The post Using “Few” vs. “A Few” vs. “Quite a Few” in a GMAT Verbal Question appeared first on Veritas Prep Blog.

]]>*“Few”*, when used without a preceding “a”, means “very few” or “none at all”. “Few” is a negative, which puts the quantity of what you are describing near zero.

On the other hand, “*a few”* is used to indicate “not a large number”. “*A few”* also indicates a small approximate number, but it is positive nonetheless.

The difference between the two is subtle, yet there are instances where the two can mean completely opposite things. For example, “I have a few friends” is the same as saying “I have some friends”. “I have few friends”, however, implies that I have *only* very few friends (as opposed to many). It can also imply that I don’t feel very well about it, and I wish I had more friends.

Also, note that there is a very common expression, “quite a few”, which looks like it could mean “rather few” or “very few”, but it does not. It actually means the exact opposite: “a large or significant number” or “many”. So saying, “I have quite a few friends,” is the same as saying “I have quite a lot of friends”.

Here are a few other simple examples:

- A few people think that red wine is healthy.
- This implies some people think that red wine is healthy.

- Few people think that red wine is healthy.
- This implies only very few people, a very small number, think that red wine is healthy; most think that it is not.

- Quite a few people think that red wine is healthy.
- This implies many people, a large number, think that red wine is healthy.

Let’s examine an official Critical Reasoning question in which confusion among these terms could lead to an incorrect answer:

*Until now, only injectable vaccines against influenza have been available. They have been primarily used by older adults who are at risk for complications from influenza. A new vaccine administered in a nasal spray form has proven effective in preventing influenza in children. Since children are significantly more likely than adults to contract and spread influenza, making the new vaccine widely available for children will greatly reduce the spread of influenza across the population. *

*Which of the following, if true, most strengthens the argument?*

*(A) If a person receives both the nasal spray and the injectable vaccine, they do not interfere with each other. *

*(B) The new vaccine uses the same mechanism to ward off influenza as injectable vaccines do. *

*(C) Government subsidies have kept the injectable vaccines affordable for adults. *

*(D) Of the older adults who contract influenza, relatively few contract it from children with influenza. *

*(E) Many parents would be more inclined to have their children vaccinated against influenza if it did not involve an injection. *

Let’s break down the argument of this passage first. We are given following premises:

- Until now, only injections of the influenza vaccine were available.
- These injections were primarily used by older adults.
- Now nasal sprays are available that prevent influenza in children.
- Children are more likely to contract and spread influenza.
- Conclusion: If nasal sprays are made available for children, it will greatly reduce the spread of influenza across the population.

Does something come to mind when you read this conclusion? What initially came to my mind was that if children are most likely to contract and spread influenza, they should have just been given the injections and that would have prevented the spread of disease across the population. Why is it that the availability of a nasal spray will prevent the spread of influenza but injections have not been able to do this?

We need to strengthen the argument, so we should focus on our conclusion and find out what will strengthen it the most. Let’s go through each of the answer choices:

*(A) If a person receives both the nasal spray and the injectable vaccine, they do not interfere with each other.*

If a person has already been given an injection, he or she is immune to influenza – taking the nasal spray on top of this will not have any impact on his or her immunity. This option is irrelevant to the argument, thus A cannot be our answer.

*(B) The new vaccine uses the same mechanism to ward off influenza as injectable vaccines do.*

This answer choice only says that the nasal sprays work in the same way the injections do. We are not told exactly why injections could not prevent the spread of influenza while the nasal spray will, so this option is also not correct.

*(C) Government subsidies have kept the injectable vaccines affordable for adults.*

This option tells us that the subsidies have kept injections affordable for all older adults, but it doesn’t say anything about the cost of the nasal spray. If, instead, this option stated, “Injections are very expensive but nasal spray is a cheap alternative”, it might have made a stronger contender, however we do not know whether cost is a factor that parents consider at all when getting their children vaccinate (to make this option the correct answer, we might even have to add something like, “Parents are not willing to get their kids immunized if the vaccine is very expensive”). As is, however, this answer choice is not correct.

*(D) Of the older adults who contract influenza, relatively few contract it from children with influenza.*

Here is the trick – many test takers feel that this option is like an assumption, and hence, it certainly strengthens the conclusion. “Few” is assumed to be “some”, so it seems to them that this option is saying, “Some older adults do contract influenza from children”. It certainly seems to be an assumption, since that is how the spread of influenza will reduce across the population of older adults.

We know, however, that “few” actually means “hardly any” or “near zero”. If few (near zero) older adults catch flu from children, it doesn’t strengthen the conclusion. If anything, it has the opposite effect since the older adults will be unaffected, and hence, it is unlikely that the spread of influenza will reduce across the population. Because of this, option D is not correct.

*(E) Many parents would be more inclined to have their children vaccinated against influenza if it did not involve an injection.*

Now this is what we are looking for – a reason why parents don’t give influenza shots to their kids but will be willing to give them nasal sprays. Parents don’t like to give shots to their kids (could be due pain associated with a shot or whatever, the reason why doesn’t really matter here), but now that a nasal spray version of the vaccine is available, they will be more inclined to get their kids vaccinated. This will probably help prevent the spread of influenza across the population. The correct answer, therefore, is E.

*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 How to Quickly Interpret Ranges of Variables in GMAT Questions appeared first on Veritas Prep Blog.

]]>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.

Our answer is C.

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.

*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: Ignore the Diagram in That GMAT Geometry Question! appeared first on Veritas Prep Blog.

]]>But sometimes, the GMAT Testmakers give such diagrams that we wish we were not given the diagram at all. In fact, the addition of a diagram – something that often simplifies our questions – can take the difficulty of the question to a whole new level. By now you are probably thinking that I am surely exaggerating, so I will proceed with an example.

Try to figure this out: when the figure given below is cut along the solid lines, folded along the dashed lines, and then taped along the solid lines, the result is a model of a geometric solid.

Now, can you use your imagination and figure out what kind of a geometric solid you will get in this case? Don’t go ahead just yet – first, give it a shot for a few minutes:

To be honest, I have given it a try and it is certainly not easy. I will know for sure only when I actually carry out the aforementioned steps – cut the paper along the solid lines, fold along the dashed lines and then tape up along the solid lines. Without carrying out the steps I am not sure exactly what kind of a figure I will get.

So the test maker comes to our rescue here. Here is the complete question:

*When the figure above is cut along the solid lines, folded along the dashed lines, and taped along the solid lines, the result is a model of a geometric solid. This geometric solid consists of two pyramids each with a square base that they share. What is the sum of number of edges and number of faces of this geometric solid?*

*(A) 10*

*(B) 18*

*(C) 20*

*(D) 24*

*(E) 25*

The Testmaker specifies what kind of a figure we get – two pyramids, each with a square base that they share. Figuring this out in one minute without an actual paper and scissor at hand would need extraordinary skill. Many test-takers spend precious minutes trying to make sense of the given diagram, but in problems like this, it should be completely ignored because we already know what it will look like – two pyramids with a common square base.

This, we understand! We know what a pyramid looks like – triangular faces converge to a single point at the top with a polygon (often a square) base. We need two pyramids joined together at the base.

This is what the solid will look like:

Just the 4 triangular faces of each of the two pyramids (8 triangles total) will be visible. Since they will share the square base, the base will not be visible. Hence, the figure will have 8 faces.

Now let’s see how many edges there will be: to make the top pyramid, four triangular faces join to give four edges. To make the bottom pyramid, another four triangular faces join to give four more edges. The two pyramids join on the square base to give yet another four edges.

So all in all, we have 4 + 4 + 4 = 12 edges

When we sum up the faces and edges, we get 8 + 12 = 20

The question is much more manageable now. All we had to do was ignore the diagram given to us!

*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: The 3-Step Method to Solving Complex GMAT Algebra Problems appeared first on Veritas Prep Blog.

]]>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.

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]]>The post Which is Worse to Encounter on a GMAT Question: Median or Mean? appeared first on Veritas Prep Blog.

]]>Median is the value at a point – to be precise, the point which divides the increasing data set into two equal halves. You don’t care what is on the left and what is on the right of this point, so an outlier will do nothing to the median. The mean, however depends on every value in the set. If you increase one element of data, the mean of the set changes – outliers can drastically change the value of the mean. Hence, every element has to be kept in mind! With the median, there is a lot less to worry about.

Let’s illustrate this with an example data sufficiency question:

**Question on Median:**

*At a bakery, cakes are sold every day for a certain number of days. If 6 or more cakes were sold for 20% of the total number of days, is the median number of cakes sold less than 4?*

*Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.*

*Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.*

The following is the number of cakes sold on any of the days mentioned in the question:

Say there were 100 days (since all figures are in terms of percentages, we can assume a number to simplify our understanding).

The question stem tells us that 6 or more cakes were sold for 20% of the days, so for 20 days, 6 or more cakes were sold. Then for 80 days, 1/2/3/4/5 cakes were sold.

With this information in mind, is the median number of cakes sold in one day less than 4?

We know how to get the median. When we arrange all figures in increasing order, the median will be the average of the 50th and the 51st terms. We need to know if the average of the 50th and 51st term is less than 4. Let’s tackle the statements one at a time:

*Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.*

The number of days that less than 6 cakes were sold = 80. 75% of these 80 days will be 60 days. In 60 days, less than 4 cakes were sold. So the 50th and 51st terms will be less than 4 and so will their average. Hence, the median will be less than 4. This statement alone is sufficient.

*Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.*

In 20 days, 6 or more cakes were sold. This constitutes 50% of the days during which 4 or more cakes were sold, so in another 20 days, 4 or 5 cakes were sold. Hence, during the leftover 60 days, less than 4 cakes were sold. The 50th and 51st terms will be less than 4 and so will their average. Hence, the median will be less than 4. This statement alone is also sufficient, so our answer is D.

All we needed to worry about here were the 50th and 51st terms, however the whole problem changes when we talk about mean instead of median.

**Same Question on Mean:**

*At a bakery, cakes are sold every day for a certain number of days. If 6 or more cakes were sold for 20% of the total number of days, is the average number of cakes sold less than 4?*

*Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4.*

*Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more.*

Again, the question stem tells us that 6 or more cakes were sold for 20% of the days, so for 20 days, 6 or more cakes were sold. Then for 80 days, 1/2/3/4/5 cakes were sold.

We now need to ask ourselves is the average number of cakes sold in one day less than 4?

This question asks us about the average. – that is far more complicated than the median. Every value matters when we talk about the average. We need to know the number of cakes sold on each of these 100 days to get the average.

6 or more cakes were sold in 20 days. Note that the number of cakes sold during these 20 days could be any number greater than 6, such as 20 or 50 or 120, etc. The minimum number of cakes sold on these 20 days would be 6*20 = 120. There is no limit to the maximum number of cakes sold.

With this in mind, let’s examine the statements:

In 80 days, less than 6 cakes were sold. Of this number, 75% is 60 days. In 60 days, less than 4 cakes were sold.

So in 60 days, you have a minimum of 1*60 = 60 cakes sold and a maximum of 3*60 = 180 cakes sold. During the leftover 20 days 4 or 5 cakes were sold, so you have a minimum of 4*20 = 80 cakes and a maximum of 5*20 = 100 cakes.

The minimum value of the average is (120 + 60 + 80)/ 100 = 2.6 cakes, but the maximum average could be anything. Therefore, this statement alone is not sufficient.

The 20 days when 6 or more cakes were sold make up 50% of the days when 4 or more cakes were sold. So for another 20 days, 4 or 5 cakes were sold. This gives us a minimum of 4*20 = 80 cakes and a maximum of 5*20 = 100 cakes. For 60 days, 1/2/3 cakes were sold. So in 60 days, you have minimum of 1*60 = 60 cakes sold and a maximum of 3*60 = 180 cakes sold.

The minimum value of the average is (120 + 60 + 80)/ 100 = 2.6 cakes, but again, the maximum average could be anything. This statement alone is also not sufficient.

Note that both statements give you the same information, so if they are not sufficient independently, they are not sufficient together. The answer of this modified question would be E.

Here, we had to assume the minimum and maximum value for each data point to get the range of the average – we couldn’t just rely on one or two data points. Finding the mean during a GMAT question requires much more information than finding the median!

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]]>The post An Interesting Right Triangle Property You’ll Need to Know for the GMAT appeared first on Veritas Prep Blog.

]]>**Property: The circumcenter of a right triangle is the mid point of the hypotenuse.**

Let’s prove this first and then we will see its application.

Say, we have a right triangle ABC right angled at B. Let’s draw the perpendicular bisector of AB which intersects AB at its mid point M. Say this line intersects the hypotenuse AC at N. We need to prove that AN = CN. Note that triangle AMN and triangle ABC are similar triangles using the AA property (angle AMN = angle ABC = 90 degrees and angle A is common to both triangles). So the ratio of the sides of the two triangle is the same. Since MN is the perpendicular bisector of line AB, AM = MB which means that AM is half of AB.

So AM/AB = 1/2 = AN/AC

Hence AN = NC

So N is the mid point of AC.

Using the exact same logic for side BC, we will see that its perpendicular bisector also bisects the hypotenuse. So N would be the circumcenter of triangle ABC and the mid point of AC.

Using an official question, let’s see how this property can be useful to us:

*In the rectangular coordinate system shown above, points O, P, and Q represent the sites of three proposed housing developments. If a fire station can be built at any point in the coordinate system, at which point would it be equidistant from all three developments?*

*(A) (3,1)*

*(B) (1,3)*

*(C) (3,2)*

*(D) (2,2)*

*(E) (2,3)*

First, let’s see how we will solve this question without knowing this property and using co-ordinate geometry instead.

Method 1:

Points O and Q lie on the X axis and are 4 units apart. We need a point equidistant from both O and Q. All such points will lie on the line lying in the middle of O and Q and perpendicular to the X axis. The equation of such a line will be x = 2. The fire station should be somewhere on this line.

Points O and P lie on the Y axis and are 6 units apart. We need a point equidistant from both O and P. All such points will lie on the line lying in the middle of O and P and perpendicular to the Y axis. The equation of such a line will be y = 3. The fire station should be somewhere on this line too.

Any two lines on the XY plane intersect at most at one point (if they are not overlapping). Since the fire station must lie on both these lines, it must be on their intersection i.e. at (2, 3).

This point (2,3) will be equidistant from O, Q and P. Therefore, the answer is E.

Method 2:

Think of the question in terms of the perpendicular bisectors of triangle OPQ. Their point of intersection will be equidistant from all three vertices.

We know that the circumcenter lies on the mid point of the hypotenuse. The end points of the hypotenuse are (4, 0) and (0, 6). The mid point will be

x = (4 + 0)/2 = 2

y = (0 + 6)/2 = 3

As in Method 1, the point (2, 3) will be equidistant from all three points, O, P and Q. Again, the answer is E.

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]]>The post Tackling GMAT Critical Reasoning Boldface Questions appeared first on Veritas Prep Blog.

]]>We have often found that one strategy, which is very helpful in other question types too, helps sort out most questions of this type, though not in the same way. That strategy is – ‘find the conclusion(s)’

The conclusion of the argument is the position taken by the author.

Boldface questions (and others too) sometimes have more than one conclusion – One would be the conclusion of the argument i.e. the author’s conclusion. The argument could mention another conclusion which could be the conclusion of a certain segment of people/ some scientists/ some researchers/ a politician etc. We need to segregate these two and how each premise supports/opposes the various conclusion. Once this structure is in place, we automatically find the answer. Let’s see how with an example.

**Question:**** Recently, motorists have begun purchasing more and more fuel-efficient economy and hybrid cars that consume fewer gallons of gasoline per mile traveled. There has been debate as to whether we can conclude that these purchases will actually lead to an overall reduction in the total consumption of gasoline across all motorists.** The answer is no, since motorists with more fuel-efficient vehicles are likely to drive more total miles than they did before switching to a more fuel-efficient car, negating the gains from higher fuel-efficiency.

Which of the following best describes the roles of the portions in bold?

(A)The first describes a premise that is accepted as true; the second introduces a conclusion that is opposed by the argument as a whole.

(B)The first states a position taken by the argument; the second introduces a conclusion that is refuted by additional evidence.

(C)The first is evidence that has been used to support a position that the argument as a whole opposes; the second provides information to undermine the force of that evidence.

(D)The first is a conclusion that is later shown to be false; the second is the evidence by which that conclusion is proven false.

(E)The first is a premise that is later shown to be false; the second is a conclusion that is later shown to be false.

**Solution**: As our first step, let’s try to figure out the conclusion of the argument:

The author’s view is that “purchases of fuel efficient vehicles will NOT lead to an overall reduction in the total consumption of gasoline across all motorists.”

This is the position the argument (and author) takes.

The argument gives us another conclusion: these purchases will actually lead to an overall reduction in the total consumption of gasoline across all motorists.

Some people take this position (implied by the use of “there has been debate”)

This is our second bold statement. It introduces the opposing conclusion.

Let’s look at our options now.

(A) The first describes a premise that is accepted as true; the second introduces a conclusion that is opposed by the argument as a whole.

The first bold statement: Recently, motorists have begun purchasing more and more fuel-efficient economy and hybrid cars that consume fewer gallons of gasoline per mile traveled.

This is a premise and has been accepted as true. We know it has been accepted as true since the last line ends with – “…negating the gains from higher fuel-efficiency”

We have seen above that the second bold statement tells us about a conclusion that the argument opposes.

So (A) is correct. We have found our answer but let’s look at the other options too.

(B) The first states a position taken by the argument; the second introduces a conclusion that is refuted by additional evidence.

The first bold statement is a premise. It is not the position taken by the argument. Let’s move on.

(C) The first is evidence that has been used to support a position that the argument as a whole opposes; the second provides information to undermine the force of that evidence.

This option often confuses test-takers.

The evidence is – “Recently, motorists have begun purchasing more and more fuel-efficient economy and hybrid cars that consume fewer gallons of gasoline per mile traveled.”

That is, “the motorists have begun purchasing fuel efficient cars that give better mileage.”

The second bold statement does not undermine this evidence at all. In fact, it builds up on it with – “This brings up a debate on whether it will lead to overall decreased fuel consumption?”

Hence (C) is not correct.

(D)The first is a conclusion that is later shown to be false; the second is the evidence by which that conclusion is proven false.

The first bold statement is not a conclusion. So no point dwelling on this option.

(E)The first is a premise that is later shown to be false; the second is a conclusion that is later shown to be false.

The premise is taken to be true. The argument ends with “… the gains from higher fuel-efficiency”. Hence, this option doesn’t stand a chance either.

We hope you see how easy it is to break down the options once we identify the conclusion(s).

Keep practicing!

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]]>*Twenty-four men can complete a job in sixteen days. Thirty-two women can complete the same job in twenty-four days. Sixteen men and sixteen women started working on the job for twelve days. How many more men must be added to complete the job in 2 days?*

*(A) 16*

* (B) 24*

* (C) 36*

* (D) 48*

* (E) 54*

Here, we are dealing with two groups of people: men and women. These two groups have different rates of completing a job. We are also told that a certain number of men and women do a part of the job, and we are asked to find the number of additional “men” required to finish the job in a shorter amount of time.

Recall that we have already come across questions where workers start some work and then more workers join in to complete the work before time.

The problem with this question is that we have two types of workers, not just one. So let’s try to simplify the question to a form that we know how to easily solve.

We’ll start by finding the relation between the rate of work done by men and the rate of work done by women. Let’s make the number of men and women the same to find the number of days it will take each group to complete 1 job.

Given: 24 men complete 1 job in 16 days

Given: 32 women complete 1 job in 24 days

So how many days will 24 women take to complete 1 work? (Why 24 women? Because we know how many days 24 men take)

We know how to solve this problem. (It has already been discussed in a past post).

32 women ……………. 1 work ………………. 24 days

24 women ……………. 1 work ………………. ?? days

No. of days taken = 24 * (32/24) = 32 days

Now this is what we have: 24 men take 16 days while 24 women take 32 days

So women take twice the time taken by men to do the same work (32 days vs 16 days). This means the rate of work of women is half the rate of work of men. This means 2 women are equivalent to 1 man i.e. 2 women will do the same work as 1 man does in the same time.

So now, let us replace all women by men so that we have only one type of worker.

Now this is our regular work rate question –

Given: 24 men complete the work in 16 days

Given: 16 men and 16 women work for 12 days

This means that we have 16 men and 8 men work for 12 days

which implies 24 men work for 12 days

We know that 24 men complete the work in 16 days. If they work for 12 days, there are 4 more days to go. But the work has to be completed in 2 days.

24 men …………… 4 days

?? men ……………. 2 days

No of men needed = 24 * (4/2) = 48

So we need 24 additional men to complete the work in 2 days.

Or looking at it another way, 24 men need 16 days to complete the work, so they need another 4 days to complete. But if we want them to complete the work in half the time (2 days), we will need twice the work force. So we need another 24 men.

Answer (B)

Basically, the question involved solving two smaller work-rate problems. Doesn’t seem daunting now, right?

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]]>*In 2009, a private school spent $200,000 on a building which housed classrooms, offices, and a library. In 2010, the school was unable to turn a profit. Therefore, the principal should be fired.*

*Each of the following, if true, weakens the author’s conclusion EXCEPT:*

*(A) The principal was hired primarily for her unique ability to establish a strong sense of community, which many parents cited as a quality that kept children enrolled in the school longer.*

*(B) The new library also features a seating area big enough for all students to participate in cultural arts performances, which the head of school intends to schedule more frequently now.*

*(C) The principal was hired when the construction of the new building was almost completed.*

*(D) A significant number of families left the school in 2010 because a favourite teacher retired.*

*(E) More than half of the new families who joined the school in 2010 cited the beautiful new school facility as an important factor in their selection of the school.*

This is a weaken/exception question, so four of the five answer choices will weaken the argument, while the fifth option (which will be the correct answer) will either not have any impact on the argument or it might even strengthen it. As we know, such questions require a bit more effort to answer, since four of the five options will definitely be relevant to the argument. The important thing is to focus on what we are given and not assume what the various answer options may or may not lead to. Let’s understand this:

The gist of the argument:

- Last year, a lot of money was spent to construct a new building with many amenities.
- This year, the school did not see a profit.
- Hence, fire the principal.

Based on the two given facts – “a lot of money was spent to make the building in 2009” and “the school did not see a profit in 2010” – the author has decided to fire the principal. Many pieces of information could weaken his stance. For example:

- It was not the principal’s decision to construct the building.
- The school’s revenue in 2010 took a hit because of some other factor.
- The school’s losses reduced by a huge amount in 2010 and the probability of it seeing a profit in 2011 is high.

Information such as this could improve the principal’s case to stay. We know that for this particular question, there will only be one option that does not help the principal.

You will have to choose the answer choice which, with the given information, does not help the principal’s case. Let’s look at the options now:

*(A) The principal was hired primarily for her unique ability to establish a strong sense of community, which many parents cited as a quality that kept children enrolled in the school longer.*

With this answer choice, we see that the principal was hired not to increase school profits, but for another critical purpose. Perhaps the school’s finance department is in charge of worrying about profits, and so the head of that department needs to be fired! This answer choice makes a strong case for keeping the principal, and hence, weakens the author’s argument.

*(B) The new library also features a seating area big enough for all students to participate in cultural arts performances, which the head of school intends to schedule more frequently now.*

If true, this statement would have no impact on whether or not the principal should be fired. It describes an amenity provided by the new building and how it will be used – it neither strengthens nor weakens the principal’s case to stay, hence, this is the correct answer choice. But let’s look at the rest of the options too, just to be safe:

*(C) The principal was hired when the construction of the new building was almost completed.*

This tells us that the new building was not her decision. So if it did not have the desired effect, she cannot be blamed for it. So it again helps her case.

(D) A significant number of families left the school in 2010 because a favourite teacher retired.

This answer choice shows that there was another reason behind the school’s loss in profit. The construction of the building could still be a good idea that leads to future profits, which the principal’s case and weakens the author’s argument.

*(E) More than half of the new families who joined the school in 2010 cited the beautiful new school facility as an important factor in their selection of the school.*

For some reason, this is the answer choice that often trips up students. They feel that it doesn’t help the principal’s case – that because the new building attracts students, if there are losses, it means that the loss is due to a fault with the new building, and thus, the principal is at fault. But note that we are assuming a lot to arrive at that conclusion. All we are told is that the new building is attracting students. This means the new building is serving its purpose – it is generating extra revenue. The fact that the school is still experiencing losses could be explained by many different reasons.

Since the author’s decision to fire the principal is based solely on the premise that a lot of money was spent to construct the new building, which now seems to serve no purpose (because the school experienced losses), this answer choice certainly weakens the argument. The option tells us that the principal’s decision to make the building was justified, so it helps her case to stay with the school.

After examining each answer choice, we can see that the answer is clearly B. Remember, in Critical Reasoning questions it is crucial to come to conclusions only based on the facts that are given – creating assumptions based on information that is not given can lead you to fall in a Testmaker trap.

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]]>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:

- Division by 0 is not allowed, hence you cannot divide by a variable until and unless we know that it cannot be 0.
- 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.

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