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Math Workout for the Gre, 4th Edition_ 275+ Practice Questions with Detailed Answers and Explanations
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Knowing algebraic manipulation is important for your success on the GRE. But, the truth is, unless you study and work with algebra regularly, you’re probably much more comfortable with arithmetic. Using algebra when you are unfamiliar with it is likely to lead to errors that could have been avoided.
A basic appreciation of algebra is crucial for a good quantitative score on the GRE. But whenever you have the option, you should use arithmetic instead of algebra, for a number of reasons.
Some of you out there might think your algebra skills are passable, and that you’re not worried about making mistakes. That may well be, but the possibility of a mistake is greatly heightened by the manipulation of algebra and greatly diminished by the use of arithmetic.
The other bad thing about algebra is that often when you’re performing it, you can mess something up and not even know it. GRE question writers are experts at predicting common mistakes that students will make using algebra. They’ll use that prediction to create incorrect answer choices to trap those students who made that mistake.
So, what should you do to try to avoid algebra as much as possible? You should plug in actual numbers! Plugging In is a lot like writing your own novel; rather than wonder how many candies Phil has in his hand, assume the power to write the narrative. Decide for yourself how many he has.
There are two dead giveaways when you’re identifying problems that you can solve using your own numbers.
(B) 2(5) − 14 = −4 Nope.
(C) 2(5) + 13 = 23 Nope.
(D) 2(5) + 16 = 26 Correct!
(E) 2(5) + 26 = 36 Nope.
Since (D) is the only answer that yielded the target number of 26, that’s the correct answer.
If you had solved this algebraically, you might have added j and 3, then doubled it to 2( j + 3), then distributed it to become 2 j + 6, then added 10, and gotten the right answer. But say you forgot to distribute the 2, and 2( j + 3) became 2 j + 3, and you added 10 to get 2 j + 13. You could have chosen (C) and moved merrily along, unaware of your mistake, because ETS anticipated it.
Instead, you solved the problem using only arithmetic that you can double-check with your calculator if need be, and you avoided all of the traps. Try another.
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Here’s How to Crack It There are variables in the answer choices, so Plug In. Choose an easy number to plug in. There are 60 minutes in an hour, so make the math a little easier by choosing either a factor of 60, such as 30, or a multiple of 60. Plug in y = 30.
In 60 minutes, one machine makes 3,000 golf balls. In that case, in 30 minutes one machine would make 1,500 golf balls. The question asks how many golf balls four machines can make in 30 minutes, which is 1,500 × 4 = 6,000 golf balls. Write down 6,000 golf balls and circle it, as it is the target number. Now, check all the answer choices to see which one matches the target answer.
Replace x with 30 in each of the answer choices; the one that gives you an answer of 6,000 is the correct answer.
(A) 200(30) = 6,000 Correct!
(B) = 6.66 Nope.
(C) 750(30) = 22,500 Nope.
(D) = 250 Nope.
(E) = 24,000 Nope.
The correct answer is (A).
You might be asking yourself, “If I got a match right away, why did I have to spend that time checking all of the
others?” And that’s a good question, because your ultimate goal is to find the correct answer and scoot off to the next
question as quickly as possible. Sometimes we may choose a number that works with multiple answer choices. For
instance, say we had plugged in y = 60 instead. In that case, the four machines would have made 12,000 golf balls.
Choice (A) still works: 200(60) = 12,000. However, look at (E): = 12,000. Had we picked 60, we would have
had two answer choices that yielded the target number. If that happens, just pick a different number and check the
remaining answers.
When you plug numbers into a question, it’s perfectly fine to choose almost any number you want, with a couple of exceptions that we will outline later, as long as the number doesn’t violate restrictions that the problem stipulates. Usually, the first integer that pops into your head will work just fine, although as you get better at it you’ll get a feel for picking numbers that make your math easier.
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Here’s How to Crack It There are variables in the answer choices, so Plug In. Choose an easy number for the variable. This question has a lot of fractions in it, and fractions mean division. The fastest way to come up with a good number is to multiply the denominators, so try m = 12.
Now work through the problem in bite-sized pieces. Since of the 12 faculty members live on campus,
= 8 people live on campus. The problem then states that of those 8 people living on campus,
own a car. = 6 people on campus own a car. The question asks how many people who live on campus do
not own a car, which means that out of the 8 people on campus, 2 of them don’t own a car. The target number is 2.
easier to solve.
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Here’s How to Crack It There are no variables here, but the problem would make a little more sense if you knew the price of a share of Amalgamedia when Johanna bought into it. So Plug In for the price of Johanna’s shares. Because the question involves percents, it’s best to plug in $100 for the price of the shares. Now solve the problem.
If the stock dropped by 20% during the first month, then the price dropped by × 100, or $20, and the price went
from $100 to $80. After a 40% increase, the stock moved up by × 80, or $32, to $112. This represents a 12%
increase from the original $100, so the correct answer is (B).
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Louella’s age, then k = 3 + (2 × 5), or 13. Joshua is three times as old as Kali, so j = 3 × 13, or 39.
Louella’s age (5) is the target number. Plug in j = 39 into the answer choices to find that (E), = 5, is
the only one that matches the target number and is the correct answer.
problem involves percents, plug in 100 for the number of books. If 60% are fiction, then the other 40 are
nonfiction. Because 30% of those 40 books are about politics, then 12 books are about politics. That means
the other 28 books are neither fiction nor about politics, and reduces to. The correct answer is (B).
3 Plug In for the length and width of the smaller playground. If the length and width of the small playground are 8 feet and 10 feet, respectively, then the area of that playground is 8 × 10, or 80 square feet. Therefore, M = 80. The larger playground is twice as long (2 × 8 = 16) and five times as wide (5 × 10 = 50), so its area is 16 × 50, or 800 square feet. The difference in these areas is 800 − 80, or 720 square feet, which is the target number. Plug in M = 80 to each answer choice, to find that the only answer choice that matches the target number is (C).
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Here’s How to Crack It This is a Must Be problem, so Plug In more than once. Begin with an easy number, such as c = 2. If c = 2, then d = 3(2) − 2 = 6 − 2 = 4. Now check each answer. Choice (A) works, because −2 ≤ 4 < 10, and (B) works, because 4 ≠ −2. Eliminate (C), because 4 is not less than 0. Keep (D) because 2 < 4, and (E), because −14 < 2. There are still four answer choices remaining, so Plug In using a FROZEN number, such as c = 0, and check the remaining answer choices. If c = 0, then d = 3(0) − 2 = −2. Keep (A) because d is equal to −2. Eliminate (B) because d does equal −2. Eliminate (D), because 0 < −2 is not true. Keep (E), because −14 < −2. Plug in again using another FROZEN number to try to eliminate either (A) or (E). Since there’s an absolute value in the problem, try a negative number for c. If c = −3, then d = 3(−3) − 2 = −9 − 2 = −11. Eliminate (A) because −11 is less than −2. The only answer left is (E), which is correct because −14 is less than −11. The correct answer is (E).
Here’s what your scratch paper could look like:
FROZEN numbers are the types of numbers people normally don’t think of when working on GRE math problems. However, there may be numbers other than the FROZEN numbers that can work to eliminate answers on certain problems. For instance, a problem about even and odd numbers may require plugging in an even number and then an odd number. A problem about factoring may best be solved by plugging in prime numbers or perfect squares. FROZEN numbers will work most of the time, so stick with it unless you definitely see a different type of number
that may be better suited to eliminate answer choices in the problem.
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than once using FROZEN numbers until three of the answer choices have been eliminated.
Before we begin Plugging In on Quant Comp questions, let’s review the answer choices for Quant Comp questions first.
(A) means that Quantity A is always greater than Quantity B. Quantity B is never greater than Quantity A, and they are never the same. (B) means that Quantity B is always greater than Quantity A. Quantity A is never greater than Quantity B, and they are never the same. (C) means the two quantities are always equal. Quantity A is never greater than Quantity B, and Quantity B is never greater than Quantity A. (D) means sometimes Quantity A is greater than Quantity B, sometimes Quantity B is greater than Quantity A, or sometimes the values of the quantities are the same.
After Plugging In the first time, some information about the problem will be available and you will most likely be able to eliminate two answer choices. If Quantity A is greater than Quantity B, then eliminate (B) and (C), as Quantity B is not larger than Quantity A, and the two values are not equal. If Quantity B is greater than Quantity A, then eliminate (A) and (C) for the same reasons. If the two quantities are equal, then eliminate (A) and (B) because neither quantity is always greater than the other.
After Plugging In once and eliminating two answer choices, Plug In again using FROZEN numbers to try to eliminate another answer choice.
If there are no variables in the question, then eliminate (D). Choice (D) is correct only when there is more than one possible value for either or both of Quantity A and Quantity B. If there is only one possible value for each, then it is always possible to determine if one quantity is greater than the other, or if they are equal.
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At first glance, you might think instinctively that Quantity B must always be greater, because 3 is greater than 2. The GRE is engineered to take advantage of these instincts, however. Follow our steps and you’ll see what we mean.
This is a Quant Comp question with variables, so Plug In more than once. Plug in an easy number for the variable, such as x = 3. If x = 3, then Quantity A is 2(3) + 1 = 6 + 1 = 7. Quantity B is 3(3) + 1 = 9 + 1 = 10. Quantity B is greater than Quantity A, so eliminate (A) and (C).
Now Plug In again using FROZEN numbers to try to make Quantity B less than Quantity A. Try zero. If x = 0, then Quantity A is 2(0) + 1 = 0 + 1 = 1. Quantity B is 3(0) + 1 = 0 + 1 = 1. Now Quantity A equals Quantity B, so Quantity B is not always greater. Eliminate (B). Because two different numbers gave two different answers to the problem, the correct answer is (D).
Your scratch paper should look something like this:
Try another:
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Here’s How to Crack It
This is a Quant Comp question with variables, so Plug In more than once. The problem states that 42 < m < 49, so
plug in a number between 42 and 49, such as m = 45. If m = 45, then Quantity A is ≈ 0.803. This is greater than
0.75, so Quantity A is greater than Quantity B, so eliminate (B) and (C).
Now, Plug In again using FROZEN numbers. However, notice that because of the restrictions in the problem, not all FROZEN numbers are allowable. Because the goal is to find a way to make Quantity A less than Quantity B, the value for m needs to be as small as possible. Try an extreme value for m. The most extreme value for m possible is m = 42. Although the problem states that m is a value between 42 and 49, if the value of m is set at 42, then it is necessary that the least possible value of Quantity A has to be greater than if m = 42. If m = 42, then the value of Quantity A is 0.75, which is equivalent to Quantity B. Because m has to be greater than 42, it follows that Quantity A has to be greater than 0.75. Therefore, Quantity A is always greater than Quantity B. The correct answer is (A).
Some problems have multiple variables. In that case, Plug In for one variable at a time and try to solve for the other variables. If it’s not possible to solve for the other variables, then Plug In again for the next variable to see if it’s possible to solve for the remaining variables. If not, just continue Plugging In until all variables are assigned a number, or it’s possible to solve for the remaining variables.
4 a = 12 b 2 b = 10 c
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Here’s How to Crack It This is a Quant Comp question with variables, so Plug In more than once. But first, begin by looking at the equations in the problem to determine if there is a good number to Plug In. To solve for the variables b and c , it will be necessary to divide by both 12 and 10, so begin by equating them to their product, 120. Set 4 a = 120, so a = 30. If a = 30, then 4(30) = 12 b and b = 10. If b = 10, then 2(10) = 10 c , and c = 2. Now evaluate the quantities. Quantity A is 30, and Quantity B is also 30. The two quantities are equal, so eliminate (A) and (B). Now, Plug In again using FROZEN numbers. Try a negative number such as a = −15. In this case, b = −5, and c = −1. Quantity A is −15 and Quantity B is also −15. The quantities are still equal. In fact, no matter what number is plugged in for the quantities, the result is that the quantities are always equal. The correct answer is (C).
When you have to Plug In for multiple variables, make sure to plug in different numbers for each variable than from the first Plug In.
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Here’s How to Crack It This is a Quant Comp question with variables, so Plug In more than once. Plug in an easy number for a , such as a = 2. This does not give any further information about the value of b , so plug in an easy number for b as well, such as b = 3. If a = 2 and b = 3, then Quantity A is 2 × 3 = 6. Quantity B is 2 + 3 = 5. Quantity A is greater, so eliminate (B) and (C). Now Plug In again using FROZEN numbers. Try to make one number negative and one number positive. Plug in numbers such as a = −4 and b = 5. Now Quantity A is (−4)(5) = −20, and Quantity B is (−4) + (5) = 1, and Quantity B is greater, so eliminate (A). The correct answer is (D).
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the correct answer is (C). This problem can be solved using Plugging In, but it always helps to know basic algebraic manipulation. In this case, if the common quadratic was recognized, this problem can be solved in a matter of seconds, as opposed to Plugging In, which may have taken a couple minutes.
for heights does not mean the heights need to be realistic or average heights. Plug in numbers that make
the math easy, such as a = 11 and b = 13. Therefore, c is 12. Quantity A, the combined height of all the
people, is 11 + 13 + 12 = 36 inches. Convert this to feet by dividing by 12 to find the combined height in
feet is 3. Quantity B is 3 as well. The two quantities are equal, so eliminate (A) and (B). Plug In again
using FROZEN numbers. Combine three FROZEN numbers and plug in repeats, one, and extremes by
plugging in a = 1 and b = 1. The value of c is also 1. Quantity A is , and Quantity B is. The
quantities are still equal. The correct answer is (C).
There is one more kind of Plugging In strategy. It is called PITA, which stands for Plugging In the Answers. It is one of the most powerful types of Plugging In because it can take some of the hardest problems on the GRE and turn them into simple arithmetic. The hardest thing about this technique, however, is remembering when to use it.
The steps for PITA are as follows:
1. Recognize the opportunity. There are ways to know if a certain problem is a good candidate to Plug In the Answers: - the question asks for a specific amount that the answer choices represent - the question asks “how many” or “how much” - it seems appropriate to write and solve an algebraic equation 2. Write out the answer choices on your scratch paper. Write the numbers for each answer choice as well as the usual vertical A B C D E. 3. Label your answer choices. If the question asks for the number of hats David has, label the column of answer choices as “David’s hats.” If it’s asking for the value of x , label the column of answer choices with an x on top. 4. Plug In (C). Since the answer choices are listed in ascending or descending order, it will be helpful to start with (C). If you have more or fewer than five answer choices, use the middle answer choice. 5. Work the problem in bite-sized pieces, making a new column for each new step. Use (C) to work through the problem in bite-sized pieces. As you come up with values for each step, make a new column next to your answer choices. 6. POE. If you end up with numbers that don’t match up, then (C) is incorrect. This is why starting with (C) is so useful. Because the answer choices are listed in either ascending or descending numerical order, if it’s possible to determine the value of (C) is either greater or less than needed, it’s also possible to eliminate the answer choices that are greater or less than (C). Continue checking all the answer choices until one choice works.
If you’re not sure whether your answer was too great or not, then pick another answer choice at random, preferably either (B) or (D). Once you’ve tried another answer choice, you’ll be able to determine if that answer was further or closer away from the desired value than (C). This will inform you that the answer needs to be greater or less than the choice you just tried.
Finally, because you are using the answer choices as your number to plug in, for multiple-choice questions, once you have found one answer choice that works you do not need to check any other choices. If you Plug In the Answers and a choice works, that is the correct answer. Circle it and move on to the next problem.
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Here’s How to Crack It The question asks for a specific amount that the answer choices represent, and it seems reasonable to write an algebraic equation, so Plug In the Answers. Begin with (C). The problem states that there are three consecutive integers and the answer choices represent the greatest of those integers. If the greatest integer is 26, then the other two are 24 and 25.