The post Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient appeared first on Veritas Prep Blog.
]]>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:
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.
We start with what we know:
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:
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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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]]>The post How to Solve “Unsolvable” Equations on the GMAT appeared first on Veritas Prep Blog.
]]>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.
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: A GMAT Quant Question That Troubles Many! appeared first on Veritas Prep Blog.
]]>In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.
This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.
We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:
† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?
(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089
The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.
The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?
As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”. Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.
For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:
Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?
Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:
58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.
Along those same lines:
AB = 10A + B
BA = 10B + A
Going back to our original question:
(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)
We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.
We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.
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: Try to Answer This GMAT Challenge Question! appeared first on Veritas Prep Blog.
]]>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
Think about how you might answer this question:
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:
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
The answer is D.
But this calculation is rather tedious and time consuming. Can’t we use the deviation method we discussed for averages and weighted averages, instead? 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:
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!
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: 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:
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:
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:
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:
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:
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!
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]]>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:
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:
(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!
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]]>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, 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 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:
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, 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 Quarter Wit, Quarter Wisdom: How to Negate Assumption Answer Choices on the GMAT appeared first on Veritas Prep Blog.
]]>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, 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 Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT! appeared first on Veritas Prep Blog.
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