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

12 Practice Tests for the SAT
Practice Test Eleven Answers and Explanations I
997
Getting to the Answer:
Use x = 4 when y = 6 to solve for k:
6 = k(4)
k=:l 2
Use k = % and x = 16 to solve for y:
y=2(16) 2
y=24
5. A
Difficulty: Medium
Strategic Advice: When you are unsure of how to write the
equation of a line, remember that the answer choices are
right in front of you. Pick points off of the graph and then
plug them into the given equations.
Getting to the Answer:
Use the point (2, 0) to test out the given equations:
y= l Ox-20
0 = 10(2) - 20
0=20-20
0=0
Choice (A) works.
6. E
Difficulty: Medium
Strategic Advice: This problem may look a little
intimidating, but what it really tests is your knowledge of
positive and negative numbers. Notice that the products of
the x- and y-coordinates of the points in quadrants I and Ill
(the upper right and lower left) are positive. On the other
hand, notice that the products of the x- and y-coordinates
of the points in quadrants II and IV are negative, because in
II and IV one coordinate is positive and the other negative.
Positive numbers are larger than negative numbers, so
either (C) or (E) is correct.
Getting to the Answer:
Point Eis much farther from the origin than point C, so the
product of the coordinates is larger in Ethan in C.
7. D
Difficulty: High
Strategic Advice: This problem asks for a in terms of b.
You're given a in terms of c, so to get a in terms of b, you
need to find c in terms of b. Then plug this expression into
the first equation for c.
Getting to the Answer:
Finding c in terms of b:
b=7 + 3c
b-7 =3c
-b-7
3
- = C
Finding a in terms of b:
a= 9c2 + 4
a= 9(-b-7
3 -)2 + 4
a=9( b2 -14b +49 )+4
9
a = b2 - 14b + 49 + 4
a = b2 - 14b + 53
This is also a great problem to try Picking Numbers. Try
picking something really simple, like c = 0 . This means
b = 7 and a= 4 , and you can quickly test all the answer
choices.
8. c
Difficulty: High
Strategic Advice: There are several ways to approach
this problem. For instance, you could begin by finding the
length of the sides of the large squares, then dividing
the figure into areas based on that length and calculating
the area of each zone. This is a perfectly good way to
solve the problem, but it will most likely take more time
than the method described below, which relies on adding
a few lines to the original diagram.
Getting to the Answer:
You're given that the large squares are identical and that
DECF and FCKJ are identical. E and K must therefore be
the centers of ABHF and FHML, respectively. So each
large square in the figure can be divided into four smaller
identical squares as shown here:
The area of a large square is 100 square units, so the area
of a small square is 1 O = 25 units. The figure is made up
of 10 small squares, so its total area is 250 square units.