, since 1 remains 1^2:

(2^8 + 1)(2^4 + 1)(2^4 – 1)

And that ultimately you could do it again with 2^4 – 1 if you wanted, but you should know that 2^4 is 16 so you can now get to work on smaller numbers. 2^8 is 256 and 2^4 is 16, so you have:

257 * 17 * 15

And what really happens now is that you have to factor out 257 to see if you can break it into anything smaller than 17 as a factor (since, if not, you can select “greater than 18”). Since you can’t, you know that 257 must have a prime factor greater than 18 (it turns out that it’s prime) and correctly select E.

The lesson here? This problem directly tests the Difference of Squares (you don’t want to try to calculate 2^16, then subtract 1, then try to factor out that massive number) but it does so more subtly, layering it inside the obvious “prime factor” problem like a rapper might embed a secondary rhyme scheme in the middle of each bar.

But in really hard problems, the testmaker goes full-on Greatest of All Time rapper, testing several things at the same time and rewarding only the really astute for recognizing the game being played. Consider:

The size of a television screen is given as the length of the screen’s diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

(A) 2
(B) 4
(C) 16
(D) 38
(E) 40

Now here you KNOW you’re dealing with a geometry problem, and it also looks like a word problem given the television backstory. As you start calculating, you’ll know that you have to take the diagonal of each square TV and use that to determine the length of each side, using the 45-45-90 triangle ratio, where the diagonal = x√2. So the length of a side of the smaller TV is 19/√2 and the length of a side of the larger TV is 21/√2.

Then you have to calculate the area, which is the side squared, so the area of the smaller TV is (19/√2)^2 and the area of the larger TV is (21/√2)^2. This is starting to look messy (Who knows the squares for 21 and 19 offhand? And radicals in denominators never look fun…) UNTIL you realize that you have to subtract the two areas. Which means that your calculation is:

(21/√2)^2 – (19/√2)^2

This fits perfectly in the Difference of Squares formula, meaning that you can express x^2 – y^2 as (x + y)(x – y). Doing that, you have:

[(21 + 19)/√2][(21-19)/√2]

Which is really convenient because the math in the numerators is easy and leaves you with:

(40/√2) * (2/√2)

And when you multiply them, the √2 terms in the denominators square out to 2, which factors with the 2 in the numerator of the right-side fraction, and everything simplifies to 40. And then, in classic “oh this guy’s effing GOOD” hip-hop style (like in the Eminem lyric “you’re witnessing a massacre like you’re watching a church gathering take place” and you realize that he’s using “massacre” and “mass occur” – the church gathering taking place – simultaneously), you realize that you should have seen it coming all along. Because when you subtract the area of one square minus the area of another square you’re LITERALLY taking the DIFFERENCE of two SQUARES.

So what’s the point?

Too often people study for the GMAT like they’d listen to 1980s rap. They expect the Difference of Squares to pair nicely at the end of an Algebra-with-Exponents bar, and the Isosceles Right Triangle formula to pair nicely with a Triangle question. They learn skills in distinct silos, memorize their flashcards in nice, tidy sets, and then go into the test and realize that they’re up against an exam that looks a lot more like a 2017 mixtape with layers of rhyme schemes and motives.

You need to be prepared to use skills where they don’t seem to obviously belong, to jot down and rearrange your scratchwork, label your unknowns, etc., looking for how you might reposition the math you’re given to help you bring in a skill or concept that you’ve used countless times, just in totally different contexts. The GMAT testmaker has a much more sophisticated flow than the one you’re likely studying for, so pay attention to that nuance when you study and you’ll have a much better chance of keeping your score 800.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on Facebook, YouTubeand Twitter!

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: When a Little Information is Enough to Solve a GMAT Problem

We have reviewed what standard deviation is in a past post. We know what data is necessary to calculate the standard deviation of a set, but in some cases, we could actually do with a lot less information than the average test-taker may think they need.

Let’s explore this idea through an example GMAT data sufficiency question:

What is the standard deviation of a set of numbers whose mean is 20?

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.
Statement 2: The sum of the squares of the differences from the mean is greater than 100.

We need to determine whether the information we have been given is sufficient to get us the exact value of the standard deviation of a particular set of numbers. To find the standard deviation of a set, we need to know the deviation of each term from the mean so that we can square those deviations, sum the squares, divide them by the number of terms, and then find the square root.

Essentially, to find the standard deviation we either need to know each element of the set, or we need to know the deviation of each element from the mean (which will also give us the number of terms), or we need to know the sum of the square of deviations and the number of terms in the set.

The question stem here tells us that the mean of the set is 20. We have no other information about any of the actual elements of the set or the number of elements. With this in mind, let’s examine each of the statements:

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.

With this statement, we don’t actually know what the absolute value of the difference is. We also don’t know how many elements there are. The set could be something like:

19, 21 (each term is exactly 1 away from the mean 20)
or
18, 18, 22, 22 (each term is exactly 2 away from the mean 20)
etc.

The standard deviation in each case will be different. We don’t know the elements of the set and we don’t know the number of elements in the set. Because of this, there is no way for us to know the value of the standard deviation – this statement alone is not sufficient.

Statement 2: The sum of the squares of the differences from the mean is greater than 100.

“Greater than 100” encompasses a large range of numbers – it could be any value larger than 100. Again, we cannot find the exact standard deviation of the set, so this statement is also not sufficient alone.

Using both statements together, we still do not have any idea of what the elements of the set are or what the sum of the squares of the differences from the mean is. We also still don’t know the number of elements. Hence, both statements together are not sufficient, so the answer is E.

Now, let us add just one more piece of information to the problem in this similar question:

What is the standard deviation of a set of 7 numbers whose mean is 20?

Statement 1: The absolute value of the difference of each number in the set from the mean is equal.
Statement 2: The sum of the squares of the differences from the mean is greater than 100.

What would you expect the answer to be? Still E, right? The sum of the deviations are still unknown and the exact elements of the set are still unknown – all we know is the number of elements. Actually, this information is already too much. All we need to know is that the number of elements is odd and suddenly we can find the standard deviation.

Here is why:

Statement 1 is quite tricky.

If we have an odd number of elements, in which case can the absolute values of the differences of each number in the set from the mean be equal?

Think about it – the mean of the set is 20. What could a possible set look like such that the mean is 20 and the absolute values of the differences of each number in the set from the mean are equal. Try to think of such a set with just 3 elements. Can you come up with one?

19, 19, 21? No, the mean is not 20

19, 20, 21? No, the absolute value of the difference of each number in the set from the mean is not equal. 19 is 1 away from mean but 20 is 0 away from mean.

Note that in this case, the only possible set that could fit the given criteria is one consisting of just an odd number of 20s (all elements in this set must be 20). Only then can each number be equidistant from the mean, i.e. each number would be 0 away from mean. If the numbers of the set all have equal elements, then obviously the standard deviation of the set is 0. It doesn’t matter how many elements it has; it doesn’t matter what the mean is! In this case, Statement 1 alone is sufficient so the answer would be A.

Takeaway:
If a set has an even number of distinct terms, the absolute values of the distances of each term from the mean could be equal. But if a set has an odd number of terms and the absolute values of the distances of each term from the mean are equal, all the terms in the set must be the same and will be equal to the mean.

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

# Quarter Wit Quarter Wisdom: How to Read GMAT Questions Carefully

We all know that we need to be very careful while reading GMAT questions – that every word is important. Even small oversights can completely change an answer for you. This is what happens with many test takers who try to tackle this official question. Even though the question looks very simple, the way it is worded causes test-takers to often ignore one word, which changes the solution entirely. Let’s look at this question now:

Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month. The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save. If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Let’s consider the question stem sentence by sentence:

“Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month.”

Say Alice’s take-home pay last year was \$100 each month. She saves a fraction of this every month – let the amount saved be x.

“The total amount of money that she had saved at the end of the year was 3 times the amount of that portion of her monthly take-home pay that she did NOT save.”

What would be “the total amount of money that she had saved at the end of the year”? Since Alice saves x every month, she would have saved 12x by the end of the year.

What would be “the amount of that portion of her monthly take-home pay that she did NOT save”? Note that this is going to be (100 – x). Many test takers end up using (100 – x)*12, however this equation is not correct. The key word here is “monthly” – we are looking for how much Alice does not save each month, not how much she does not save during the whole year.

The total amount of money that Alice saved at the end of the year is 3 times the amount of that portion of her MONTHLY take-home pay that she did not save. Now we know we are looking for:

12x = 3*(100 – x)
x = 20

“If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?”

From our equation, we have determined that Alice saved \$20 out of every \$100 she earned every month, so she saved 20/100 = 1/5 of her take-home pay.

Often, test-takers make the mistake of writing the equation as:

12x = 3*(100 – x)*12
x = 300/4

However, this will give them the fraction (300/4)/100 = 3/4, and that’s when they will wonder what went wrong.

Be extra careful when reading GMAT questions so that precious minutes are not wasted on such avoidable errors.

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

# Quarter Wit Quarter Wisdom: Solving the Weighing Puzzle (Part 2)

A couple of weeks back, we discussed how to handle puzzles involving a two pan balance. In those problems, we learned how to tackle problems that ask you to measure items against one another.

Today, we will look at some puzzles that require the use of a traditional weighing scale. When we put an object on this scale, it shows us the weight of the object.

This is what such a scale looks like:

Puzzles involving a weighing scale can be quite tricky! Let’s take a look at a couple of examples:

You have 10 bags with 1000 coins in each. In one of the bags, all of the coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 grams.

If you have an accurate weighing scale, which you can use only once, how can you identify the bag with the forgeries?

We are allowed only a single weighing, so we cannot weigh all 10 bags on the scale individually to measure which one has counterfeit coins. We need to find the bag in only one weighing, so we need to somehow make the coins in the bags distinctive.

How do we do that? We can take out one coin from the first bag, two coins from the second bag, three coins from the third bag and so on. Finally, we will have 1 + 2 + 3 + … + 10 = 10*11/2 = 55 coins.

Let’s weigh these 55 coins now.

If all coins were true, the total weight would have been 55 grams. But since some coins are counterfeit, the total weight will be more. Say, the total weight comes out to be 55.2 grams. What can we deduce from this? We can deduce that there must be two counterfeit coins (because each counterfeit coin weighs 0.1 gram extra). So the second bag must be the bag of counterfeit coins.

Let’s try one more:

A genuine gummy bear has a mass of 10 grams, while an imitation gummy bear has a mass of 9 grams. You have 7 cartons of gummy bears, 4 of which contain real gummy bears while the others contain imitation bears.

Using a scale only once and the minimum number of gummy bears, how can you determine which cartons contain real gummy bears?

Now this has become a little complicated! There are three bags with imitation gummy bears. Taking a cue from the previous question, we know that we should take out a fixed number of gummy bears from each bag, but now we have to ensure that the sum of any three numbers is unique. Also, we have to keep in mind that we need to use the minimum number of gummy bears.

So from the first bag, take out no gummy bears.

From the second bag, take out 1 gummy bear.

From the third bag, take out 2 gummy bears (if we take out 1 gummy bear, the sum will be the same in case the second bag has imitation gummy bears or in case third bag has imitation gummy bears.

From the fourth bag, take out 4 gummy bears. We will not take out 3 because otherwise 0 + 3 and 1 + 2 will give us the same sum. So we won’t know whether the first and fourth bags have imitation gummy bears or whether second and third bags have imitation gummy bears.

From the fifth bag, take out 7 gummy bears. We have obtained this number by adding the highest triplet: 1 + 2 + 4 = 7. Note that anything less than 7 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 6 = 7 and 1 + 2 + 4 = 7
or
0 + 1 + 5 = 6 and 0 + 2 + 4 = 6

But we need the sum to be obtainable in only one way so that we can find out which three bags contain the imitation gummy bears.

At this point, we have taken out 0, 1, 2, 4, and 7 gummy bears.

From the sixth bag, take out 13 gummy bears. We have obtained this number by adding the highest triplet: 2 + 4 + 7 = 13. Note that anything less than 13 will, again, give us a sum that can be made in multiple ways, such as:

12 + 1 + 0 = 13 and 2 + 4 + 7 = 13
or
0 + 1 + 9 = 10 and 1 + 2 + 7 = 10
…etc.

Note that this way, we are also ensuring that we measure only the minimum number of gummy bears, which is what the question asks us to do.

From the seventh bag, take out 24 gummy bears. We have obtained this number by adding the highest triplet again: 4 + 7 + 13 = 24. Again, anything less than 24 will give us a sum that can be made in multiple ways, such as:

0 + 1 + 15 = 16 and 1 + 2 + 13 = 16
or
0 + 1 + 19 = 20 and 0 + 7 + 13 = 20
or
0 + 1 + 23 = 24 and 4 + 7 + 13 = 24
…etc.

Thus, this is the way we will pick the gummy bears from the 7 bags: 0, 1, 2, 4, 7, 13, 24.

In all, 51 gummy bears will be weighed. Their total weight should be 510 grams (51*10 = 510) but because three bags have imitation gummy bears, the weight obtained will be less.

Say the weight is less by 8 grams. This means that the first bag (which we pulled 0 gummy bears from), the second bag (which we pulled 1 gummy bear from) and the fifth bag (which we pulled 7 gummy bears from) contain the imitation gummy bears. This is because 0 + 1 + 7 = 8 – note that we will not be able to make 8 with any other combination.

We hope this tricky little problem got you thinking. Work those grey cells and the GMAT will not seem hard at all!

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!

# GMAT Tip of the Week: Keep Your GMAT Score Safe from the Bowling Green Massacre

The hashtag of the day is #bowlinggreenmassacre, inspired by an event that never happened. Whether intentionally or accidentally (we’ll let you and your news agency of choice decide which), White House staffer Kellyanne Conway referenced the “event” in an interview, inspiring an array of memes and references along the way.

Whatever Ms. Conway’s intentions (or lack thereof; again we’ll let you decide) with the quote, she is certainly guilty of inadvertently doing one thing: she didn’t likely intend to help you avoid a disaster on the GMAT, but if you’re paying attention she did.

Your GMAT test day does not have to be a Bowling Green Massacre!

Here’s the thing about the Bowling Green Massacre: it never happened. But by now, it’s lodged deeply enough in the psyche of millions of Americans that, to them, it did. And the same thing happens to GMAT test-takers all the time. They think they’ve seen something on the test that isn’t there, and then they act on something that never happened in the first place. And then, sadly, their GMAT hopes and dreams suffer the same fate as those poor souls at Bowling Green (#thoughtsandprayers).

Here’s how it works:

The Quant Section’s Bowling Green Massacre
On the Quant section, particularly with Data Sufficiency, your mind will quickly leap to conclusions or jump to use a rule that seems relevant. Consider the example:

What is the perimeter of isosceles triangle LMN?

(1) Side LM = 4
(2) Side LN = 4√2

A. Statement (1) ALONE is sufficient, but statement (2) alone is insufficient
B. Statement (2) ALONE is sufficient, but statement (1) alone is insufficient
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient

When people see that square root of 2, their minds quickly drift back to all those flash cards they studied – flash cards that include the side ratio for an isosceles right triangle: x, x, x√2. And so they then leap to use that rule, inferring that if one side is 4 and the other is 4√2, the other side must also be 4 to fit the ratio and they can then calculate the perimeter. With both statements together, they figure, they can derive that perimeter and select choice C.

But think about where that side ratio comes from: an isosceles right triangle. You’re told in the given information that this triangle is, indeed, isosceles. But you’re never told that it’s a right triangle. Much like the Bowling Green Massacre, “right” never happened. But the mere suggestion of it – the appearance of the √2 term that is directly associated with an isosceles, right triangle – baits approximately half of all test-takers to choose C here instead of the correct E (explanation: “isosceles” means only that two sides match, so the third side could be either 4, matching side LM, or 4√2, matching side LN).

Your mind does this to you often on Data Sufficiency problems: you’ll limit the realm of possible numbers to integers, when that wasn’t defined, or to positive numbers, when that wasn’t defined either. You’ll see symptoms of a rule or concept (like √2 leads to the isosceles right triangle side ratio) and assume that the entire rule is in play. The GMAT preys on your mind’s propensity for creating its own story when in reality, only part of that story really exists.

The Verbal Section’s Bowling Green Massacre
This same phenomenon appears on the Verbal section, too – most notably in Critical Reasoning. Much like what many allege that Kellyanne Conway did, your mind wants to ascribe particular significance to events or declarations, and it will often exaggerate on you. Consider the example:

About two million years ago, lava dammed up a river in western Asia and caused a small lake to form. The lake existed for about half a million years. Bones of an early human ancestor were recently found in the ancient lake-bottom sediments that lie on top of the layer of lava. Therefore, ancestors of modern humans lived in Western Asia between two million and one-and-a-half million years ago.

Which one of the following is an assumption required by the argument?

A. There were not other lakes in the immediate area before the lava dammed up the river.
B. The lake contained fish that the human ancestors could have used for food.
C. The lava that lay under the lake-bottom sediments did not contain any human fossil remains.
D. The lake was deep enough that a person could drown in it.
E. The bones were already in the sediments by the time the lake disappeared.

The key to most Critical Reasoning problems is finding the conclusion and knowing EXACTLY what the conclusion says – nothing more and nothing less. Here the conclusion is the last sentence, that “ancestors of modern humans lived” in this region at this time. When people answer this problem incorrectly, however, it’s almost always for the same reason. They read the conclusion as “the FIRST/EARLIEST ancestors of modern humans lived…” And in doing so, they choose choice C, which protects against humans having come before the ones related to the bones we have.

“First/earliest” is a classic Bowling Green Massacre – it’s a much more noteworthy event (“scientists have discovered human ancestors” is pretty tame, but “scientists have discovered the FIRST human ancestors” is a big deal) that your brain wants to see. But it’s not actually there! It’s just that, in day to day life, you’d rarely ever read about a run-of-the-mill archaeological discovery; it would only pop up in your social media stream if it were particularly noteworthy, so your mind may very well assume that that notoriety is present even when it’s not.

In order to succeed on the GMAT, you need to become aware of those leaps that your mind likes to take. We’re all susceptible to:

• Assuming that variables represent integers, and that they represent positive numbers
• Seeing the symptoms of a rule and then jumping to apply it
• Applying our own extra superlatives or limits to conclusions

So when you make these mistakes, commit them to memory – they’re not one-off, silly mistakes. Our minds are vulnerable to Bowling Green Massacres, so on test day #staywoke so that your score isn’t among those that are, sadly, massacred.

By Brian Galvin.

# Quarter Wit, Quarter Wisdom: Solving the Hourglass Puzzle

Let’s continue our puzzles discussion today with another puzzle type – time measurement using an hourglass. (Before you continue reading this article, check out our posts on how to solve pouring water puzzles and weighing and balancing puzzles)

First, understand what an hourglass is – it is a mechanical device used to measure the passage of time. It is comprised of two glass bulbs connected vertically by a narrow neck that allows a regulated trickle of sand from the upper bulb to fall into the lower one. The sand also takes a fixed amount of time to fall from the upper bulb to the lower bulb. Hourglasses may be reused indefinitely by inverting the bulbs once the upper bulb is empty.

This is what they look like:

Say a 10-minute hourglass will let us measure time in intervals of 10 minutes. This means all of the sand will flow from the upper bulb to the lower bulb in exactly 10 minutes. We can then flip the hourglass over – now sand will start flowing again for the next 10 minutes, and so on. We cannot measure, say, 12 minutes using just a 10-minute hourglass, but we can measure more time intervals when we have two hourglasses of different times. Let’s look at this practice problem to see how this can be done:

A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He used a 7-minute and an 11-minute hourglass. During the whole time, he turned the hourglasses only 3 times (turning both hourglasses at once counts as one flip). Explain how the teacher measured out 15 minutes.

Here, we have a 7-minute hourglass and an 11-minute hourglass. This means we can measure time in intervals of 7 minutes as well as in intervals of 11 minutes. But consider this: if both hourglasses start together, at the end of 7 minutes, we will have 4 minutes of sand leftover in the top bulb of the 11-minute hourglass. So we can also measure out 4 minutes of time.

Furthermore, if we flip the 7-minute hourglass over at this time and let it flow for that 4 minutes (until the sand runs out of the top bulb of the 11-minute hourglass), we will have 3 minutes’ worth of sand leftover in the 7-minute hourglass. Hence, we can measure a 3 minute time interval, too, and so on…

Now, let’s see how we can measure out 15 minutes of time using our 7-minute and 11-minute hourglasses.

First, start both hourglasses at the same time. After the top bulb of the 7-minute hourglass is empty, flip it over again. At this time, we have 4 minutes’ worth of sand still in the top bulb of the 11-minute hourglass. When the top bulb of the 11-minute hourglass is empty, the bottom bulb of 7-minute hourglass will have 4 minutes’ worth of sand in it. At this point, 11 minutes have passed

Now simply flip the 7-minute hourglass over again and wait until the sand runs to the bottom bulb, which will be in 4 minutes.

This is how we measure out 11 + 4 = 15 minutes of time using a 7-minute hourglass and an 11-minute hourglass.

Let’s look at another problem:

Having two hourglasses, a 7-minute one and a 4-minute one, how can you correctly time out 9 minutes?

Now we need to measure out 9 minutes using a 7-minute hourglass and a 4-minute hourglass. Like we did for the last problem, begin by starting both hourglasses at the same time. After 4 minutes pass, all of the sand in the 4-minute hourglass will be in the lower bulb. Now flip this 4-minute hourglass back over again. In the 7-minute hourglass, there will be 3 minutes’ worth of sand still in the upper bulb.

After 3 minutes, all of the sand from the 7-minute hourglass will be in the lower bulb and 1 minute’s worth of sand will be in the upper bulb of the 4-minute hourglass.

This is when we will start our 9-minute interval.

The 1 minute’s worth of sand will flow to the bottom bulb of the 4-minute hourglass. Then we just need to flip the 4-minute hourglass over and let all of the sand flow out (which will take 4 minutes), and then flip the hourglass over to let all of the sand flow out again (which will take another 4 minutes).

In all, we have measured out a 1 + 4 + 4 = 9-minute interval, which is what the problem has asked us to find.

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

# Quarter Wit, Quarter Wisdom: Solving the Weighing and Balancing Puzzle

Let’s continue the discussion on puzzles that we began last week. Today we look at another kind of puzzle – weighing multiple objects using a two-pan balance while we are given a limited number of times to weight the objects against each other.

First of all, do we understand what a two-pan balance looks like?

Here is a picture.

As you can see, it has two pans that will be even if the weights in them are equal. If one pan has heavier objects in it, that pan will go down due to the weight. With this in mind, let’s try our first puzzle:

One of twelve coins is a bit lighter than the other 11 (which have the same weight). How would you identify this lighter coin if you could use a two-pan balance scale only 3 times? (You can only balance one set of coins against another, i.e. you have no weight measurements.)

There are various ways in which we can solve this.

We are given 12 coins, all of same weight, except one which is a bit lighter.

Let’s split the coins into two groups of 6 coins each and put them in the two pans. Since there is one lighter coin, one pan will be lighter than the other and will rise higher. So now we know that one of these 6 coins is the lighter coin.

Now split these 6 coins into another two groups of 3 coins each. Again, one pan will rise higher since it will have the lighter coin on it. Now we know that one of these three coins is the lighter coin.

Now what do we do? We have 3 coins and we cannot split them equally. What we can do is put one coin in each pan. What happens if the pans are not balanced? Then we know the pan that rises higher has the lighter coin on it (and thus, we have identified our coin). But what if both pans are balanced? The catch is that then the leftover coin is the lighter one! In any case, we would be able to identify the lighter coin using this strategy.

We hope you understand the logic here. Now let’s try another puzzle:

One of 9 coins is a bit lighter than the other 8. How would you identify this lighter coin if you could use a two-pan balance scale only 2 times?

Now we can use the balance only twice, and we are given an odd number of coins so we cannot split them evenly. Recall what we did in the first puzzle when we had an odd number of coins – we put one coin aside. What should we do here? Can we try putting 1 coin aside and splitting the rest of the 8 coins into two groups of 4 each? We can but once we have a set of 4 coins that contain the lighter coin, we will still need 2 more weighings to isolate the light coin, and we only have a total of 2 weighings to use.

Instead, we should split the 9 coins into 3 groups of 3 coins each. If we put one group aside and put the other two groups into the two pans of the scale, we will be able to identify the group which has the lighter coin. If one pan rises up, then that pan is holding the lighter coin; if the pans weight the same, then the group put aside has the lighter coin in it.

Now the question circles back to the strategy we used in the first puzzle. We have 3 coins, out of which one is lighter than the others, and we have only one chance left to weigh the coins. Just like in the first puzzle, we can put one coin aside and weigh the other two against each other – if one pan rises, it is holding the lighter coin, otherwise the coin put aside is the lighter coin! Thus, we were able to identify the lighter coin in just two weighings. Can you use the same method to answer the first puzzle now?

We will leave you with a final puzzle:

On a Christmas tree there were two blue, two red, and two white balls. All seemed the same, however in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls weighed the same. Using a 2-pan balance scale only twice, identify the lighter balls.

Can you solve this problem using the strategies above? Let us know in the comments!

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 Answer GMAT Questions That are About an Unfamiliar Topic

Usually, GMAT questions that are based on your field of work or interests are simple to comprehend and relatively easy to answer correctly. But what happens when, say, an engineer gets a question based on psychiatry? Is he or she bound to fail? No.

Remember that the GMAT offers a level playing field for test takers from different backgrounds – it doesn’t matter whether your major was literature or physics. If you feel lost on a question about renaissance painters, remember that the guy next to you is lost on the problem involving planetary systems.

So how can you successfully handle GMAT questions on any topic? By sticking to the basics. The logic and reasoning required to answer these questions will stay the same no matter which field the information in the question stem comes from.

To give an example of this, let’s today take a look at a GMAT question involving psychoanalysis:

Studies in restaurants show that the tips left by customers who pay their bill in cash tend to be larger when the bill is presented on a tray that bears a credit-card logo. Consumer psychologists hypothesize that simply seeing a credit-card logo makes many credit-card holders willing to spend more because it reminds them that their spending power exceeds the cash they have immediately available.

Which of the following, if true, most strongly supports the psychologists’ interpretation of the studies?

(A) The effect noted in the studies is not limited to patrons who have credit cards.
(B) Patrons who are under financial pressure from their credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.
(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.
(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.
(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

Let’s break down the argument:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo.

Why would that be? Why would there be a difference in customer behavior when the tray has no logo from when the tray has a credit card logo? Psychologists’ hypothesize that seeing a credit-card logo reminds people of the spending power given by the credit card they carry (and that their spending power exceeds the actual cash they have right now).

The question asks us to support the psychologists’ interpretation. And what is the psychologists’ interpretation of the studies? It is that seeing a logo reminds people of their own credit card status. Say we change the argument a little by adding a line:

Argument: Studies show that cash tips left by customers are larger when the bill is presented on a tray that bears a credit-card logo. Patrons under financial pressure from credit-card obligations tend to tip less when presented with a restaurant bill on a tray with credit-card logo than when the tray has no logo.

Now, does the psychologists’ interpretation make even more sense? The psychologists’ interpretation is only that “seeing a logo reminds people of their own credit card status.” The fact “that their spending power exceeds the cash they have right now” explains the higher tips. If we are given that some customers tip more upon seeing that card logo and some tip less upon seeing it, it makes sense, right? Different people have different credit card obligation status, hence, people are reminded of their own card obligation status and they tip accordingly.

Answer choice B increases the probability that the psychologists’ interpretation is true because it tells you that in the cases of very high credit card obligations, customers tip less. This is what you would expect if the psychologists’ interpretation were correct.

In simpler terms, the logic here is similar to the following situation:

A: After 12 hours of night time sleep, I can’t study.
B: Yeah, because your sleep pattern is linked to your level of concentration. After a long sleep, your mind is still muddled and lazy so you can’t study.
A: After only 4 hrs of night time sleep, I can’t study either.

Does B’s theory make sense? Sure! B’s theory is that “sleep pattern is linked to level of concentration.” If A sleeps too much, her concentration is affected. If she sleeps too little, again her concentration is affected. So B’s theory certainly makes more sense.

(E) The percentage of restaurant bills paid with given brand of credit card increases when that credit card’s logo is displayed on the tray with which the bill is prepared.

This option supports the hypothesis that credit card logos remind people of their own card – not of their card obligations. The psychologists’ interpretation talks about the logo reminding people of their card status (high spending power or high obligations). Hence, this option is not correct.

Now let’s examine the rest of the answer choices to see why they are also incorrect:

(A) The effect noted in the studies is not limited to patrons who have credit cards.

This argument is focused only on credit cards, not on credit cards and their logos, so this is irrelevant.

(C) In virtually all of the cases in the studies, the patrons who paid bills in cash did not possess credit cards.

This option questions the validity of the psychologists’ interpretation. Hence, this is also not correct.

(D) In general, restaurant patrons who pay their bills in cash leave larger tips than do those who pay by credit card.

This argument deals with people who have credit cards but are tipping by cash, hence this is also irrelevant.

We hope you see that if you approach GMAT questions logically and stick to the basics, it is not hard to interpret and solve them, even if they include information from an unfamiliar field.

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!

# GMAT Tip of the Week: Taking the Least Amount of Time to Solve “At Least” Probability Problems

In its efforts to keep everyone from getting perfect 800s, the GMAT has two powerful tools to stop you from perfection. For one, it can bait you into wrong answers (with challenging content, tempting trap answers, or a combination thereof). And secondly, it can waste your time, making it look like you need to do a lot of work when there’s a much simpler way.

Fortunately, and contrary to popular belief, the GMAT isn’t “pure evil.” Wherever it provides opportunities for less-savvy examinees to waste their time, it also provides a shortcut for those who have put in the study time to learn it or who have the patience to look for the elevator, so to speak, before slogging up the stairs. And one classic example of that comes with the “at least one” type of probability question.

To illustrate, let’s consider an example:

In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?

(A) 5/17
(B) 12/17
(C) 25/81
(D) 56/81
(E) 4/9

Here, you can first streamline the process along the lines of one of those “There are two types of people in the world: those who _______ and those who don’t _______” memes. Your goal is to determine whether you get a yellow marble, so you don’t care as much about “blue” and “black”…those can be grouped into “not yellow,” thereby giving you only two groups: 8 yellow marbles and 10 not-yellow marbles. Fewer groups means less ugly math!

But even so, trying to calculate the probability of every sequence that gives you one or two yellow marbles is labor intensive. You could accomplish that “not yellow” goal several ways:

First marble: Yellow; Second: Not Yellow
First: Not Yellow; Second: Yellow
First: Yellow; Second: Yellow

That’s three different math problems each involving fractions and requiring attention to detail. There ought to be an easier way…and there is. When a probability problem asks you for the probability of “at least one,” consider the only situation in which you WOULDN’T get at least one: if you got none. That’s a single calculation, and helpful because if the probability of drawing two marbles is 100% (that’s what the problem says you’re doing), then 100% minus the probability of the unfavorable outcome (no yellow) has to equal the probability of the favorable outcome. So if you determine “the probability of no yellow” and subtract from 1, you’re finished. That means that your problem should actually look like:

PROBABILITY OF NO YELLOW, FIRST DRAW: 10 non-yellow / 18 total
PROBABILITY OF NO YELLOW, SECOND DRAW: 9 remaining non-yellow / 17 remaining total

10/18 * 9/17 reduces to 10/2 * 1/17 = 5/17. Now here’s the only tricky part of using this technique: 5/17 is the probability of what you DON’T want, so you need to subtract that from 1 to get the probability you do want. So the answer then is 12/17, or B.

More important than this problem is the lesson: when you see an “at least one” probability problem, recognize that the probability of “at least one” equals 100% minus the probability of “none.” Since “none” is always a single calculation, you’ll always be able to save time with this technique. Had the question asked about three marbles, the number of favorable sequences for “at least one yellow” would be:

Yellow Yellow Yellow
Yellow Not-Yellow Not-Yellow
Yellow Not-Yellow Yellow
Yellow Yellow Not-Yellow
Not-Yellow Yellow Yellow

(And note here – this list is not yet exhaustive, so under time pressure you may very well forget one sequence entirely and then still get the problem wrong even if you’ve done the math right.)

Whereas the probability of No Yellow is much more straightforward: Not-Yellow, Not-Yellow, Not-Yellow would be 10/18 * 9/17 * 8/16 (and look how nicely that last fraction slots in, reducing quickly to 1/2). What would otherwise be a terrifying slog, the “long way” becomes quite quick the shorter way.

So, remember, when you see “at least one” probability on the GMAT, employ the “100% minus probability of none” strategy and you’ll save valuable time on at least one Quant problem on test day.

By Brian Galvin.

# Solving GMAT Geometry Problems That Involve Infinite Figures

Sometimes, we come across GMAT geometry questions that involve figures inscribed inside other figures. One shape inside of another shape may not be difficult to work with, but how do we handle problems that involve infinite figures inscribed inside one another? Such questions can unsettle even the most seasoned test takers. Let’s take a look at one of them today:

A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in this way and this process is continued indefinitely. If a side of the first square is 4 cm. Determine the sum of areas of all squares?

(A) 18
(B) 32
(C) 36
(D) 64
(E) None

Now the first thing that might come to our mind is this – how do we mathematically, in the time limit of approximately 2 minutes, calculate areas of infinite squares?

There has to be a formula for this. Recall that we do, in fact, have a formula that calculates the sum of infinite terms – the geometric progression formula! Let’s see if we can use that to find the areas of the squares mentioned in this problem.

First, we’ll see if we can find a pattern in the areas of the squares:

Say the side of the outermost square is “s“. The area of the outermost square will be s^2 and half of the side will be s/2. The side of the next square inside this outermost square (the second square) forms the hypotenuse of a right triangle with legs of length s/2 each. Using the Pythagorean Theorem:

Hypotenuse^2 = (s/2)^2 + (s/2)^2 = s^2/2
Hypotenuse = s/√(2)

So now we know the sides of the second square will each equal s/√(2), and the area of the second square will be s^2/2.

Our calculations will be far easier if we note that the diagonal of the second square will be the same length as the side of the outer square. We know that area of a square given diagonal d is d^2/2, so that would directly bring us to s^2/2 as the area of the second square.

The second square and the square inscribed further inside it (the third square) will have the same relation. The area of the third square will be (s^2/2)*(1/2) = s^2/4.

Now we know the area of every subsequent square will be half the area of the outside square. So the total area of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …Each term is half the previous term.

Therefore, the sum of an infinite Geometric Progression where the common ratio is less than 1 is:

Total Sum = a/(1-r)
a: First Term
r: Common Ratio

Sum of areas of all squares = s^2 + s^2/2 + s^2/4 + s^2/8 + …
Sum of areas of all squares = s^2/(1 – 1/2)
Sum of areas of all squares = 2s^2

Since s is the length of the side of the outermost square, and s = 4 (this fact is given to us by the questions stem), the sum of the areas of all the squares = 2*4^2 = 32 cm^2. Therefore, our answer is B.

We hope you understand how we have used the geometric progression formula to get to our answer. To recap, the sum of an infinite geometric progression is a/(1 – r).

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 Find the Maximum Distance Between Points on a 3D Object

How do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?

A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)

There are various different diagonals in a rectangular solid. Look at the given figure:

BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?

The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.

Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.

In our question we know that:

l = 10 inches
w = 10 inches
h = 5 inches

Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.

DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:

DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125

Going back to triangle DHB, we can now say that:

BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15

Thus, our answer to this question is A.

Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:

The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?

(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15

Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:

The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:

Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5

In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.

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!

# GMAT Sentence Correction: How to Tackle Inverted Sentence Structures

One of the challenges test-takers encounter on Sentence Correction questions is the tendency of question writers to structure sentences in a way that departs from the way we typically write or speak. Take a simple example: “My books are on the table,” could also be written as “On the table are my books.” If you’re like me, you cringe a little bit with the second option – it sounds starchy and pretentious, but it’s a perfectly legitimate sentence, and an example of what’s called “inverted structure.”

In a standard structure, the subject will precede the verb. In an inverted structure, the subject comes after the verb. The tipoff for such a construction is typically a prepositional phrase – in this case, “on the table,” followed by a verb. It is important to recognize that the object of the prepositional phrase, “table,” cannot be the subject of the verb, “are,” so we know that the subject will come after the verb.

Let’s look at an example from an official GMAT question:

The Achaemenid empire of Persia reached the Indus Valley in the fifth century B.C., bringing the Aramaic script with it, from which was derived both northern and southern Indian alphabets.

(A) the Aramaic script with it, from which was derived both northern and
(B) the Aramaic script with it, and from which deriving both the northern and the
(C) with it the Aramaic script, from which derive both the northern and the
(D) with it the Aramaic script, from which derives both northern and
(E) with it the Aramaic script, and deriving from it both the northern and

The first thing you might notice is the use of the relative pronoun “which.” We’d like for “which” to be as close to as possible to its referent. So what do we think the alphabets were derived from? From the Aramaic script.

Notice that in options A and B, the closes referent to “which” is “it.” There are two problems here. One, it would be confusing for one pronoun, “which,” to have another pronoun, “it,” as its antecedent. Moreover, “it” here seems to refer to the Achaemenid Empire. Do we think that the alphabets derived from the empire? Nope. Eliminate A and B. Though E eliminates the “which,” this option also seems to indicate that the alphabets derived from the empire, so E is out as well.

We’re now down to C and D. Notice that our first decision point is to choose between “from which derive” and “from which derives.” This is an instance of inverted sentence structure. We have the prepositional phrase “from which,” followed immediately by a verb “derive or “derives.” Thus, we know that the subject for this verb is going to come later in the sentence, in this case, the northern and southern alphabets.  If we were to rearrange the sentences so that they had a more conventional structure, our choice would be between the following options:

C) Both the northern and the southern Indian alphabets derive from [the empire.]

or

D) Both northern and southern Indian alphabets derives from [the empire.]

Because “alphabets” is plural, we want to pair this subject with the plural verb, “derive.” Therefore, the correct answer is C.

Takeaway: anytime we see the construction “prepositional phrase + verb,” we are very likely looking at a sentence with an inverted sentence structure. In these cases, make sure to look for the subject of the sentence after the verb, rather than before.

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

# Advanced Number Properties on the GMAT – Part VII

We have seen a number of posts on divisibility, odd-even concepts and perfect squares. Individually, each topic has very simple concepts but when they all come together in one GMAT question, it can be difficult to wrap one’s head around so many ideas. The GMAT excels at giving questions where multiple concepts are tested. Let’s take a look at one such Data Sufficiency question today:

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

1) When p is divided by 8, the remainder is 5.
2) x – y = 3

This Data Sufficiency question has a lot of information in the question stem.  First, we need to sort through this information before we move on to the statements.

We know that p, x and y are positive integers. y is an unknown odd number, so it can be written in the form 2n + 1. We also know that p = x^2 + y^2.

Because y is written in the form 2n + 1, y^2 can be written as:

y^2 =(2n + 1)^2

y^2 = 4n^2 + 4n + 1

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

An interesting thing to note here is that one case of n and (n+1) will be odd and the other will be even. In every case, n(n + 1) is even. Therefore, y^2 is 1 more than a multiple of 8. In other words, we can write it as y^2 = 8m + 1.

Now we can say p = x^2 + 8m + 1.

With this in mind, is x divisible by 4? Let’s examine the statements to find out:

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

Because y^2 = 8m + 1, we can see that when y^2 is divided by 8, the remainder will be 1. Therefore, to get a remainder of 5 when p is divided by 8, when x^2 is divided by 8, we should get a remainder of 4.

Now we know that x^2 can be written in the form 8a + 4 (i.e. we can make “a’” groups of 8 each and have 4 leftover).

x^2 = 4*(2a + 1)

So x = 2 * √(an odd number)

Note that square root of an odd number will be an odd number only. If there is no 2 in the perfect square, obviously there was no 2 in the number, too.

So, x = 2 * some other odd number, which means x will be a multiple of 2, but not of 4 definitely. This statement alone is sufficient.

Now let’s look at the next statement:

Statement 2: x – y = 3

Since y is odd, we can say that x will be even (an even – an odd = an odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since we have no constraints on p.

This statement alone is not sufficient to answer the question. Therefore, our answer is A.

Test takers might feel that not every step in this solution is instinctive. For example, how do we know that we should put y^2 in the form 4n(n+1) + 1? Keep the target in mind – we know that we need to find whether x is divisible by 4. Hence, try to get everything in terms of multiples of 4 + a remainder.

See you next week!

(For more advanced number properties on the GMAT, check out Parts IIIIIIIV, V and VI of this series.)

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!

# GMAT Tip of the Week: 3 Guiding Principles for Exponent Problems

If you’re like many GMAT examinees, you’ve found yourself in this familiar situation. You KNOW the rules for exponents. You know them cold. When you’re multiplying the same base and different exponents, you add the exponents. When you’re taking one exponent to another power, you multiply those exponents. A negative exponent? Flip that term into the denominator. A number to the zero power? You’ve got yourself a 1.

But as thoroughly and quickly as you know those rules, this exponent-based problem in front of you has you stumped. You know what you need to KNOW, but you’re not quite sure what you need to DO. And that’s an ever-important part about taking the GMAT – it’s necessary to know the core rules, facts, and formulas, but it’s also every bit as important to have action items for how you’ll apply that knowledge to tricky problems.

For exponents, there are three “guiding principles” that you should keep in mind as your action items. Any time you’re stuck on an exponent-based problem, look to do one (or more) of these things:

1) Find Common Bases
Most of the exponent rules you know only apply when you’re dealing with two exponents of the same base. When you multiply same-base exponents, you add the exponents; when you divide two same-base exponents, you subtract. And if two exponents of the same base are set equal, then you know that the exponents are equal. But keep in mind – these major rules all require you to be using exponents with the same base! If the GMAT gives you a problem with different bases, you have to find ways to make them common, usually by factoring them into their prime bases.

So for example, you might see a problem that says that:

2^x * 4^2x = 8^y. Which of the following must be true?

(A) 3x = y
(B) x = 3y
(C) y = (3/5)x
(D) x = (3/5)y
(E) 2x^2 = y

In order to apply any rules that you know, you must get the bases in a position where they’ll talk to each other. Since 2, 4, and 8 are all powers of 2, you should factor them all in to base 2, rewriting as:

2^x * (2^2)^2x = (2^3)^y

Which simplifies to:

2^x * 2^4x = 2^3y

Now you can add together the exponents on the left:

2^5x = 2^3y

And since you have the same base set equal with two different exponents, you know that the exponents are equal:

5x = 3y

This means that you can divide both sides by 5 to get x = (3/5)y, making answer choice D correct. But more importantly in a larger context, heed this lesson – when you see an exponent problem with different bases for multiple exponents, try to find ways to get the bases the same, usually by prime-factoring the bases.

2) Factor to Create Multiplication
Another important thing about exponents is that they represent recurring multiplication. x^5, for example, is x * x * x * x * x…it’s a lot of x’s multiplied together. Naturally, then, pretty much all exponent rules apply in cases of multiplication, division, or more exponents – you don’t have rules that directly apply to addition or subtraction. For that reason, when you see addition or subtraction in an exponent problem, one of your core instincts should be to factor common terms to create multiplication or division so that you’re in a better position to leverage the rules you know. So, for example, if you’re given the problem:

2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

(A) 18
(B) 20
(C) 21
(D) 22
(E) 24

You should see that in order to do anything with the left-hand side of the equation, you’ll need to factor the common 2^x in order to create multiplication and be in a position to divide and cancel terms from the right. Doing so leaves you with:

2^x(1 + 2^3) = (6^2)(2^18)

Here, you can simplify the 1 + 2^3 parenthetical: 2^3 = 8, so that term becomes 9, leaving you with:

9(2^x) = (6^2)(2^18)

And here, you should heed the wisdom from above and find common bases. The 9 on the left is 3^2, and the 6^2 on the right can be broken into 3^2 * 2^2. This gives you:

(3^2)(2^x) = (3^2)(2^2)(2^18)

Now the 3^2 terms will cancel, and you can add the exponents of the base-2 exponents on the right. That means that 2^x = 2^20, so you know that x = 20. And a huge key to solving this one was factoring the addition into multiplication, a crucial exponent-based action item on test day.

3) Test Small Numbers and Look For Patterns
Remember: exponents are a way to denote repetitive, recurring multiplication. And when you do the same thing over and over again, you tend to get similar results. So exponents lend themselves well to finding and extrapolating patterns. When in doubt – when a problem involves too much abstraction or too large of numbers for you to get your head around – see what would happen if you replaced the large or abstract terms with smaller ones, and if you find a pattern, then look to extrapolate it. With this in mind, consider the problem:

What is the tens digit of 11^13?

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

Naturally, calculating 11^13 without a calculator is a fool’s errand, but you can start by taking the first few steps and seeing if you establish a pattern:

11^1 = 11 –> tens digit of 1

11^2 = 121 –> tens digit of 2

11^3 = 1331 –> tens digit of 3

And depending on how much time you have you could continue:

11^4 = 14641 –> tens digit of 4

But generally feel pretty good that you’ve established a recurring pattern: the tens digit increases by 1 each time, so by 11^13 it will be back at 3. So even though you’ll never know exactly what 11^13 is, you can be confident in your answer.

Remember: the GMAT is a test of how well you apply knowledge, not just of how well you can memorize it. So for any concept, don’t just know the rules, but also give yourself action items for what you’ll do when problems get tricky. For exponent problems, you have three guiding principles:

1) Find Common Bases
2) Factor to Create Multiplication
3) Test Small Numbers to Find a Pattern

By Brian Galvin.

# How NOT to Write the Equation of a Line on the GMAT

A question brought an interesting situation to our notice. Let’s start by asking a question: How do we write the equation of a line? There are two formulas:

y = mx + c (where m is the slope and c is the y-intercept)
and
yy1 = m * (xx1) [where m is the slope and (x1,y1) is a point on the line]

We also know that m = (y2y1)/(x2x1) – this is how we find the slope given two points that lie on a line. The variables are x1, y1 and x2, y2, and they represent specific values.

But think about it, is m = (y2y)/(xx1) really the equation of a line? Let’s further clarify this idea using a GMAT practice question:

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x + y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

We have been given that the line passes through (0, 0) and has a slope of 2. We can find the equation of the line from this information.

y = mx + c
y = 2x + 0 (Since the line passes through (0, 0), its y-intercept is 0 – when x is 0, y is also 0.)
y = 2x

Since we are given two other points, (3, y) and (x, 4), on the line and we have a slope of 2, many test-takers will be tempted to make another equation for the line using this information.

(4 – y)/(x – 3) = 2
(4 – y) = 2*(x – 3)
Thus, 2+ y = 10

Here, test-takers will use the two equations to solve for x and y and get x = 5/2 and y = 5.

After adding x and y together, they then wonder why 7.5 is not one of the answer choices. If this were an actual GMAT question, it is quite likely that 7.5 would have been one of the options. So all in all, the test-taker would not even have realized that he or she made a mistake, and would choose 7.5 as the (incorrect) answer.

The error is conceptual here. Note that the equation of the line, 2x + y = 10, is not the same as the equation we obtained above, y = 2x. They represent two different lines, but we have only a single line in the question. So which is the actual equation of that line?

To get the second equation, we have used m = (y2y)/(xx1). But is this really the equation of a line? No. This formula doesn’t have y and x, the generic variables for the x– and y-coordinates in the equation of a line.

To further clarify, instead of x and y, try using the variables a and b in the question stem and see if it makes sense:

“In the coordinate plane, line k passes through the origin and has slope 2. If points (3, a) and (b, 4) are on line k, then a + b =”

You can write (4 – a)/(b – 3) = 2 and this would be correct. But can we solve for both a and b here? No – we can write one of them in terms of the other, but we can’t get their exact values.

We know a and b must have specific values. (3, a) is a point on the line y = 2x. For x = 3, the value of of the y-coordinate, a, will be y = 2*3 = 6. Therefore, a = 6.

(b, 4) is also on the line y = 2x. So if the y-coordinate is 4, the x-coordinate, b, will be 4 = 2b, i.e. b = 2. Thus, a + b = 6 + 2 = 8, and our answer is C.

This logic remains the same even if the variables used are x and y, although test-takers often get confused because of it. Let’s solve the question in another way using the variables as given in the original question.

Recall what we have learned about slope in the past. If the slope of the line is 2 and the point (0, 0) lies on the line, the value of y – if point (3, y) also lies on the line – will be 6 (a slope of 2 means a 1-unit increase in x will lead to a 2-unit increase in y).

Again, if point (x, 4) lies on the line too, an increase of 4 in the y-coordinate implies an increase of 2 in the x-coordinate. So x will be 2, and again, x + y = 2 + 6 = 8.

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!

# 3 Formats for GMAT Inequalities Questions You Need to Know

As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

1. Must Be True
2. Could Be True
3. Complete Range

Case 1: Must Be True
If |-x/3 + 1| < 2, which of the following must be true?
(A) x > 0
(B) x < 8
(C) x > -4
(D) 0 < x < 3
(E) None of the above

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2
(1/3) * |x – 3| < 2
|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0
Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8
Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4
Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3
Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Case 2: Could Be True
If −1 < x < 5, which is the following could be true?
(A) 2x > 10
(B) x > 17/3
(C) x^2 > 27
(D) 3x + x^2 < −2
(E) 2x – x^2 < 0

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10
x > 5
No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3
x > 5.67
No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27
x^2 – 27 > 0
x > 3*√(3) or x < -3*√(3)
√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

(Details on how to solve such inequalities are discussed here.)

(D) 3x + x^2 < −2
x^2 + 3x + 2 < 0
(x + 1)(x + 2) < 0
-2 < x < -1
No values of x will lie between -2 and -1, so this also cannot be true.

(E) 2x – x^2 < 0
x * (x – 2) > 0
x > 2 or x < 0
If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E.

Case 3: Complete Range
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?
(A) 0 < |x| < ½
(B) |x| > ½
(C) -½ < x < 0 or ½ < x
(D) x < -½ or 0 < x < ½
(E) x < -½ or x > 0

We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice.

We are given that x^3 – 4x^5 < 0. This inequality can be solved to:

x^3 ( 1 – 4x^2) < 0
x^3*(2x + 1)*(2x – 1) > 0
> 1/2 or -1/2 < x < 0

This is our universe of the values of x. It is given that all values of x lie in this range.

Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer.

We hope these practice problems will help you become able to distinguish between the three cases 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!

# Using Special Formats on GMAT Variable Problems

In today’s post, we will discuss some special formats when we assume variables on the GMAT. These will allow us to minimize the amount of manipulations and calculations that are required to solve certain Quant problems.

Here are some examples:

An even number: 2a
Logic: It must be a multiple of 2.

An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.

Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).

Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.

Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.

A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.

Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.

Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.

Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:

The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Let’s have the four consecutive odd numbers be the following, where “a” is any integer: (2a – 3), (2a – 1), (2a + 1), (2a + 3)

The sum of these numbers is: (2a – 3) + (2a – 1) + (2a + 1) + (2a + 3) = 8a

Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)

The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b

Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.

We are given that the sum 8a is equal to the sum 6b.

8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.

With this in mind, possible solutions for a and b are:

a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
etc.

We are also given that the middle term of the even numbers is greater than 101 and less than 200.

So 101 < 2b < 200, i.e. 50.5 < b < 100.

B must be an integer, hence, 51 ≤ b ≤ 99.

Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96

The number of values b can take = (Last term – First term)/Common Difference + 1 = (96 – 52)/4 + 1 = 12

For each of these 12 values of b, there will be a corresponding value of a and, hence, we will get 12 such sequences. Therefore, the answer to our question 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!

# Assumption vs. Strengthen Critical Reasoning Questions: What’s the Difference?

I had a discussion with a tutoring student the other day about the distinction between Assumption and Strengthen questions in the Critical Reasoning section. The two categories feel similar, after all. They are different, however, and the difference, as with most Critical Reasoning questions, lies mainly in the texture of the language that would be most appropriate for a correct answer in either category.

To illustrate, let’s take a simple argument: Dave opens a coffee shop in Veritasville called Dave’s Blends. According to surveys, Dave’s Blends has the best tasting coffee in the city. Therefore, Dave’s Blends will garner at least 50% the local market.

First, imagine that this is a simple Strengthen question. In order to strengthen this somewhat fanciful conclusion, we’re going to want strong language. For example: Virtually all coffee drinkers in Veritasville buy coffee daily from Dave’s. That’s a pretty good strengthener. The statement increases the likelihood that Dave’s Blends will dominate the local market. But an answer choice such as, “Some people buy coffee at Dave’s,” would be a lousy choice, as the fact that Dave’s has at least one customer is hardly a compelling reason to conclude that it will get to at least a 50% market share.

Now imagine that we take the same argument and make it an Assumption question. The first aforementioned answer choice is now much less appealing. Can we really assume that virtually everyone in town will get their coffee at Dave’s? Not really. If Dave’s has 51% of the market share, it doesn’t mean that virtually everyone gets their coffee there. But now consider the second answer choice – if we’re concluding that Dave’s will get at least half of the local market, we are assuming that some people will purchase coffee there, so now this would be a good answer.

The difference is that in a Strengthen question, we’re looking for new information that will make the conclusion more likely. In an Assumption question, we’re looking for what is true based on the conclusion.  Put another way, strong language (“virtually everyone”) is often desirable in a Strengthen question, whereas softer language (“some people”) is usually more desirable in an Assumption question.

Let’s see this in action with a GMAT practice question:

For most people, the left half of the brain controls linguistic capabilities, but some people have their language centers in the right half. When a language center of the brain is damaged, for example by a stroke, linguistic capabilities are impaired in some way. Therefore, people who have suffered a serious stroke on the left side of the brain without suffering any such impairment must have their language centers in the right half.

Which of the following is an assumption on which the reasoning in the argument above depends?

(A) No part of a person’s brain that is damaged by a stroke ever recovers.
(B) Impairment of linguistic capabilities does not occur in people who have not suffered any damage to any language center of the brain.
(C) Strokes tend to impair linguistic capabilities more severely than does any other cause of damage to language centers in the brain.
(D) If there are language centers on the left side of the brain, any serious stroke affecting that side of the brain damages at least one of them.
(E) It is impossible to determine which side of the brain contains a person’s language centers if the person has not suffered damage to either side of the brain.

First, let’s break this argument down:

Conclusion: People who suffer a stroke on the left side of the brain and don’t’ suffer language impairment have language centers in the right half of the brain.

Premises: Most people have language centers on the left side of the brain, while some have them on the right. Damage impairs linguistic capabilities.

This is an Assumption question, so we’re looking for what is be true based on the way the premises lead to the conclusion. Put another way, softer language might be preferable here. Now let’s examine each of the answer choices:

(A) Notice the extreme language, “No part…ever recovers“. Can we really assume that? Of course not – some portion might recover. No good.

(B) We don’t know this. Imagine someone has a part of his or her brain removed and this part of the brain doesn’t contain a language center. Surely we can’t assume that this person will have no language impairment at all. No good.

(C) Again, notice the extreme language, “…more severely than other cause. Can we assume that a stroke is worse than every other kind of brain trauma? Of course not. No good.

(D) Now we’re talking. Here, we are given more generous language: damages at least one of them. “At least one” is a pretty low bar. Remember that the conclusion is that someone who suffers a left-brain stroke and doesn’t have language impairment must have language centers on the right side. Well, that only makes sense if there’s some damage somewhere on the left. This answer choice looks good.

(E) Notice again the extreme language, “…it is impossible“. There may be some other way to assess where the language centers are. No good.

Takeaway: Strengthen questions and Assumption questions are not identical. In a Strengthen question, we want a strong answer choice that will make a conclusion more likely. In an Assumption question we want a soft answer that is indisputable based on how the premises lead to the conclusion. Attention to details in the language (some vs. most vs. all) is the key.

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

# Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT!

Test-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that  there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

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 Absolute Values – Part V

We will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)

Let’s look at the following GMAT question:

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

In this question, we are given the inequality |x – 6| > 3*|x + 2|

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

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

# The Holistic Approach to Absolute Values – Part IV

Last week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

• |x – 3| will be the distance covered by the friend at 3 to reach x.
• |x + 1| will be the distance covered by the friend at -1 to reach x.
• |x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.

Today, let’s look at how to solve a more advanced GMAT question using the same logic:

For how many integer values of x, is |x – 5| + |x + 1| + |x|  + |x – 7| < 15?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x|  + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

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

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

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 Patterns to Solve GMAT Questions with Reversed-Digit Numbers – Part II

, I wrote about the GMAT’s tendency to ask questions regarding the number properties of two two-digit numbers whose tens and units digits have been reversed.

The biggest takeaways from that post were:

1. Anytime we add two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 11.
2. Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.

For the hardest GMAT questions, we’re typically mixing and matching different types of number properties and strategies, so it can be instructive to see how the above axioms might be incorporated into such problems.

Take this challenging Data Sufficiency question, for instance:

When the digits of two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

(1) The integer (M –N) has 12 unique factors.

(2) The integer (M –N) is a multiple of 9.

The average test-taker looks at Statement 1, sees that it will be very difficult to simply pick numbers that satisfy this condition, and concludes that this can’t possibly be enough information. Well, the average test-taker also scores in the mid-500’s, so that’s not how we want to think.

First, let’s concede that Statement 1 is a challenging one to evaluate and look at Statement 2 first. Notice that Statement 2 tells us something we already know – as we saw above, anytime you have two two-digit numbers whose tens and units digits are reversed, the difference will be a multiple of 9. If Statement 2 is useless, we can immediately prune our decision tree of possible correct answers. Either Statement 1 alone is sufficient, or the statements together are not sufficient, as Statement 2 will contribute nothing. So right off the bat, the only possible correct answers are A and E.

If we had to guess, and we recognize that the average test-taker would likely conclude that Statement 1 couldn’t be sufficient, we’d want to go in the opposite direction – this question is significantly more difficult (and interesting) if it turns out that Statement 1 gives us considerably more information than it initially seems.

In order to evaluate Statement 1, it’s helpful to understand the following shortcut for how to determine the total number of factors for a given number. Say, for example, that we wished to determine how many factors 1000 has. We could, if we were sufficiently masochistic, simply list them out (1 and 1000, 2 and 500, etc.). But you can see that this process would be very difficult and time-consuming.

Alternatively, we could do the following. First, take the prime factorization of 1000. 1000 = 10^3, so the prime factorization is 2^3 * 5^3. Next, we take the exponent of each prime base and add one to it. Last, we multiply the results. (3+1)*(3+1) = 16, so 1000 has 16 total factors. More abstractly, if your number is x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Now let’s apply this process to Statement 1. Imagine that the difference of M and N comes out to some two-digit number that can be expressed as x^a * y^b. If we have a total of 12 factors, then we know that (a+1)(b+1) = 12. So, for example, it would work if a = 3 and b = 2, as a + 1 = 4 and b + 1 = 3, and 4*3 =12. But it would also work if, say, a = 5 and b = 1, as a + 1 = 6 and b + 1 = 2, and 6*2 = 12. So, let’s list out some numbers that have 12 factors:

1. 2^3 * 3^2 (3+1)(2+1) = 12
2. 2^5 * 3^1 (5+1)(1+1) = 12
3. 2^2 * 3^3 (2+1)(3+1) = 12

Now remember that M – N, by definition, is a multiple of 9, which will have at least 3^2 in its prime factorization. So the second option is no longer a candidate, as its prime factorization contains only one 3. Also recall that we’re talking about the difference of two two-digit numbers. 2^2 * 3^3 is 4*27 or 108. But the difference between two positive two-digit numbers can’t possibly be a three-digit number! So the third option is also out.

The only possibility is the first option. If we know that the difference of the two numbers is 2^3 * 3^2, or 8*9 = 72, then only 91 and 19 will work. So Statement 1 alone is sufficient to answer this question, and the answer is A.

Algebraically, if M = 10x + y, then N = 10y + x.

M – N = (10x + y) – (10y + x) = 9x – 9y = 9(x – y).

If 9(x – y) = 72, then x – y = 8. If the difference between the tens and units digits is 8, the numbers must be 91 and 19.

Takeaway: the hardest GMAT questions will require a balance of strategy and knowledge. In this case, we want to remember the following:

• Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.
• If one statement is easier to evaluate than the other, tackle the easier one first. If it’s the case that one statement gives you absolutely nothing, and the other is complex, there is a general tendency for the complex statement alone to be sufficient.
• For the number x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

# The Holistic Approach to Absolute Values – Part III

A while back, we discussed some holistic approaches to answering absolute value questions. Today, we will enhance our understanding of absolute values with some variations that you might see on the GMAT.

Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.

A quick review:

• |x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.
• |x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or  -3.
• |x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.

Let’s move ahead now and see how we can use these concepts to solve inequalities:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:

What will be the total distance at any value of x between these two points?

Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4

Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7

In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).

Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.

Along the same lines, consider a slight variation of this question:

For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.

We hope this logic is clear. We will look at some other variations of this concept 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!

# Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient

Today we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?

Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.

A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?

(1) The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday

We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.

There are four variables to consider here:

1. Number of compact cars rented (this is what we need to find)
2. Number of luxury cars rented
3. Daily rental rate of compact cars
4. Daily rental rate of luxury cars

Let’s examine the information given to us by the statements:

Statement 1: The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.

This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:

Statement 2: The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday.

This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.

Now, let’s try to tackle both statements together:

The daily rate for luxury cars is \$15 higher than it is for compact cars, and the total rental rates for luxury cars is \$105 higher than it is for compact cars. What constitutes this \$105? It is the higher rental cost of each luxury car (the extra \$15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more  luxury cars were rented, 105 would account for the \$15 higher rent of each luxury car and also for the rent of the extra luxury cars.

Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.

The total extra money collected by renting luxury cars is \$105.

105/15 = 7

Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is \$0 (theoretically), the rent of luxury cars is \$15 and the extra rent charged will be \$105 (7*15 = 105) – this is a valid case.

Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.

Let’s make a slight change to our current numbers to see if they still fit:

Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be \$15*8 = \$120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is \$105 – the \$15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be \$15/9 = \$5/3.

This would mean that the daily rental rate of each luxury car is \$5/3 + \$15 = \$50/3

The total rental cost of luxury cars in this case would be 8 * \$50/3 = \$400/3

The total rental cost of compact cars in this case would be 17 * \$5/3 = \$85/3

The difference between the two total rental costs is \$400/3 – \$85/3 = 315/3 = \$105

Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.

The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:

Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be 10*\$15 = \$150 extra, but it is actually only \$105 extra, a difference of \$45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be \$45/5 = \$9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, \$105. This is a valid case, too.

Hence, there are two strategies we saw in action today:

• Tweak the numbers slightly to see if you will get the same results
• Go for the easy split when choosing numbers to plug in

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!

# GMAT Tip of the Week: Taking The GMAT Like It’s Nintendo Switch

The non-election trending story of the day is the announcement of the forthcoming Nintendo Switch gaming system, a system that promises to help you take the utmost advantage of your leisure time…but that may help you maximize the value of your GMAT experience, too.

How?

The main feature of the Switch (and the driving factor behind its name) is its flexibility. It can be an in-home gaming system attached to a fixed TV set, but then immediately Switch to a hand-held portable system that allows you to continue your game on the go. Nintendo’s business plan is primarily based on offering flexibility…and on the GMAT, your plan should be to prove to business schools that you can offer the same.

The GMAT, of course, tests algebra skills and critical thinking skills and grammar skills, but beneath the surface it also has a preference for testing flexibility. Many problems will punish those with pure tunnel vision, but reward those who can identify that their first course of action isn’t working and who can then Switch to another plan. This often manifests itself in:

• Math problems that seem to require algebra…but halfway through beg to be back-solved using answer choices.
• Sentence Correction problems that seem to ask you to make a decision about one major difference…but for which the natural choices leave you with clearer-cut errors elsewhere.
• Critical Reasoning answer choices that seem out of scope at first, but reward those who read farther and then see their relevance.
• Data Sufficiency problems for which you’ve made a clear, confident decision on one statement…but then the other statement shows you something you hadn’t considered before and forces you to reconsider.
• The overall concept that if you’re a one-trick pony – you’re a master of plugging in answer choices, for example – you’ll find questions that just won’t reward that strategy and will force you to do something else.

Flexibility matters on the GMAT! As an example, consider the following Data Sufficiency question:

Is x/y > 3?

1) 3x > 9y
2) y > 3y

If you’re like many, you’ll confidently address the algebra in Statement 1, divide both sides by 3 to get x > 3y, and then see that if you divide both sides by y, you can make it look exactly like the question stem: x/y > 3. And you may very well say, “Statement 1 is sufficient!” and confidently move on to Statement 2.

But when you look at Statement 2 – either conceptually or algebraically – something should stand out. For one, there’s no way that it’s sufficient because it doesn’t help you determine anything about x. And secondly, it brings up the point that “y is negative” (algebraically you’d subtract y from both sides to get 0 > 2y, then divide by 2 to get 0 > y). And here’s where, if it hasn’t already, your mind should Switch to “positive/negative number properties” mode. If you weren’t thinking about positive vs. negative properties when you considered Statement 1, this one gives you a chance to Switch your thinking and reconsider – what if y were negative? Algebraically, you’d then have to flip the sign when you divide both sides by y:

3x > 9y : Divide both sides by 3

x > 3y : Now divide both sides by y, but remember that if y is positive you keep the sign (x/y > 3), and if y is negative you flip the sign (x/y < 3).

With this in mind, Statement 1 doesn’t really tell you anything. x/y can be greater than 3 or less than 3, so all Statement 1 does is eliminate that x/y could be exactly 3. Now you have the evidence to Switch your answer. If you initially thought Statement 1 was sufficient, Statement 2 has given you a chance to reassess (thereby demonstrating flexibility in thinking) and realize that it’s not, until you know whether y is negative or positive.

Statement 2 supplies that missing piece, and the answer is thus C. But more important is the lesson – because the GMAT so values mental flexibility, it will often provide you with clues that can help you change your mind if you’re paying attention. So on the GMAT, take a lesson from Nintendo Switch: flexibility is an incredibly marketable skill, so look for clues and opportunities to Switch your line of thinking and save yourself from trap answers.

By Brian Galvin.

# How to Solve “Unsolvable” Equations on the GMAT

The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Statement 2: x^2 – x – 2 < 0

This, we can easily solve:

x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0

We know how to solve this inequality using the method discussed here.

This this will give us -1 < x < 2.

Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.

To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some 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!

# Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question!

Today, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.

Let’s take a look at the question stem:

 Date of Transaction Type of Transaction June 11 Withdrawal of \$350 June 16 Withdrawal of \$500 June 21 Deposit of x dollars

For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was \$1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was \$1,000, what was the amount of the deposit on June 21?

(A) \$1,000
(B) \$1,150
(C) \$1,200
(D) \$1,450
(E) \$1,600

The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000

Now we have been given the only three transactions that took place:

• A withdrawal of \$350 on June 11 – so on June 11, the account balance goes down to \$650.
• A withdrawal of \$500 on June 16 – so on June 16, the account balance goes down to \$150.
• A deposit of \$x on June 21 – So on June 21, the account balance goes up to 150 + x.

Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) +  150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000

One might then end up doing this calculation to find the value of x:

[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = \$1,450

But this calculation is rather tedious and time consuming. Can’t we use the deviation method ? After all, we are dealing with large values here! How?

Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:

• The amount from June 11 to June 30 is 350 less.
• The amount from June 16 to June 30 is another 500 less.
• The amount from June 21 to June 30 is x in excess.

Through the deviation method, we can see the shortfall = the excess:

350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)

This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!

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

# Quarter Wit, Quarter Wisdom: Evaluating Nasty GMAT Answer Choices

In some Quant questions, we are given big nasty numbers in the answer choices and little else in the question stem. Often in such cases, the starting point is difficult for the test-taker to find, so today, we will discuss how to handle such questions.

The first and only rule with these types of problems is that familiarity helps. Evaluate the answer choices that make sense to you first.

Let’s look at a few questions to understand how to do that:

Which of the following is NOT prime?

(A) 1,556,551
(B) 2,442,113
(C) 3,893,257
(D) 3,999,991
(E) 9,999,991

The first thing that comes to mind when we consider how to find prime numbers should be to “check the number N for divisibility by all prime factors until we get to the √N.” But note that here, we have four numbers that are prime and one number that is not. Also, the numbers are absolutely enormous and, hence, will be very difficult to work with. So, let’s slide down to a number that seems a bit more sane: 3,999,991 (it is very close to a number with lots of 0’s).

3,999,991 = 4,000,000 – 9
= (2000)^2 – 3^2

This is something we recognise! It’s a difference of squares, which can be written as:

= (2000 + 3) * (2000 – 3)
= 2003 * 1997

Hence, we see that 3,999,991 is a product of two factors other than 1 and itself, so it is not a prime number. We have our answer! The answer is D.

Let’s try another problem:

Which of the following is a perfect square?

(A) 649
(B) 961
(C) 1,664
(D) 2,509
(E) 100,000

Here, start by looking at the answer choices. The first one that should stand out is option E, 100,000, since multiples of 10 are always easy to handle. However, we have an odd number of zeroes here, so we know this cannot be a perfect square.

Next, we look at the answer choices that are close to the perfect squares that we intuitively know, such as 30^2 = 900, 40^2 = 1600, 50^2 = 2500. The only possible number whose perfect square could be 961 is 31 – 31^2 will end with a 1 and will be a bit greater than 900 (32^2 will end with a 4, so that cannot be the square root of 961, and the perfect squares of other greater numbers will be much greater than 900).

31^2 = (30 + 1)^2 = 900 + 1 + 2*30*1 = 961

So, we found that 961 is a perfect square and is, hence, the answer!

In case 961 were not a perfect square, we would have tried 1,664 since it is just 64 greater than 1,600. It could be the perfect square of 42, as the perfect square of 42 will end in a 4.

If 1,664 were also not a perfect square (it is not), we would have looked at 2,509. We would have known immediately that 2,509 cannot be a perfect square because it is too close to 2,500. 2,509 ends in a 9, so we may have considered 53 to be its square root, but the difference between consecutive perfect squares increases as we get to greater numbers.

(4^2 is 16 while 5^2 is 25 – the difference between them is 9. The difference between 5^2 and 6^2 will be greater than 9, and so will the difference between the perfect squares of any pair of consecutive integers greater than 6. Hence, the difference between the squares of 50 and 53 certainly cannot be 9.)

Therefore, our answer is B. Let’s try one more question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

(A) 1,296
(B) 1,369
(C) 1,681
(D) 1,764
(E) 2,500

This question is, again, on perfect squares. We can use the same concepts here, too.

30^2 = 900
31^2 = 961 (=(30+1)^2 = 900 + 1 + 2*30)

40^2= 1,600
41^2 = 1,681 (=(40+1)^2 = 1,600 + 1 + 2*40)

50^2 = 2,500
51^2 = 2,601 (=(50+1)^2 = 2,500 + 1 + 2*50)

We know that the difference between consecutive squares increases as we go to greater numbers: going from 30^2 to 31^2 is a difference of 61, while jumping from 40^2 to 41^2 is a difference of 81.

All the answer choices lie in the range from 900 to 2500. In this range, the difference between consecutive squares is between 60 and 100. So, when you add 148 to a perfect square to get another perfect square in this range, we can say that the numbers must be 2 apart, such as 33 and 35 or 42 and 44, etc. Also, the numbers must lie between 30 and 40 because twice 61 is 122 and twice 81 is 162 – 148 lies somewhere in between 122 and 162.

A and B are the only two possible options.

Consider option A – it ends in a 6, so the square root must end in a 6, too. If you add 148, then it will end with a 4 (the perfect square of a number ending in 8 will end in 4). So this answer choice works.

Consider option B – it ends in a 9, so the square root must end in a 3 or a 7. When you add 148, it ends in 7. No perfect square ends in 7, so this option is out. Our answer is, therefore, A.

We hope you see how a close evaluation of the answer choices can help you solve questions of this type. Go get ’em!

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!

# Evaluating “Useful to Evaluate” Critical Reasoning Questions on the GMAT

In today’s post, we will look at how to answer “useful to evaluate” Critical Reasoning questions in the Verbal section of the GMAT. Arguably, this is one of the toughest question types for test-takers to tackle (perhaps right after boldfaced questions).

To answer this type of question, all you will need to do is follow these six simple steps:

1) Identify the conclusion.
3) Answer it with a “yes” and figure out whether it affects the conclusion.
4) Answer it with a “no” and figure out whether it affects the conclusion.
5) Repeat this for all other answer choices.
6) Only one option will affect the conclusion differently in the two cases – that is your answer.

Let’s illustrate this concept with a problem:

In a certain wildlife park, park rangers are able to track the movements of many rhinoceroses because those animals wear radio collars. When, as often happens, a collar slips off, it is put back on. Putting a collar on a rhinoceros involves immobilizing the animal by shooting it with a tranquilizer dart. Female rhinoceroses that have been frequently re-collared have significantly lower fertility rates than uncollared females. Probably, therefore, some substance in the tranquilizer inhibits fertility.

In evaluating the argument, it would be most useful to determine which of the following?

(A) Whether there are more collared female rhinoceroses than uncollared female rhinoceroses in the park.
(B) How the tranquilizer that is used for immobilizing rhinoceroses differs, if at all, from tranquilizers used in working with other large mammals
(C) How often park rangers need to use tranquilizer darts to immobilize rhinoceroses for reasons other than attaching radio collars
(D) Whether male rhinoceroses in the wildlife park lose their collars any more often than the park’s female rhinoceroses do
(E) Whether radio collars are the only practical means that park rangers have for tracking the movements of rhinoceroses in the park

First, we need to break down the argument to find the premises and the conclusion:

• Many rhinoceroses wear radio collars.
• Often, collars slip.
• When a collar slips, the animal is shot with a tranquilizer to re-collar.
• The fertility of frequently re-collared females is less than the fertility of uncollared females.
• Conclusion: Some substance in the tranquilizer inhibits fertility.

Let’s take a look at each answer choice:

(A) Whether there are more collared female rhinoceroses than uncollared female rhinoceroses in the park.

Even if there are more collared female rhinoceroses than uncollared females, this does not affect the argument’s conclusion. This answer choice talks about collared females vs. uncollared females; we are comparing the fertility of re-collared females with that of uncollared females. Anyway, how many of either type there are doesn’t matter. So, whether you answer “yes” or “no” to this question, it is immaterial.

(B) How the tranquilizer that is used for immobilizing rhinoceroses differs, if at all, from tranquilizers used in working with other large mammals.

This option is comparing the tranquilizers used for rhinoceroses with the tranquilizers used for other large mammals. What the conclusion does, however, is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer “very different” or “not different at all” to this question, in the end, it doesn’t matter.

(C) How often park rangers need to use tranquilizer darts to immobilize rhinoceroses for reasons other than attaching radio collars.

This answer choice can be evaluated in two ways:

• Very Often – Tranquilizers are used very often for uncollared females, too. In this case, can we still say that “tranquilizers inhibit fertility”? No! If they did, fertility in uncollared females would have been low, too.
• Rarely – This would strengthen our conclusion. If tranquilizers are not used on uncollared females, it is possible that something in these tranquilizers inhibits fertility.

(D) Whether male rhinoceroses in the wildlife park lose their collars any more often than the park’s female rhinoceroses do.

This answer choice is comparing the frequency of tranquilizers used on male rhinoceroses with the frequency of tranquilizers used on female rhinoceroses. What the conclusion actually does is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer this question with “more frequently” or “not more frequently,” it doesn’t matter.

(E) Whether radio collars are the only practical means that park rangers have for tracking the movements of rhinoceroses in the park.

This option is comparing radio collars with other means of tracking. What the conclusion does is compare collared female rhinoceroses with uncollared female rhinoceroses. Hence, whether you answer this question with “there are other means” or “there are no other means,” again, it does not matter.

Note that only answer choice C affects the conclusion – if you answer the question it raises differently, it affects the conclusion differently. Option C would be good to know to evaluate the conclusion of the argument, therefore, the answer must be C.

Now try this question on your own:

Following several years of declining advertising sales, the Greenville Times reorganized its advertising sales force two years ago. Before the reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ businesses by having each sales representative deal with only one type of industry or of retailing. After the reorganization, advertising sales increased.

In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be helpful to find out each of the following EXCEPT:

(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?
(D) Before the reorganization, had sales representatives found it difficult to keep up with relevant developments in all types of businesses to which they are assigned?
(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?

We hope you will find this post useful to evaluate the “useful to evaluate” 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!

# How to Simplify Complicated Combination and Permutation Questions on the GMAT

When test-takers first learn how to tackle combination and permutation questions, there’s typically a moment of euphoria when the proper approach really clicks.

If, for example, there are 10 people in a class, and you wish to find the number of ways you can form a cabinet consisting of a president, a vice president, and a treasurer, all you need to do is recognize that if you have 10 options for the president, you’ll have 9 left for the vice president, and 8 remaining for the treasurer, and the answer is 10*9*8. Easy, right?

But on the GMAT, as in life, anything that seems too good to be true probably is. An easy question can be tackled with the type of mechanical thinking illustrated above. A harder question will require a more sophisticated approach in which we consider disparate scenarios and perform calculations for each.

Take this question, for example:

Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

A) 84
B) 91
C) 100
D) 105
E) 243

It’s natural to see this problem and think, “All I have to do is reason out how many options I have for each digit. So for the hundreds digit, I have 3 options (7, 8, or 9); the tens digit has to be different from the hundreds digit, and it must be non-zero, so I’ll have 8 options here; then the last digit has to be odd, so…”

Here’s where the trouble starts. The number of eligible numbers in the 700’s will not be the same as the number of eligible numbers in the 800’s -if the digits must all be different, then a number in the 700’s can’t end in 7, but a number in the 800’s could. So, we need to break this problem into separate cases:

First Case: Numbers in the 700’s
If we’re dealing with numbers in the 700’s, then we’re calculating how many ways we can select a tens digit and a units digit. 7___ ___.

Let’s start with the units digit. Well, we know that this number needs to be odd. And we know that it must be different from the hundreds and the tens digits. This leaves us the following options, as we’ve already used 7 for the hundreds digit: 1, 3, 5, 9. So there are 4 options remaining for the units digit.

Now the tens digit must be a non-zero number that’s different from the hundreds and units digit. There are 9 non-zero digits. We’re using one of those for the hundreds place and one of those for the units place, leaving us 7 options remaining for the tens digit. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, there are 4*7 = 28 options in the 700’s.

Second Case: Numbers in the 800’s
Same logic: 8 ___ ___. Again, this number must be odd, but now we have 5 options for the units digit, as every odd number will obviously be different from the hundreds digit, which is even (1, 3, 5, 7, or 9). The tens digit logic is the same – 9 non-zero digits total, but it must be different from the hundreds and the units digit, leaving us 7 options remaining. If there are 5 ways we can select the units digit and 7 ways we can select the tens digit, there are 5*7 = 35 options in the 800’s.

Third Case: Numbers in the 900’s
This calculation will be identical to the 700’s scenario: 9___ ___. For the units digit, we want an odd number that is different from the hundreds digit, giving us (1, 3, 5, 7), or 4 options. We’ll have 7 options again for the tens digit, for the same reasons that we’ll have 7 options for the tens digit in our other cases. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, then there are 4*7 = 28 options in the 900’s.

To summarize, there are 28 options in the 700’s, 35 options in the 800’s, and 28 options in the 900’s. 28 + 35 + 28 = 91. Therefore, B is the correct answer.

Takeaway: for a simpler permutation question, it’s fine to simply set up your slots and multiply. For a more complicated problem, we’ll need to work case-by-case, bearing in mind that each individual case is, on its own, actually not nearly as hard as it looks, sort of like the GMAT itself.

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

# How to Solve “Hidden” Factor Problems on the GMAT

One of the interesting things to note about newer GMAC Quant questions is that, while many of these questions test our knowledge of multiples and factors, the phrasing of these questions is often more subtle than earlier versions you might have seen. For example, if I ask you to find the least common multiple of 6 and 9, I’m not being terribly artful about what topic I’m testing you on – the word “multiple” is in the question itself.

But if tell you that I have a certain number of cupcakes and, were I so inclined, I could distribute the same number of cupcakes to each of 6 students with none left over or to each of 9 students with none left over, it’s the same concept, but I’m not telegraphing the subject in the same conspicuous manner as the previous question.

This kind of recognition comes in handy for questions like this one:

All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
(2) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.

Initially, we have stacks of 12 boxes with no boxes left over, meaning we could have 12 boxes or 24 boxes or 36 boxes, etc. This is when you want to recognize that we’re dealing with a multiple/factor question. That first sentence tells you that the number of boxes is a multiple of 12. After 60 more boxes were added, the boxes were arranged in stacks of 14 with none left over – after this change, the number of boxes is a multiple of 14.

Because 60 is, itself, a multiple of 12, the new number must remain a multiple of 12, as well. [If we called the old number of boxes 12x, the new number would be 12x + 60. We could then factor out a 12 and call this number 12(x + 5.) This number is clearly a multiple of 12.] Therefore the new number, after 60 boxes are added, is a multiple of both 12 and 14. Now we can find the least common multiple of 12 and 14 to ensure that we don’t miss any possibilities.

The prime factorization of 12: 2^2 * 3

The prime factorization of 14: 2 * 7

The least common multiple of 12 and 14: 2^2 * 3 * 7 = 84.

We now know that, after 60 boxes were added, the total number of boxes was a multiple of 84. There could have been 84 boxes or 168 boxes, etc. And before the 60 boxes were added, there could have been 84-60 = 24 boxes or 168-60 = 108 boxes, etc.

A brief summary:

After 60 boxes were added: 84, 168, 252….

Before 60 boxes were added: 24, 108, 192….

That feels like a lot of work to do before even glancing at the statements, but now look at how much easier they are to evaluate!

Statement 1 tells us that there were fewer than 110 boxes before the 60 boxes were added, meaning there could have been 24 boxes to start (and 84 once 60 were added), or there could have been 108 boxes to start (and 168 once 60 were added). Because there are multiple potential solutions here, Statement 1 alone is not sufficient to answer the question.

Statement 2 tells us that there were fewer than 120 boxes after 60 boxes were added. This means there could have been 84 boxes – that’s the only possibility, as the next number, 168, already exceeds 120. So we know for a fact that there are 84 boxes after 60 were added, and 24 boxes before they were added. Statement 2 alone is sufficient, and the answer is B.

Takeaway: questions that look strange or funky are always testing concepts that have been tested in the past – otherwise, the exam wouldn’t be standardized. By making these connections, and recognizing that a verbal clue such as “none left over” really means that we’re talking about multiples and factors, we can recognize even the most abstract patterns on the toughest of GMAT questions.