- 3 Draw a line with points P, Q, R, andSon the
line in that order. You are given that PS
—
= 2 PR
—
and
that PS—
= 4 PQ
—
, so choose values for those lengths, like
PS—
=12, PR
—
=6, and PQ—
=3.
This means that QS—
=9, so QS—
/PQ
—
=9/3 =3.
(Chapter 6, Lesson 2: Analyzing Problems)CHAPTER 16 / PRACTICE TEST 3 761
- 750 25% of $600 is $150. Therefore, the club
earned $150 more in 2007 than it did in 2006, or $600 +
$150 =$750. Remember, also, that increasing any
quantity by 25% is the same as multiplying that quan-
tity by 1.25.
(Chapter 7, Lesson 5: Percents) - 3 Set up equations: x+y= 4
x−y= 2
Add straight down: 2 x= 6
Divide by 2: x= 3
Plug in 3 for x:3 +y= 4
Subtract 3: y= 1
Final product: (x)(y) = (3)(1) = 3
(Chapter 8, Lesson 2: Systems) - 32 Let LM=x,and let LO=y.Since xis twice the
length of y, x= 2 y.
x+x+y+y= P
Substitute for x: 2 y+ 2 y+y+y= P
Combine terms: 6 y= P
Plug in 48 for P: 6 y= 48
Divide by 6: y= 8
Solve for x: x= 2 y=2(8) = 16
To find the area of the shaded region, you might notice
that if PMis the base of the shaded triangle, then LOis
the height, so area =^1 ⁄ 2 (base)(height) =^1 ⁄ 2 (8)(8) =32.
If you don’t notice this, you can find the shaded area
by finding the area of the rectangle and subtracting
the areas of the two unshaded triangles.
Area of rectangle =(length)(width)
Area of rectangle =(x)(y) =(16)(8) = 128Area of triangle PLO=^1 ⁄ 2 (base)(height)
Area of triangle PLO=^1 ⁄ 2 (8)(8) = 32
Area of triangle MNO=^1 ⁄ 2 (base)(height)
Area of triangle MNO=^1 ⁄ 2 (16)(8) = 64
Area of triangle OPM= 128 − 64 − 32 = 32
(Chapter 10, Lesson 5: Areas and Perimeters)
- 9 643 = 4 x
Substitute 4^3 for 64: (4^3 )^3 = 4 x
Simplify: 49 = 4 x
Equate the exponents: x= 9
(Chapter 8, Lesson 3: Working with Exponentials)
LP
16
8
88
8 M
ON
33 6
PQRSyx(–1, 6) (4, 6)(m,n)O15.1.5 Since the graph is a parabola, it has a vertical
axis of symmetry through the vertex. The points (−1, 6)
and (4, 6) have the same y-coordinate, so each one is the
reflection of the other over the axis of symmetry. This
axis, therefore, must be halfway between the two
points. Since the average of −1 and 4 is (− 1 +4)/2 =1.5,
the axis of symmetry must be the line x=1.5, and there-
fore m=1.5.
(Chapter 11, Lesson 2: Functions)- 18 Since these numbers are “evenly spaced,”
their mean (average) is equal to their median (middle
number). The average is easy to calculate: 110/5 =22.
Therefore, the middle number is 22, so the numbers
are 18, 20, 22, 24, and 26.
Alternatively, you can set up an equation to find the
sum of five consecutive unknown even integers,
where xis the least of these:
x+(x+2) +(x+4) +(x+6) +(x+8) = 110
Combine like terms: 5 x+ 20 = 110
Subtract 20: 5 x= 90
Divide by 5: x= 18
So the five integers are 18, 20, 22, 24, and 26.
(Chapter 9, Lesson 2: Mean/Median/Mode Problems) - 20 Use the percent change formula:
(Chapter 7, Lesson 5: Percents)24 000 20 000
20 000
100 20
,,
,
%%
−
×()=
Final Original
Original