SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (43 page)

Now let’s take a look at what we might get if we decide to diagram this. First, let’s get a rough idea of where point
B
is:

When we try to plot point
A
, we start to see the difficulty in the question: we have one coordinate for
A
, but not

the other, so we can’t plot it. All we can do is say that
A
must lie somewhere on the line of
y
= 3, because its
y
-value is 3.

So now the issue is that
A
is 17 units away from
B
, and
A
is also somewhere on the dashed line in the diagram. So we have to figure out exactly where
A
can go. At this point, I would draw a circle (or part of a circle) showing the points that are 17 units away from
B
.
A
must be on a point where this circle intersects the line
y
= 3 above.

Now, all we have to do is figure out the
x
-value of either one of those two spots where
A
could be. For this, we’ll want to use the Pythagorean theorem.  The separation between the
y
-values of the two points will be one leg in our right triangle. The separation between the
x
-values of the two points will be the other leg. And the hypotenuse will be the 17-unit straight-line distance between the two points:

From here, we can apply the Pythagorean theorem as we did in the other approach, and realize that the horizontal leg in the triangle must have a length of 8 units, which means that the
x
-value for point
A
must be either 2 or 18.

Again, we could have done this in a more formal way by using the distance formula:

d
= √((
y
2
-
y
1
)
2
+ (
x
2
-
x
1
)
2
)

But, in general, I advise you to avoid formulaic thinking on the SAT, because there are so many instances on the average SAT Math section in which a formula might seem appropriate when it actually isn’t. And even when a formula
might work on a particular question, it’s almost always going to take more time and energy than a non-formulaic approach, and it will also typically increase the likelihood of a mistake on your part.

Also, in the case of the distance formula, I find that it’s much easier for most people to remember and apply the Pythagorean theorem than to recall every detail of the distance formula correctly. And since the SAT doesn’t care how you arrive at the answer, I generally advise people to work in whichever way seems easiest at the time.

For this question,
the key thing is understanding what the idea of modeling the data means. We're looking for the line that provides the best match to the general trend of the data points. Since the data doesn't increase or decrease as you move from left to right, the only thing that works is (A).

Note that (B), (C), and (D) would all have to go through the point (0,0), which clearly isn’t appropriate here. Lastly, (E) would be increasing as it went to the right—it would pass through (0,44) and (5,49). We don’t want that kind of positive slope because if we look at the plotted points we can see that the values are not increasing overall as we move left-to-right. The highest value is in the second trial, and one of the lowest values is in the 5th.

This is one more example of a question that’s fairly easy if we read carefully and pay attention, but that many test-takers will miss anyway.

As with many SAT Math questions, there are a lot of ways to go about this one. The easiest approach, I think, is to convert either 12
L
or 10
L
to
W
, and then figure out the number of
L
x
W
rectangles, since that’s what the question asks for.

The only way to convert L to W is to notice that 2
L
= 3
W
, which we know because 2
L
and 3
W
both correspond to the height of the big rectangle in the diagram.

If 2
L
is 3
W
, then 12
L
is 18
W
.

That means the area being covered
can be expressed as 18
W
x 10
L
, or 180
LW
. Since
LW
is the size of each tile, that means we would need 180 tiles. So the answer is (E).

Notice that one of the wrong answers, (B), is 1/5 of the right answer. This makes sense, because a lot of people might accidentally determine the number of times that the big rectangle in the diagram could fit into the 12
L
x 10
W
region.

Most students I’ve worked with on this question have ultimately arrived at the right answer, but they often don’t see the easiest way to get there.

Most of them try to figure out the median slope by determining the slop
e of every single line segment in the question, and then trying to arrange them from least to greatest (which they usually do by converting them all to a common denominator). It’s quite time-consuming.

The much easier thing is to pay attention to the diagram and to think about what the concept of slope means in the first place. The slope of a line is a measurement of how slanted it is, in a manner of speaking. So the line segment with the median slope will be the one whose ‘slantiness’ puts it in the middle of all the other segments.

In other words, we can tell that line segment
OC
is the one with the median slope just by looking at the diagram. Now all we have to do is determine the slope of
OC
, and we’re done. Since
OC
starts at the origin and point
C
is at (4,3), that means it goes up 3 units and over 4, for a slope of 3/4. So (C) is right.

Notice that all the other answer choices are the slopes of the line segments, which allows test-takers to make all kinds of mistakes and still find a wrong answer that they’ll like. Remember to pay attention to details!

This question is one more excellent example of how we always have to be ready to apply basic math ideas in non-traditional ways. In this case, we’re asked to find equivalent proportions even though every proportion involves 4 variables and we’re never told the values of any of those variables. How is that possible?

In situations like this, it’s often helpful to fall back on the idea of trying to identify concepts related to the things that appear in the question. In this case, pretty much the only usable concept we have in front of us is the idea of cross-multiplying.

As it turns out, if we cross-multiply each answer choice, we’ll
see that choice (A) gives us
ac
=
bf
, but all the other choices give us
af
=
bc
.

That means choice (A) is the one that’s not equivalent to the others.

So, in the end, we never learned the values of any of these variables, and we never did anything more complicated than cross-multiplying to answer the question. Many, many test-takers must have missed this question, but when I talk to students who’ve missed it I’ve never met one who didn’t know what cross-multiplying was. Remember that the challenge on the SAT Math section is to identify the basic concepts that will solve the problem.

Remember that these questions with roman numerals are typically going to be based on some kind of abstract property of the concepts in the question. If we can figure out what that property is, answering the question will usually be pretty straightforward.

For instance, the question tells us that roman numeral I will work out to
ab
-
b
. So we have to ask ourselves if there are any conditions in which that might be equal to zero.

That should raise the question, “when can subtracting one thing from another give us zero?”

The answer is that subtracting one value from another can only result in zero when the two values were equal to start with. This is a property of zero.

So
ab
-
b
can only equal zero if
ab
and
b
are equal to each other.

And that leads to another question: when can
ab
and
b
be equal? When can we multiply something by
a
and have it equal our original starting value, in this case
b
?

We can only multiply something by
a
without changing its value if
a
equals 1. This is a special property of the number 1.

So, in the end,
ab
-
b
can only equal zero when
a
equals 1, because if
a
is 1 then the expression
ab
-
b
is the same as the expression
b
-
b
, which must be zero.

So that means that the expressions in each of the other answer choices can only come out to be zero if the part of the expression before the box can possibly equal 1, like
a
could in roman numeral I.

For roman numeral III, then, it’s pretty clear that the value can work out to be zero if we just make
a
be 1 again.

But what about roman numeral II?

The (
a
+
b
) expression at the beginning of roman numeral II could work out to be 1 if
a
and
b
could be fractions or even if one could be negative. But the question tells us that
a
and
b
must be positive integers, which means the lowest possible value for
a
+
b
is 2. And that can’t work.

So the correct answer here is (E): only roman numerals I and III can work.

Notice that this question has really no resemblance to anything that ever happens in the average math class in school. Notice that with this solution we never really needed a calculator, or even a pencil. Notice, also, that there’s no other question in the entire book that calls for the same specific approach to that this one calls for. In other words, the lesson to learn from this question is NOT the idea of specifically attacking roman numeral questions by looking to see what kinds of subtraction can result in zero, because you’ll probably never see a question that tests that concept in this exact way again. Instead, the lesson to take away from this is, as always, that we need to practice thinking about SAT Math questions in terms of basic principles, definitions, and properties. We need to read everything carefully. And we have to learn to be flexible in our thinking in the way that the SAT Math section rewards over and over again.