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.
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.
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 =
and triangle area =
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
. 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.
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
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.
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
Step 3: Mine the Math. Adding the word
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.
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
Now some basic arithmetic is in order, making this a multiple-concept
question. We can use the Magic X to simplify our equation to
, 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.
h + y = 10 + 6 = 16 No good.
Not even close.
That’s the one. E
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.