Practice Problems
3.1 PS X-Ray
3.3 PS Step Method
3.4 Practice Problems
Practice Problems
This question type is called Problem Solving, after all, so you may as well get some practice solving some problems. There may be more than one effective way to solve any given problem, so remember to use Step 2 of the PS step method and give some thought as to the approach that might work best for you before jumping in. See how you make out with the problems in the set, and then review the guided explanations that follow.

Directions: The following questions have five answer choices. Select the best of the answer choices given.

1. Working together at a constant rate, machines L, M, and N can produce a total of 200 bolts in 4 hours. If machine L can produce 20 bolts in 30 minutes, and machines M and N work at the same constant rate as each other, how many bolts can machine M produce in 1 hour?
(A) 5
(B) 10
(C) 20
(D) 30
(E) 50
2. If the area of the circle above with center O is 36p, what is the area of the shaded region?
(B) 9π – 18
(C) 3p +
(D) 9π
(E) 36π – 18
3. The average (arithmetic mean) of the first five numbers in a group of ten positive integers is 110. The sum of the remaining numbers in the group is 250. What is the average (arithmetic mean) of all ten numbers?
(A) 22
(B) 36
(C) 50
(D) 80
(E) 180
4. What is .5 percent of 55?
(A) .275
(B) 2.75
(C) 27.5
(D) 50
(E) 110
5. A coin flipped a number of times landed on heads y more times than twice the number of times it landed on tails. If h is the number of times the coin landed on heads, how many times was the coin flipped, expressed in terms of h and y?
(A) h + y
Guided Explanations
1. A
Step 1: Get the Specs. That’s sure an awful lot of words for a math question, which tells us we’ve entered word problem territory. Specifically, we’re asked to determine the number of bolts that can be made in an hour, so it’s a work kind of word problem. (It doesn’t hurt that the first word of the question is working, another clue as to what kind of problem we’re up against.) The numbers in the choices seem fairly manageable, so we’ll take that into account as we move on to Step 2.
Step 2: Plan the Attack. Since there are straightforward numbers in the choices, you may have been tempted to work backward. But since there’s really only one basic formula for work problems, we’ll go for the standard approach and see how that pans out.
Step 3: Mine the Math. The only real piece of math we need is the work formula: rate × time = amount. We will, however, need to be careful to keep the units straight, since time in the problem is expressed both in minutes and in hours.
Step 4: Power Through. Applying the work formula from Step 3 to the first sentence of the problem gives rate × 4 hours = 200 bolts for all three machines working together. We can then solve for rate by dividing both sides of the equation by 4, yielding rate = 50 bolts per hour for all three machines working together. The next most concrete piece of information concerns machine L: It alone can produce 20 bolts in 30 minutes. Since we’ve been dealing with hours so far, let’s keep it that way and multiply both sides by 2 to convert L’s rate to 40 bolts per hour. Now we’re getting somewhere: If all three machines can turn out 50 bolts in an hour, and machine L alone can turn out 40 in one hour, then machines M and N together must account for 50 – 40 = 10 bolts per hour. Since we’re told that machines M and N have identical rates, each must produce 10 ÷ 2 = 5 bolts per hour, choice A.
2. B
Step 1: Get the Specs. The diagram pretty much gives it away: Geometry is the name of this game, and a mish-mash challenge at that given the overlap between the circle, right triangle, and shaded region. The choices are pretty intimidating, so it doesn’t appear as if it will be helpful to work backward from them.
Step 2: Plan the Attack. Looks like we’ll just go the traditional route and work our way through the question, keeping in mind that mish-mash problems often require us to recognize which pieces of which geometric figures overlap with the other figures.
Step 3: Mine the Math. Area is the key element here, so at the very least circle area = p r 2 and triangle area = base × height should spring to mind to get you heading in the right direction. Since this is a multiple-concept question, a few other bits and pieces of math knowledge may be needed, but we’ll cross those bridges when we come to them.
Step 4: Power Through. We’re asked for the area of the shaded region, but as is often the case in mish-mashes, we have to step back and consider how the shaded area might be the result of simpler shapes. Even though it looks like a complicated shape, the right angle at the center tells us the shaded region is really just a quarter circle with a right triangle subtracted from it. Since the area of the whole circle is 36π, the area of the quarter circle must be .
Now we need the area of the right triangle. Since the area of the whole circle is 36π, we can use the trusty area formula recalled in Step 3 above to solve for r:
pr2 = 36p
r2 = 36
r = 6
Both of the lines that intersect O are radii, so each must have a length of 6. Since the base and height of the right triangle are 6, its area is . Subtracting this area from the area of the quarter circle yields the area of the shaded region, 9π – 18, which is choice B.
Before we leave this one, we’d like to make a quick point about approximating. We discussed the importance of approximating answers in our chapter 1 introduction to GRE math. If you were stuck for a guess in this problem, estimating the answer would have been an excellent approach. The area of the entire circle is 36π, or about 120 (recall that π is roughly equal to 3.14). The shaded region is just a small portion of that, and it looks to be less than one-tenth of the whole figure. That means that the correct answer should be less than 12, eliminating C, D, and E. To decide between A and B, keep in mind that the correct answer to a problem involving a circle often contains π. This makes B the best guess, even if you blanked on the problem otherwise.
3. D
Step 1: Get the Specs. The problem clearly concerns averages and contains basic numbers in the choices. We’ll get to the specifics soon enough, but there’s not much more to notice at this point.
Step 2: Plan the Attack. You can certainly go with a straightforward application of the arithmetic mean formula, but let’s, for the sake of practice, use our alternative working-backward approach.
Step 3: Mine the Math. The problem involves averages, so we’ll need to call to mind the standard formula .
Step 4: Power Through. Since the choices are in ascending order, we’ll start with C, the middle one. This, as we demonstrated earlier, allows us to knock off three choices by testing just one—or, of course, get the answer straightaway if C happens to be correct. Let’s work with the info in the first two sentences of the problem and then see whether 50 works as the final answer.
Manipulating the formula from Step 3 results in sum of terms=(average)×(# of terms). If five numbers average out to 110, then the sum of those five numbers is 110 × 5 = 550. We’re told that the sum of the remaining five numbers is 250, which means the sum of all ten numbers must be 550 + 250 = 800. Now let’s see if choice C fits with this calculation. If the average of all ten numbers is 50, then the sum of all ten would be 50 × 10 = 500, which doesn’t match the 800 figure we just calculated. So not only is C incorrect, but we can also see that it’s too small, which means choices A and B, containing even smaller values, must be incorrect as well. All we have to do now is try one of the two remaining choices; either it will work, or the other choice will be correct. D works. An average of 80 for all ten numbers yields a sum of 80 × 10 = 800, which does match the total of 800 we calculated initially. D is therefore correct.
You may have gone with a traditional application of the arithmetic mean formula all the way through, instead of stopping to work backward from the choices. That’s fine if it worked for you. Still, give some thought to the working-backward strategy demonstrated above to see if you may have been able to get the answer faster and with less risk.
4. A
Step 1: Get the Specs. Not much to it, is there? Straight arithmetic, nothing too crazy.
Step 2: Plan the Attack. A straightforward approach is the way to go. We’ll dig out the math concept we need, and use it to do the math.
Step 3: Mine the Math. Adding the word percent to a number means taking that number two places to the left when converting it to a decimal. For example, 20 percent = .2. In this example, .5 percent = .005. The other thing you need to know is that the word of means multiplication; whenever we take a certain percent of something, we multiply the figures.
Step 4: Power Through. Let’s do the math: .005 × 55 = .275, which is choice A. Maybe you just multiplied it out by hand, or instead used approximations to get into the ballpark. Here’s one way you can work through it without actually multiplying: 10 percent of 55 is 5.5, so to get 1 percent of 55, we just move the decimal place back one more place, giving us .55. Now, .5 percent is half of 1 percent, so we have to divide .55 by 2, which gives us .275.
As for the distractors, 2.75 (B) is what you get if you ignore the decimal point and take 5 percent of 55. 27.5 (C) is what you get if you ignore the word percent and simply take .5 of 55. 50 (D) is what you get if you subtract 5 from 55, and 110 (E) is what you get if you double 55, perhaps by mistakenly translating the problem into 55 ÷ .5. So while the question is a straightforward test of your arithmetic knowledge, there are still a few concepts you need to know, some steps you need to perform, and some traps that could potentially trip you up.
5. E
Step 1: Get the Specs. The presence of a coin in the problem might initially suggest a data analysis probability problem, but that’s not how it plays out—it’s actually an algebra question. We’re given a word problem that requires some English-to-math translation, and some scary-looking expressions containing variables in the choices. These are the basic aspects of the problem that should catch your eye.
Step 2: Plan the Attack. Word problems containing variables lend themselves naturally to algebraic solutions, so if you’re comfortable with algebra, you could power through this one by setting up and solving some basic equations. Alternatively, the variables scattered across the answer choices suggest that this problem may also lend itself well to our making-up-numbers strategy, so you may have chosen that route instead. Tell you what—for practice, we’ll do both.
Step 3: Mine the Math. For the algebraic approach, you’ll need to draw on your knowledge of constructing and solving multiple equations. For both approaches, you’ll need to correctly translate the English into math.
Step 4: Power Through, Algebra Style. First, the English-to-math translation: If we designate tails as t, then “twice the number of times it landed on tails” is 2t. y more times than “twice the number of times it landed on tails” is therefore 2t + y. Using h for heads as instructed, our equation becomes h = 2t + y. The question is looking for the total number of flips, which is the number of heads plus the number of tails, or h + t. However, the question asks for the total expressed in terms of h and y. No problem—we can solve for t in terms of y in our first equation and substitute that into the equation representing the total:
Now replace t with this new value in the total equation:
Now some basic arithmetic is in order, making this a multiple-concept question. We can use the Magic X to simplify our equation to
or , choice E
Step 5: Power Through by Making Up Numbers. Okay, so what if all those equations in the previous solution look like Egyptian hieroglyphics to you? Then the “make up numbers” alternative approach is the way to go. We start the same way by translating the wording of the question into the expression h = 2t + y. Variable h is the dependent variable because its value is determined by the values of t and y, so when it comes to picking numbers, we’ll want to save h for last. In this case, we’ll only need to use our imagination for t and y, since h will follow from those. And remember to go with simple numbers—why make your life more difficult than it needs to be? t = 2 is friendly enough; a nice small number that’s easy to double. 2t is 4, so if we set y = 6, then h will be an even 10, giving us:
t = 2
y = 6
h = 10
With all the variables set, we can focus on what the question is after. The total number of flips equals the number of heads plus the number of tails. In our imaginary world, that’s 10 + 2 = 12. Now we simply need to determine which combination of hs and ys in the choices calculates to 12. Let’s try them out.
A: h + y = 10 + 6 = 16 No good.
B: 11 Nope.
C: Uh-uh.
D: Not even close.
E: That’s the one. E is correct, no matter which approach we take. We presented two solutions to this final problem, both to give you extra practice and to remind you of the old, somewhat sadistic saying “There are many ways to skin a cat”—not that we can figure out why anyone would want to. While there are definite math concepts to know and specific options for applying them, you should always attempt to determine the approach that works best for you in each situation.
You’ll get more practice with Problem Solving in the practice test at the end of the book. For now, let’s move on to our next, slightly more complicated math question type, Quantitative Comparisons.
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