The post Quarter Wit, Quarter Wisdom: Evaluating Nasty GMAT Answer Choices appeared first on Veritas Prep Blog.

]]>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!

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!*

*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**!*

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]]>The post Evaluating “Useful to Evaluate” Critical Reasoning Questions – Part II appeared first on Veritas Prep Blog.

]]>*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 organised 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?*

Let’s first break down what the argument says:

- Advertising sales were declining.
- The paper reorganized the advertising sales team two years back.
- Advertising sales increased after reorganisation.

Now, we want to figure out whether the increase actually happened due to the reorganization; in other words, we need to evaluate what else could have caused the increase in sales, if not the reorganization. Say the lead of the sales team changed two years back – it is possible that he is responsible for the increase in revenue. Four of the five answer choices will raise similar questions, while the leftover option (which will be our answer) will not. Let’s take a look at each of the answer choices:

*(A) Two years ago, what proportion of the Greenville Times’ total revenue was generated by the advertising sales?*

The proportion of advertising sales as a part of the total revenue is immaterial to us – we only need to evaluate why the advertising sales have increased. It is possible that the revenue from other sources has increased much more than the revenue from advertising sales, and hence, advertising sales could be a smaller proportion of the overall revenue now, however this doesn’t matter at all. This option has nothing to do with the increase in advertising sales, and hence, is the correct answer.

Let’s take a look at all the other options too, just to be safe:

*(B) Has the circulation of the Greenville Times increased substantially in the last two years?*

This answer choice can be evaluated in two ways:

- Yes, it has increased – If the circulation increased substantially in the last two years, that could have led to the increase in advertising sales.
- No, it has not increased – If the circulation hasn’t increased substantially, then there must be another reason for the increase in advertising sales. In that case, the reorganization could be the reason.

These two answers affect the argument differently, and hence, this option will be useful in evaluating the argument.

*(C) Has there been a substantial turnover in personnel in the advertising sales force over the last two years?*

Again, the answer choice can be evaluated in two ways:

- Yes, there has – If there has been a substantial turnover in personnel, it is possible that more capable people have been hired, which could have led to higher advertising sales.
- No, there hasn’t – If there hasn’t been a substantial turnover in personnel, then there would need to be another reason for the increased advertising sales. In that case, the reorganization could be the reason.

The two answers affect the argument differently, so this option will also be useful in evaluating the argument.

*(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?*

This option can also be evaluated in two ways:

- Yes, they did find it difficult – Did reorganization make it easier to keep track of relevant developments? If yes, then the reorganization could be responsible for the increase in sales.
- No, they did not find it difficult – If they did not find it difficult to keep up with relevant developments, then we cannot say whether the reorganization was responsible for the increase in sales or not.

These two responses affect the argument differently. Hence, this option will be useful in evaluating the argument.

*(E) Has the economy in Greenville and the surrounding regions been growing rapidly over the last two years?*

Answer choice E can also be evaluated in two ways:

- Yes, it has – If the economy has been growing rapidly over the past two years, it could be the reason for higher advertising sales. Then we may not be able to attribute the improvement in advertising sales to the reorganization.
- No it has not – If there has been no such growth in the economy, then reorganization could be the reason for higher advertising sales.

Again, the two responses affect the argument differently, so this option will also be useful in evaluating the argument.

We see that B, C, D and E are all useful in evaluating the argument. Therefore, our answer is A. We hope you will find it easier to handle such questions in the future!

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!*

*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 post Evaluating “Useful to Evaluate” Critical Reasoning Questions – Part II appeared first on Veritas Prep Blog.

]]>The post Evaluating “Useful to Evaluate” Critical Reasoning Questions on the GMAT appeared first on Veritas Prep Blog.

]]>To answer this type of question, all you will need to do is follow these six simple steps:

1) Identify the conclusion.

2) Ask yourself the question raised by answer choice A.

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!

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!*

*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 post Evaluating “Useful to Evaluate” Critical Reasoning Questions on the GMAT appeared first on Veritas Prep Blog.

]]>The post Quarter Wit, Quarter Wisdom: Using the Deviation Method for Weighted Averages appeared first on Veritas Prep Blog.

]]>The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:

*What is the average of 452, 452, 453, 460, 467, 480, 499, 499, 504?*

What would you say the average is here? Perhaps, around 470?

**Shortfall:**

We have two 452s – 452 is 18 less than 470.

453 is 17 less than 470.

460 is 10 less than 470.

467 is 3 less than 470.

Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.

**Excess:**

480 is 10 more than 470.

We have two 499s – 499 is 29 more than 470.

504 is 34 more than 470.

Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.

The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.

So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.

Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)

This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.

We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:

*Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?*

*(A) y + 3z*

*(B) (y +z) / 4*

*(C) 2y + 3z*

*(D) 3y + z*

*(E) 3y + 4.5z*

Grade 1 milk contains 1% fat. Grade 2 milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.

Shortfall = x*(1.5 – 1)

Excess = y*(2 – 1.5) + z*(3 – 1.5)

Since 1.5 is the actual average, the shortfall = the excess.

x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)

x/2 = y/2 + 3z/2

x = y + 3z

And there you have it – the answer is A.

We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your GMAT toolkit.

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!*

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

The post Quarter Wit, Quarter Wisdom: Using the Deviation Method for Weighted Averages appeared first on Veritas Prep Blog.

]]>The post Quarter Wit, Quarter Wisdom: How to Negate Assumption Answer Choices on the GMAT appeared first on Veritas Prep Blog.

]]>We already know that many sentences are invalidated by negating the verb of the dominant clause. For example:

*There has been a corresponding increase in the number of professional companies devoted to other performing arts.*

becomes

*There has not been a corresponding increase in the number of professional companies devoted to other performing arts.*

Recently, we got a query on how to negate various modifiers such as “most” and “a majority”. So today, we will examine how to negate the most popular modifiers we come across:

- All -> Not all
- Everything -> Not everything
- Always -> Not always
- Some -> None
- Most -> Half or less than half
- Majority -> Half or less than half
- Many -> Not many
- Less than -> Equal to or more than
- Element A -> Not element A
- None -> Some
- Never -> Sometimes

Let’s take a look at some examples with these determiners:

1) “All of the 70 professional opera companies are commercially viable options.”

This becomes, “Not all of the 70 professional opera companies are commercially viable options.”

2) “There were fewer than 45 professional opera companies that had been active 30 years ago and that ceased operations during the last 30 years.”

This becomes, “There were 45 or more professional opera companies that had been active 30 years ago and that ceased operations during the last 30 years.”

3) “No one who is feeling isolated can feel happy.”

This becomes, “Some who are feeling isolated can feel happy.”

4) “Anyone who is able to trust other people has a meaningful emotional connection to at least one other human being.”

This becomes, “Not everyone who is able to trust other people has a meaningful emotional connection to at least one other human being.”

5) “The 45 most recently founded opera companies were all established as a result of enthusiasm on the part of a potential audience.”

This becomes, “The 45 most recently founded opera companies were not all established as a result of enthusiasm on the part of a potential audience.”

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

This becomes, “Not many of the vehicles that were ticketed for exceeding the speed limit were ticketed more than once in the time period covered by the report.”

7) “The birds of prey capture and kill every single Spotted Mole that comes above ground.”

This becomes, “Not every single Spotted Mole that comes above ground is captured and killed by the birds of prey.”

8) “At least some people who do not feel isolated are happy.”

This becomes, “No people who do not feel isolated are happy.”

9) “Some land-based mammals active in this region, such as fox, will also hunt and eat the Spotted Mole on a regular basis.”

This becomes, “None of the land-based mammals active in this region, such as fox, will also hunt and eat the Spotted Mole on a regular basis.”

10) “No other animal could pose as significant a threat to the above-ground fruits as could the Spotted Mole.”

This becomes, “Some other animals could pose as significant a threat to the above-ground fruits as could the Spotted Mole.”

We hope the next time you come across an assumption question, you will not face any trouble negating the answer choices!

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!*

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

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]]>The post Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT! appeared first on Veritas Prep Blog.

]]>*The last digit of 12^12 + 13^13 – 14^14 × 15^15 =*

*(A) 0*

*(B) 1*

*(C) 5*

*(D) 8*

*(E) 9*

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12

The units digit of 12 is 2.

2 has a cyclicity of 2 – 4 – 8 – 6.

The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13

The units digit of 13 is 3.

3 has a cyclicity of 3 – 9 – 7 – 1.

A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)

This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

**(i) 100-29**

100

-29

071

**(ii) 29-100**

100

-29

071

(But since the sign of 100 is negative, your answer is actually -71.)

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0

– pq9

ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

*Getting ready to take the GMAT? We have **free online GMAT seminars** **running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!*

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

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]]>The post Quarter Wit, Quarter Wisdom: Linear Relations in GMAT Questions appeared first on Veritas Prep Blog.

]]>We know the equation of a line: it is **y = mx + c**, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

*A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?*

*(A) 20*

*(B) 36*

*(C) 48*

*(D) 60*

*(E) 84*

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)

60 = 24m + c ……(II)

(II) – (I)

30 = 18m

m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c

c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20

R = 48

48 (answer choice C) is our answer.

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer.

*Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!*

*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!*

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]]>The post Advanced Number Properties on the GMAT – Part VI appeared first on Veritas Prep Blog.

]]>Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

**Say N = 3**

The numbers are 5, 6, 7 (any three consecutive numbers)

Their sum is 5 + 6 + 7 = 18

Their product is 5*6*7 = 210

Note that both the sum and the product are divisible by 3 (i.e. N).

**Say N = 5**

The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)

Their sum is 2 + 3 + 4 + 5 + 6 = 20

Their product is 2*3*4*5*6 = 720

Again, note that both the sum and the product are divisible by 5 (i.e. N)

**Say N = 4**

The numbers are 3, 4, 5, 6 (any five consecutive numbers)

Their sum is 3 + 4 + 5 + 6 = 18

Their product is 3*4*5*6 = 360

Now note that the sum is not divisible by 4, but the product is divisible by 4.

**If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.**

Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

(Multiple of N),

(Multiple of N) +1,

(Multiple of N) + 2,

… ,

(Multiple of N) + (N-2),

(Multiple of N) + (N-1)

In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:

(Multiple of 3) + 2 = 5

(Multiple of 3) = 6

(Multiple of 3) + 1 = 7

In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:

(Multiple of 4) + 3 = 3

(Multiple of 4) = 4

(Multiple of 4) + 1 = 5

(Multiple of 4) + 2 = 6

etc.

What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will always add up in pairs to give the sum of N:

1 + (N – 1) = N

2 + (N – 2) = N

3 + (N – 3) = N

…

So when you add up all the integers, you will get a multiple of N.

What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will add up to give integers of N but one will be leftover:

1 + (N – 1) = N

2 + (N – 2) = N

3 + (N – 3) = N

…

The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.

For example, let’s reconsider the previous example in which we had four consecutive integers:

(Multiple of 4) = 4

(Multiple of 4) + 1 = 5

(Multiple of 4) + 2 = 6

(Multiple of 4) + 3 = 3

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

*Which of the following must be true?*

*1) The sum of N consecutive integers is always divisible by N.*

*2) If N is even then the sum of N consecutive integers is divisible by N.*

*3) If N is odd then the sum of N consecutive integers is divisible by N.*

*4) The Product of K consecutive integers is divisible by K.*

*5) The product of K consecutive integers is divisible by K!*

*(A) 1, 4, 5*

*(B) 3, 4, 5*

*(C) 4 and 5*

*(D) 1, 2, 3, 4*

*(E) only 4*

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

*5) The product of K consecutive integers is divisible by K!*

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts **I**, **II**, **III**, **IV** and **V** of this series.)

*Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!*

*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!*

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]]>The post Quarter Wit, Quarter Wisdom: Divisibility by Powers of 2 appeared first on Veritas Prep Blog.

]]>For 2 – If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2.

For 4 – If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4.

For 8 – If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8.

A similar rule applies to all powers of 2:

For 16 – If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16.

For 32 – If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32.

and so on…

Let’s figure out why:

The generic rule can be written like this: **A number M is divisible by 2^n if the last n digits of M are divisible by 2^n.**

Take, for example, a division by 8 (= 2^3), where M = 65748048 and n = 3.

Our digits of interest are the last three digits, 048.

48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.

A valid question here is, “What about the remaining five digits? Why do we ignore them?”

Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).

Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.

All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.

In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.

In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.

We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:

*What is the remainder when 1990990900034 is divided by 32 ?*

*(A) 16*

*(B) 8*

*(C) 4*

*(D) 2*

*(E) 0*

Breaking down our given number, 1990990900034 = 1990990900000 + 00034.

1990990900000 ends in five 0’s so it is divisible by 32. 34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. Our answer is D.

*Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!*

*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!*

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]]>The post Quarter Wit Quarter Wisdom: What is Your Favorite Number? appeared first on Veritas Prep Blog.

]]>*“ The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”*

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this number.

*Which of the following is a factor of 1001^(32) – 1 ?*

*(A) 768*

*(B) 819*

*(C) 826*

*(D) 858*

*(E) 924*

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Let me explain:

1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)

1001^32 = 7^32 * 11^32 * 13^32

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

**free online GMAT seminars** running all the time. And, be sure to follow us on **Facebook**, **YouTube**, **Google+**, and **Twitter**!

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

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