12.Practice.Tests.for.the.SAT_2015-2016_1128p

12 Practice Tests for the SAT
Practice Test Twelve Answers and Explanations I
1081
8. D
Difficulty: High
Strategic Advice: The first thing you need to do is find what
the tens digit can be. You are given that the sum of the
other two digits must equal the square of the tens digit and
that the hundreds digit must be 3, 2, or 1.
Getting to the Answer:
Therefore, the greatest possible sum of the hundreds and
units digits is 3 + 9, or 12. The square of the tens digit must
be 12 or less, and the only perfect squares less than 12 are
0, 1, 4, and 9. Since there is no way the hundreds digit can
equal 0, you can eliminate O as an option. The tens digit
can either be vT = 1, \14 = 2, or V9 = 3. Now count the
combinations of numbers between 100 and 400 in which
the sum of the hundreds digit and the units digit equals 1,
4, or 9. The easiest way to do this is to list the possibilities:
110 123 222 32 1 138 237 336
There are 7 possibilities, (D).
15
9. 4 or 3.75
Difficulty: Low
Strategic Advice: This is a pretty straightforward algebra
problem that involves solving for one variable. Work
carefully to avoid any mathematical errors.
Getting to the Answer:
-6x + 8 = -2x - 7
1
10. 10 or .1
-4x=-15
x = --1.2
4
Difficulty: Medium
Strategic Advice: You can either solve this problem
algebraically, or pick numbers, starting with z.
Getting to the Answer:
x=-l-r=-l-(iz) = 2
2o z=-kz
11. 12
Difficulty: Medium
Strategic Advice: Here, you have a figure that is not drawn
to scale, so eyeballing won't help much. The area of the top
of the solid is much greater than the areas of the sides, so
the solid probably doesn't look much like the figure anyway.
Don't let the figure mislead you.
Getting to the Answer:
You know that volume is length x width x height, so label
the edges accordingly. Assume that the vertical lines
represent the height of the solid and label all the vertical
lines h. You can also assume that the horizontal lines
represent the width and label them w, and the diagonal
lines are the length, so label them /. So the area of face I is
wl, the area of face II is hw, and the area of face Ill is lh . This
means that you can write down three algebraic equations;
wl = 24, hw = 2, and lh = 3. Now you have three
equations and three unknowns, and you should be able to
solve. It's not obvious how to solve, however, so you'll have
to play around with these equations. Adding or subtracting
them doesn't get you very far, but if you multiply any two of
them together, something interesting happens. For example,
if you multiply the first two equations together, you get a
new equation; wl x hw = 24 x 2 or w2 /h = 48. Notice that
the left side of the new equation is w2 times lh, which is also
w2 times the left side of the third equation. So, if you divide
the new equation by the third equation, you get /h = 8 ,
or w2 = 16. That means that w must be 4, so I is 6 and h is
" and the volume is 4 x 6 x " or 12.
12. 30
Difficulty: Medium
Strategic Advice: Many of the more difficult geometry
questions on the SAT can be simplified by noticing
relationships between different parts of the diagram. In this
case, knowing the relationship between the area of the
shaded sector and the area of the circle as a whole gives us
the measure of the angle between AD and CF. This in turn
gives us the measure of q.
Getting to the Answer:
Since the shaded area is i the area of the circle, the angle
between AD and CF is i of 360 degrees, or 60 degrees.
The angles along CF add up to 180 degrees, so
60 + 90 + q = 180 and q = 30.